Fundamental Actuarial Mathematics/Present Value Random Variables for Long-Term Insurance Coverages

Learning objectives

The Candidate will be able to perform calculations on the present value random variables associated with benefits and expenses for long term insurance coverages.

Learning outcomes

The Candidate will be able to:

1. Identify the present value random variables associated with life insurance, endowment, and annuity payments for single lives, based on annual, 1/m-thly and continuous payment frequency.
2. Calculate probabilities, means, variances and covariances for the random variables in 1., using fractional age or claims acceleration approximations where appropriate.
3. Understand the relationships between the insurance, endowment, and annuity present value random variables in 1., and between their expected values.
4. Calculate the effect of changes in underlying assumptions (e.g., mortality and interest).
5. Identify and apply standard actuarial notation for the expected values of the random variables in 1.

Introduction to life insurance

In a life insurance policy, when the insured dies at a time within the coverage of the life insurance, the insurance company needs to pay a benefit for the insured at a certain time. The amount and the time of benefit payment depends on the terms stated in the contract, and the type of the life insurance. In this chapter, we will discuss some models for the payments of different types of life insurance, and perform some calculations related to them. Also, we will introduce some related actuarial notations.

Insurances payable at the moment of death

In this section, we will model some types of insurances payable at the moment of death, which is associated with the random variable $T_{x}$, and all insurances mentioned in this section are payable at the moment of death unless stated otherwise. In practice, the insurance is not payable at the moment of death, since it is infeasible. However, with this theoretical assumption, some calculations can be more convenient.

Before modeling the insurances, we need to introduce a fundamental random variable: present-value random variable.

Definition. (Present-value random variable) The present-value random variable $Z=b_{T}v_{T}$, in which $b_{T}$ and $v_{T}$ are benefit function and discount function of $T$ ($T=T_{x}$), respectively.

In the models discussed in this section, they differ in only the definitions of $b_{T}$ and $v_{T}$, and the model for $n$-year term life insurance is the basic one, and other models are just some modifications on this model. Because of this, it is important to understand the $n$-year term life insurance model clearly.

We often want to know the expected value of the present-value random variable so that we would somehow predict how much benefit will be paid, in the present value term. As a result, we have a special name for such expected value:

Definition. (Actuarial present value) The actuarial present value (APV) is $\mathbb {E} [Z]$, the expected value of the present-value random variable.

For simplicity, we will assume the benefit amount is one in the following models unless otherwise specified. If we want to change the benefit amount to other values, then we can just multiply it accordingly. For the insurances for which the benefit amount is varying, we can also change the benefit amounts "together" through appropriate multiplication.

n-year term life insurance

Definition. ($n$-year term life insurnace) A $n$-year term life insurance provides a payment if the insured dies within the $n$-year term, starting from the time of policy issue, and does not provide any payment otherwise.

For $n$-year term life insurance, we have $b_{T}={\begin{cases}1,&T\leq n;\\0,&T>n,\end{cases}}$ and $v_{T}=v^{T},\quad T\geq 0$ ($v$ is the discount factor which is ${\frac {1}{1+i}}$, in which $i$ is the (annual) effective interest rate). Hence, $Z={\begin{cases}v^{T},&T\leq n\\0,&T>n\end{cases}}$.

In this case, the actuarial present value, denoted by ${\bar {A}}_{x:{\overline {n}}|}^{1}$, is $\mathbb {E} [Z]=\int _{0}^{n}v^{t}f_{x}(t)\,dt=\int _{0}^{n}v^{t}{}_{t}p_{x}\mu _{x+t}\,dt$. Here, we use the result $f_{x}(t)={}_{t}p_{x}\mu _{x+t}$ from the Fundamental Actuarial Mathematics/Mortality Models chapter.

The notation for the APV may seem weird at first, but after knowing the meaning of each of symbols involved, the notation will make more sense. Indeed, the other notations for the APV in other types of life insurance are constructed in a similar way. The meaning of the symbols involved is as follows:

• $A$ means "a life insurance".
• ${\bar {}}$ (on top of $A$) means this APV is "in a continuous manner" (since the benefit is paid at the moment of death, the APV is "somehow continuous").
• $x$ means the insured is aged $x$ at the time of policy issue.
• $:{\overline {n}}|$ means the term of the life insurance is $n$ years.
• $1$ on top of $x$ (and "before" $:{\overline {n}}|$) means the benefit of amount 1 is paid to the insured (aged $x$ at the time of policy issue), if he dies before $n$ years passed from the time of policy issue.

Example. Consider a 10-year term life insurance with benefit of 1, issued to a life aged 60. Suppose the pdf of $T_{x}$ is $f_{x}(t)={\frac {1}{100}},\quad 0\leq t\leq 100$, and the annual force of interest (the force of interest in the following will be annual unless otherwise specified) is (a constant) $\delta =0.05$. Then,

• The APV is denoted by ${\bar {A}}_{60:{\overline {10|}}}^{1}$
• The discount function is $v^{t}=e^{-\delta t}=e^{-0.05t}$ (here we use a result from Financial Math FM)
• The value of APV is thus $\int _{0}^{10}e^{-0.05t}\cdot {\frac {1}{100}}\,dt={\frac {1}{100(-0.05)}}[e^{-0.05t}]_{0}^{10}=-{\frac {1}{5}}(e^{-0.5}-1)=0.0787$.

• This means it is "expected" that benefit with present value of 0.0787 will be paid for each insurance issued.
• It may be intuitive that the insurance should also cost 0.0787 when the life aged 60 purchases it immediately after reaching age 60. However, this may not necessarily be the case, since there is also some cost for handling the purchase of insurance, and also the insurance company selling it will also want to make some profits. As a result, the resultant cost may be higher than this APV.

Exercise.

1 What is the value of APV if the benefit of the insurance is 1 million instead of 1?

	0.0787
	787
	78700
	787000

2 Will the APV increase, decrease, or remain unchanged when the annual (constant) force of interest increases to a higher value?

	Increase.
	Decrease.
	Remain unchanged.

3 Will the APV increase, decrease, or remain unchanged when the term of the insurance is longer?

	Increase.
	Decrease.
	Remain unchanged.

Remark.

• If the force of interest is not constant, the calculation will be more complicated.
• Thus, for simplicity, the force of interest will be assumed to be constant unless otherwise specified in the following.

Example. Amy, aged 30, purchases a 20-year term life insurance with benefit 1,000,000 from XYZ life insurance company. Suppose the survival distribution for Amy at aged $x$ is $S_{x}(t)=e^{-0.05t}$, and the force of interest is 7% (constant).

(a) What is the expression for the APV of this life insurance?

(b) Suppose the premium of this policy is the APV of this life insurance plus 1000. Calculate the premium of this policy.

Solution:

(a) $1000000{\bar {A}}_{30:{\overline {20}}|}^{1}$.

(b) First, given the survival function, we know that $_{t}p_{30}=e^{-0.05t}$. Also, the force of mortality is $\mu =0.05$ which is constant (so we are actually assuming constant force of mortality). The premium is

$$1000000{\bar {A}}_{30:{\overline {20}}|}^{1}+1000=1000000\int _{0}^{20}e^{-0.07t}{}_{t}p_{30}\mu _{30+t}\,dt+1000=50000\int _{0}^{20}e^{-0.12t}\,dt+1000={\frac {50000}{-0.12}}[e^{-0.12t}]_{0}^{20}+1000\approx 379867.52.$$

Exercise.

1. Suppose that XYZ life insurance company finds out that Amy is actually a smoker, and thus the survival function for Amy at aged $x$ is adjusted to $S_{x}(t)=e^{-0.08t}$. Following the assumptions made above, calculate the new premium of this policy.

Solution

The new premium is

$$80000\int _{0}^{20}e^{-0.07t}e^{-0.08t}\,dt+1000={\frac {80000}{-0.15}}[e^{-0.15t}]_{0}^{20}+1000\approx 507780.23.$$

Notice that the premium increases quite a lot.

2. Following the assumption in the previous question, calculate the probability that no benefit is paid to Amy within these 20 years.

Solution

No benefit is paid to Amy within these 20 years is equivalent to Amy lives for these 20 years. Thus, the probability is $_{20}p_{30}=e^{-0.05(20)}\approx 0.368$.

3. After 10 years, Amy still survives and she decides to withdraw from the insurance. According to the withdrawal policy of XYZ life insurance company, when the policyholder withdraws from a life insurance policy, the company will return 50% of the premium of the remaining part of the insurance concerned, that is, the premium charged when that policyholder purchases a life insurance in that length (length of the remaining part) at the withdrawal time point. Calculate the return amount in this case, following the assumption in previous questions.

Solution

The return amount is

$$0.5{\bar {A}}_{40:{\overline {10}}|}=0.5\cdot 80000\int _{0}^{10}e^{-0.07t}e^{-0.08t}\,dt={\frac {40000}{-0.15}}[e^{-0.15t}]_{0}^{10}\approx 207165.29.$$

4. Suppose that Amy unfortunately dies 1 year after the withdrawal from insurance, so no benefit is paid to Amy. Calculate the value of the loss of Amy, at the time point where Amy is at age 30, defined by the sum of the value of her net cash outflow, and the value of the benefit she will have if she does not withdraw from the insurance.

Solution

The value of the loss is

$$507780.23-207165.29e^{-0.07(10)}+1000000e^{-0.07(11)}\approx 867918.046.$$

Whole life insurance

For whole life insurance, it can be interpreted as a "$\infty$-year term life insurance", and so payment will be eventually provided since no one lives for eternity. To be more precise, its definition is as follows:

Definition. (Whole life insurance) A whole life insurance provides a payment following the death of the insured (at any time in the future).

For whole life insurance, we have $b_{T}=1,\quad t\geq 0$, $v_{T}=v^{T},\quad T\geq 0$, and thus $Z=v^{T},\quad t\geq 0$. The APV, denoted by ${\bar {A}}_{x}$, is $\int _{0}^{\infty }v^{t}{}_{t}p_{x}\mu _{x+t}\,dt$.

In the notation, we omit the "$1$" and "$:{\overline {n}}|$" appearing in the notation of APV for $n$-year term life insurance., since for whole life insurnace, the "term" does not exist (or is "infinite"). Also, the benefit must be paid eventually, and so there is no need to emphasize the benefit payment, unlike above, where the benefit may or may not be paid, depending on how long the insured live.

Example. (Formula of ${\bar {A}}_{x}$ with constant force of interest and force of mortality) Given the force of interest is $\delta$ and the force of mortality is $\mu$, ${\bar {A}}_{x}={\frac {\mu }{\mu +\delta }}$.

Proof. When the force of mortality is $\mu$, $_{t}p_{x}={\frac {S_{0}(x+t)}{S_{0}(x)}}={\frac {e^{-\int _{0}^{x+t}\mu \,ds}}{e^{-\int _{0}^{x}\mu \,ds}}}={\frac {e^{-(x+t)\mu }}{e^{-x\mu }}}=e^{-\mu t}$. Thus,

$${\bar {A}}_{x}=\int _{0}^{\infty }v^{t}{}_{t}p_{x}\mu _{x+t}\,dt=\int _{0}^{\infty }e^{-\delta t}{}_{t}p_{x}(\mu )\,dt=\mu \int _{0}^{\infty }e^{-\delta t}e^{-\mu t}\,dt=\mu \int _{0}^{\infty }e^{(-\delta +\mu )t}\,dt={\frac {\mu }{-(\delta +\mu )}}[e^{(-\delta +\mu )t}]_{0}^{\infty }={\frac {\mu }{-(\delta +\mu )}}(0-1)={\frac {\mu }{\mu +\delta }}.$$

$\Box$

Remark.

• We can see that when $\delta =0$, then ${\bar {A}}_{x}=1$ for each nonzero $\mu$ ($\mu$ should not be zero, or else the life will never die (the survival function is always equal to 1)).
• This is intuitive, since for whole life insurance, the benefit must be paid eventually (since the life will die eventually), and if the interest rate is always zero, then the present value of the benefit is always one. Thus, its expected value is also one.

n-year pure endowment

The $n$-year pure endowment is similar to the $n$-year term life insurance in some sense, but the death benefit in the $n$-year term life insurnace is replaced by the survival benefit (i.e. benefit is paid when the insured survives for, but not dies within, $n$ years) in the $n$-year pure endowment.

Definition. ($n$-year pure endowment) The $n$-year pure endowment provides a payment at the end of the $n$ years when the insured survives for $n$ years from the time of policy issue, and does not provide any payment otherwise.

For $n$-year pure endowment, we have $b_{T}={\begin{cases}0,&T\leq n;\\1,&T>n\end{cases}}$, $v_{T}=v^{n},\quad T\geq 0$ . Hence, $Z={\begin{cases}0,&T\leq n;\\v^{n},&T>n.\end{cases}}$. The APV, denoted by $A_{x:{\overline {n}}|}^{\;\;1}$, is $\int _{n}^{\infty }v^{n}f_{x}(t)\,dt=v^{n}\int _{n}^{\infty }f_{x}(t)\,dt=v^{n}\mathbb {P} (T_{x}>n)=v^{n}{}_{n}p_{x}$.

Since this notation may look a bit clumsy, there is an alternative notation: $_{n}E_{x}$.

In the notation, there is no ${\bar {}}$ on top of $A$, since the benefit is paid at a fixed timing, so the APV is "not quite in a continuous manner". Also, the "$1$" is placed on top of "$:{\overline {n}}|$", since the benefit (if exists) is paid at the end of $n$th year (or time $n$).

n-year endowment insurance

The $n$-year endowment insurance is not purely endowment. Instead, it is a mixture between $n$-year term life insurance and $n$-year endowment. That is, both death and survival benefits exist. Because of this, this type of insurance is similar to to whole life insurance, in the sense that benefit must be paid.

Definition. ($n$-year endowment insurance) A $n$-year endowment insurance provides an payment either following the death of the insured when the insured dies within $n$ years, or at the end of $n$ years when the insured survives for $n$ years.

For $n$-year endowment insurance, we have $b_{T}=1,\quad T\geq 0$, $v_{T}={\begin{cases}v^{T},&T\leq n;\\v^{n},&T>n\end{cases}}$, and thus $Z={\begin{cases}v^{T},&T\leq n;\\v^{n},&T>n.\end{cases}}$ We can observe that $Z$ in this case is the sum of $Z$ for $n$-year term life insurance and $Z$ for $n$-year pure endowment. It follows that the APV is also the sum of the two corresponding APV's, that is, the APV for $n$-year endowment insurance is ${\bar {A}}_{x:{\overline {n}}|}^{1}+A_{x:{\overline {n}}|}^{\;\;1}$. Such APV is denoted by ${\bar {A}}_{x:{\overline {n}}|}$.

In the notation, we can see that there is a ${\bar {}}$ on top of $A$, since the benefit may be paid at the moment of the death of the insured, so the APV is somehow "in a continuous manner". Also, the "$1$" is omitted, with the same reason for whole life insurance: the benefit must be paid.

m-year deferred whole life insurance

As suggested by its name, this type of insurance is a deferred whole life insurance, i.e. the "start" of the insurance takes place in some years after the policy issue. To be more precise, we have the following definition.

Definition. ($m$-year deferred whole life insurance) A $m$-year deferred whole life insurance provides a payment when the insured dies at least $m$ years following policy issue.

For $m$-year deferred insurance, we have $b_{T}={\begin{cases}0,&T\leq m;\\1,&T>m\end{cases}}$, $v_{T}=v^{T},\quad T\geq 0$ , and thus $Z={\begin{cases}0,&T\leq m;\\v^{T},&T>m.\end{cases}}$

The APV, denoted by $_{m|}{\bar {A}}_{x}$, is $\int _{m}^{\infty }v^{t}{}_{t}p_{x}\mu _{x+t}\,dt$, which is similar to the one for whole life insurance, except that the lower bound of the integral is replaced by $m$. The "$m|$" in the notation refers to deferring $m$ years, and since the insurance does not have a term, "$:{\overline {n}}|$" is omitted.

m-year deferred n-year term life insurance

In a similar manner, we can also defer a $n$-year term life insurance.

Definition. ($m$-year deferred $n$-year term life insurance) A $m$-year deferred $n$-year term life insurance provides a payment when the insured survives for at least $m$ years following policy issue, and then dies within the coming $n$ years. The insurance does not provide any payment otherwise.

Remark.

• That is, the insurance provides payment when the insured dies between $m$ years and $m+n$ years following policy issue, and does not provide payment otherwise.

For such insurance, we have $b_{T}={\begin{cases}0,&T\leq m;\\1,&m<T\leq m+n;\\0,&T>m+n\end{cases}}$, $v_{T}=v^{T},\quad T\geq 0$ . Thus, $Z={\begin{cases}0,&T\leq m;\\v^{T},&m<T\leq m+n;\\0,&T>m+n.\end{cases}}$

The APV, denoted by $_{m|n}{\bar {A}}_{x}$, is $\int _{m}^{m+n}v^{t}{}_{t}p_{x}\mu _{x+t}\,dt$, which is similar to the one for the $n$-year term life insurance, except that both the lower and upper bound is added by $m$. In the notation, $m|n$ means that $m$ years are deferred and the term is $n$ years after the deferral, i.e. the insurance lasts for $n$ years after deferred by $m$ years. This has the similar meaning as the "$m|n$" in "$_{m|n}q_{x}$". Indeed, this is how "$m|n$" means in actuarial notations.

Annually increasing whole life insurance

Starting from this section, we will discuss the life insurances for which the benefit amount is varying.

Definition. (Annually increasing whole life insurance) An annually increasing whole life insurance provides payment of 1, 2, ... at the moment of death during the first, second year, etc., respectively.

For the annually increasing whole life insurance, we have $b_{T}=\lfloor T\rfloor +1,\quad T\geq 0$ , $v_{T}=v^{T},\quad T\geq 0$, and thus $Z=(\lfloor T\rfloor +1)v^{T},\quad T\geq 0$. Graphically, the value of benefit function $b_{T}$ is illustrated below:

b_T  =       1    2     3
            /\    /\    /\
           /  \  /  \  /  \
          /    \/    \/    \
        --*-----*-----*-----*-----
          0     1     2     3  ...

The APV, denoted by $(I{\bar {A}})_{x}$, is $\int _{0}^{\infty }(\lfloor t\rfloor +1)v^{t}{}_{t}p_{x}\mu _{x+t}\,dt$. In practice, since it is difficult to integrate "$\lfloor t\rfloor$" directly, we will split the integration interval to $[0,1),[1,2),\dotsc$, and multiple integrals are created from this, so that in each of these intervals, $\lfloor t\rfloor$ equals an integer only. That is, we split the integral like this: $\int _{0}^{\infty }(\lfloor t\rfloor +1)v^{t}{}_{t}p_{x}\mu _{x+t}\,dt=\int _{0}^{1}v^{t}{}_{t}p_{x}\mu _{x+t}\,dt+\int _{1}^{2}2v^{t}{}_{t}p_{x}\mu _{x+t}\,dt+\int _{2}^{3}3v^{t}{}_{t}p_{x}\mu _{x+t}\,dt+\dotsb$, which is an infinite sum.

In the notation, the "$I$" stands for "increasing" (annually), and we add a bracket so that the notation is not to be confused with $I\cdot {\bar {A}}_{x}$.

Annually increasing n-year term life insurance

When we "strip off" the benefit starting from year $n+1$ and onward, we obtain an annually increasing $n$-year insurance. To be more precise, we have the following definition:

Definition. (Annually increasing $n$-year term life insurance) An annually increasing $n$-year insurance provides payment of 1, 2, ..., $n$ at the moment of death during the first, second, ..., $n$th year respectively.

For this insurance, we have a slightly different benefit function:

• $b_{T}={\begin{cases}\lfloor T\rfloor +1,&T\leq n;\\0,&T>n\end{cases}}$ and
• $v_{T}=v^{T},\quad t\geq 0$.

Thus, $Z={\begin{cases}(\lfloor T\rfloor +1)v^{T},&T\leq n;\\0,&T>n.\end{cases}}$

Graphically, the benefit function looks like:

b_T  =       1    2     3    ...       n
            /\    /\    /\            / \
           /  \  /  \  /  \          /   \
          /    \/    \/    \        /     \
        --*-----*-----*-----*-------*------*-----
          0     1     2     3  ... n-1     n

The APV, denoted by $(I{\bar {A}})_{x:{\overline {n}}|}^{1}=\int _{0}^{n}(\lfloor t\rfloor +1)v^{t}{}_{t}p_{x}\mu _{x+t}\,dt$. We have "$x:{\overline {n}}|$" instead of $x$ in the notation, since this is a term life insurance.

Annually decreasing n-year term life insurance

Similarly, we have annually decreasing $n$-year insurance, but the benefit does not "start" at 1.

Definition. (Annually decreasing $n$-year term life insurance) An annually decreasing $n$-year term life insurance provides payment of $n$, $n-1$, ... , 1 at the moment of death during the first, second year, ..., $n$th year.

The benefit function is $b_{T}={\begin{cases}n-\lfloor T\rfloor ,&T\leq n;\\0,&T>n.\end{cases}}$, the discount function is $v_{T}=v^{T},\quad T\geq 0$. Thus, $Z={\begin{cases}(n-\lfloor T\rfloor )v^{T},&T\leq n;\\0,&T>n.\end{cases}}$ Graphically, the benefit function looks like:

b_T  =       n    n-1   n-2  ...       1
            /\    /\    /\            / \
           /  \  /  \  /  \          /   \
          /    \/    \/    \        /     \
        --*-----*-----*-----*-------*------*-----
          0     1     2     3  ... n-1     n

The APV, denoted by $(D{\bar {A}})_{x:{\overline {n}}|}^{1}$, equals $\int _{0}^{n}(n-\lfloor t\rfloor )v^{t}{}_{t}p_{x}\mu _{x+t}\,dt$. We have "$D$" instead of "$I$" in the notation to reflect that the insurance is decreasing rather than increasing.

m-thly increasing whole life insurance

This is a "more frequent" version of the annually increasing whole life insurance.

