Financial Math FM/Stochastic Interest

Stochastic interest

In this book, we have mainly discussed deterministic (i.e. non-random) interest, and we will briefly introduce stochastic (i.e. random) interest, by regarding the interest rate as a random variable. We use the following notations:

• $I_{t}$: interest rate random variable for the period $t-1$ to $t$
• $\mu _{t}$: mean of $I_{t}$
• $\sigma _{t}^{2}$: variance of $I_{t}$

Accumulation of single investment

Example. (some simple formulas) For the accumulation of a unit sum of money over the period $t-1$ to $t$,

• the mean is $\mathbb {E} [1+I_{t}]=1+\mathbb {E} [I_{t}]=1+\mu _{t}$ (for non-random interest, it is $1+i_{t}$)
• the variance is $\operatorname {Var} (1+I_{t})=\operatorname {Var} (I_{t})=\sigma _{t}^{2}$
• the 2nd moment is $\mathbb {E} [(1+I_{t})^{2}]=\operatorname {Var} (1+I_{t})+(\mathbb {E} [1+I_{t}])^{2}=\sigma _{t}^{2}+(1+\mu _{t})^{2}$

Accumulation of single payment over several time periods

Assume that $I_{t}$ are independent for $t=1,\ldots ,n$. Let $S_{n}$ be the accumulation of a single unit sum of money invested for $n$ years, i.e.

$$S_{n}=(1+I_{1})\cdots (1+I_{n}).$$

Then, by independence,

$${\begin{aligned}\mathbb {E} [S_{n}]&=(1+\mu _{1})\cdots (1+\mu _{n})\\\mathbb {E} [S_{n}^{2}]&=\mathbb {E} [(1+I_{1})^{2}\cdots (1+I_{n})^{2}]\\&=\mathbb {E} [(1+I_{1})^{2}]\cdots \mathbb {E} [(1+I_{n})^{2}]\\&=\left(\sigma _{1}^{2}+(1+\mu _{1})^{2}\right)\cdots \left(\sigma _{n}^{2}+(1+\mu _{n})^{2}\right)\\\operatorname {Var} (S_{n})&=\mathbb {E} [S_{n}^{2}]-(\mathbb {E} [S_{n}])^{2}\\&=\left(\sigma _{1}^{2}+(1+\mu _{1})^{2}\right)\cdots \left(\sigma _{n}^{2}+(1+\mu _{n})^{2}\right)-(1+\mu _{1})^{2}\cdots (1+\mu _{n})^{2}\end{aligned}}$$

For simplicity, further assume that $I_{t}$'s are i.i.d. (identically and independently distributed), with mean $\mu$ and variance $\sigma ^{2}$. Then,

$${\begin{aligned}\mathbb {E} [S_{n}]&=\underbrace {(1+\mu )\cdots (1+\mu )} _{n{\text{ copies}}}=(1+\mu )^{n}\\\mathbb {E} [S_{n}^{2}]&=(\sigma ^{2}+(1+\mu )^{2})^{n}=(1+2\mu +\mu ^{2}+\sigma ^{2})^{n}\\\operatorname {Var} (S_{n})&=\mathbb {E} [S_{n}^{2}]-(\mathbb {E} [S_{n}])^{2}=(1+2\mu +\mu ^{2}+\sigma ^{2})^{n}-(1+\mu )^{2n}\end{aligned}}$$

Accumulation of investments with log-normal distribution

Some information about log-normal distribution

If $Y=\ln X$ has a normal distribution with mean $\mu$ and variance $\sigma ^{2}$, then $X$ has a log-normal distribution with parameters (not mean/variance generally) $\mu$ and $\sigma ^{2}$. The following are some properties of random variables following log-normal distribution with parameters $\mu$ and $\sigma ^{2}$:

• probability density function (pdf): $f(x)={\frac {1}{x\sigma {\sqrt {2\pi }}}}\exp \left(-{\frac {1}{2}}\left({\frac {\ln x-\mu }{\sigma }}\right)^{2}\right)$
• mean: $\mathbb {E} [X]=e^{\mu +{\frac {\sigma ^{2}}{2}}}$
• variance: $\operatorname {Var} (X)=e^{2\mu +\sigma ^{2}}\left(e^{\sigma ^{2}}-1\right)=(\mathbb {E} [X])^{2}\left(e^{\sigma ^{2}}-1\right)$

Motivation of using log-normal distribution

Let's apply log-normal distribution to stochastic interest. If $1+I_{t}$ follows a log-normal distribution with parameters $\mu$ and $\sigma ^{2}$, then $\ln(1+I_{t})$ will be normally distributed with mean $\mu$ and variance $\sigma ^{2}$.

