Geometric Distribution
edit
Geometric
Probability mass function
Cumulative distribution function
Parameters
0
<
p
≤
1
{\displaystyle 0<p\leq 1}
real )
Support
k
∈
{
1
,
2
,
3
,
…
}
{\displaystyle k\in \{1,2,3,\dots \}\!}
PMF
(
1
−
p
)
k
−
1
p
{\displaystyle (1-p)^{k-1}\,p\!}
CDF
1
−
(
1
−
p
)
k
{\displaystyle 1-(1-p)^{k}\!}
Mean
1
p
{\displaystyle {\frac {1}{p}}\!}
Median
⌈
−
1
log
2
(
1
−
p
)
⌉
{\displaystyle \left\lceil {\frac {-1}{\log _{2}(1-p)}}\right\rceil \!}
−
1
/
log
2
(
1
−
p
)
{\displaystyle -1/\log _{2}(1-p)}
Mode
1
{\displaystyle 1}
Variance
1
−
p
p
2
{\displaystyle {\frac {1-p}{p^{2}}}\!}
Skewness
2
−
p
1
−
p
{\displaystyle {\frac {2-p}{\sqrt {1-p}}}\!}
Ex. kurtosis
6
+
p
2
1
−
p
{\displaystyle 6+{\frac {p^{2}}{1-p}}\!}
Entropy
−
(
1
−
p
)
log
2
(
1
−
p
)
−
p
log
2
p
p
{\displaystyle {\tfrac {-(1-p)\log _{2}(1-p)-p\log _{2}p}{p}}\!}
MGF
p
e
t
1
−
(
1
−
p
)
e
t
{\displaystyle {\frac {pe^{t}}{1-(1-p)e^{t}}}\!}
for
t
<
−
ln
(
1
−
p
)
{\displaystyle t<-\ln(1-p)\!}
CF
p
e
i
t
1
−
(
1
−
p
)
e
i
t
{\displaystyle {\frac {pe^{it}}{1-(1-p)\,e^{it}}}\!}
There are two similar distributions with the name "Geometric Distribution".
The probability distribution of the number X of Bernoulli trials needed to get one success, supported on the set { 1, 2, 3, ...} The probability distribution of the number Y = X − 1 of failures before the first success, supported on the set { 0, 1, 2, 3, ... } These two different geometric distributions should not be confused with each other. Often, the name shifted geometric distribution is adopted for the former one. We will use X and Y to refer to distinguish the two.
The shifted Geometric Distribution refers to the probability of the number of times needed to do something until getting a desired result. For example:
How many times will I throw a coin until it lands on heads ?
How many children will I have until I get a girl?
How many cards will I draw from a pack until I get a Joker? Just like the Bernoulli Distribution , the Geometric distribution has one controlling parameter: The probability of success in any independent test.
If a random variable X is distributed with a Geometric Distribution with a parameter p we write its probability mass function as:
P
(
X
=
i
)
=
p
(
1
−
p
)
i
−
1
{\displaystyle P\left(X=i\right)=p\left(1-p\right)^{i-1}}
With a Geometric Distribution it is also pretty easy to calculate the probability of a "more than n times" case. The probability of failing to achieve the wanted result is
(
1
−
p
)
k
{\displaystyle \left(1-p\right)^{k}}
Example: a student comes home from a party in the forest, in which interesting substances were consumed. The student is trying to find the key to his front door, out of a keychain with 10 different keys. What is the probability of the student succeeding in finding the right key in the 4th attempt?