Definition. ($m$-thly increasing whole life insurance) A $m$-thly increasing whole life insurance provides payment of $1/m$, $2/m$, ... at the moment of death during the first $1/m$ of year, second $1/m$ of year, etc., respectively.

The benefit function is $b_{T}={\frac {\lfloor mT\rfloor +1}{m}},\quad T\geq 0$, the discount function is $v_{T}=v^{T},\quad T\geq 0$, and thus $Z={\frac {\lfloor mT\rfloor +1}{m}}v^{T},\quad T\geq 0$. Graphically, the benefit function looks like:

b_T  =     1/m   2/m   3/m
            /\    /\    /\
           /  \  /  \  /  \
          /    \/    \/    \
        --*-----*-----*-----*-----
          0    1/m   2/m   3/m ...

The APV, denoted by $(I^{(m)}{\bar {A}})_{x}$, is $\int _{0}^{\infty }{\frac {\lfloor mt\rfloor +1}{m}}v^{t}{}_{t}p_{x}\mu _{x+t}\,dt$. The "$I^{(m)}$" in the notation reflects that the insurance is increasing $m$-thly.

Continuously increasing whole life insurance

We can further increasing the frequency and obtain a "continuous" version of the annually increasing whole life insurance. This version is a bit theoretical, and there may not be such kind of insurance in real life. However, the calculation is simpler using this continuous model, as we will see.

Definition. (Continuously increasing whole life insurance) A continuously increasing whole life insurance provides a payment of $t$ when the time of death is $t$.

• The benefit function is $b_{T}=T,\quad T\geq 0$, and
• the discount function is $v_{T}=v^{T},\quad T\geq 0$, and
• thus, $Z=Tv^{T},\quad T\geq 0$.

Its APV, denoted by $({\bar {I}}{\bar {A}})_{x}$, is $\int _{0}^{\infty }tv^{t}{}_{t}p_{x}\mu _{x+t}\,dt$. There is a "${\bar {}}$" in the notation, since the increase is continuous.

Example. (Alternative formula for APV of continuously increasing whole life insurance) Show that $({\bar {I}}{\bar {A}})_{x}=\int _{0}^{\infty }{}_{s|}{\bar {A}}_{x}\,ds$.

Proof.

$${\begin{aligned}({\bar {I}}{\bar {A}})_{x}&=\int _{0}^{\infty }{\color {darkgreen}t}v^{t}{}_{t}p_{x}\mu _{x+t}\,dt\\&=\int _{0}^{\infty }v^{t}{}_{t}p_{x}\mu _{x+t}{\color {darkgreen}\int _{0}^{t}\,ds}\,dt\\&=\int _{0}^{\infty }\int _{0}^{t}v^{t}{}_{t}p_{x}\mu _{x+t}\,ds\,dt&{\text{since }}v^{t}{}_{t}p_{x}\mu _{x+t}{\text{ is independent from }}s\\&=\int _{0}^{\infty }{\color {blue}\int _{s}^{\infty }v^{t}{}_{t}p_{x}\mu _{x+t}\,dt}\,ds&{\text{by changing the expression of the integration region}}\\&=\int _{0}^{\infty }{}_{s|}{\bar {A}}_{x}\,ds.\\\end{aligned}}$$

$\Box$

m-thly and continuously increasing n-year term life insurance

We can modify slightly $m$-thly/continuously increasing whole life insurance to $m$thly/continuously increasing $n$-year term life insurance, which is similar to the previous case for modifying annually increasing whole life insurance to annually increasing $n$-year term life insurance.

Despite their definitions are quite similar to the whole life insurance counterparts, we still include their definition as follows for completeness.

Definition. ($m$-thly increasing $n$-year term life insurance) A $m$-thly increasing $n$-year term life insurance provides payment of $1/m$, $2/m$, ... at the moment of death during the first $1/m$ of year, second $1/m$ of year, ... , $mn$th $1/m$ of year respectively.

Definition. (Continuously increasing $n$-year term life insurance) A continuously increasing $n$-year term life insurance provides a payment of $t$ when the time of death is $t\leq n$.

For $m$-thly increasing $n$-year term life insurance, the benefit function is $b_{T}={\begin{cases}{\frac {\lfloor mT\rfloor +1}{m}},&T\leq n;\\0,&T>n\end{cases}}$, the discount function is $v_{T}=v^{T},\quad T\geq 0$, and thus $Z={\begin{cases}{\frac {\lfloor mT\rfloor +1}{m}}v^{T},&T\leq n;\\0,&T>n.\end{cases}}$ Hence, the APV, denoted by $(I^{(m)}{\bar {A}})_{x:{\overline {n}}|}$, is $\int _{0}^{n}{\frac {\lfloor mt\rfloor +1}{m}}v^{t}{}_{t}p_{x}\mu _{x+t}\,dt$ ("$:{\overline {n}}|$" is added to the notation to represent the $n$-year term).

For continuously increasing $n$-year whole life insurance, the benefit function is $b_{T}={\begin{cases}T,&T\leq n;\\0,&T>n\\\end{cases}}$, the discount function is $v_{T}=v^{T},\quad T\geq 0$. Thus, $Z={\begin{cases}Tv^{T},&T\leq n;\\0,&T>n.\end{cases}}$ Hence, the APV, denoted by $({\bar {I}}{\bar {A}})_{x:{\overline {n}}|}$, is $\int _{0}^{n}tv^{t}{}_{t}p_{x}\mu _{x+t}\,dt.$

Of course, we can also have $m$-thly/continuously decreasing whole/$n$-year term life insurance, and their APV notations are constructed in a similar manner. But, to avoid being repetitive, these insurances are omitted here.

Rule of moments

We have discussed how to calculate the actuarial present value, which is $\mathbb {E} [Z]$. How about the higher moments: $\mathbb {E} [Z^{2}],\mathbb {E} [Z^{3}],\dotsc$? We will provide a convenient way (for some type of insurance) below to calculate those moments by using the actuarial present value ($\mathbb {E} [Z]$), namely rule of moments.

Theorem. (Rule of moments) For insurances with one payment with unit amount (i.e. payment of 1), then $\mathbb {E} [Z^{j}]$ when the force of interest is $\delta _{t}$ (a function of $t$) equals $\mathbb {E} [Z]$ when the force of interest is $j\delta _{t}$ ($j=2,3,\dotsc$) .

Proof. We assume the insurance concerned is a whole life insurance of benefit 1. For other types of insurance (that satisfy the conditions, i.e. the benefit is a single payment of 1), we can just change the integration region below suitably, and we will get the same result.

For whole life insurance, with force of interest $\delta _{t}$, we have

$$\mathbb {E} [Z^{j}]=\mathbb {E} [v^{jT}]=\int _{0}^{\infty }{\big (}v^{T}{\big )}^{j}{}_{t}p_{x}\mu _{x+t}\,dt=\int _{0}^{\infty }\left(e^{-\int _{0}^{t}\delta _{s}\,ds}\right)^{\color {darkgreen}j}{}_{t}p_{x}\mu _{x+t}\,dt=\int _{0}^{\infty }e^{-{\color {darkgreen}j}\int _{0}^{t}\delta _{s}\,ds}{}_{t}p_{x}\mu _{x+t}\,dt,=\int _{0}^{\infty }e^{-\int _{0}^{t}{\color {darkgreen}j}\delta _{s}\,ds}{}_{t}p_{x}\mu _{x+t}\,dt,$$

which is $\mathbb {E} [Z]$ when the force of interest is $j\delta _{t}$.

$\Box$

Remark.

• When the benefit payment is not 1, we cannot apply this rule directly. We can see from the proof that the benefit payment needs to satisfy $b_{T}^{j}=b_{T}$ for this rule to hold. In this case, we have $Z^{j}=(b_{T}v_{T})^{j}=b_{T}v_{T}^{j}$. When $b_{T}$ is nonzero (benefit payment should not be zero for an insurance), when $b_{T}=1$, this is satisfied, and for other (positive) values of $b_{T}$, this is not satisfied.
• For values of benefit payment other than one, we need to perform some procedure so that we can still use the rule of moments, as follows:

• When the benefit payment is $b_{T}$ ($\neq 1$), we have $\mathbb {E} [Z^{j}]=\mathbb {E} [(b_{T}v_{T})^{j}]=b_{T}^{j}\mathbb {E} [v_{T}^{j}]$.
• After that, the random variable inside the expectation is $Z$ with payment 1, and we can use the rule of moments.

Because of this result, we introduce some special notation for the higher moment: we add a "$j$" in the upper left corner of the notation when we are discussing the $j$th moment of $Z$, rather than actuarial present value of the concerning insurance. For example, we use $^{2}{\bar {A}}_{x}$ to denote $\mathbb {E} [Z^{2}]$ in which $Z$ is the present-value random variable for the whole life insurance .

Example. Recall that ${\bar {A}}_{x}={\frac {\mu }{\mu +\delta }}$ under the constant force of interest $\delta$ and constant force of mortality $\mu$. By rule of moments, we have $^{2}{\bar {A}}_{x}={\frac {\mu }{\mu +2\delta }}$.

Exercise.

Let $Z$ be the present-value random variable corresponding to the above actuarial present value. What is $\operatorname {Var} (Z)$?

	${\frac {\mu }{\mu +2\delta }}$
	${\frac {\mu ^{2}}{\mu +2\delta }}$
	${\frac {\mu (\mu +\delta )^{2}-\mu ^{2}(\mu +2\delta )}{(\mu +2\delta )(\mu +\delta )^{2}}}$
	${\frac {\mu (\mu +\delta )^{2}+\mu ^{2}(\mu +2\delta )}{(\mu +2\delta )(\mu +\delta )^{2}}}$

Summary of insurance types

Summary

Insurance name	benefit function $b_{T}$	discount function $v_{T}$	present-value random variable $Z$	APV notation
Whole life insurance	$1,\quad T\geq 0$	$v^{T},\quad T\geq 0$	$v^{T},\quad T\geq 0$	${\bar {A}}_{x}$
$n$-year term life insurance	${\begin{cases}1,&T\leq n;\\0,&T>n\\\end{cases}}$	$v^{T},\quad T\geq 0$	${\begin{cases}v^{T},&T\leq n;\\0,&T>n\end{cases}}$	${\bar {A}}_{x:{\overline {n}}|}^{1}$
$n$-year pure endowment	${\begin{cases}0,&T\leq n;\\1,&T>n\\\end{cases}}$	$v^{n},\quad T\geq 0$	${\begin{cases}0,&T\leq n;\\v^{n},&T>n\end{cases}}$	${\bar {A}}_{x:{\overline {n}}|}^{\;\;1}$ or $_{n}E_{x}$
$n$-year endowment insurance	$1,\quad T\geq 0$	${\begin{cases}v^{T},&T\leq n;\\v^{n},&T>n\\\end{cases}}$	${\begin{cases}v^{T},&T\leq n;\\v^{n},&T>n\end{cases}}$	${\bar {A}}_{x:{\overline {n}}|}$
$m$-year deferred whole life insurance	${\begin{cases}0,&T\leq n;\\1,&T>n\\\end{cases}}$	$v^{T},\quad T\geq 0$	${\begin{cases}v^{T},&T\leq n;\\0,&T>n\end{cases}}$	$_{m|}{\bar {A}}_{x}$
$m$-year deferred $n$-year term life insurance	${\begin{cases}0,&T\leq m;\\1,&m<T\leq m+n;\\0,&T>m+n,\\\end{cases}}$	$v^{T},\quad T\geq 0$	${\begin{cases}0,&T\geq 0;\\v^{T},&m<T\leq m+n;\\0,&T>m+n\\\end{cases}}$	$_{m|n}{\bar {A}}_{x}$
Annually increasing whole life insurance	$\lfloor T\rfloor +1,\quad T\geq 0$	$v^{T},\quad T\geq 0$	$(\lfloor T\rfloor +1)v^{T},\quad T\geq 0$	$(I{\bar {A}})_{x}$
Annually increasing $n$-year term life insurance	${\begin{cases}\lfloor T\rfloor +1,&T\leq n\;\\0,&T>n\\\end{cases}}$	$v^{T},\quad T\geq 0$	${\begin{cases}(\lfloor T\rfloor +1)v^{T},&T\leq n\;\\0,&T>n\\\end{cases}}$	$(I{\bar {A}})_{x:{\overline {n}}|}^{1}$
Annually decreasing $n$-year term life insurance	${\begin{cases}n-\lfloor T\rfloor ,&T\leq n\;\\0,&T>n\\\end{cases}}$	$v^{T},\quad T\geq 0$	${\begin{cases}(n-\lfloor T\rfloor )v^{T},&T\leq n\;\\0,&T>n\\\end{cases}}$	$(D{\bar {A}})_{x:{\overline {n}}|}^{1}$
$m$-thly increasing whole life insurance	${\frac {\lfloor mT\rfloor +1}{m}},\quad T\geq 0$	$v^{T},\quad T\geq 0$	${\frac {\lfloor mT\rfloor +1}{m}}v^{T},\quad T\geq 0$	$(I^{(m)}{\bar {A}})_{x}$
$m$-thly increasing $n$-year term life insurance	${\begin{cases}{\frac {\lfloor mT\rfloor +1}{m}},&T\leq n;\\0,&T>n\end{cases}}$	$v^{T},\quad T\geq 0$	${\begin{cases}{\frac {\lfloor mT\rfloor +1}{m}}v^{T},&T\leq n;\\0,&T>n\end{cases}}$	$(I^{(m)}{\bar {A}})_{x:{\overline {n}}|}$
Continuously increasing whole life insurance	$T,\quad T\geq 0$	$v^{T},\quad T\geq 0$	$Tv^{T},\quad T\geq 0$	$({\bar {I}}{\bar {A}})_{x}$
Continuously increasing $n$-year term life insurance	${\begin{cases}T,&T\leq n;\\0,&T>n\end{cases}}$	$v^{T},\quad T\geq 0$	${\begin{cases}Tv^{T},&T\leq n;\\0,&T>n\end{cases}}$	$({\bar {I}}{\bar {A}})_{x:{\overline {n}}|}$

Insurance payable at the end of the year of death

In practice, it is uncommon that the benefit is paid at the moment of death. Instead, it is more common to pay the benefit at the end of the year of death (or may be in other time unit, e.g. at the end of month of death, etc.). In this case, we can just modify the benefit function and discount function in the previous section slightly, and then we can calculate the actuarial present value under this assumption similarly.

Under this case, the present-value random variable is defined using $K$ ($K=K_{x}$) instead of $T$. Since the payment is made at the end of the year of death, but $K$ gives the time at the beginning of the year of death. Because of this, the input of benefit function and discount function is $K+1$, rather than $K$. Then, we have the present-value random variable $Z=b_{K+1}v_{K+1}$.

For example, when a life aged 50 dies at the middle between age 53 and 54, then benefit is paid at age 54. Graphically,

                    death  benefit paid
                         | |
                         v v
---*-----*-----*-----*-----*----
  50    51    52     53    54

Basically, the different insurance models under this assumption are quite similar to the corresponding one under previous assumption, except that the present-value random variable $Z$ is now discrete rather than continuous (since $K$ is discrete), and thus to calculate the actuarial present value (expected value of $Z$) here, we need to use summation instead of integration.

For notations, the notations here are highly similar to the one under previous assumption, except that the ${\bar {}}$ on top of $A$ (if exists) is removed, since the APV is now "in a discrete manner". Again, we assume the amount of benefit is one unless otherwise specified.

Let us discuss the model for $n$-year term life insurance under this assumption. For other models, we can just build them from this model, in a similar way as in the previous section.

Before the discussion, we need to recall that the pmf $\mathbb {P} (K_{x}=k)={}_{k}p_{x}q_{x+k}$.

For $n$-year term life insurance,

• the benefit function is $b_{K+1}={\begin{cases}1,&K=0,1,\dotsc ,n-1;\\0,&K=n,n+1,\dotsc \end{cases}}$, and
• the discount function is $v_{K+1}=v^{K+1},\quad K=0,1,\dotsc$ .

Thus, the present-value random variable $Z={\begin{cases}v^{K+1},&K=0,1,\dotsc ,n-1;\\0,&K=n,n+1,\dotsc \end{cases}}$

As a result, the APV is $\mathbb {E} [Z]=\sum _{k=0}^{n-1}v^{k+1}\mathbb {P} (K_{x}=k)=\sum _{k=0}^{n-1}v^{k+1}{}_{k}p_{x}q_{x+k}$, and as we mentioned, its notation is basically the same as previous one, except that the ${\bar {}}$ is removed. That is, the notation is $A_{x:{\overline {n}}|}^{1}$.

Remark.

• You may notice that a notation for $n$-year pure endowment is $A_{x:{\overline {n}}|}^{\;\;1}$, in which there is no "${\bar {}}$" under previous assumption already. So, if we follow our convention here, then the notation is exactly the same under this discrete assumption.

• Thus, this may cause some ambiguity when only the notation is presented. Hence, in this case, we usually specify clearly which assumption is to be used, and then use the notation, which can help us to differentiate between these two APV's.

For other types of insurance, they are defined in a similar manner. Since you should be now quite familiar with those types of insurance, we simply summarize most of the definitions for different types of insurance (under the discrete assumption) in the following table:

Summary

Insurance name	benefit function $b_{K+1}$	discount function $v_{K+1}$	present-value random variable $Z$	APV notation
Whole life insurance	$1,\quad K=0,1,\dotsc$	$v^{K+1},\quad K=0,1,\dotsc$	$v^{K+1},\quad K=0,1,\dotsc$	$A_{x}$
$n$-year term life insurance	${\begin{cases}1,&K=0,1,\dotsc ,n-1;\\0,&K=n,n+1,\dotsc \\\end{cases}}$	$v^{K+1},\quad K=0,1,\dotsc$	${\begin{cases}v^{K+1},&K=0,1,\dotsc ,n-1;\\0,&K=n,n+1,\dotsc \end{cases}}$	$A_{x:{\overline {n}}|}^{1}$
$n$-year pure endowment	${\begin{cases}0,&K=0,1,\dotsc ,n-1;\\1,&K=n,n+1,\dotsc \\\end{cases}}$	$v^{n},\quad K=0,1,\dotsc$	${\begin{cases}0,&K=0,1,\dotsc ,n-1;\\v^{n},&K=n,n+1,\dotsc \end{cases}}$	$A_{x:{\overline {n}}|}^{\;\;1}$ or $_{n}E_{x}$ (the latter one is the same as the continuous one)
$n$-year endowment insurance	$1,\quad K=0,1,\dotsc$	${\begin{cases}v^{K+1},&K=0,1,\dotsc ,n-1;\\v^{n},&K=n,n+1,\dotsc \\\end{cases}}$	${\begin{cases}v^{K+1},&K=0,1,\dotsc ,n-1;\\v^{n},&K=n,n+1,\dotsc \end{cases}}$	$A_{x:{\overline {n}}|}$
$m$-year deferred whole life insurance	${\begin{cases}0,&K=0,1,\dotsc ,n-1;\\1,&K=n,n+1,\dotsc \\\end{cases}}$	$v^{K+1},\quad K=0,1,\dotsc$	${\begin{cases}v^{K+1},&K=0,1,\dotsc ,n-1;\\0,&K=n,n+1,\dotsc \end{cases}}$	$_{m|}A_{x}$
$m$-year deferred $n$-year term life insurance	${\begin{cases}0,&K=0,1,\dotsc ,m-1;\\1,&K=m,m+1,\dotsc ,m+n-1;\\0,&K=m+n,m+n+1,\dotsc \\\end{cases}}$	$v^{K+1},\quad K=0,1,\dotsc$	${\begin{cases}0,&K=0,1,\dotsc ,m-1;\\v^{K+1},&K=m,m+1,\dotsc ,m+n-1;\\0,&K=m+n,m+n+1,\dotsc \\\end{cases}}$	$_{m|n}A_{x}$
Annually increasing whole life insurance	$K+1,\quad K=0,1,\dotsc$	$v^{K+1},\quad K=0,1,\dotsc$	$(K+1)v^{K+1},\quad K=0,1,\dotsc$	$(IA)_{x}$
Annually increasing $n$-year term life insurance	${\begin{cases}K+1,&K=0,1,\dotsc ,n-1\;\\0,&K=n,n+1,\dotsc \\\end{cases}}$	$v^{K+1},\quad K=0,1,\dotsc$	${\begin{cases}(K+1)v^{K+1},&K=0,1,\dotsc ,n-1\;\\0,&K=n,n+1,\dotsc \\\end{cases}}$	$(IA)_{x:{\overline {n}}|}^{1}$
Annually decreasing $n$-year term life insurance	${\begin{cases}n-K,&K=0,1,\dotsc ,n-1\;\\0,&K=n,n+1,\dotsc \\\end{cases}}$	$v^{K+1},\quad K=0,1,\dotsc$	${\begin{cases}(n-K)v^{K+1},&K=0,1,\dotsc ,n-1\;\\0,&K=n,n+1,\dotsc \\\end{cases}}$	$(DA)_{x:{\overline {n}}|}^{1}$

We should notice that there is no continuously increasing whole life/ $n$-year term life insurance under this discrete assumption, since the payment is made discretely, and hence the payment cannot be continuously increasing.