Then, considering the natural logarithm of accumulation of a single investment of one unit for a period of $n$ time units, we have

$$\ln S_{n}=\ln((1+I_{1})\cdots (1+I_{n}))=\ln(1+I_{1})+\cdots +\ln(1+I_{n}).$$

Assuming $I_{t}$'s are independent, $\ln(1+I_{t})$ will also be independent. If we further assume that $1+I_{t}$'s are also log-normally distributed with parameters $\mu _{t}$ and $\sigma _{t}^{2}$, then $\ln(1+I_{t})$'s are normally distributed with mean $\mu$ and variance $\sigma ^{2}$x, and the sum of independent normal random variables $\ln(1+I_{1})+\cdots +\ln(1+I_{n})$ is normally distributed with mean $\mu _{1}+\cdots +\mu _{n}$ and variance $\sigma _{1}^{2}+\cdots +\sigma _{n}^{2}$ (which is a well-known result about normal distribution). That is,

$$\ln S_{n}=\ln(1+I_{1})+\cdots +\ln(1+I_{n})\sim N(\mu _{1}+\cdots +\mu _{n},\sigma _{1}^{2}+\cdots +\sigma _{n}^{2}).$$

Thus, if we apply log-normal distribution to stochastic interest, we can obtain this nice result ($\ln S_{n}$ follows a simple normal distribution).

Examples

Example. (a) It is given that $1+I$ follows log-normal distribution with parameters $\mu$ and $\sigma ^{2}$, $\mathbb {E} [I]=0.1$ and $\operatorname {Var} (I)=0.005$. Compute $\mu$ and $\sigma ^{2}$.

Solution: (a) Based on the given mean and variance, we have $\mathbb {E} [1+I]=1+0.1=1.1$ and $\operatorname {Var} (1+I)=0.005.$ So,

$${\begin{aligned}&&&{\begin{cases}e^{\mu +{\frac {\sigma ^{2}}{2}}}=1.1\\e^{2\mu +\sigma ^{2}}\left(e^{\sigma ^{2}}-1\right)=0.005\\\end{cases}}\\&\Rightarrow &1.1^{2}\left(e^{\sigma ^{2}}-1\right)&=0.005\\&\Rightarrow &\sigma ^{2}&=\ln \left(1+{\frac {0.005}{1.1^{2}}}\right)\approx 0.004124\\&\Rightarrow &e^{\mu +0.004124/2}&=1.1\\&\Rightarrow &\mu &\approx \ln(1.1)-0.004124/2\approx 0.09325.\end{aligned}}$$

(b) It is further given that $I_{t}$ is the annual yield rate for the $t$th year, $I_{t}$'s are i.i.d., and follow the distribution mentioned above. Compute the mean and variance of the accumulation of $100$ for $10$ years, and the probability that its accumulated value will be less than $220$.

Solution: (b) Since $\ln(1+I_{t})\sim N(0.09325,0.004124)$,

$$\ln(S_{1}0)\sim N(10(0.09325)+10(0.004124))=N(0.9325,0.04124).$$

So, $S_{1}0$ follows log-normal distribution with parameters $\mu =0.9325$ and $\sigma ^{2}=0.04124$. Thus, its mean and variance are as follows:

• mean: $\mathbb {E} [100S_{10}]=100e^{0.9325+0.04124/2}\approx 259.3790$
• variance: $\operatorname {Var} (100S_{10})=100^{2}e^{2(0.9325+0.04124)}(e^{0.04124}-1)\approx 2832.5273$

Then, we can compute the probability by

$${\begin{aligned}\mathbb {P} (100S_{10}<220)&=\mathbb {P} (\ln S_{10}<\ln 2.2)\\&=\mathbb {P} \left(Z<{\frac {\ln 2.2-0.9325}{\sqrt {0.04124}}}\right)\qquad {\text{in which }}Z\sim N(0,1)\\&=\mathbb {P} (Z<-0.709303)\\&\approx 0.23885\qquad {\text{from standard normal table}}\end{aligned}}$$