P
(
X
=
4
)
=
1
10
(
1
−
1
10
)
4
−
1
=
1
10
(
9
10
)
3
=
0.0729
{\displaystyle P\left(X=4\right)={\frac {1}{10}}\left(1-{\frac {1}{10}}\right)^{4-1}={\frac {1}{10}}\left({\frac {9}{10}}\right)^{3}=0.0729}
Unshifted
edit
The probability mass function is defined as:
f
(
x
)
=
p
(
1
−
p
)
x
{\displaystyle f(x)=p(1-p)^{x}\,}
x
∈
{
0
,
1
,
2
,
…
}
{\displaystyle x\in \{0,1,2,\dots \}}
E
[
X
]
=
∑
i
f
(
x
i
)
x
i
=
∑
0
∞
p
(
1
−
p
)
x
x
{\displaystyle \operatorname {E} [X]=\sum _{i}f(x_{i})x_{i}=\sum _{0}^{\infty }p(1-p)^{x}x}
Let q=1-p
E
[
X
]
=
∑
0
∞
(
1
−
q
)
q
x
x
{\displaystyle \operatorname {E} [X]=\sum _{0}^{\infty }(1-q)q^{x}x}
E
[
X
]
=
∑
0
∞
(
1
−
q
)
q
q
x
−
1
x
{\displaystyle \operatorname {E} [X]=\sum _{0}^{\infty }(1-q)qq^{x-1}x}
E
[
X
]
=
(
1
−
q
)
q
∑
0
∞
q
x
−
1
x
{\displaystyle \operatorname {E} [X]=(1-q)q\sum _{0}^{\infty }q^{x-1}x}
E
[
X
]
=
(
1
−
q
)
q
∑
0
∞
d
d
q
q
x
{\displaystyle \operatorname {E} [X]=(1-q)q\sum _{0}^{\infty }{\frac {d}{dq}}q^{x}}
We can now interchange the derivative and the sum.
E
[
X
]
=
(
1
−
q
)
q
d
d
q
∑
0
∞
q
x
{\displaystyle \operatorname {E} [X]=(1-q)q{\frac {d}{dq}}\sum _{0}^{\infty }q^{x}}
E
[
X
]
=
(
1
−
q
)
q
d
d
q
1
1
−
q
{\displaystyle \operatorname {E} [X]=(1-q)q{\frac {d}{dq}}{1 \over 1-q}}
E
[
X
]
=
(
1
−
q
)
q
1
(
1
−
q
)
2
{\displaystyle \operatorname {E} [X]=(1-q)q{1 \over (1-q)^{2}}}
E
[
X
]
=
q
1
(
1
−
q
)
{\displaystyle \operatorname {E} [X]=q{1 \over (1-q)}}
E
[
X
]
=
(
1
−
p
)
p
{\displaystyle \operatorname {E} [X]={(1-p) \over p}}
Variance
edit
We derive the variance using the following formula:
Var
[
X
]
=
E
[
X
2
]
−
(
E
[
X
]
)
2
{\displaystyle \operatorname {Var} [X]=\operatorname {E} [X^{2}]-(\operatorname {E} [X])^{2}}
We have already calculated E[X ] above, so now we will calculate E[X2 ] and then return to this variance formula:
E
[
X
2
]
=
∑
i
f
(
x
i
)
⋅
x
2
{\displaystyle \operatorname {E} [X^{2}]=\sum _{i}f(x_{i})\cdot x^{2}}
E
[
X
2
]
=
∑
0
∞
p
(
1
−
p
)
x
x
2
{\displaystyle \operatorname {E} [X^{2}]=\sum _{0}^{\infty }p(1-p)^{x}x^{2}}
Let q=1-p
E
[
X
2
]
=
∑
0
∞
(
1
−
q
)
q
x
x
2
{\displaystyle \operatorname {E} [X^{2}]=\sum _{0}^{\infty }(1-q)q^{x}x^{2}}
We now manipulate x2 so that we get forms that are easy to handle by the technique used when deriving the mean.