For the $m$-thly increasing $n$-year term/whole life insurance, it is quite different from the previous deviation for the continuous one. Let us illustrate the deviation for $m$-thly increasing whole life insurance as follows. For the $m$-thly increasing $n$-year term life insurance, the deviation can be done in a similar manner.

Before discussing $m$-thly increasing whole life insurance, we should discuss similar insurances where payment is made $m$-thly, and the payment is not varying first.

In financial mathematics, you may have encountered similar situations: annuities where payments are made $m$-thly, instead of annually. In that case, we can simply calculate the equivalent $m$-thly interest rate corresponding to the annual interest rate, and then treat such annuities as ordinary one (i.e. annuities payable annually), so that the calculation process is the same as the ordinary annuities.

However, in the current context, we also incorporate the survival probabilities in the calculation, and we define the time-until-death random variable using years as units. Thus, $K$ is in the unit of years, and we cannot convert $K$ to some other "equivalent $m$-thly $K$" using simple ways. As a result, we need to develop some methods and formulas for the insurance payable $m$-thly.

Remark.

• We should notice that for the insurances where benefit is paid at the moment of death, there is no such "insurance payable $m$-thly", since those insurances are by assumption payable continuously already.

Consider the following diagram:

          death benefit
              |  |
              v  v
--*--...--*------*-----...-----*-----...-----*--
  x  ... x+k  x+k+1/m  ...  x+k+j/m  ...   x+k+1

In this type of insurance, the benefit is paid at the end of the $m$thly interval where death occurs, instead of at the end of the year where death occurs.

We can observe that when death occurs in any $m$-thly interval within one year, say from year $k$ to $k+1$, the value of $K$ is the same, which is $k$. Because of this, to perform calculations related to this kind of insurance, we need to introduce another random variable, say $J$, to consider which interval within one year does death occur.

Recall that $K$ represents the number of complete years lived before death. Hence, it is natural to define $J$ in a similar manner, namely the number of complete $m$-th of a year lived in the year of death, before death. Using this definition of $J$, we know that $J$ can only take values of $0,1,\dotsc ,m-1$.

Exercise. You may notice that $J$ cannot take the value of $m$. Why?

Solution

When $J=m$, it means that the person lives for $m$ "$m$-th of a year" in his year of death, i.e. he lives for the whole year in his year of death. This means that he does not even die in his "year of death"! Hence, this cannot be the case.

Remark.

• Suppose the person lives for the whole year $k$, and then die in the following $m$-thly interval in year $k+1$. Then, in this case, we do not have $K=k$ and $J=m$. Instead, we have $K=k+1$ and $J=0$.

Using these definitions, we can define, for example, the benefit, discount function, present-value random variable of whole life insurance payable $m$-thly as follows:

• the benefit function is $1,\quad K=0,1,\dotsc ,{\text{ and }}J=0,1,\dotsc ,m-1$;
• the discount function is $v^{K+{\frac {J+1}{m}}},\quad K=0,1,\dotsc {\text{ and }}J=0,1,\dotsc ,m-1$;
• the present-value random variable is $v^{K+{\frac {J+1}{m}}},\quad K=0,1,\dotsc ,{\text{ and }}J=0,1,\dotsc ,m-1$.

Since there are two random variables involved, the calculation of APV is more complicated. We first consider the joint pmf of $K$ and $J$: since $\mathbb {P} (K=k{\text{ and }}J=j)=\mathbb {P} \left({\text{the person dies between time }}k+{\frac {j}{m}}{\text{ and }}k+{\frac {j+1}{m}}\right)=\mathbb {P} \left(k+{\frac {j}{m}}\leq T\leq k+{\frac {j+1}{m}}\right)={}_{k+{\frac {j}{m}}}p_{x}\;_{\frac {1}{m}}q_{x+k+{\frac {j}{m}}}$, the joint pmf of $K$ and $J$ is $f_{K,J}(k,j)={}_{k+{\frac {j}{m}}}p_{x}\;_{\frac {1}{m}}q_{x+k+{\frac {j}{m}}}$. It follows by the definition of expectation that the APV is

$$A_{x}^{(m)}=\sum _{k=0}^{\infty }\sum _{j=0}^{m-1}v^{k+{\frac {j+1}{m}}}{}_{k+{\frac {j}{m}}}p_{x}\;_{\frac {1}{m}}q_{x+k+{\frac {j}{m}}}.$$

Notice that the APV notation has an extra "$(m)$" at the upper-right corner of "$A$" to represent that this insurance is payable $m$-thly.

Exercise. (Another expression of $A_{x}^{(m)}$) Show that $A_{x}^{(m)}=\sum _{k=0}^{\infty }v^{k}{}_{k}p_{x}\left(\sum _{j=0}^{m-1}v^{\frac {j+1}{m}}{}_{\frac {1}{m}}q_{x+k+{\frac {j}{m}}}\;_{\frac {j}{m}}p_{x+k}\right).$

Solution

Proof.

$$A_{x}^{(m)}=\sum _{k=0}^{\infty }\sum _{j=0}^{m-1}{\color {blue}v^{k+{\frac {j+1}{m}}}}{\color {red}{}_{k+{\frac {j}{m}}}p_{x}}\;_{\frac {1}{m}}q_{x+k+{\frac {j}{m}}}=\sum _{k=0}^{\infty }\sum _{j=0}^{m-1}{\color {blue}v^{k}v^{\frac {j+1}{m}}}{\color {red}{}_{k}p_{x}{}_{\frac {j}{m}}p_{x+k}}\;_{\frac {1}{m}}q_{x+k+{\frac {j}{m}}}=\sum _{k=0}^{\infty }\underbrace {{\color {blue}v^{k}}{\color {red}{}_{k}p_{x}}} _{{\text{independent from }}j}\left(\sum _{j=0}^{m-1}{\color {blue}v^{\frac {j+1}{m}}}{\color {red}{}_{\frac {j}{m}}p_{x+k}}\;_{\frac {1}{m}}q_{x+k+{\frac {j}{m}}}\right)$$

$\Box$

Remark.

• Intuitively, we can interpret this formula as follows:

• For each time $k$, only considering the upcoming year, i.e. from time $k$ to $k+1$, the APV at time $k$ is $\sum _{j=0}^{m-1}v^{\frac {j+1}{m}}{}_{\frac {1}{m}}q_{x+k+{\frac {j}{m}}}\;_{\frac {j}{m}}p_{x+k}$
• Graphically,

v^{(j+1)/m}
     <-----------------*
                 death |
                    |  | benefit
                    v  |    
-----*----...----*-----*--------...---------*
     k    ...  k+j/m  k+(j+1)/m       k+1
     ------------>  q
      _{j/m} p _{x+k}

• To consider the "whole" APV, we need to sum this term over all $k$, discount these APV's to time 0, and also consider the probabilities for surviving to time $k$. As a result, the APV is $\sum _{k=0}^{\infty }v^{k}{}_{k}p_{x}\left(\sum _{j=0}^{m-1}v^{\frac {j+1}{m}}{}_{\frac {1}{m}}q_{x+k+{\frac {j}{m}}}\;_{\frac {j}{m}}p_{x+k}\right)$.

After understanding the construction of this insurance, we can construct other types of insurance that are also payable $m$-thly similarly. Also, their APV notations also have an extra "$(m)$" added at the upper-right corner of "$A$".

Besides, we can apply this "$m$-thly concept" to the frequency of increasing/decreasing payment. To illustrate this, let us consider the $m$-thly increasing $n$-year term life insurance. The benefit function is $K+{\frac {J+1}{m}},\quad K=0,1,\dotsc ,n-1,{\text{ and }}J=0,1,\dotsc ,m-1$, the discount function is $v^{K+1},\quad K=0,1,\dotsc$, and thus the present-value random variable is $\left(K+{\frac {J+1}{m}}\right)v^{K+1},\quad K=0,1,\dotsc ,n-1,{\text{ and }}J=0,1,\dotsc ,m-1$. It follows that the APV is

$$(I^{(m)}A)_{x:{\overline {n}}|}=\sum _{k=0}^{n-1}\sum _{j=0}^{m-1}\left(k+{\frac {j+1}{m}}\right)v^{k+1}{}_{k+{\frac {j}{m}}}p_{x}\;_{\frac {1}{m}}q_{x+k+{\frac {j}{m}}},$$

where there is an additional "$(m)$" at the upper-right corner of "$I$" to represent that the increase is $m$-thly.

Example. (Alternative formula of $(DA)_{x:{\overline {n}}|}^{1}$) From the above table, we can deduce that

$$(DA)_{x:{\overline {n}}|}^{1}=\sum _{k=0}^{n-1}(n-k)v^{k+1}{}_{k}p_{x}q_{x+k}.$$

We can also express $(DA)_{x:{\overline {n}}|}^{1}$ as:

$$(DA)_{x:{\overline {n}}|}^{1}=\sum _{j=0}^{n-1}A_{x:{\overline {n-j}}|}^{1}.$$

An intuitive explanation for this expression is illustrated in the following diagrams:

              benefit amount at different time point
n          1
n-1        1     1
.          .     .  .
.          .     .     .
.          .     .        .
2          1     1           1
1          1     1           1     1
      *----*-----*----...----*-----*--
      0    1     2    ...   n-1    n
"sum up" (considering discounting and survival probabilities) each "vertical bar" gives APV

n          1                            A x:1
n-1        1     1                      A x:2
.          .     .  .                     .
.          .     .     .                  .
.          .     .        .               .
2          1     1           1         A x:n-1
1          1     1           1     1   A x:n
      *----*-----*----...----*-----*--
      0    1     2    ...   n-1    n
sum up all A (which consider discounting and survival probabilities already) ("horizontal bar") gives APV.

Exercise. Show that

$$(DA)_{x:{\overline {n}}|}^{1}=\sum _{j=0}^{n-1}A_{x:{\overline {n-j}}|}^{1}$$

formally and mathematically.

Solution

Proof.

$${\begin{aligned}(DA)_{x:{\overline {n}}|}^{1}&=\sum _{k=0}^{n-1}{\color {darkgreen}(n-k)}v^{k+1}{}_{k}p_{x}q_{x+k}\\&=\sum _{k=0}^{n-1}{\color {darkgreen}\left(\sum _{j=0}^{n-k-1}1\right)}v^{k+1}{}_{k}p_{x}q_{x+k}&({\text{trick}})\\&=\sum _{j=0}^{n-1}\sum _{k=0}^{n-j-1}v^{k+1}{}_{k}p_{x}q_{x+k}&({\text{changing summation order }}{\color {darkgreen}{\textit {suitably}}})\\&=\sum _{j=0}^{n-1}A_{x:{\overline {n-j}}|}^{1}&({\text{definition}}).\end{aligned}}$$

$\Box$

Recursion relations of APV's

For the insurances payable at the end of the year of death, we can develop recursion relations for them. These recursion relations can be useful when we need to calculate some APV's, given some other APV's and related terms only. Also, these recursion relations can give some insights about the relationship between different APV's in some sense. The following proposition includes some recursion relations, where the their deviations contain similar idea.

Proposition. (Recursion relations of APV's)

1. $A_{x}=vq_{x}+vp_{x}A_{x+1}$.
2. $A_{x:{\overline {n}}|}^{1}=vq_{x}+vp_{x}A_{x+1:{\overline {n-1}}|}^{\;\;1}$.
3. $A_{x:{\overline {n}}|}=vq_{x}+vp_{x}A_{x+1:{\overline {n-1}}|}$ (a simple corollary from 2.).
4. $_{m|n}A_{x}=vp_{x}{}_{m-1|n}A_{x+1}$.
5. $(IA)_{x:{\overline {n}}|}^{1}=A_{x:{\overline {n}}|}^{\;\;1}+vp_{x}(IA)_{x+1:{\overline {n-1}}|}^{\;\;1}$.
6. $(DA)_{x:{\overline {n}}|}^{1}=nvq_{x}+vp_{x}(DA)_{x+1:{\overline {n-1}}|}^{\;\;1}$.
7. $(IA)_{x}=A_{x}+vp_{x}(IA)_{x+1}$.

Proof. We will only prove 1. and 5., and the remaining relations will be explained in an intuitive manner after this proof.

1.:

$${\begin{aligned}A_{x}&=\sum _{k=0}^{\infty }v^{k+1}{}_{k}p_{x}q_{x+k}\\&=v\cdot \underbrace {{}_{0}p_{x}} _{=1}\cdot q_{x}+\sum _{k=1}^{\infty }v^{k+1}{}_{k}p_{x}q_{x+k}&({\text{split the `}}k=0{\text{' term out, and }}{}_{0}p_{x}=\mathbb {P} (T_{x}\geq 0)=1)\\&=vq_{x}+\sum _{k'=0}^{\infty }{\color {blue}v^{k'+2}}{\color {red}{}_{k'+1}p_{x}}q_{x+1+k'}&(k'=k-1)\\&=vq_{x}+{\color {blue}v}{\color {red}p_{x}}\sum _{k'=0}^{\infty }{\color {blue}v^{k'+1}}{\color {red}{}_{k'}p_{x+1}}q_{x+1+k'}\\&=vq_{x}+vp_{x}A_{x+1}&({\text{definition}}).\end{aligned}}$$

5.:

$${\begin{aligned}(IA)_{x:{\overline {n}}|}^{1}&=\sum _{k=0}^{n-1}(k+1)v^{k+1}{}_{k}p_{x}q_{x+k}\\&=\sum _{k=0}^{n-1}v^{k+1}{}_{k}p_{x}q_{x+k}+\sum _{k=0}^{n-1}k{\color {blue}v^{k+1}}{\color {red}{}_{k}p_{x}}q_{x+k}&({\text{split the sum}})\\&=A_{x:{\overline {n}}|}^{1}+{\color {blue}v}{\color {red}p_{x}}\sum _{k=0}^{n-1}k{\color {blue}v^{k}}{\color {red}{}_{k-1}p_{x+1}}q_{x+k}\\&=A_{x:{\overline {n}}|}^{1}+vp_{x}\sum _{k=1}^{n-1}kv^{k}{}_{k-1}p_{x+1}q_{x+k}&({\text{`}}k=0{\text{' term is zero}})\\&=A_{x:{\overline {n}}|}^{1}+vp_{x}\sum _{k'=0}^{n-2}(k'+1)v^{k'+1}{}_{k'}p_{x+1}q_{x+k}&(k'=k-1)\\&=A_{x:{\overline {n}}|}^{1}+vp_{x}(IA)_{x+1:{\overline {n-1}}|}^{\;\;1}&({\text{definition}}).\end{aligned}}$$

$\Box$

Actuarial discounting

One of the key ideas in the intuitive explanations to these recursive relations is the concept of actuarial discounting. To understand this, let us consider the intuitive explanation to the above recursion relation 2. Graphically, the idea in the recursion looks like

"actuarial discounting"
   *---
   |   \
   v    \
         |----------------|      A x+1:n-1
   |-----|                       A x:1= vq_x
   |----------------------|      A x:n
---*-----*----...----*----*---
   x    x+1   ...  x+n-1  x+n   age

We "split" the $n$-year term life insurance issued to a person aged $x$ into two parts:

• 1-year term life insurance issued to a person aged $x$, and
• $n-1$-year term life insurance issued to a person aged $x+1$,

as illustrated in the above graph. The corresponding APV's to these two insurances are $A_{x:{\overline {1}}|}^{1}$ and $A_{x+1:{\overline {n-1}}|}^{\;\;1}$ respectively.

Of course, we cannot simply regard the APV of $n$-year term life insurance issued to the life aged $x$ as $A_{x:{\overline {1}}|}^{1}+A_{x+1:{\overline {n-1}}|}^{\;\;1}$ directly, since $A_{x+1:{\overline {n-1}}|}^{\;\;1}$ is not for a life aged $x$. Instead, it is for a life aged $x+1$. Hence, adjustment needs to be made on this insurance, and the process of doing this adjustment is called "actuarial discounting".

In financial mathematics, discounting means multiplying the discounting factor, where the effect of interest is considered. On the other hand, in this context, apart from the interest effect, we also have "survival effect", that is, we also need to consider the survival probabilities of a life.

To be more precise, "$n-1$-year term life insurance issued to a person aged $x+1$" only takes into effect given that the person aged $x$ actually lives to age $x+1$. Otherwise, if the person dies within the first year, there is not any "life aged $x+1$". Thus, apart from multiplying the discounting factor, we also need to multiply the "survival factor".

In this case, the discounting factor is $v$ since 1 year is discounted back, and the "survival factor" is $p_{x}$ since the person aged $x$ is required to live for 1 year for the $n-1$-year term life insurance to take into effect, and we need to multiply this probability to ensure that this happens (the multiplication of probability is related to the "conditional" concept in probability). Since we need to multiply both $v$ and $p_{x}$ in this case, the term $vp_{x}$ is called "actuarial discount factor" (notice that this is actually the APV of a 1-year pure endowment).

Of course, it may not be convincing if we just say multiplying such probability can do this. Thus, let us explain the theory behind this intuition in the following. Recall that $A_{x:{\overline {n}}|}^{1}=\mathbb {E} [Z]$, where $Z$ is the present-value random variable for the $n$-year term life insurance issued to the person aged $x$. Now, we apply a result in probability ("generalized" law of total probability) to get the following equation:

$$\mathbb {E} [Z]=\mathbb {E} [Z|K=0]\mathbb {P} (K=0)+\mathbb {E} [Z|K=1,2,\dotsc ]\mathbb {P} (K=1,2,\dotsc ).$$

Notice that the event $\{K=0\}$ means the person dies within first year, and the event $\{K=1,2,\dotsc \}$ means the person lives to age $x+1$. After noticing these, we know that $\mathbb {E} [Z|K=0]=A_{x:{\overline {1}}|}^{1}$, since given that $K=0$, it is impossible to have the benefit after age $x+1$, that is, the only possibility for getting benefit is that a life aged $x$ dies within first year. So, $\mathbb {E} [Z|K=0]$ is essentially the same as the APV of a 1-year term life insurance issued to a person aged $x$. Also, we know that $\mathbb {P} (K=0)=q_{x}$ since this is the probability for the person aged $x$ to die within first year. On the other hand, $\mathbb {E} [Z|K=1,2,\dotsc ]$ is essentially the same as the APV of a $n-1$-year term life issued to a person aged $x+1$, since given that $K=1,2,\dotsc$, the benefit can only possibly be made at age $x+2,x+3,\dotsc ,x_{n}$. These are time 1,2,...,$n-1$ with respect to a life aged $x+1$. Since the benefits in this APV are made with respect to a life aged $x+1$, to convert this APV to the APV for a person aged $x$, we need to discount back the benefit for 1-year, i.e. multiplying $v$. Now, consider the probability $\mathbb {P} (K=1,2,\dotsc )$. Since this probability is the probability for the person aged $x$ to live for 1 year, i.e. live to age $x+1$, we have $\mathbb {P} (K=1,2,\dotsc )=p_{x}$. Thus, we can obtain the recursion relation

$$A_{x:{\overline {n}}|}^{1}=vq_{x}+vp_{x}A_{x+1:{\overline {n-1}}|}^{\;\;1}.$$

In financial mathematics, when we want to discount back $n$ years, we multiply the discount factor to the power $n$. However, in this case, when we want to "actuarially" discount back $n$ years, we are not multiplying $vp_{x}$ to the power $n$. The reason for this is due to the "survival part" of the actuarial discount factor: the probability for a person to survive for $n$ years is not $p_{x}^{n}$ (unless constant force assumption is assumed). Instead, the probability is given by $_{n}p_{x}$. Hence, when we want to "actuarially" discount back $n$ years, we are multiplying $v^{n}{}_{n}p_{x}$, which is actually the APV of $n$-year pure endowment (in other words, the APV of a particular payment made at time $n$ can be obtained by "actuarially" discount it back to time 0. This idea will be useful when we discuss life annuities).

Intuitive explanations

Now, we are ready to explain the recursion relations intuitively.

• recursion relation 2: we first extract 1-year term life insurance ($vq_{x}$) from the $n$-year term life insurance ($A_{x:{\overline {n}}|}^{1}$). Then, the remaining part of insurance is a $n-1$-year term life insurance with respect to $(x+1)$ ($A_{x+1:{\overline {n-1}}|}^{\;\;1}$). After that, we actuarially discount by 1 year (multiply $vp_{x}$) the remaining part of insurance. .
• recursion relation 3: notice that we can split the endowment insurance to term life insurance and pure endowment. For the term life insurance part, it follows from relation 2. For the pure endowment part, the pure endowment at RHS ($_{n-1}E_{x+1}$) is with respect to $(x+1)$, and thus we actuarially discount it by 1 year (multiply $vp_{x}$) to get the pure endowment at LHS ($_{n}E_{x}$). Thus, we have $_{n}E_{x}=vp_{x}{}_{n-1}E_{x+1}$, and the relation follows.
• recursion relation 4: we consider the $m$-year deferred $n$-year term life insurance issued to a life aged $x$ ($_{m|n}A_{x}$) with respect to a life aged $x+1$. Then, from this point of view, the APV is $_{m-1|n}A_{x+1}$ (since with respect to a life aged $x+1$, such insurance is just deferred for $m-1$ years, and its term is still $n$ years (from age $x+1+m-1$ to $x+m+n$). After that, we actuarially discount this insurance back 1 year (multiply $vp_{x}$).
• recursion relation 6: we first extract the "first year part" ($nvq_{x}$) out. Then, the remaining part is an annually decreasing $n-1$-year term life insurance with respect to $(x+1)$ ($(DA)_{x+1:{\overline {n-1}}}^{\;\;1}$). After that, we actuarially discount the remaining part back 1 year (multiply $vp_{x}$).
• recursion relation 7: we first extract benefit of 1 from each original benefit to get a whole life insurance (with benefit of 1) ($A_{x}$). Then, the remaining part of insurance is an annually increasing whole life insurance with respect to $(x+1)$ ($(IA)_{x+1}$). After that, we actuarially discount the remaining part back 1 year (multiply $vp_{x}$).

Exercise. Try to explain recursion relation 1 and 5 intuitively.

Solution

• recursion relation 1: we first extract 1-year term life insurance ($vq_{x}$) from the whole life insurance, and then the remaining part of the insurance is a whole life insurance with respect to ($x+1)$ ($A_{x+1}$). After that, we actuarially discount the remaining part back 1 year (multiply $vp_{x}$).
• recursion relation 5: we extract benefit of 1 from each original benefit to get a $n$-year term life insurance (with benefit of 1) ($A_{x:{\overline {n}}|}$). Then, the remaining part of insurance is an annually increasing $n-1$-year term life insurance with respect to $(x+1)$ ($(IA)_{x+1:{\overline {n-1}}|}^{\;\;1}$). After that, we actuarially discount the remaining part back 1 year (multiply $vp_{x}$).

Exercise. Using similar arguments, we can obtain many recursion relations other than those mentioned above. Can you construct another recursion relation for $A_{x}$ that involves $A_{x:{\overline {n}}|}^{1}$, and explain it intuitively?

Solution

$$A_{x}=A_{x:{\overline {n}}|}^{1}+v^{n}{}_{n}p_{x}A_{x+n}.$$

In this case, we extract $n$-year, instead of 1-year, term life insurance from the whole life insurance. Then, the remaining part of the insurance is a whole life insurance with respect to $(x+n)$. After that, we need to actuarially discount this part of insurance back $n$ years (multiply $v^{n}{}_{n}p_{x}$).

Example.

(a) Show that $(IA)_{x:{\overline {n}}|}=vq_{x}+vp_{x}(A_{x+1:{\overline {n-1}}|}^{\;\;1}+(IA)_{x+1:{\overline {n-1}}|}^{\;\;1})$.

(b) Given that $i=0.04,p_{x}=0.9,(IA)_{x:{\overline {n}}|}=3.52,A_{x+2:{\overline {n-2}}|}=0.125$, calculate $(IA)_{x+1:{\overline {n-1}}|}^{\;\;1}$.

Solution:

(a)

Proof. Since $A_{x:{\overline {n}}|}=vq_{x}+vp_{x}A_{x+1:{\overline {n-1}}|}$, we have

$$(IA)_{x:{\overline {n}}|}=A_{x:{\overline {n}}|}+vp_{x}(IA)_{x+1:{\overline {n-1}}|}^{\;\;1}=vq_{x}+vp_{x}A_{x+1:{\overline {n-1}}|}+vp_{x}(IA)_{x+1:{\overline {n-1}}}^{\;\;1}=vq_{x}+vp_{x}(A_{x+1:{\overline {n-1}}|}^{\;\;1}+(IA)_{x+1:{\overline {n-1}}|}^{\;\;1}).$$

$\Box$

(b) Notice that we are given $A_{x+2:{\overline {n-2}}|}$ instead of $A_{x+1:{\overline {n-1}}}$. Thus, we cannot apply the equation derived in (a) directly. However, it is simple to deduce the value of $A_{x+1:{\overline {n-1}}}$ from the given information, by considering the following recursion relation:

$$A_{x+1:{\overline {n-1}}}=vq_{x}+vp_{x}A_{x+2:{\overline {n-1}}|}\implies A_{x+1:{\overline {n-1}}}=(1.04)^{-1}(1-0.9)+(1.04)^{-1}(0.9)(0.125)\implies A_{x+1:{\overline {n-1}}}\approx 0.2043.$$

Now, we can apply the equation in (a) to get

$$3.52=1.04^{-1}(0.1)+1.04^{-1}(0.9)(0.2043+(IA)_{x+1:{\overline {n-1}}}^{\;\;1})\implies (IA)_{x+1:{\overline {n-1}}|}^{\;\;1}\approx 3.752.$$

Exercise. Given that $i=0.07,\ell _{x}=100-x\quad (0\leq x\leq 100),\;_{5|5}A_{30}=0.063$, calculate $_{10|5}A_{25}$.

Solution

Since $\ell _{x}=100-x$, ${}_{t}p_{x}={\frac {100-x-t}{100-x}}=1-{\frac {t}{100-x}}$ where $0\leq x\leq 100-t$ and $0\leq t\leq 100-x$.

We can develop a new recursion relation as follows:

$$_{10|5}A_{25}=v^{5}{}_{5}p_{25}{}_{5|5}A_{30}$$

(possibly using some intuitive arguments).

Hence,

$$_{10|5}A_{25}=(1.07)^{-5}\left(1-{\frac {5}{75}}\right)(0.063)\approx 0.0419.$$

Exercise.

(a) Given that $A_{x}=0.012,A_{x+1}=0.01,v=0.9$, calculate $p_{x}$.

Solution

By the recursion relation, we have

$$A_{x}=vq_{x}+vp_{x}A_{x+1}\implies 0.012=0.9(1-p_{x})+0.9p_{x}(0.01)\implies p_{x}\approx 0.9966.$$

(b) Given that $A_{x-1}=0.012,A_{x}=0.01,v=0.9,{}_{2}p_{x-1}=0.994$, calculate $p_{x}$.

Solution

By the recursion relation, we have

$$A_{x-1}=vq_{x-1}+vp_{x-1}A_{x}\implies 0.012=0.9(1-p_{x-1})+0.9p_{x-1}(0.01)\implies p_{x-1}\approx 0.9966.$$

Thus,

$$p_{x-1}p_{x}={}_{2}p_{x-1}\implies 0.9966p_{x}=0.994\implies p_{x}\approx 0.997$$

Now, you may ask that there are some similar recursion relations holds for continuous insurances. The answer is yes, but since the insurances are continuous, the relations may get more complicated. Also, such relations are generally not quite useful, since often there are more than one type of insurance involved in the relation, and also we often are not given the values of APV of continuous insurances directly, which is theoretical.

For instance, we have ${\bar {A}}_{x}={\bar {A}}_{x:{\overline {1}}|}+vp_{x}{\bar {A}}_{x+1}$, which is analogous to the recursion relation 1 for the discrete insurance above. But, this relation is not very helpful actually, since we need to have the values of ${\bar {A}}_{x}$, ${\bar {A}}_{x:{\overline {1}}|}$, $p_{x}$ and $v$ to get ${\bar {A}}_{x+1}$, but often we do not have such information.

Indeed, instead of such relationship of APV between integer ages ($x$ vs. $x+1$), which is applicable to discrete insurances. To have the idea about relationship of APV between different ages for continuous insurances, it is better for us to consider the rate of change of the APV.

Recursion relations in continuous case: differential equations

We can also develop "recursion relations" of APV's of continuous insurances. Since the insurances are continuous, we can use differential equations to have some "recursions". To understand this more clearly, let us consider the following example.

Example. (Differential equation for ${\bar {A}}_{x}$) Recall that we can write

$${\bar {A}}_{x}=\int _{0}^{\infty }v^{t}{}_{t}p_{x}\mu _{x+t}\,dt.$$

(a) Show that ${\bar {A}}_{x}={\frac {1}{{}_{x}p_{0}v^{x}}}\int _{x}^{\infty }v^{y}{}_{y}p_{0}\mu _{y}\,dy$.

(b) Show that $\int _{x}^{\infty }v^{y}{}_{y}p_{0}\mu _{y}\,dy=\int _{-\infty }^{-x}v^{-u}{}_{-u}p_{0}\mu _{-u}\,du$.

(c) Hence, show that ${\frac {d}{dx}}{\bar {A}}_{x}=(\mu _{x}+\delta ){\bar {A}}_{x}-\mu _{x}$.

Solution:

(a)

Proof.

$${\begin{aligned}{\bar {A}}_{x}&=\int _{0}^{\infty }{\color {blue}v^{t}}{\color {red}{}_{t}p_{x}}\mu _{x+t}\,dt\\&={\frac {1}{{\color {red}{}_{x}p_{0}}{\color {blue}v^{x}}}}\int _{0}^{\infty }{\color {blue}v^{x+t}}{\color {red}{}_{x+t}p_{0}}\,dt&({\color {red}{}_{x+t}p_{0}={}_{x}p_{0}{}_{t}p_{x}})\\&={\frac {1}{{}_{x}p_{0}v^{x}}}\int _{x}^{\infty }v^{y}{}_{y}p_{0}\mu _{y}\,dy&(y=x+t).\\\end{aligned}}$$

$\Box$

(b)

Proof.

$${\begin{aligned}\int _{x}^{\infty }v^{y}{}_{y}p_{0}\mu _{y}\,dy&=-\int _{-x}^{-\infty }v^{-u}{}_{-u}p_{0}\mu _{-u}\,du&(u=-y)\\&=\int _{-\infty }^{-x}v^{-u}{}_{-u}p_{0}\mu _{-u}\,du.\end{aligned}}$$

$\Box$

(c)

Proof.

$${\begin{aligned}{\frac {d}{dx}}{\bar {A}}_{x}&={\frac {1}{{}_{x}p_{0}v^{x}}}\cdot {\frac {d}{dx}}\int _{-\infty }^{-x}v^{-u}{}_{-u}p_{0}\mu _{-u}\,du+\int _{x}^{\infty }v^{y}{}_{y}p_{0}\mu _{y}\,dy\cdot {\frac {d}{dx}}{\frac {1}{{}_{x}p_{0}v^{x}}}\\&={\frac {1}{{}_{x}p_{0}v^{x}}}\cdot (-1)v^{x}{}_{x}p_{0}\mu _{x}+\int _{x}^{\infty }v^{y}{}_{y}p_{0}\mu _{y}\,dy\cdot {\frac {d}{dx}}{\frac {1}{{}_{x}p_{0}v^{x}}}&({\text{fundamental theorem of calculus, chain rule}})\\&=-\mu _{x}+\int _{x}^{\infty }v^{y}{}_{y}p_{0}\mu _{y}\,dy\cdot -{\frac {{}_{x}p_{0}v^{x}\ln v+v^{x}(-{}_{x}p_{0}\mu _{x})}{({}_{x}p_{0}v^{x})^{2}}}&\left({\frac {d}{dx}}{}_{x}p_{0}={\frac {d}{dx}}S_{0}(x)=-f_{0}(x)=-{}_{x}p_{0}\mu _{x}\right)\\&=-\mu _{x}+\int _{x}^{\infty }v^{y}{}_{y}p_{0}\mu _{y}\,dy\cdot -\left({\frac {-\delta }{{}_{x}p_{0}v^{x}}}-{\frac {\mu _{x}}{{}_{x}p_{0}v^{x}}}\right)\\&=-\mu _{x}+{\frac {1}{{}_{x}p_{0}v^{x}}}\int _{x}^{\infty }v^{y}{}_{y}p_{0}\mu _{y}\,dy\cdot (\mu _{x}+\delta )\\&=-\mu _{x}+{\bar {A}}_{x}\cdot (\mu _{x}+\delta )&({\text{by (a)}})\\\end{aligned}}$$

The result follows.

$\Box$

Incorporating selection ages to insurances

We have discussed the idea of selection ages in the previous chapter. We can also use the selection ages for the ages in the insurances. For example, we can have $A_{[x]},{\bar {A}}_{[x]+1:{\overline {n}}|},{}_{n}E_{[x]},$ etc. However, it is more common to calculate the discrete APV's involving selection ages, since the life table is usually given in a discrete form, so it is often only possible to calculate the discrete, rather than continuous APV's. To calculate the values of these APV's, we usually obtain the value of different terms related to the survival probability from the life table, as discussed in the previous chapter.

Example. The following is a select-and-ultimate table with 3-year select period:

$${\begin{array}{ccccc}x&\ell _{[x]}&\ell _{[x]+1}&\ell _{[x]+2}&\ell _{x+3}\\\hline 50&9700&9690&9670&9640\\51&9660&9650&9625&9600\\52&9640&9615&9585&9570\\\end{array}}$$

Also, it is given that the annual interest rate is 4%.

Then, based on this table, we have

$${\begin{aligned}A_{[50]:{\overline {3}}|}&=\sum _{k=0}^{2}v^{k+1}{}_{k}p_{[50]}q_{[50]+k}\\&=(1.04)^{-1}(1)(1-\ell _{[50]+1}/\ell _{[50]})+(1.04)^{-2}(\ell _{[50]+1}/\ell _{[50]})(1-\ell _{[50]+2}/\ell _{[50]})+(1.04)^{-3}(\ell _{[50]+2}/\ell _{[50]})(1-\ell _{53}/\ell _{[50]})\\&=1.04^{-1}(1-9690/9700)+1.04^{-2}(9690/9700)(1-9670/9700)+(1.04)^{-3}(9670/9700)(1-9640/9700)\\&\approx 0.0093297.\\\end{aligned}}$$

Exercise.

1 Based on above, which of the following APV's can be calculated?

	$A_{[51]:{\overline {3}}|}$
	$A_{[52]:{\overline {3}}|}$
	$A_{[53]:{\overline {3}}|}$
	$_{1|1}A_{[50]}$
	$_{1|1}A_{[51]}$
	$_{1|1}A_{[52]}$
	$_{1|1}A_{[53]}$

2 Based on above, which of the following APV's can be calculated?

	$A_{[51]:{\overline {1}}|}^{(2)}$
	$(IA)_{[52]:{\overline {3}}|}$
	$(DA)_{[51]+1:{\overline {3}}|}$
	$_{3}E_{[52]}$ (discrete one)
	$A_{[50]:{\overline {3}}|}$

Connections between continuous and discrete insurances

We have discussed continuous insurances where the benefit is payable at the moment of death ("${\bar {A}}$") and discrete insurances where the benefit is payable at the end of year of death ("$A$"), and we know that in practice, discrete insurances are more common than continuous insurances. However, in terms of the calculations, the calculations of APV for continuous insurances (which involve integral) are often more convenient and simple than the calculations of APV for discrete insurances (which involve summation), particularly in the case that there is no special "nice" formula to calculate the summation, where we may actually need to calculate the terms one by one and sum them up, which can be quite complicated and tedious.

This motivates us to find some relationships between the APV of continuous and discrete insurances, so that we can use the APV of continuous insurances (which is often more convenient to calculate) to calculate the APV of discrete insurances, which makes the calculation of the APV of discrete insurances simpler.

Of course, the following relationship has some limitations. Otherwise, if the relationship holds for every insurance with no further assumptions made, then we can simply calculate the APV's of continuous insurances and then apply the relationship, and therefore we do not need to develop the formula for APV's of discrete insurances at all!

Indeed, we need to have the UDD assumption (from previous chapter) to develop such relationship for some insurances.

Theorem. (Relationship between ${\bar {A}}$ and $A$) Suppose for a continuous insurance, the benefit function $b_{T}$ is a function of $\lfloor T\rfloor =K$, and the discount function $v_{T}=v^{T}$. Then, under UDD assumption, the APV of the continuous insurance is the product of ${\frac {i}{\delta }}$ and the APV of its discrete counterpart, where $i$ is the annual interest rate and $\delta$ is the constant annual force of interest.

Proof. Suppose for the continuous insurances, the benefit function is a function of $K$, so we can write $b_{T}=b(K)$ where $b$ is a function of $K$, and the discount function is $v^{T}$. Also, assume UDD. Then, the present-value random variable for the continuous insurance

$$Z=b_{T}v_{T}=b(K)v^{T}=b(K)v^{K+1}v^{S-1}$$

where $T=K+S$. Under UDD assumption, $K$ and $S$ are independent. Hence, the function of $K$ and the function of $S$ are also independent. Thus, applying expectation on $Z$, the APV of the continuous insurance equals

$$\mathbb {E} [Z]=\mathbb {E} [b(K)v^{K+1}]\mathbb {E} [v^{S-1}]$$

by independence. Observe that the function $b(K)$ is the benefit function of the corresponding discrete insurance (this is an important observation) . Hence, $\mathbb {E} [b(K)v^{K+1}]$ is the APV of the discrete counterpart. Now, it suffices to show that $\mathbb {E} [v^{S-1}]={\frac {i}{\delta }}$, which is actually true since

$${\begin{aligned}\mathbb {E} [v^{S-1}]&=\int _{0}^{1}v^{s-1}\cdot {\frac {1}{1-0}}\,ds&(S{\text{ follows the uniform distribution on }}(0,1))\\&=\int _{0}^{1}(1+i)^{1-s}\,ds\\&=-\int _{1}^{0}(1+i)^{u}\,du&(u=1-s)\\&=-\left[{\frac {(1+i)^{u}}{\ln(1+i)}}\right]_{u=1}^{u=0}\\&=-{\frac {1}{\delta }}+{\frac {1+i}{\delta }}&(\delta =\ln(1+i))\\&={\frac {i}{\delta }}.\end{aligned}}$$

$\Box$

Example. Under UDD assumption,

• ${\bar {A}}_{x}={\frac {i}{\delta }}A_{x}$ since for $A_{x}$, the benefit function $b_{T}$ is a function of $\lfloor T\rfloor$, and the discount function $v_{T}$ is $v^{T}$;
• $(I{\bar {A}})_{x:{\overline {n}}|}={\frac {i}{\delta }}(IA)_{x:{\overline {n}}|}$ since $b_{T}=\lfloor T\rfloor +1$ and $v_{T}=v^{T}$.
• ${\bar {A}}_{x:{\overline {n}}|}^{1}={\frac {i}{\delta }}A_{x:{\overline {n}}|}^{1}$ since $b_{T}$ is a function of $\lfloor T\rfloor$ and $v_{T}=v^{T}$.

Exercise. Do we have ${\bar {A}}_{x:{\overline {n}}|}={\frac {i}{\delta }}A_{x:{\overline {n}}|}$? Explain why.

Solution

No. This is because the discount function is $v_{T}={\begin{cases}v^{T},&T\leq n;\\v^{n},&T>n.\end{cases}}$. Thus, the discount function is not $v^{T}$ when $T>n$.

Remark.

• However, since the discount function is indeed $v^{T}$ when $T\leq n$ and the benefit function is a function of $\lfloor T\rfloor$, we can use the relationship when $T\leq n$:

• We can write ${\bar {A}}_{x:{\overline {n}}|}={\bar {A}}_{x:{\overline {n}}|}^{1}+{}_{n}E_{x}={\frac {i}{\delta }}A_{x:{\overline {n}}|}^{1}+{}_{n}E_{x}$ to "use the relationship when $T\leq n$" (where $_{n}E_{x}$ refers to the continuous $n$-year pure endowment, since it is originated from continuous $n$-year endowment insurance).
• Further question: can we apply the relationship on this continuous $n$-year pure endowment, and write something like $_{n}E_{x}={\frac {i}{\delta }}{}_{n}E_{x}$ where $_{n}E_{x}$ is continuous at LHS and $_{n}E_{x}$ is discrete at RHS?

Introduction to life annuities

In the previous discussion about life insurance, the benefit payment is contingent on death. Now, we will discuss another type of product, namely life annuity, which is contingent on survival.

For life annuity, a series of payments are made continuously (for continuous life annuities), or at equal intervals (for discrete life annuities), where the equal intervals may be years, quarters, months, etc., while the given life (called annuitant) survives, that is, the life only receives these payments when he survives. Hence, these payments can be regarded as survival benefits.

Similar to life insurance, life annuity may or may not have a term. For life annuity with term, we call such life annuities as temporary life annuities (since the payments are only made temporarily for a certain interval, and even if the life still survives after that interval, no payments will be made). On the other hand, for life annuities without term, they are similarly called whole life annuities.

In financial mathematics, we have learnt about various types of annuities, where the payments are certainly made (not contingent on survival). Hence, they can be called annuities-certain. Many concepts and formulas developed there can apply to life annuities. For example, there are life annuity-immediate, lie annuity-due, continuous life annuity, etc.

Even if life annuity is not a life insurance itself, it is still quite important in life insurance operations. For example, we will later see that life insurances are often purchased using a life annuity of premium, instead of a single premium. Also, in retirement plans for workers, deferred life annuities are often involved, where the payments begin at retirement, and to purchase such annuities, the workers contribute payments, in the form of temporary life annuity, while they are actively working. Such annuities can ensure that the workers still have some stable incomes even after retirement.

Continuous life annuities

Let us first discuss some life annuities payable continuously at the rate of 1 per year (level payment), similar to the case of life insurance. Recall that we have discussed continuously paying $n$-year annuities-certain in financial mathematics $\left({\bar {a}}_{{\overline {n}}|}={\frac {1-v^{n}}{\delta }}\right)$. This will be useful for the following discussions related to the present-value random variable.

For life annuities, the present-value random variable is usually denoted by $Y$, instead of $Z$.

Whole life annuity

Definition. (Whole life annuity) A whole life annuity provides payments to a life until death.

The present-value random variable for whole life annuity is thus

$$Y={\bar {a}}_{{\overline {T}}|},\quad T\geq 0.$$

Remark.

• When the annual interest rate is $i$ and the force of interest is a constant $\delta$, we can also write

$$Y={\frac {1-v^{T}}{\delta }},$$

from the formula developed for annuities payable continuously in financial mathematics.

• Recall that the force of interest will be assumed to be constant unless otherwise specified. Thus, in the following discussions, we may use this expression in place of ${\bar {a}}_{{\overline {T}}|}$.

Hence, the APV of this annuity, denoted by ${\bar {a}}_{x}$ (we use "$a$" for life annuity, and similarly we add "$j$" to the upper-left corner of the notation when we are discussing the $j$th moment of $Y$), is

$$\mathbb {E} [Y]=\int _{0}^{\infty }{\bar {a}}_{{\overline {t}}|}{}_{t}p_{x}\mu _{x+t}\,dt.$$

Apparently, computing this integral can be somewhat complicated since "${\bar {a}}_{{\overline {t}}|}={\frac {1-v^{t}}{\delta }}$" is involved in the integral. Fortunately, there is an alternative, and often more convenient formula for calculating this APV, as shown in the following example.

Example. (Alternative formula of ${\bar {a}}_{x}$) Show that ${\bar {a}}_{x}=\int _{0}^{\infty }v^{t}{}_{t}p_{x}\,dt$.

Proof. We use integration by parts.

$${\begin{aligned}{\bar {a}}_{x}&=\int _{0}^{\infty }{\bar {a}}_{{\overline {t}}|}{}_{t}p_{x}\mu _{x+t}\,dt\\&=\int _{0}^{\infty }{\bar {a}}_{{\overline {t}}|}\,d({\color {darkgreen}-}{}_{t}p_{x})&\left({\frac {d}{dt}}{\big (}-{}_{t}p_{x}{\big )}={\frac {d}{dt}}{\big [}-(1-F_{x}(t)){\big ]}=f_{x}(t)={}_{t}p_{x}\mu _{x+t}\right)\\&=\left[-{\bar {a}}_{{\overline {t}}|}{}_{t}p_{x}\right]_{t=0}^{t=\infty }{\color {darkgreen}+}\int _{0}^{\infty }{}_{t}p_{x}\,d\left({\bar {a}}_{{\overline {t}}|}\right)\\&=\left[-{\bar {a}}_{{\overline {t}}|}{}_{t}p_{x}\right]_{t=0}^{t=\infty }{\color {darkgreen}+}\int _{0}^{\infty }{}v^{t}{}_{t}p_{x}\,dt&\left({\frac {d}{dt}}{\bar {a}}_{{\overline {t}}|}={\frac {d}{dt}}{\frac {1-v^{t}}{\delta }}=-{\frac {v^{t}\overbrace {\ln v} ^{=-\delta }}{\delta }}=v^{t}\right)\\&={\bigg [}-\underbrace {{\bar {a}}_{{\overline {\infty }}|}} _{=1/\delta }\underbrace {{}_{\infty }p_{x}} _{=0}-\underbrace {{\bar {a}}_{{\overline {0}}|}} _{=0}\underbrace {{}_{0}p_{x}} _{=1}{\bigg ]}+\int _{0}^{\infty }{}v^{t}{}_{t}p_{x}\,dt&({\text{abuse of notations}})\\&=\int _{0}^{\infty }{}v^{t}{}_{t}p_{x}\,dt.\end{aligned}}$$

$\Box$

Exercise. (Interpretation of the above formula) We have shown that the above formula holds. Now, let us consider its interpretation.

(a) Use an actuarial notation to describe $v^{t}{}_{t}p_{x}$.

(b) Interpret $v^{t}{}_{t}p_{x}$ intuitively (Hint: you should have seen something similar before).

(c) Hence, interpret $\sum _{t=1}^{\infty }v^{t}{}_{t}p_{x}$ intuitively.

(d) Apply the idea in (c) to the continuous case (i.e. interpret the above formula intuitively).

Solution

(a) ${}_{n}E_{x}$.

(b) It is the term we multiply for "actuarially" discount back $t$ years.

(c) We have a series of payment of 1 payable at the end of each year, and we "actuarially" discount back to time 0 for every payment, to get the APV of each payment. Then, summing the APV of every payment yields the APV of the whole series of payment (linearity of expectation).

(d) We can consider that there is a payment $1\,dt$ at time $t$. Then, multiplying $v^{t}{}_{t}p_{x}$ "actuarially" discounts the payment back to time zero. Hence, integrating over the APV of every payment yields the APV of the whole series of payment.

Remark.

• The payment discussed in part (d) is called "current payment".
• This gives rise to the name "current payment technique", which is discussed in the following.

We can apply the idea in the previous exercise to general cases, and such formula derived from the idea is called current payment technique (CPT). In general, the current payment technique is given by

$${\text{APV}}=\int _{0}^{\infty }\underbrace {v^{t}\times \mathbb {P} ({\text{payments are being made at time }}t)} _{\text{actuarial discounting to time zero}}\times \underbrace {({\text{payment rate at time }}t)\,dt} _{{\text{`amount' of payment at time }}t}.$$

Example. Consider a continuous whole life annuity payable at the annual rate of 100, which is issued to a life aged $x$. Suppose the force is interest is $\delta =0.06$, and the force of mortality of the life is $\mu =0.03$. Calculate the actuarial present value of this life annuity.

Solution. Since the force of mortality is $\mu =0.03$, we have ${}_{t}p_{x}=e^{-0.04}$. Thus, $f_{x}(t)=0.03e^{-0.03t}$. Also, the force of interest is $\delta =0.06$. Thus, $v^{t}=e^{-0.06t}$, and hence ${\bar {a}}_{{\overline {t}}|}={\frac {1-e^{-0.06t}}{0.06}}$. Then, we have two methods to calculate the APV.

Method 1: Definition

The APV is

$$\int _{0}^{\infty }{\color {blue}100}{\bar {a}}_{{\overline {t}}|}f_{x}(t)\,dt=\int _{0}^{\infty }100\cdot {\frac {1-e^{-0.06t}}{\color {blue}0.06}}\cdot {\color {blue}0.03}e^{-0.03t}\,dt={\color {blue}50}\int _{0}^{\infty }e^{-0.03t}-e^{-0.09t}\,dt=50\left[{\frac {e^{-0.03t}}{-0.03}}-{\frac {e^{-0.09t}}{-0.09}}\right]_{0}^{\infty }\approx 1111.1111.$$

Method 2: Current payment technique

The APV is

$$\int _{0}^{\infty }100v^{t}{}_{t}p_{x}\,dt\int _{0}^{\infty }100e^{-0.06t}e^{-0.03t}\,dt={\frac {100}{-0.09}}\left[e^{-0.09t}\right]_{0}^{\infty }\approx 1111.1111.$$

(Notice how the current payment technique simplifies the calculation)

Exercise. Let $Y$ be the present-value random variable for this life annuity. Calculate $\operatorname {Var} (Y)$ (Hint: the APV, i.e. $\mathbb {E} [Y]$, is calculated above. Hence, it suffices to calculate $\mathbb {E} [Y^{2}]$).

Solution

Let us calculate $\mathbb {E} [X^{2}]$. Notice that the current payment technique does not work in this case, since we are calculating $\mathbb {E} [Y^{2}]$, instead of $\mathbb {E} [Y]$ here.

Hence, we consider the definition:

$${\begin{aligned}\mathbb {E} [Y^{2}]&=\int _{0}^{\infty }\left(100{\bar {a}}_{{\overline {t}}|}\right)^{2}f_{x}(t)\,dt\\&=\int _{0}^{\infty }10000\cdot \left({\frac {1-e^{-0.06t}}{0.06}}\right)^{2}\cdot 0.03e^{-0.03t}\,dt\\&={\frac {250000}{3}}\int _{0}^{\infty }(1-2e^{-0.06t}+e^{-0.12t})\cdot e^{-0.03t}\,dt\\&={\frac {250000}{3}}\int _{0}^{\infty }e^{-0.03t}-2e^{-0.09t}+e^{-0.15t}\,dt\\&={\frac {250000}{3}}\left[{\frac {e^{-0.03t}}{-0.03}}-2{\frac {e^{-0.09t}}{-0.09}}+{\frac {e^{-0.15t}}{-0.15}}\right]_{0}^{\infty }\\&=1481481.481481\end{aligned}}$$

It follows that the variance

$$\operatorname {Var} (Y)=\mathbb {E} [Y^{2}]-(\mathbb {E} [Y])^{2}\approx 1481481.481481-(1111.1111)^{2}\approx 246913.6.$$

From the previous exercise, we have calculated the variance of the present-value random variable $Y$, and the calculation process is a bit complicated. Indeed, there are some alternative methods to calculate the variance (and also expectation) of the present-value random variable $Y$, by relating $Y$ with $Z$ (present-value random variable for life insurance).

Proposition. (Relationships between life annuity and life insurance) Let $Y$ be the present-value random variable for the whole life annuity payable at the rate of 1 per year. Then,

• ${\bar {a}}_{x}={\frac {1-{\bar {A}}_{x}}{\delta }}$ and
• $\operatorname {Var} (Y)={\frac {^{2}{\bar {A}}_{x}-({\bar {A}}_{x})^{2}}{\delta }}$.

Proof. We have

• ${\bar {a}}_{x}=\mathbb {E} \left[{\frac {1-v^{T}}{\delta }}\right]={\frac {\mathbb {E} [1-v^{T}]}{\delta }}={\frac {1-{\color {darkgreen}\mathbb {E} [v^{T}]}}{\delta }}={\frac {1-{\color {darkgreen}{\bar {A}}_{x}}}{\delta }}$ (${\bar {A}}_{x}=\mathbb {E} [v^{T}]$ by definition);
• $\operatorname {Var} (Y)=\operatorname {Var} \left({\frac {1-v^{T}}{\delta }}\right)={\frac {\operatorname {Var} (1-v^{T})}{\delta ^{2}}}={\frac {\operatorname {Var} (-v^{T})}{\delta ^{2}}}={\frac {(-1)^{2}{\color {darkgreen}\operatorname {Var} (v^{T})}}{\delta ^{2}}}={\frac {\color {green}^{2}{\bar {A}}_{x}-({\bar {A}}_{x})^{2}}{\delta ^{2}}}$.

$\Box$

Exercise. Calculate the $\operatorname {Var} (Y)$ in the previous exercise using this proposition.

Solution

First, we have ${\bar {A}}_{x}=\int _{0}^{\infty }e^{0.06t}\cdot 0.03e^{0.03t}\,dt=0.03\int _{0}^{\infty }e^{-0.09t}\,dt=-{\frac {0.03}{0.09}}\left[e^{-0.09t}\right]_{0}^{\infty }={\frac {1}{3}}$, and $^{2}{\bar {A}}_{x}=\int _{0}^{\infty }e^{0.12t}\cdot 0.03e^{0.03t}\,dt=0.03\int _{0}^{\infty }e^{-0.15t}\,dt=-{\frac {0.03}{0.15}}\left[e^{-0.15t}\right]_{0}^{\infty }={\frac {1}{5}}$. Let $Y'$ be the present-value random variable for the same life annuity, but with the annual payment rate of 1. Then, the variance is

$$\operatorname {Var} (Y)=\operatorname {Var} (100Y')=100^{2}\operatorname {Var} (Y')=100^{2}\cdot {\frac {1/5-(1/3)^{2}}{0.06^{2}}}\approx 246913.6.$$

Exercise. Under the constant force of mortality $\mu$ and constant force of interest $\delta$, propose a formula for ${\bar {a}}_{x}$ (not in integral form), and prove it.

Solution

Proposition: ${\bar {a}}_{x}={\frac {1}{\mu +\delta }}$.

Proof. Recall that under these assumptions, we have

$${\bar {A}}_{x}={\frac {\mu }{\mu +\delta }}.$$

Applying the relationship between life annuity and life insurance, we have

$${\bar {a}}_{x}={\frac {1-{\bar {A}}_{x}}{\delta }}={\frac {1-{\frac {\mu }{\mu +\delta }}}{\delta }}={\frac {\frac {\delta }{\mu +\delta }}{\delta }}={\frac {1}{\mu +\delta }}.$$

$\Box$

Example. (Interpretation of rate of change of APV of whole life annuity with respect to age) Show that ${\frac {d}{dx}}{\bar {a}}_{x}=\delta {\bar {a}}_{x}+\mu _{x}{\bar {a}}_{x}-1$, and interpret this result verbally.

Proof. By the current payment technique, we have

$${\bar {a}}_{x}=\int _{0}^{\infty }v^{t}{}_{t}p_{x}\,dt.$$

Using this formula, we have

$${\begin{aligned}{\frac {d}{dx}}{\bar {a}}_{x}&={\frac {\partial }{\partial x}}\int _{0}^{\infty }v^{t}{}_{t}p_{x}\,dt&\left({\text{changing }}{\frac {d}{dx}}{\text{ to }}{\frac {\partial }{\partial x}}{\text{ does not affect the result here}}\right)\\&=\int _{0}^{\infty }{\frac {\partial }{\partial x}}(v^{t}{}_{t}p_{x})\,dt&({\text{we can do this by some mathematical results}})\\&=\int _{0}^{\infty }v^{t}{\frac {\partial }{\partial x}}{}_{t}p_{x}\,dt\\&=\int _{0}^{\infty }v^{t}{\frac {\partial }{\partial x}}{}\left({\frac {S_{0}(x+t)}{S_{0}(x)}}\right)\,dt\\&=\int _{0}^{\infty }v^{t}\cdot {\frac {S_{0}(x){\frac {\partial }{\partial x}}S_{0}(x+t)-S_{0}(x+t){\frac {\partial }{\partial x}}S_{0}(x)}{(S_{0}(x))^{2}}}\,dt\\&=\int _{0}^{\infty }v^{t}{\frac {{\cancel {S_{0}(x)}}(-S_{0}(x+t)\mu _{x+t})-S_{0}(x+t)(-{\cancel {S_{0}(x)}}\mu _{x})}{(S_{0}(x))^{\cancel {2}}}}\,dt&\left({\text{Let }}u=x+t.{\text{ Then }}{\frac {du}{dx}}=1,{\text{ and hence }}{\frac {\partial }{\partial x}}S_{0}(x+t)={\frac {\partial }{\partial u}}S_{0}(u)\cdot {\frac {du}{dx}}=-S_{0}(u)\mu _{u}=-S_{0}(x+t)\mu _{x+t}\right)\\&=\int _{0}^{\infty }v^{t}\cdot {\frac {S_{0}(x+t)(\mu _{x}-\mu _{x+t})}{S_{0}(x)}}\,dt\\&=\int _{0}^{\infty }v^{t}{}_{t}p_{x}(\mu _{x}-\mu _{x+t})\,dt\\&=\mu _{x}\underbrace {\int _{0}^{\infty }v^{t}{}_{t}p_{x}\,dt} _{{\bar {a}}_{x}{\text{ by CPT}}}-\underbrace {\int _{0}^{\infty }v^{t}{}_{t}p_{x}\mu _{x+t}\,dt} _{{\bar {A}}_{x}}\\&=\mu _{x}{\bar {a}}_{x}-{\bar {A}}_{x}\\&=\mu _{x}{\bar {a}}_{x}-(1-\delta {\bar {a}}_{x})&\left({\bar {a}}_{x}={\frac {1-{\bar {A}}_{x}}{\delta }}\iff {\bar {A}}_{x}=1-\delta {\bar {a}}_{x}\right)\\&=\delta {\bar {a}}_{x}+\mu _{x}{\bar {a}}_{x}-1.\end{aligned}}$$

$\Box$

We can interpret the future potential payments in the whole life annuity to be "owned" by the payer for now (future payments from the payer have not yet been made), and the APV of the whole life annuity is ${\bar {a}}_{x}$ for now. Then, we can interpret the rate of change of the APV with respect to age as the sum of

• rate of interest income on the APV: $\delta {\bar {a}}_{x}$ (under the constant force of interest, the APV ${\bar {a}}_{x}$ grows in value, so its increase is "earned")
• rate of mortality benefit: $\mu _{x}{\bar {a}}_{x}$ ($\mu _{x}$ is the probability for a life to die "instantaneously" at age $x$ (given the life survives to age $x$), and when the life dies, there will not be future payments with APV ${\bar {a}}_{x}$ made to the life. So, the payer "save" the payments with APV ${\bar {a}}_{x}$ with probability $\mu _{x}$, and hence this expected value is "earned")

and then subtracted by the rate of payment made: $1$ (we subtract it since from the payer is making this payment at this rate "currently" (i.e. at age $x$)).

n-year temporary life annuity

Definition. ($n$-year temporary life annuity) A $n$-year temporary life annuity provides payments to a life until the end of $n$ years, or until the death of the life, whichever occurs first.

From this definition, we know that the present-value random variable for $n$-year temporary life annuity is

$$Y={\begin{cases}{\bar {a}}_{{\overline {T}}|},&T<n;\\{\bar {a}}_{{\overline {n}}|},&T\geq n.\end{cases}}$$

It follows that the APV of this annuity, denoted by ${\bar {a}}_{x:{\overline {n}}|}$, is

$$\mathbb {E} [Y]=\int _{0}^{n}{\bar {a}}_{{\overline {t}}|}{}_{t}p_{x}\mu _{x+t}\,dt+\int _{n}^{\infty }{\bar {a}}_{{\overline {n}}|}{}_{t}p_{x}\mu _{x+t}\,dt.$$

Since

$$\int _{n}^{\infty }{\bar {a}}_{{\overline {n}}|}{}_{t}p_{x}\mu _{x+t}\,dt={\bar {a}}_{{\overline {n}}|}\int _{n}^{\infty }f_{x}(t)\,dt={\bar {a}}_{{\overline {n}}|}\mathbb {P} (T>n)={\bar {a}}_{{\overline {n}}|}{}_{n}p_{x},$$

we can also write the APV as

$$\int _{0}^{n}{\bar {a}}_{{\overline {t}}|}{}_{t}p_{x}\mu _{x+t}\,dt+{}_{n}p_{x}{\bar {a}}_{{\overline {n}}|}.$$

But we often use the current payment technique for calculating the APV instead:

$${\bar {a}}_{x:{\overline {n}}|}=\int _{0}^{n}v^{t}{}_{t}p_{x}\,dt.$$

(there are payments in the first $n$ years only, so from the current payment technique, the integration region is from $t=0$ to $t=n$)

For whole life annuity, we have some results that relate it with life insurance. Naturally, one may ask whether there are some similar results for $n$-year temporary life annuity. Indeed, there are similar results.

Proposition. (Relationships between life annuity and life insurance) Let $Y$ be the present-value random variable for the $n$-year temporary life annuity payable at the rate of 1 per year. Then,

• ${\bar {a}}_{x:{\overline {n}}|}={\frac {1-{\bar {A}}_{x:{\overline {n}}|}}{\delta }}$ and
• $\operatorname {Var} (Y)={\frac {^{2}{\bar {A}}_{x:{\overline {n}}|}-({\bar {A}}_{x:{\overline {n}}|})^{2}}{\delta }}$.

Exercise. Prove the above proposition (Hint: Let $Z$ be the present-value random variable of the $n$-year endowment insurance (with benefit of 1). Then, relate $Y$ with $Z$ by considering their definitions).

Solution

Proof. Notice that the present-value random variable of the $n$-year endowment insurance is

$$Z={\begin{cases}v^{T},&T<n;\\v^{n},&T\geq n,\end{cases}}$$

while the present-value random variable of the $n$-year temporary life annuity is

$$Y={\begin{cases}{\frac {1-v^{T}}{\delta }},&T<n;\\{\frac {1-v^{n}}{\delta }},&T\geq n.\end{cases}}$$

Comparing the definitions of $Y$ and $Z$, we can observe that $Y={\frac {1-Z}{\delta }}$. It follows that

$$\mathbb {E} [Y]=\mathbb {E} \left[{\frac {1-Z}{\delta }}\right]={\frac {1-\mathbb {E} [Z]}{\delta }}={\frac {1-{\bar {A}}_{x:{\overline {n}}|}}{\delta }},$$

and

$$\operatorname {Var} (Y)=\operatorname {Var} \left({\frac {1-Z}{\delta }}\right)={\frac {\operatorname {Var} (Z)}{\delta ^{2}}}={\frac {^{2}{\bar {A}}_{x:{\overline {n}}|}-({\bar {A}}_{x:{\overline {n}}|})^{2}}{\delta ^{2}}}.$$

$\Box$

Recall that we have the following relationship between ${\bar {a}}_{{\overline {n}}|}$ and ${\bar {s}}_{{\overline {n}}|}$ in financial mathematics:

$${\bar {s}}_{{\overline {n}}|}={\bar {a}}_{{\overline {n}}|}(1+i)^{n}={\frac {{\bar {a}}_{{\overline {n}}|}}{v^{n}}}$$

(we accumulates the present value ${\bar {a}}_{{\overline {n}}|}$ to the end of $n$ years (multiply $(1+i)^{n}$) to get the accumulated value at the end of the $n$th year ${\bar {s}}_{{\overline {n}}|}$). Naturally, we will expect a similar kind of relationship should hold for $n$-year temporary life annuity. This is indeed true, but of course, there is a difference between the discounting in financial mathematics and actuarial discounting.

In this context, we call the value of APV at the end of $n$th year as actuarial accumulated value (AAV). For the AAV of the $n$-year temporary life annuity, it is denoted by ${\bar {s}}_{x:{\overline {n}}|}$, and is given by

$${\bar {s}}_{x:{\overline {n}}|}={\frac {{\bar {a}}_{x:{\overline {n}}|}}{v^{n}{}_{n}p_{x}}}$$

(Notice that this equation is similar to the above relationship between ${\bar {a}}_{{\overline {n}}|}$ and ${\bar {s}}_{{\overline {n}}|}$). We can interpret the above equation as "actuarially discounting (multiply $v^{n}{}_{n}p_{x}$) the AAV (${\bar {s}}_{x:{\overline {n}}|}$) to time zero gives the APV (${\bar {a}}_{x:{\overline {n}}|}$)". Graphically,

    actuarial discounting: v^n _n p_x
      *-------------*
     /               \
    |                 \
    v                  |
  -                  -
  a x:n              s x:n
----*------------------*----
    0                  n   time

Example.

(a) Propose a relationship between ${\bar {a}}_{x}$ and $A_{x}$ under UDD assumption, and prove it.

(b) (recursion relation) Show that ${\bar {a}}_{x}={\bar {a}}_{x:{\overline {n}}|}+v^{n}{}_{n}p_{x}{\bar {a}}_{x+n}$ ($n$ is a positive integer), and explain it intuitively.

(c) Hence, propose a relationship between ${\bar {a}}_{x:{\overline {n}}|}$, $A_{x}$ and $A_{x+n}$ under UDD assumption, and prove it.

(d) Given that $A_{x}=0.01$, $A_{x+1}=0.009$, the annual interest rate is $i=0.04$, calculate ${\bar {a}}_{x:{\overline {1}}|}$ (Hint: calculate $p_{x}$ based on the given information first).

Solution.

(a) Proposition: we have ${\bar {a}}_{x}={\frac {1}{\delta }}\left(1-{\frac {i}{\delta }}A_{x}\right)$.

Proof. Under UDD assumption, we have ${\bar {A}}_{x}={\frac {i}{\delta }}A_{x}$. Since ${\bar {a}}_{x}={\frac {1-{\bar {A}}_{x}}{\delta }}$, we have ${\bar {a}}_{x}={\frac {1}{\delta }}\left(1-{\frac {i}{\delta }}A_{x}\right)$.

$\Box$

(b)

Proof. By current payment technique, we have

$${\begin{aligned}{\bar {a}}_{x}=\int _{0}^{\infty }v^{t}{}_{t}p_{x}\,dt&=\underbrace {\int _{0}^{n}v^{t}{}_{t}p_{x}\,dt} _{{\bar {a}}_{x:{\overline {n}}|}{\text{ by CPT}}}+\int _{n}^{\infty }v^{t}{}_{t}p_{x}\,dt\\&={\bar {a}}_{x:{\overline {n}}|}+v^{n}{}_{n}p_{x}\int _{n}^{\infty }v^{t-n}{}_{t-n}p_{x+n}\,dt\\&={\bar {a}}_{x:{\overline {n}}|}+v^{n}{}_{n}p_{x}\int _{0}^{\infty }v^{u}{}_{u}p_{x+n}\,dt&\left({\text{let }}u=t-n\right)\\&={\bar {a}}_{x:{\overline {n}}|}+v^{n}{}_{n}p_{x}{\bar {a}}_{x+n}.\end{aligned}}$$

$\Box$

(c) Proposition: we have ${\frac {1}{\delta }}\left(1-{\frac {i}{\delta }}A_{x}\right)={\bar {a}}_{x:{\overline {n}}|}+v^{n}{}_{n}p_{x}{\frac {1}{\delta }}\left(1-{\frac {i}{\delta }}A_{x+n}\right)$.

Proof. Using the relationship in (a), we have also ${\bar {a}}_{x+n}={\frac {1}{\delta }}\left(1-{\frac {i}{\delta }}A_{x+n}\right)$ (the "$x$" there just indicates an age, which is arbitrary). Hence, using (b), we have

$${\bar {a}}_{x}={\bar {a}}_{x:{\overline {n}}|}+v^{n}{}_{n}p_{x}{\bar {a}}_{x+n}\implies {\frac {1}{\delta }}\left(1-{\frac {i}{\delta }}A_{x}\right)={\bar {a}}_{x:{\overline {n}}|}+v^{n}{}_{n}p_{x}{\frac {1}{\delta }}\left(1-{\frac {i}{\delta }}A_{x+n}\right).$$

$\Box$

(d) First, we have the following recursion relation:

$$A_{x}=vq_{x}+vp_{x}A_{x+1}.$$

From the given information, we then have

$$0.01=(1.04)^{-1}(1-p_{x})+(1.04)^{-1}p_{x}(0.009)\iff 0.952885p_{x}=0.951538\iff p_{x}\approx 0.998586.$$

Since $i=0.04\iff \delta =\ln(1.04)\approx 0.03922$, using the relationship developed in (c), we have

$${\frac {1}{0.03922}}\left(1-{\frac {0.04}{0.03922}}(0.01)\right)={\bar {a}}_{x:{\overline {1}}|}+(1.04)^{-1}(0.998586)\cdot {\frac {1}{0.03922}}\left(1-{\frac {0.04}{0.03922}}(0.009)\right)\iff {\bar {a}}_{x:{\overline {1}}|}\approx 0.98.$$

Example. An insurance company offers two kinds of product to a life aged $x$: a 10-year term life insurance with benefit of 1,000,000 payable at the moment of death and a 10-year temporary life annuity payable continuously at a rate of $r$ per year. It is given that the actuarial present value of these two kinds of product both equals to 95000, and the annual force of interest is 3%. Calculate $r$.

Solution. First, since the actuarial present value of the insurance is 95000, we have

$$1000000{\bar {A}}_{x:{\overline {10}}|}=95000\iff {\bar {A}}_{x:{\overline {10}}|}=0.095.$$

Since the actuarial present value of the life annuity is also 95000, we have

$$95000=r{\bar {a}}_{x:{\overline {10}}|}=r\cdot {\frac {1-{\bar {A}}_{x:{\overline {10}}|}}{\delta }}={\frac {r(1-0.095)}{0.03}}\iff r\approx 3149.171.$$

Exercise. Suppose we replace the 10-year term life insurance with whole life insurance and the 10-year temporary life annuity with whole life annuity in the above example. But the other information remains unchanged (in particular, the actuarial present value of both of them is still 95000). Calculate $r$ again.

Solution

In this case, we can replace the "${\bar {A}}_{x:{\overline {10}}|}$" by "${\bar {A}}_{x}$", and "${\bar {a}}_{x:{\overline {10}}|}$" by "${\bar {a}}_{x}$" in the above calculation. We can observe that the same argument still holds, and hence the value of $r$ is still (approximately) 3149.171.

n-year deferred whole life annuity

Similar to insurances, we have deferred life annuities. The idea involved is basically the same: the "start" of the life annuity starts some time later than the issue of the life annuity. This kind of life annuity is indeed quite common in real life. For instance, we may purchase a deferred whole life annuity before our retirement, so that we can receive income from the life annuity after our retirement, while we survive. This can ensure a stable income even after the retirement, which should be quite desirable.

Definition. ($n$-year deferred whole life annuity) A $n$-year deferred whole life annuity provides payments to a life until death, starting at the end of $n$th year.

The present-value random variable is hence

$$Y={\begin{cases}0,&T<n;\\v^{n}{\bar {a}}_{{\overline {T-n}}|},&T\geq n.\\\end{cases}}$$

Thus, its actuarial present value, denoted by ${}_{n|}{\bar {a}}_{x}$, is

$$\mathbb {E} [Y]=\int _{n}^{\infty }v^{n}{\bar {a}}_{{\overline {t-n}}|}{}_{t}p_{x}\mu _{x+t}\,dt.$$

Using current payment technique, we then have

$$_{n|}{\bar {a}}_{x}=\int _{n}^{\infty }v^{t}{}_{t}p_{x}\,dt$$

(there are only payments starting from time $n$ (the end of $n$th year)).

Example. Amy, aged 30, plans to retire after 30 years, i.e. at age 60. Hence, she purchases a 30-year deferred whole life annuity payable continuously at the rate of 10000 per year. Her mortality follows the de Moivre's law with limiting age $\omega =120$. The annual force of interest is 5%.

(a) What is the probability that Amy will (unfortunately) not receive anything from this life annuity?

(b) Calculate the actuarial present value of this life annuity.

Solution.

(a) When Amy dies within the deferral period, she will not receive anything from the life annuity (also consider the above definition of $Y$). Hence, the probability is $_{30}q_{30}=1-{\frac {S_{0}(60)}{S_{0}(30)}}=1-{\frac {1-60/120}{1-30/120}}={\frac {1}{3}}$ (this is quite high actually!).

(b) The actuarial present value is

$$10000_{30|}{\bar {a}}_{30}=\int _{30}^{90}e^{-0.05t}{}_{t}p_{30}\,dt=10000\int _{30}^{90}e^{-0.05t}{\frac {1-(30+t)/120}{1-30/120}}\,dt=10000\int _{30}^{90}e^{-0.05t}{\frac {90-t}{90}}\,dt=10000\int _{30}^{90}e^{-0.05t}\,dt-{\frac {10000}{90}}\int _{30}^{90}te^{-0.05t}\,dt.$$

For the first integral, we have

$$10000\int _{30}^{90}e^{-0.05t}\,dt={\frac {1}{-0.05}}\left[e^{-0.05t}\right]_{30}^{90}\approx 42404.23272$$

For the second integral, we have

$${\frac {10000}{90}}\int _{30}^{90}te^{-0.05t}\,dt={\frac {10000}{90(-0.05)}}\int _{30}^{90}t\,d(e^{-0.05t})={\frac {10000}{90(-0.05)}}\left([te^{-0.05t}]_{30}^{90}-\int _{30}^{90}e^{-0.05t}\,dt\right)\approx {\frac {10000}{90(-0.05)}}(-5.694095-42404.2372/10000)\approx 22076.70827.$$

Hence, the actuarial present value is $42404.23272+22076.70827\approx 64480.90499.$

Exercise. Suppose Amy needs to purchase this life annuity by 10 end-of-year level payments, starting at age 30, where the present value of these 10 payments (at age 30) equals the actuarial present value of the life annuity (at age 30). Calculate the amount of each payment.

Solution

Let $P$ be the amount of each payment. Then, we have

$$Pa_{{\overline {10}}|}=64480.90499\iff P\left({\frac {1-v^{10}}{i}}\right)=64480.90499\iff P\left({\frac {1-e^{-0.05(10)}}{e^{0.05}-1}}\right)=64480.90499\iff P\approx 8402.201.$$

Remark.

• Notice that Amy can die before making all of the 10 payments, and then the payments for the life annuity of course stop.
• We will have more discussions about this when we introduce premiums in the next chapter.

n-year certain and life annuity

Let us discuss a new kind of concept that does not appear for insurances, and applies to annuities. For this kind of annuity, it is a "mixture" of annuity-certain (studied in financial mathematics) and life annuity.

Definition. ($n$-year certain and life annuity) A $n$-year certain and life annuity is a whole life annuity where payments in the first $n$ years are certainly made.

The present-value random variable of the $n$-year certain and life annuity is

$$Y={\begin{cases}{\bar {a}}_{{\overline {n}}|},&T\leq n;\\{\bar {a}}_{{\overline {T}}|},&T>n.\end{cases}}$$

Thus, the APV of this annuity, denoted by ${\bar {a}}_{\overline {x:{\overline {n}}|}}$, is

$$\mathbb {E} [Y]=\int _{0}^{n}{\bar {a}}_{{\overline {n}}|}{}_{t}p_{x}\mu _{x+t}\,dt+\int _{n}^{\infty }{\bar {a}}_{{\overline {t}}|}{}_{t}p_{x}\mu _{x+t}\,dt={\bar {a}}_{{\overline {n}}|}\int _{0}^{n}{}_{t}p_{x}\mu _{x+t}\,dt+\int _{n}^{\infty }{\bar {a}}_{{\overline {t}}|}{}_{t}p_{x}\mu _{x+t}\,dt={}_{n}q_{x}{\bar {a}}_{{\overline {n}}|}+\int _{n}^{\infty }{\bar {a}}_{{\overline {t}}|}{}_{t}p_{x}\mu _{x+t}\,dt.$$

Using current payment technique, we have

$${\bar {a}}_{\overline {x:{\overline {n}}|}}=\underbrace {{\bar {a}}_{{\overline {n}}|}} _{\text{certainly made}}+\int _{n}^{\infty }v^{t}{}_{t}p_{x}\,dt.$$

In particular, there is no survival probability involved for the "certain" part of this annuity: ${\bar {a}}_{{\overline {n}}|}$.

From the definition, we can observe that when $T>n$, there are some extra payments contingent on survival, and it appears that those extra payments form a $n$-year deferred whole life annuity. This is indeed the case, as shown in the following exercise.

Exercise.

(a) Show that ${\bar {a}}_{{\overline {T}}|}={\bar {a}}_{{\overline {n}}|}+v^{n}{\bar {a}}_{{\overline {T-n}}|}$.

(b) Hence, propose a relationship between the present-value random variable of $n$-year certain and life annuity (denoted by $Y_{1}$), that of $n$-year deferred whole life annuity (denoted by $Y_{2}$), and ${\bar {a}}_{{\overline {n}}|}$. Then, prove it.

(c) Hence, propose a relationship between ${\bar {a}}_{\overline {x:{\overline {n}}|}}$, $_{n|}{\bar {a}}_{x}$ and ${\bar {a}}_{{\overline {n}}|}$. Then, prove it.

Solution

(a)

Proof.

$${\bar {a}}_{{\overline {T}}|}={\frac {1-v^{T}}{\delta }}={\frac {1-v^{n}}{\delta }}+{\frac {v^{n}-v^{T}}{\delta }}={\bar {a}}_{{\overline {n}}|}+v^{n}\cdot {\frac {1-v^{T-n}}{\delta }}={\bar {a}}_{{\overline {n}}|}+v^{n}{\bar {a}}_{{\overline {T-n}}|}.$$

$\Box$

(b) Proposition: $Y_{1}=Y_{2}+{\bar {a}}_{{\overline {n}}|}$.

Proof. Notice that

$$Y_{1}={\begin{cases}{\bar {a}}_{{\overline {n}}|},&T\leq n;\\{\bar {a}}_{{\overline {T}}|},&T>n,\end{cases}}={\begin{cases}{\bar {a}}_{{\overline {n}}|},&T<n;\\{\bar {a}}_{{\overline {n}}|}+v^{n}{\bar {a}}_{{\overline {T-n}}||},&T\geq n,\end{cases}}$$

and

$$Y_{2}={\begin{cases}0,&T<n;\\v^{n}a_{{\overline {T-n}}|},&T\geq n.\end{cases}}$$

Then, we can observe that

$$Y_{1}=Y_{2}+{\bar {a}}_{{\overline {n}}|}.$$

$\Box$

(c) Proposition: ${\bar {a}}_{\overline {x:{\overline {n}}|}}=_{n|}{\bar {a}}_{x}+{\bar {a}}_{{\overline {n}}|}$.

Proof. Taking expectation on both side of the equation in (b) yields the desired result.

$\Box$

Example. An insurance company offers Bob, aged 60, two kinds of life annuity: life annuity V and W. For life annuity V, payments at an annual rate of 1000 are made continuously until Bob dies. For life annuity W, payments at an annual rate of $k$ are made certainly for the first 10 years. Afterward, if Bob still survives, then payments at an annual rate of 500 are made until Bob dies. Let $V$ and $W$ be the present-value random variable for life annuity V and W respectively. It is given that the annual force of interest is $\delta =0.04$ and the constant force of mortality is $\mu =0.06$.

(a) Define $V$ and $W$ based on the given information.

(b) Suppose the actuarial present value of life annuity V and W are the same. Calculate $k$.

Solution.

(a)

$$V=1000{\bar {a}}_{{\overline {T}}|},\qquad T\geq 0,\qquad {\text{ and }}\qquad W={\begin{cases}k{\bar {a}}_{{\overline {10}}|},&T\leq 10;\\k{\bar {a}}_{{\overline {10}}|}+500{\bar {a}}_{{\overline {T-10}}|},&T>10.\end{cases}}$$

(b) Since ${}_{t}p_{60}=e^{-0.06t}$, we have

$$1000{\bar {a}}_{60}=k{\bar {a}}_{{\overline {10}}|}+500{}_{10|}{\bar {a}}_{60}\iff 1000\int _{0}^{\infty }e^{-0.04t}e^{-0.06t}\,dt={\frac {k(1-e^{-0.04(10)})}{0.04}}+500\int _{10}^{\infty }e^{-0.04t}e^{-0.06t}\,dt\iff {\frac {1000}{-0.1}}\left[e^{0.1t}\right]_{0}^{\infty }=8.242k+{\frac {500}{-0.1}}\left[e^{-0.1t}\right]_{10}^{\infty }\iff 10000=8.242k+1839.3972\iff k\approx 880.124.$$

Exercise. Suppose $k=500$.

(a) Express the actuarial value of life annuity W using actuarial notations.

(b) What is the difference between the actuarial present value of life annuity V and W?

Solution

(a) The actuarial present value of life annuity W is $500{\bar {a}}_{\overline {60:{\overline {10}}|}}$ (since the life annuity W is now a 10-year certain and life annuity).

(b) The actuarial present value of life annuity V is 10000 from (b) in above example. Hence, it suffices to calculate the actuarial present value of life annuity W. Its APV is

$$500{\bar {a}}_{\overline {60:{\overline {10}}|}}=500\left({\bar {a}}_{{\overline {10}}|}+\int _{10}^{\infty }e^{-0.04t}e^{-0.06t}\,dt\right)=500\left({\frac {1-e^{-0.04(10)}}{0.04}}+{\frac {1}{-0.1}}\left[e^{-0.1t}\right]_{10}^{\infty }\right)\approx 5960.3966.$$

Hence, the difference between them is $10000-5960.3966\approx 4039.6034$.

Just like annuities-certain, there are also other types of continuous annuities, e.g. annually (continuously) increasing (decreasing) continuous annuities. We can also denote their APV's, and calculate their APV's (using current payment technique) similarly:

• $({\bar {I}}{\bar {a}})_{x}=\int _{0}^{\infty }tv^{t}{}_{t}p_{x}\,dt$
• $({\bar {I}}{\bar {a}})_{x:{\overline {n}}|}=\int _{0}^{n}tv^{t}{}_{t}p_{x}\,dt$
• $(I{\bar {a}})_{x}=\int _{0}^{\infty }\lfloor t+1\rfloor v^{t}{}_{t}p_{x}\,dt$
• $(I{\bar {a}})_{x:{\overline {n}}|}=\int _{0}^{n}\lfloor t+1\rfloor v^{t}{}_{t}p_{x}\,dt$
• $({\bar {D}}{\bar {a}})_{x:{\overline {n}}|}=\int _{0}^{n}(n-t)v^{t}{}_{t}p_{x}\,dt$
• $(D{\bar {a}})_{x:{\overline {n}}|}=\int _{0}^{n}(n-\lfloor t\rfloor )v^{t}{}_{t}p_{x}\,dt$

Of course, the payment patterns can also be a "mixture" of different kinds of patterns discussed, something that is irregular, etc. Then, for those life annuities, one can still calculate their APV using the current payment technique, by considering their present-value random variable from their definitions.

Discrete life annuities

After discussing continuous life annuities, let us discuss discrete life annuities. Just as the case for annuities-certain in financial mathematics, there are quite many types of discrete life annuities:

• $n$-year temporary/whole life annuity
• payable at the beginning/end of the year (due/immediate)
• payable annually/$m$-thly
• increasing/decreasing annually/$m$-thly

Again, for the life annuities discussed in this section, the amount of each payment is one unless otherwise specified.

In general, the present-value random variable of the discrete life annuities is a function of $K$, say $y(K)$. Then, their actuarial present values have the form of

$$\sum _{k}^{}y(k){}_{k}p_{x}q_{x+k}\quad ({\text{definition}}),$$

or using current payment technique, for the annuities with level payments of 1, their actuarial present values have the form of

$$\sum _{k}^{}v^{k}{}_{k}p_{x}.$$

Let us consider some simple cases first.

Whole life annuity-due (annuity-immediate)

Definition. (Whole life annuity-due (annuity-immediate)) A whole life annuity-due (annuity-immediate) provides a payment at the beginning (end) of each year while the annuitant still survives at the time of payment.

For whole life annuity-due, the present-value random variable $Y={\ddot {a}}_{{\overline {K+1}}|},\quad K=0,1,\dotsc$. The APV, denoted by ${\ddot {a}}_{x}$ is $\sum _{k=0}^{\infty }{\ddot {a}}_{{\overline {k+1}}|}{}_{k}p_{x}q_{x+k}$ by definition. Using current payment technique, we have ${\ddot {a}}_{x}=\sum _{k=0}^{\infty }v^{k}{}_{k}p_{x}$. We can observe that the formula developed by the current payment technique is much simpler.

One may ask that why $Y$ is defined to be ${\ddot {a}}_{{\overline {K+1}}|}$ instead of ${\ddot {a}}_{{\overline {K}}|}$. This is because the payment is made at the beginning of each year. So, even if the annuitant does not survive the whole year, he will still get the payment at the beginning, so there is a "$+1$". To be more precise, suppose the annuitant dies between year $k$ and $k+1$ (i.e. within year $k+1$). Then, $K=k$, but the annuitant gets the payment at the beginning of year $1,2,\dotsc ,k+1$, since he still survives at those time points. Therefore, the annuitant gets $k+1$ payments, and hence the present value is ${\ddot {a}}_{{\overline {k+1}}|}$.

It may appear that such annuity is advantageous to the annuitant, since the annuitant can get the payment for the whole year at the beginning of that year, even if he does not survive the whole year. In a later section, we will discuss a more "fair" life annuity-due: apportionable annuity-due, which requires the annuitant to refund some part of the payment to the payer of the life annuity if the annuitant does not survive the whole year.

Example. Show that ${\ddot {a}}_{x}={\frac {1-A_{x}}{d}}$.

Proof. Notice that the present-value random variable $Y={\ddot {a}}_{K+1}={\frac {1-v^{K+1}}{d}},\quad K=0,1,\dotsc$, and the present-value random variable of the whole life insurance payable at the end of year of death is $Z=v^{K+1},\quad K=0,1,\dotsc$. Then, we can observe that $Y={\frac {1-Z}{d}}$. Hence,

$${\ddot {a}}_{x}=\mathbb {E} [Y]=\mathbb {E} \left[{\frac {1-Z}{d}}\right]={\frac {1-\mathbb {E} [Z]}{d}}={\frac {1-A_{x}}{d}}.$$

$\Box$

For the derivations of whole life annuity-immediate, see the following exercise.

Exercise.

(a) Propose the present-value random variable of whole life annuity-immediate, $Y$, based on its definition.

(b) The APV of whole life annuity-immediate is denoted by $a_{x}$. Find the expression of $a_{x}$

(i) by definition;

(ii) by current payment technique.

(c) Propose a relationship between ${\ddot {a}}_{x}$ and $a_{x}$, and prove it.

Solution

(a) $Y=a_{{\overline {K}}|},\quad K=0,1,\dotsc$ (Notice that the annuitant needs to survive the whole year to get the payment for that year, which is made at the end of that year, in this case).

Remark.

• We can similarly notice some "unfairness" here. It appears that this annuity is advantageous to the payer of the life annuity this time, since even if the annuitant survives for a part of a year, and then die, he cannot get any payment for that year since he does not survive at the end of that year. Again, in a later section, we will discuss a way to address this issue by considering complete annuity-immediate, which requires the payer to pay some part of the payment for that year to the annuitant if he survives only part of the year.

(b) (i) $a_{x}=\sum _{k=0}^{\infty }a_{{\overline {k}}|}{}_{k}p_{x}q_{x+k}=\sum _{k=1}^{\infty }a_{{\overline {k}}|}{}_{k}p_{x}q_{x+k}$ (since $a_{{\overline {0}}|}=0$).

(ii) $a_{x}=\sum _{k=1}^{\infty }v^{k}{}_{k}p_{x}$ (the payment starts at time 1, and the annuitant needs to survive $k$ years to get the $k$th payment).

(c) Proposition: ${\ddot {a}}_{x}=1+a_{x}$.

Proof. By current payment technique, we have ${\ddot {a}}_{x}=\sum _{k=0}^{\infty }v^{k}{}_{k}p_{x}=v^{0}{}_{0}p_{x}+\sum _{k=1}^{\infty }v^{k}{}_{k}p_{x}=1+\sum _{k=1}^{\infty }v^{k}{}_{k}p_{x}$. Also, we have $a_{x}=\sum _{k=1}^{\infty }v^{k}{}_{k}p_{x}$. Hence, we have the desired result.

$\Box$

n-year temporary life annuity-due (annuity-immediate)

Definition. ($n$-year temporary life annuity-due (annuity-immediate)) A $n$-year temporary life annuity-due (annuity-immediate) provides a payment at the beginning (end) of each of the first $n$ years while the annuitant still survives at the time of payment.

Based on the definition, it should not be too hard to derive the present-value random variable and expressions of the APV of these annuities. The details are left to the following exercise.

Exercise.

(a) Fill in the missing information in the following table.

$${\begin{array}{ccc}&n{\text{-year temporary life annuity-due}}&n{\text{-year temporary life annuity-immediate}}\\\hline {\text{present-value random variable }}Y&?&?\\{\text{APV notation}}&{\ddot {a}}_{x:{\overline {n}}|}&a_{x:{\overline {n}}|}\\{\text{APV expression by definition}}&?&?\\{\text{APV expression by CPT}}&?&?\\\end{array}}$$

(b) Hence, propose a relationship between ${\ddot {a}}_{x:{\overline {n}}|}$ and $a_{x:{\overline {n}}|}$, and prove it.

Solution

(a)

$${\begin{array}{cc|c}&n{\text{-year temporary life annuity-due}}&n{\text{-year temporary life annuity-immediate}}\\\hline {\text{present-value random variable }}Y&{\begin{cases}{\ddot {a}}_{{\overline {K+1}}|},&K=0,1,\dotsc ,n-1;\\{\ddot {a}}_{{\overline {n}}|},&K=n,n+1,\dotsc \end{cases}}&{\begin{cases}{\ddot {a}}_{{\overline {K}}|},&K=0,1,\dotsc ,n-1;\\{\ddot {a}}_{{\overline {n}}|},&K=n,n+1,\dotsc \end{cases}}{\text{ OR }}{\begin{cases}{\ddot {a}}_{{\overline {K}}|},&K=0,1,\dotsc ,n;\\{\ddot {a}}_{{\overline {n}}|},&K=n+1,n+2,\dotsc \end{cases}}\\{\text{APV notation}}&{\ddot {a}}_{x:{\overline {n}}|}&a_{x:{\overline {n}}|}\\{\text{APV expression by definition}}&\sum _{k=0}^{n-1}{\ddot {a}}_{{\overline {k+1}}|}v^{k}{}_{k}p_{x}q_{x+k}+{\ddot {a}}_{{\overline {n}}|}\underbrace {\sum _{k=n}^{\infty }v^{k}{}_{k}p_{x}} _{{}_{n}p_{x}}&\sum _{k=0}^{n-1}a_{{\overline {k}}|}v^{k}{}_{k}p_{x}q_{x+k}+a_{{\overline {n}}|}\underbrace {\sum _{k=n}^{\infty }v^{k}{}_{k}p_{x}} _{{}_{n}p_{x}}{\text{ OR }}\sum _{k=0}^{n}a_{{\overline {k}}|}v^{k}{}_{k}p_{x}q_{x+k}+a_{{\overline {n}}|}\underbrace {\sum _{k=n+1}^{\infty }v^{k}{}_{k}p_{x}} _{{}_{n+1}p_{x}}\\{\text{APV expression by CPT}}&\sum _{k=0}^{n-1}v^{k}{}_{k}p_{x}&\sum _{k=1}^{n}v^{k}{}_{k}p_{x}\\\end{array}}$$

(b) Proposition: ${\ddot {a}}_{x:{\overline {n}}|}+v^{n}{}_{n}p_{x}=1+a_{x:{\overline {n}}|}$.

Proof. Consider the above APV expressions by CPT. We have

$${\ddot {a}}_{x:{\overline {n}}|}+v^{n}{}_{n}p_{x}=\sum _{k=0}^{n-1}v^{k}{}_{k}p_{x}+v^{n}{}_{n}p_{x}=\sum _{k=0}^{n}v^{k}{}_{k}p_{x},$$

and

$$1+a_{x:{\overline {n}}|}=1+\sum _{k=1}^{n}v^{k}{}_{k}p_{x}=\sum _{k=0}^{n}v^{k}{}_{k}p_{x}.$$

Hence, we have the desired result.

$\Box$

n-year deferred whole life annuity-due (annuity-immediate)

Definition. ($n$-year deferred whole life annuity-due (annuity-immediate)) A $n$-year deferred whole life annuity-due (annuity-immediate) provides a payment at the beginning (end) of each year while the annuitant still survives at the time of payment, starting at the beginning (end) of $n$th year.

Again, let's left the derivation of the present-value random variable and expressions of the APV of these annuities to the following exercise.

Exercise.

(a) Fill in the missing information in the following table (Hint: you may consider the $n$-year deferred whole life continuous annuity for some ideas).

$${\begin{array}{ccc}&n{\text{-year deferred whole life annuity-due}}&n{\text{-year deferred whole life annuity-immediate}}\\\hline {\text{present-value random variable }}Y&?&?\\{\text{APV notation}}&{}_{n|}{\ddot {a}}_{x}&_{n|}a_{x}\\{\text{APV expression by definition}}&?&?\\{\text{APV expression by CPT}}&?&?\\\end{array}}$$

(b) Hence, propose a relationship between $_{n|}{\ddot {a}}_{x}$ and $_{n|}a_{x}$, and prove it.

Solution

(a)

$${\begin{array}{cc|c}&n{\text{-year deferred whole life annuity-due}}&n{\text{-year deferred whole life annuity-immediate}}\\\hline {\text{present-value random variable }}Y&{\begin{cases}0,&K=0,1,\dotsc ,n-1;\\v^{n}{\ddot {a}}_{{\overline {K+1-n}}|},&K=n,n+1,\dotsc \end{cases}}&{\begin{cases}0,&K=0,1,\dotsc ,n-1;\\v^{n}a_{{\overline {K-n}}|},&K=n,n+1,\dotsc \end{cases}}{\text{ OR }}{\begin{cases}0,&K=0,1,\dotsc ,n;\\v^{n}a_{{\overline {K-n}}|},&K=n+1,n+2,\dotsc \end{cases}}\\{\text{APV notation}}&{}_{n|}{\ddot {a}}_{x}&_{n|}a_{x}\\{\text{APV expression by definition}}&\sum _{k=n}^{\infty }v^{n}{\ddot {a}}_{{\overline {K+1-n}}|}{}_{k}p_{x}q_{x+k}&\sum _{k=n}^{\infty }v^{n}a_{{\overline {K-n}}|}{}_{k}p_{x}q_{x+k}{\text{ OR }}\sum _{k=n+1}^{\infty }v^{n}a_{{\overline {K-n}}|}{}_{k}p_{x}q_{x+k}\\{\text{APV expression by CPT}}&\sum _{k=n}^{\infty }v^{k}{}_{k}p_{x}&\sum _{k=n+1}^{\infty }v^{k}{}_{k}p_{x}\\\end{array}}$$

(In particular, for the APV expression by CPT of $n$-year deferred whole life annuity-immediate, the payment starts at time $k+1$, since the payment starts at the end of $n$th year, which is time $k+1$.)

(b) Proposition: $_{n|}{\ddot {a}}_{x}={}_{n|}a_{x}+v^{n}{}_{n}p_{x}$.

Proof. The result follows by noticing that

$$_{n|}{\ddot {a}}_{x}=\sum _{k=n}^{\infty }v^{k}{}_{k}p_{x}=v^{n}{}_{n}p_{x}+\sum _{k=n+1}^{\infty }v^{k}{}_{k}p_{x}=v^{n}{}_{n}p_{x}+{}_{n|}a_{x}.$$

$\Box$

n-year certain and life annuity-due (annuity-immediate)

Definition. ($n$-year certain and life annuity-due (annuity-immediate)) A $n$-year certain and life annuity-due (annuity-immediate) is a whole life annuity-due (annuity-immediate) where payments in the first $n$ years are certainly made.

Exercise.

(a) Fill in the missing information in the following table (Hint: you may consider the $n$-year continuous certain and life annuity for some ideas).

$${\begin{array}{ccc}&n{\text{-year certain and life annuity-due}}&n{\text{-year certain and life annuity-immediate}}\\\hline {\text{present-value random variable }}Y&?&?\\{\text{APV notation}}&{\ddot {a}}_{\overline {x:{\overline {n}}|}}&a_{\overline {x:{\overline {n}}|}}\\{\text{APV expression by definition}}&?&?\\{\text{APV expression by CPT}}&?&?\\\end{array}}$$

(b) Hence, propose a relationship between ${\ddot {a}}_{\overline {x:{\overline {n}}|}}$ and $a_{\overline {x:{\overline {n}}|}}$, and prove it.

Solution

(a)

$${\begin{array}{cc|c}&n{\text{-year certain and life annuity-due}}&n{\text{-year certain and life annuity-immediate}}\\\hline {\text{present-value random variable }}Y&{\begin{cases}{\ddot {a}}_{{\overline {n}}|},&K=0,1,\dotsc ,n-1;\\{\ddot {a}}_{{\overline {K+1}}|},&K=n,n+1,\dotsc \end{cases}}&{\begin{cases}a_{{\overline {n}}|},&K=0,1,\dotsc ,n-1;\\a_{{\overline {K}}|},&K=n,n+1,\dotsc \end{cases}}{\text{ OR }}{\begin{cases}a_{{\overline {n}}|},&K=0,1,\dotsc ,n;\\a_{{\overline {K}}|},&K=n+1,n+2,\dotsc \end{cases}}\\{\text{APV notation}}&{\ddot {a}}_{\overline {x:{\overline {n}}|}}&a_{\overline {x:{\overline {n}}|}}\\{\text{APV expression by definition}}&{\ddot {a}}_{{\overline {n}}|}\underbrace {\sum _{k=0}^{n-1}{}_{k}p_{x}q_{x+k}} _{{}_{n}q_{x}}+\sum _{k=n}^{\infty }{\ddot {a}}_{{\overline {k+1}}|}{}_{k}p_{x}q_{x+k}&a_{{\overline {n}}|}\underbrace {\sum _{k=0}^{n-1}{}_{k}p_{x}q_{x+k}} _{{}_{n}q_{x}}+\sum _{k=n}^{\infty }a_{{\overline {k+1}}|}{}_{k}p_{x}q_{x+k}{\text{ OR }}a_{{\overline {n}}|}\underbrace {\sum _{k=0}^{n}{}_{k}p_{x}q_{x+k}} _{{}_{n+1}q_{x}}+\sum _{k=n+1}^{\infty }a_{{\overline {k+1}}|}{}_{k}p_{x}q_{x+k}\\{\text{APV expression by CPT}}&{\ddot {a}}_{{\overline {n}}|}+\sum _{k=n}^{\infty }v^{k}{}_{k}p_{x}&a_{{\overline {n}}|}+\sum _{k=n+1}^{\infty }v^{k}{}_{k}p_{x}\\\end{array}}$$

(b) Proposition: ${\ddot {a}}_{\overline {x:{\overline {n|}}}}=1-v^{n}+v^{n}{}_{n}p_{x}+a_{\overline {x:{\overline {n|}}}}$

Proof. First, we have

$${\ddot {a}}_{{\overline {n}}|}=1+a_{{\overline {n}}|}-v^{n}$$

(intuitively, after adding payment of 1 at time 0 and taking away the last payment from the $n$-year annuity-immediate, it becomes $n$-year annuity-due). Also,

$$\sum _{k=n}^{\infty }v^{k}{}_{k}p_{x}=v^{n}{}_{n}p_{x}+\sum _{k=n+1}^{\infty }v^{k}{}_{k}p_{x}.$$

Hence,

$${\ddot {a}}_{\overline {x:{\overline {n|}}}}={\ddot {a}}_{{\overline {n}}|}+\sum _{k=n}^{\infty }v^{k}{}_{k}p_{x}=(1+a_{{\overline {n}}|}-v^{n})+\left(v^{n}{}_{n}p_{x}+\sum _{k=n+1}^{\infty }v^{k}{}_{k}p_{x}\right)=1-v^{n}+v^{n}{}_{n}p_{x}+a_{\overline {x:{\overline {n|}}}}.$$

$\Box$

Example. The parents of Chris purchase a life annuity for Chris, who is aged 10, which pays $1000 at the beginning of each year, starting at age 18.

(a) This life annuity can be classified as $m$-year deferred whole life annuity-due or $n$-year deferred whole life annuity-immediate. What is $m$ and $n$?

(b) Calculate the actuarial present value of this life annuity, assuming the force of interest is $\delta =0.02$ and the force of mortality is $\mu =0.01$.

Solution.

(a) Since the beginning of year 18 can be interpreted as the end of year 17, or in general, the beginning of year $k$ can be interpreted as the end of year $k-1$, we have $m=8$ and $n=7$.

(b) The actuarial present value is

$$1000{}_{8|}{\ddot {a}}_{\overline {10|}}=1000\sum _{k=8}^{\infty }v^{k}{}_{k}p_{10}=1000\sum _{k=8}^{\infty }e^{-0.02k}{\frac {e^{-0.01(k+10)}}{e^{-0.01(10)}}}=1000\sum _{k=8}^{\infty }e^{-0.03k}=1000\cdot \underbrace {\frac {e^{-0.24}}{1-e^{-0.03}}} _{\text{geometric series}}\approx 26616.20917.$$

Life annuities-due (annuities-immediate) with m-thly payments

In practice, the payment frequency of the life annuities may not be annual. Instead, it can be semiannual, quarterly, monthly, etc. Thus, we will develop life annuities-due (annuities-immediate) payable $m$-thly here. For the annuities here, we assume the total payment is 1 per year, payable in installments of $1/m$ at the beginning (or end) of every $m$-th of the year. Graphically,

Annuity-due:
  1/m  1/m   1/m         1/m
--*-----*-----*----...----*-----*-----
  0    1/m   2/m   ...   1-1/m  1      time
Annuity-immediate:
       1/m   1/m         1/m   1/m
--*-----*-----*----...----*-----*-----
  0    1/m   2/m   ...   1-1/m  1      time

Definition. (Whole life annuity-due (annuity-immediate) with $m$-thly payment) A whole life annuity-due (annuity-immediate) with $m$-thly payment provides a payment at the beginning (end) of each $m$-th of a year while the annuitant still survives at the time of payment.

We can develop the present-value random variable and APV using similar ideas as in the insurances payable $m$-thly: introducing a random variable $J$ that represents which interval within one year does death occur. In the following, we will first develop the present-value random variable and APV for whole life annuity-due with $m$-thly payment. Then, we will relate whole life annuity-due payable $m$-thly with whole life annuity-immediate payable $m$-thly, so we can use the whole life annuity-due with $m$-thly payment as a bridge for the calculation related to whole life annuity-immediate with $m$-thly payment.

For whole life annuity-due with $m$-thly payment, the present-value random variable is

$$Y={\ddot {a}}_{{\overline {K+{\frac {J+1}{m}}}}{\big |}}^{(m)}=\sum _{j=0}^{mK+J}{\frac {1}{m}}v^{j/m}={\frac {1-v^{K+{\frac {J+1}{m}}}}{d^{(m)}}},\quad K=0,1,\dotsc ,{\text{ and }}J=0,1,\dotsc ,m-1.$$

Graphically,

Annuity-due:
                                          death: K=k, J=m-2
                                            |
1/m  ...  1/m  1/m   1/m     ...      1/m   v     
*----...--*-----*-----*------...-------*--------*----------*------
0         k   k+1/m  k+2/m   ...   k+(m-2)/m   k+(m-1)/m  k+1      time

It appears that the present-value random variable of this life annuity is quite complicated. Hence, the expression of the actuarial present value by definition will be quite complex, and therefore not useful for the actual calculation.

Thus, we will just use the current payment technique and the following method to get an expression of the APV, denoted by ${\ddot {a}}_{x}^{(m)}$.

Exercise. Propose a relationship between ${\ddot {a}}_{x}^{(m)}$ and $A_{x}^{(m)}$, and prove it.

Solution

Proposition: ${\ddot {a}}_{x}^{(m)}={\frac {1-A_{x}^{(m)}}{d^{(m)}}}$.

Proof. Notice that $Z=v^{K+{\frac {J+1}{m}}},\quad K=0,1,\dotsc ,{\text{ and }}J=0,1,\dotsc ,m-1.$ where $Z$ is the present-value random variable of whole life insurance payable $m$-thly. Hence, we can observe that $Y={\frac {1-Z}{d^{(m)}}}$ where $Y$ is the present-value random variable of whole life annuity-due with $m$-thly payment. Thus,

$${\ddot {a}}_{x}^{(m)}={\frac {1-A_{x}^{(m)}}{d^{(m)}}}.$$

$\Box$

For the current payment technique, we have

$${\ddot {a}}_{x}^{(m)}={\frac {1}{m}}\sum _{j=0}^{\infty }v^{j/m}{}_{j/m}p_{x}.$$

However, the above two expressions can also be be quite difficult to be calculated sometimes. In particular, if we are not given $A_{x}^{(m)}$, it is complicated to calculate $A_{x}^{(m)}$, and then apply the relationship between ${\ddot {a}}_{x}^{(m)}$ and $A_{x}^{(m)}$. Also, for the current payment technique, we may the term in the sum may not be "nice", and so there may not be a formula for calculating the sum efficiently. Hence, in the following, we will give a third approach of expressing ${\ddot {a}}_{x}^{(m)}$, which relates it with ${\ddot {a}}_{x}$, under UDD assumption.

Example. The following is under UDD assumption.

(a) Prove that $\mathbb {P} (J=j)={\frac {1}{m}}$ ($J$ is as defined previously) for each $j\in \{0,1,\dotsc ,m-1\}$.

(b) Hence, or otherwise, prove that $K$ and $J$ (as defined previously) are independent.

(c) Hence, or otherwise, prove that $A_{x}^{(m)}={\frac {i}{i^{(m)}}}A_{x}$. (analogous to ${\bar {A}}_{x}={\frac {i}{\delta }}A_{x}$)

Solution.

(a)

Proof. First, notice that $J=0,1,\dotsc ,m-1$ corresponds to death in the 1st,2nd,...,(m-1)th of ${\frac {1}{m}}$ of year. Also, recall that the fractional part of a year lived in the year of death ($S=T-K$) follows the uniform distribution on $(0,1)$ (the pdf is simply $f(x)=1,\quad x\in (0,1)$). Hence, the probability of death in the 1st,2nd,...,(m-1)th of ${\frac {1}{m}}$ of year is $\int _{0}^{1/m}1\,dt={\frac {1}{m}},\dotsc ,\int _{(m-1)/m}^{1}1\,dt={\frac {1}{m}}$ respectively.

$\Box$

(b)

Proof. Since for every $k\in \{0,1,\dotsc \}$ and $j\in \{0,1,\dotsc ,m-1\}$,

$$\mathbb {P} (K=k{\text{ and }}J=j)=\mathbb {P} \left(k+{\frac {j}{m}}\leq T<k+{\frac {j+1}{m}}\right)={}_{k}p_{x}\left({}_{\frac {j+1}{m}}q_{x+k}-{}_{\frac {j}{m}}q_{x+k}\right){\overset {\text{UDD}}{=}}{}_{k}p_{x}\left({\frac {j+1}{m}}q_{x+k}-{\frac {j}{m}}q_{x+k}\right)={\color {blue}{}_{k}p_{x}q_{x+k}}\cdot {\color {red}{\frac {1}{m}}}={\color {blue}\mathbb {P} (K=k)}{\color {red}\mathbb {P} (J=j)},$$

$K$ and $J$ are independent by definition.

$\Box$

(c)

Proof.

$${\begin{aligned}A_{x}^{(m)}&=\mathbb {E} \left[v^{K+{\frac {J+1}{m}}}\right]&({\text{definition}})\\&=\mathbb {E} [v^{K{\color {darkgreen}+1}}]\mathbb {E} \left[v^{{\frac {J+1}{m}}{\color {darkgreen}-1}}\right]&(K{\text{ and }}J{\text{ are independent}})\\&=A_{x}\cdot v^{{\frac {1}{m}}-1}\mathbb {E} \left[v^{\frac {J}{m}}\right]&({\text{definition, linearity of expectation}})\\&=A_{x}v^{{\frac {1}{m}}-1}\sum _{j=0}^{m-1}\left(v^{\frac {j}{m}}\cdot {\frac {1}{m}}\right)&({\text{definition}})\\&=A_{x}\cdot {\frac {v^{{\frac {1}{m}}-1}}{m}}\cdot {\frac {1-(v^{1/m})^{m}}{1-v^{1/m}}}&({\text{geometric series formula}})\\&=A_{x}\cdot {\frac {1}{m}}\cdot {\frac {v^{-1}(1-v)}{v^{-1/m}(1-v^{1/m})}}\\&=A_{x}\cdot {\frac {1}{m}}\cdot {\frac {1+i-1}{(1+i)^{1/m}-1}}\\&=A_{x}\cdot {\frac {i}{m{\big (}(1+i)^{1/m}-1{\big )}}}\\&={\frac {i}{i^{(m)}}}A_{x}.&({\text{definition}})\end{aligned}}$$

$\Box$

Exercise. Hence, or otherwise, prove that ${\ddot {a}}_{x}^{(m)}=\alpha (m){\ddot {a}}_{x}-\beta (m)$ where $\alpha (m)={\frac {id}{i^{(m)}d^{(m)}}}$ and $\beta (m)={\frac {i-i^{(m)}}{i^{(m)}d^{(m)}}}$. (Hint: apart from considering the above example, also consider the relationship between ${\ddot {a}}_{x}^{(m)}$ and $A_{x}^{(m)}$ in a previous exercise.)

Solution

Proof.

$${\begin{aligned}{\ddot {a}}_{x}^{(m)}&={\frac {1-A_{x}^{(m)}}{d^{(m)}}}&({\text{previous exercise}})\\&={\frac {1-{\frac {i}{i^{(m)}}}A_{x}}{d^{(m)}}}&({\text{(c)}})\\&={\frac {1-{\frac {i}{i^{(m)}}}(1-d{\ddot {a}}_{x})}{d^{(m)}}}\\&={\frac {1-{\frac {i}{i^{(m)}}}}{d^{(m)}}}+{\frac {d{\ddot {a}}_{x}}{d^{(m)}}}\cdot {\frac {i}{i^{(m)}}}\\&={\frac {i^{(m)}-i}{i^{(m)}d^{(m)}}}+{\frac {id}{i^{(m)}d^{(m)}}}{\ddot {a}}_{x}\\&=\alpha (m){\ddot {a}}_{x}-\beta (m).\end{aligned}}$$

$\Box$

Apart from the whole life annuity-due with $m$-thly payment, we can define $n$-year temporary life annuity-due with $m$-thly payment, $n$-year deferred whole life annuity-due with $m$-thly payment, etc., similarly. Their notations are constructed in a similar manner.

Exercise. Based on the relation that ${\ddot {a}}_{x}^{(m)}=\alpha (m){\ddot {a}}_{x}-\beta (m)$, we can deduce two corollaries:

• Corollary 1: ${\ddot {a}}_{x:{\overline {n}}|}^{(m)}=\alpha (m){\ddot {a}}_{x:{\overline {n}}|}-\beta (m){\color {darkgreen}(1-v^{n}{}_{n}p_{x})}$.
• Corollary 2: ${}_{n|}{\ddot {a}}_{x}^{(m)}=\alpha (m){}_{n|}{\ddot {a}}_{x}-\beta (m){\color {darkgreen}v^{n}{}_{n}p_{x}}$.

Prove them. (Hint: we have recursion relations for the life annuities with $m$-thly payment which can be explained intuitively:

• ${\ddot {a}}_{x}^{(m)}={\ddot {a}}_{x:{\overline {n}}|}^{(m)}+v^{n}{}_{n}p_{x}{\ddot {a}}_{x+n}^{(m)}$.
• ${}_{n|}{\ddot {a}}_{x}^{(m)}=v^{n}{}_{n}p_{x}{\ddot {a}}_{x+n}^{(m)}$.

)

Solution

Proof.

Corollary 1: Applying recursion relations, we have

$${\color {blue}{\ddot {a}}_{x}^{(m)}}=\alpha (m){\color {red}{\ddot {a}}_{x}}-\beta (m)\implies {\color {blue}{\ddot {a}}_{x:{\overline {n}}|}^{(m)}+v^{n}{}_{n}p_{x}{\ddot {a}}_{x+n}^{(m)}}=\alpha (m)({\color {red}{\ddot {a}}_{x:{\overline {n}}|}+v^{n}{}_{n}p_{x}{\ddot {a}}_{x+n}})-\beta (m)\implies {\ddot {a}}_{x:{\overline {n}}|}^{(m)}+v^{n}{}_{n}p_{x}({\color {darkgreen}{\cancel {\alpha (m){\ddot {a}}_{x+n}}}-\beta (m)})=\alpha (m)({\ddot {a}}_{x:{\overline {n}}|}{\cancel {+v^{n}{}_{n}p_{x}{\ddot {a}}_{x+n}}})-\beta (m).$$

Then, rearranging the last equation yields the desired result.

Corollary 2: Applying recursion relations, we have

$${\color {blue}{\ddot {a}}_{x+n}^{(m)}}=\alpha (m){\color {red}{\ddot {a}}_{x+n}}-\beta (m)\implies {\color {blue}{\frac {{}_{n|}{\ddot {a}}_{x}^{(m)}}{v^{n}{}_{n}p_{x}}}}=\alpha (m)\left({\color {red}{\frac {{}_{n|}{\ddot {a}}_{x}}{v^{n}{}_{n}p_{x}}}}\right)-\beta (m)$$

Then, multiplying $v^{n}{}_{n}p_{x}$ on both sides of the last equation yields the desired result.

$\Box$

Now, we discuss whole life annuity-immediate with $m$-thly payment. As you may expect, the present-value random variable is quite similar:

$$Y=a_{\overline {K+{\frac {J}{m}}{\big |}}}^{(m)},\quad K=0,1,\dotsc ,{\text{ and }}J=0,1,\dotsc ,m-1.$$

Graphically,

Annuity-immediate:
                                                  death: K=k, J=m-2
                                                    |
     1/m  ....    1/m  1/m   1/m     ...     1/m    v     
*-----*---....----*-----*-----*------...-------*--------*----------*------
0     1           k   k+1/m  k+2/m   ...   k+(m-2)/m   k+(m-1)/m  k+1      time

By current payment technique, the actuarial present value of whole life annuity-immediate with $m$-thly payment, denoted by $a_{x}^{(m)}$, is

$$a_{x}^{(m)}={\frac {1}{m}}\sum _{j=1}^{\infty }v^{j/m}{}_{j/m}p_{x}.$$

Thus, we can relate $a_{x}^{(m)}$ to ${\ddot {a}}_{x}^{(m)}$, as follows.

Example. Show that $a_{x}^{(m)}={\ddot {a}}_{x}^{(m)}-{\frac {1}{m}}$.

Proof. By current payment technique,

$$a_{x}^{(m)}={\frac {1}{m}}\sum _{j=1}^{\infty }v^{j/m}{}_{j/m}p_{x}={\frac {1}{m}}\sum _{j=0}^{\infty }v^{j/m}{}_{j/m}p_{x}-{\frac {1}{m}}\underbrace {v^{0/m}{}_{0/m}p_{x}} _{1}={\ddot {a}}_{x}^{(m)}-{\frac {1}{m}}.$$

$\Box$

Exercise. Show that $a_{x:{\overline {n}}|}^{(m)}={\ddot {a}}_{x:{\overline {n}}|}^{(m)}-{\frac {1}{m}}(1-v^{n}{}_{n}p_{x})$. (The notation $a_{x:{\overline {n}}|}^{(m)}$ means the $n$-year temporary life annuity-immediate with $m$-thly payment. This is similar to the corresponding life annuity-due.)

Solution

Proof. Applying recursion relations, we have

$$a_{x}^{(m)}={\ddot {a}}_{x}^{(m)}-{\frac {1}{m}}\implies a_{x:{\overline {n}}|}^{(m)}+v^{n}{}_{n}p_{x}{\color {darkgreen}a_{x+n}^{(m)}}={\ddot {a}}_{x:{\overline {n}}|}^{(m)}+v^{n}{}_{n}p_{x}{\ddot {a}}_{x+n}^{(m)}-{\frac {1}{m}}.\implies a_{x:{\overline {n}}|}^{(m)}+v^{n}{}_{n}p_{x}({\color {darkgreen}{\cancel {{\ddot {a}}_{x+n}^{(m)}}}-1/m})={\ddot {a}}_{x:{\overline {n}}|}^{(m)}{\cancel {+v^{n}{}_{n}p_{x}{\ddot {a}}_{x+n}^{(m)}}}-{\frac {1}{m}}.$$

Then, rearranging the last equation yields the desired result.

$\Box$

Apportionable annuities-due and complete annuities-immediate

Previously, we have mentioned that the life annuities-due are advantageous to the annuitant, and the life annuities-immediate are advantageous to the payer. These cause some "unfairness", and the essential cause of this unfairness is that the annuities payments are discrete, rather than continuous. (We do not have this issue if the payments are made continuously.) Thus, it is natural to address this issue by appropriately "converting" the discrete payments to some "equivalent" continuous payments. This is the key idea for developing the apportionable annuities-due and complete annuities-immediate.

Let us the consider the case for whole life annuities-due (annuities immediate) first. In this case, these two kinds of annuities are addressing the unfairness issues in whole life annuities-immediate and whole life annuities-due.

Let us first consider apportionable annuities-due. Consider the following diagram.

    $1
     |   die
     v    |
-----*----------*------------- time
     k    t    k+1
     \    /\    /
      \  /  \  /
       \/    \/
    earned  unearned
    portion portion

Suppose the annuitant receives an annual payment from the life annuity-due at time $K=k$ (i.e. at the beginning of year $k+1$), and then dies at time $T=t$ (within year $k+1$). In this case, the annuitant only lives for a portion of the year $k+1$, so he should just earn a "portion" of the benefit for the year $k+1$. To determine the size of the benefit earned, we assume the payments within each year are earned continuously at a certain annual rate $r$ such that $r{\bar {a}}_{{\overline {1}}|}=1$ (so that each stream of payments has the same present value (i.e. value at the beginning of the year) as the single payment made at the beginning of the year). Thus, in terms of the value at time $t$, the annuitant earns $r{\bar {s}}_{{\overline {t-k}}|}$ (retrospective), and the remaining $r{\bar {a}}_{{\overline {1-t}}|}$ is unearned (prospective). Of course, for the previous years: year 1,2,...,$k$, all continuously paid payments in each are earned. Therefore, to summarize, the annuitant can earn the payments at an annual rate $r$ continuously until death. Notice that this earning of payment is the same as the earning in whole life continuous annuity. Since the payment rate

$$r={\frac {1}{{\bar {a}}_{{\overline {1}}|}}}={\frac {\delta }{1-v}}={\frac {\delta }{d}},$$

(recall that $v=1-d$ from a result in financial mathematics) the actuarial present value of this apportionable annuity-due, denoted by ${\ddot {a}}_{x}^{\{1\}}$ is

$${\ddot {a}}_{x}^{\{1\}}={\frac {\delta }{d}}{\bar {a}}_{x}.$$

For the unearned portion, the annuitant should refund to the payer at the moment of death, where the value of the refund payment at that moment (time $t$) is

$$r{\bar {a}}_{{\overline {1-t}}|}={\frac {\delta }{d}}{\bar {a}}_{{\overline {1-t}}|},$$

as suggested above.

We can extend this idea to whole life annuity-due with $m$-thly payment.

    $1/m
     |   die
     v    |
-----*----------*------------- time
     k    t    k+1/m
     \    /\    /
      \  /  \  /
       \/    \/
    earned  unearned
    portion portion

In this case, we assume the payments within each $1/m$ of year are earned continuously at a certain $m$thly rate $r'$ such that $r'{\bar {a}}_{1/m}=1/m$ (so that each stream of payments has the same present value as the single payment of $1/m$ made at the beginning). Now it follows that in terms of the value at time $t$, the annuitant earns $r'{\bar {s}}_{{\overline {t-k}}|}$ (retrospetive), and the remaining $r'{\bar {a}}_{{\overline {1/m-t}}|}$ is unearned (prospective). Since

$$r'={\frac {1/m}{{\bar {a}}_{{\overline {1/m}}|}}}={\frac {\delta /m}{1-v^{1/m}}}={\frac {\delta }{m(1-v^{1/m})}},$$

and

$$\left(1-{\frac {d^{(m)}}{m}}\right)^{-m}=v^{-1}\implies 1-{\frac {d^{(m)}}{m}}=v^{1/m}\implies d^{(m)}=m(1-v^{1/m}),$$

we have

$$r'={\frac {\delta }{d^{(m)}}}.$$

Thus, the actuarial present value of the apportionable annuity-due (for whole life annuity-due with $m$-thly payments), denoted by ${\ddot {a}}_{x}^{\{m\}}$ is similarly

$${\ddot {a}}_{x}^{\{m\}}={\frac {\delta }{d^{(m)}}}{\bar {a}}_{x}.$$

Exercise. Determine the value of the refund payment at time $t$ in this case.

Solution

The value is ${\frac {\delta }{d^{(m)}}}{\bar {a}}_{{\overline {1/m-t}}|}$.

For the complete annuities-immediate, the situation is somehow "reversed". Consider the following diagram.

    $1/m       $1/m (cannot get)
     |   die    |    
     v    |     v 
-----*----------*------------- time
     k    t    k+1/m
     \    /\    /
      \  /  \  /
       \/    \/
    earned  unearned
    portion portion

We have similar treatments on the payments within each year. We assume the payments within each year are earned continuously at a certain rate $r$ such that $r{\bar {s}}_{{\overline {1/m}}|}=1/m$ (so that each stream of payments has the same accumulated value (i.e. value at the end of the $m$th of year) as the single payment made at the end of the $m$th of year). Then, in terms of the value at time $t$, the annuitant earns $r{\bar {s}}_{{\overline {t-k}}|}$ (retrospective), and the remaining $r{\bar {a}}_{{\overline {1/m-t}}|}$ is unearned (prospective). Since

$$r={\frac {1/m}{{\bar {s}}_{{\overline {1/m}}|}}}={\frac {\delta /m}{(1+i)^{1/m}-1}}={\frac {\delta }{m((1+i)^{1/m}-1)}}={\frac {\delta }{i^{(m)}}},$$

the actuarial present value of the complete annuity-immediate (for whole life annuity-immediate with $m$-thly payment), denoted by ${\overset {\circ }{a}}_{x}^{(m)}$ (remark: when $m=1$, the "$(m)$" is omitted in this case), is

$${\overset {\circ }{a}}_{x}^{(m)}={\frac {\delta }{i^{(m)}}}{\bar {a}}_{x}.$$

Of course, in this case, the annuitant does not receive any payment for the year of death, and hence is not responsible for "refund". Instead, the payer should be responsible to pay additional payment (the amount of earned payments yet not paid) to the annuitant, at time $t$. We call such payment as adjustment payment.

Exercise. Determine the value of the adjustment payment at time $t$.

Solution

The value of the adjustment payment is the value of the payments earned by the annuitant in the year of death. Hence, the value of the adjustment payment is

$${\frac {\delta }{i^{(m)}}}{\bar {s}}_{{\overline {t-k}}|}.$$

Example. Assume the constant force of mortality is $\mu =0.04$ and the constant force of interest is $\delta =0.06$. Calculate the actuarial present value of a complete life annuity-immediate issued to a life aged $x$ with 1000 payable at the end of each quarter.

Solution. Under the constant force of mortality and force of interest, we have

$${\bar {a}}_{x}=\int _{0}^{\infty }v^{t}{}_{t}p_{x}\,dt=\int _{0}^{\infty }e^{-0.06t}e^{-0.04t}\,dt={\frac {1}{-0.1}}[e^{-0.1t}]_{0}^{\infty }=10.$$

Notice that in each quarter, the payment is of $1000=4000/4$. Also, we have

$$\left(1+{\frac {i^{(4)}}{4}}\right)^{4}=e^{0.06}\implies i^{(4)}\approx 0.0605.$$

Hence, the required present value is

$$4000{\overset {\circ }{a}}_{x}^{(4)}\approx {\frac {4000(0.06)}{0.0605}}\cdot {\bar {a}}_{x}\approx 39669.421.$$

Exercise. Calculate the actuarial present value of the apportionable annuity-due issued to a life aged $x$ with 1000 payable at the beginning of each quarter.

Solution

Since

$$\left(1-{\frac {d^{(4)}}{4}}\right)^{-4}=e^{0.06}\implies d^{(4)}\approx 0.0596,$$

the actuarial present value is

$$4000{\ddot {a}}_{x}^{\{4\}}={\frac {4000(0.06)}{0.0596}}\cdot {\bar {a}}_{x}\approx 40268.456.$$

Similarly, we have ${\ddot {a}}_{x:{\overline {n}}|}^{\{m\}}$ and ${\overset {\circ }{a}}_{x:{\overline {n}}|}^{(m)}$, which correspond to $n$-year temporary life annuities. Using analogous arguments, we can show that

• ${\ddot {a}}_{x:{\overline {n}}|}^{\{m\}}={\frac {\delta }{d^{(m)}}}{\bar {a}}_{x:{\overline {n}}|}$ and
• ${\overset {\circ }{a}}_{x:{\overline {n}}|}^{(m)}={\frac {\delta }{i^{(m)}}}{\bar {a}}_{x:{\overline {n}}|}$.

(There are no payments after a certain time point for each of the case. So, we divide our arguments by cases. For the case where there are payments, it is similar to the above argument. For the case where there are no payments, the actuarial present value is simply zero.)

Recursion relations

Similar to the case for insurances, we can also develop recursion relations for life annuities intuitively. For life annuities, the current payment technique is quite important for the development. This is because when we use current payment technique, we can "split" the potential future payments in some ways, similar to the case for insurances where the "coverage" of the insurance is split. As a result, we can develop similar recursion relations for life annuities, e.g., (the intuitive explanations is in the brackets)

• ${\bar {a}}_{x}={\bar {a}}_{x:{\overline {n}}|}+v^{n}{}_{n}p_{x}{\bar {a}}_{x+n}$ (split the potential payments to two parts: payments in first $n$ years (${\bar {a}}_{x:{\overline {n}}|}$) and payments afterward (${\bar {a}}_{x+n}$). Then, actuarially discount (multiply $v^{n}{}_{n}p_{x}$) the payments afterward to age $x$.
• ${\ddot {a}}_{x}={\ddot {a}}_{x:{\overline {n}}|}+v^{n}{}_{n}p_{x}{\ddot {a}}_{x+n}$ (similar to the first one)
• $a_{x}^{(m)}=a_{x:{\overline {n}}|}^{(m)}+v^{n}{}_{n}p_{x}a_{x+n}^{(m)}$ (similar to the first one)
• ${\ddot {a}}_{x}^{\{m\}}={\ddot {a}}_{x:{\overline {n}}|}^{\{m\}}+v^{n}{}_{n}p_{x}{\ddot {a}}_{x+n}^{\{m\}}$ (similar to the first one)
• $(I{\ddot {a}})_{x:{\overline {n}}|}={\ddot {a}}_{x:{\overline {n}}|}+vp_{x}(I{\ddot {a}})_{x+1:{\overline {n}}|}$ ("extract" an unit payment for each of the payments to form an life annuity-due (${\ddot {a}}_{x:{\overline {n}}|}$), and then the remaining payments form "$(I{\ddot {a}})_{x+1:{\overline {n}}|}$", and we need to actuarially discount (multiply $vp_{x}$) it back to age $x$).

Incorporating selecting ages to life annuities

Of course, we can incorporate selecting ages to life annuities. We just replace some terms related to the survival probability by their values for selection ages. Also, we can similarly put a square bracket in the notation when selecting ages are involved, e.g. ${\bar {a}}_{[x]}$. The idea involved is quite similar to the case for insurances: we often determine appropriate values for selection ages from life tables.