E
[
X
2
]
=
(
1
−
q
)
∑
0
∞
q
x
[
(
x
2
−
x
)
+
x
]
{\displaystyle \operatorname {E} [X^{2}]=(1-q)\sum _{0}^{\infty }q^{x}[(x^{2}-x)+x]}
E
[
X
2
]
=
(
1
−
q
)
[
∑
0
∞
q
x
(
x
2
−
x
)
+
∑
0
∞
q
x
x
]
{\displaystyle \operatorname {E} [X^{2}]=(1-q)\left[\sum _{0}^{\infty }q^{x}(x^{2}-x)+\sum _{0}^{\infty }q^{x}x\right]}
E
[
X
2
]
=
(
1
−
q
)
[
q
2
∑
0
∞
q
x
−
2
x
(
x
−
1
)
+
q
∑
0
∞
q
x
−
1
x
]
{\displaystyle \operatorname {E} [X^{2}]=(1-q)\left[q^{2}\sum _{0}^{\infty }q^{x-2}x(x-1)+q\sum _{0}^{\infty }q^{x-1}x\right]}
E
[
X
2
]
=
(
1
−
q
)
q
[
q
∑
0
∞
d
2
(
d
q
)
2
q
x
+
∑
0
∞
d
d
q
q
x
]
{\displaystyle \operatorname {E} [X^{2}]=(1-q)q\left[q\sum _{0}^{\infty }{\frac {d^{2}}{(dq)^{2}}}q^{x}+\sum _{0}^{\infty }{\frac {d}{dq}}q^{x}\right]}
E
[
X
2
]
=
(
1
−
q
)
q
[
q
d
2
(
d
q
)
2
∑
0
∞
q
x
+
d
d
q
∑
0
∞
q
x
]
{\displaystyle \operatorname {E} [X^{2}]=(1-q)q\left[q{\frac {d^{2}}{(dq)^{2}}}\sum _{0}^{\infty }q^{x}+{\frac {d}{dq}}\sum _{0}^{\infty }q^{x}\right]}
E
[
X
2
]
=
(
1
−
q
)
q
[
q
d
2
(
d
q
)
2
1
1
−
q
+
d
d
q
1
1
−
q
]
{\displaystyle \operatorname {E} [X^{2}]=(1-q)q\left[q{\frac {d^{2}}{(dq)^{2}}}{1 \over 1-q}+{\frac {d}{dq}}{1 \over 1-q}\right]}
E
[
X
2
]
=
(
1
−
q
)
q
[
q
2
(
1
−
q
)
3
+
1
(
1
−
q
)
2
]
{\displaystyle \operatorname {E} [X^{2}]=(1-q)q\left[q{2 \over (1-q)^{3}}+{1 \over (1-q)^{2}}\right]}
E
[
X
2
]
=
2
q
2
(
1
−
q
)
2
+
q
(
1
−
q
)
{\displaystyle \operatorname {E} [X^{2}]={2q^{2} \over (1-q)^{2}}+{q \over (1-q)}}
E
[
X
2
]
=
2
q
2
+
q
(
1
−
q
)
(
1
−
q
)
2
{\displaystyle \operatorname {E} [X^{2}]={2q^{2}+q(1-q) \over (1-q)^{2}}}
E
[
X
2
]
=
q
(
q
+
1
)
(
1
−
q
)
2
{\displaystyle \operatorname {E} [X^{2}]={q(q+1) \over (1-q)^{2}}}
E
[
X
2
]
=
(
1
−
p
)
(
2
−
p
)
p
2
{\displaystyle \operatorname {E} [X^{2}]={(1-p)(2-p) \over p^{2}}}
We then return to the variance formula
Var
[
X
]
=
[
(
1
−
p
)
(
2
−
p
)
p
2
]
−
(
1
−
p
p
)
2
{\displaystyle \operatorname {Var} [X]=\left[{(1-p)(2-p) \over p^{2}}\right]-\left({1-p \over p}\right)^{2}}
Var
[
X
]
=
(
1
−
p
)
p
2
{\displaystyle \operatorname {Var} [X]={(1-p) \over p^{2}}}
External links
edit
![{\displaystyle \operatorname {E} [X]=\sum _{i}f(x_{i})x_{i}=\sum _{0}^{\infty }p(1-p)^{x}x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/14772fe3fb1aac3aeda043b8844d8e44ea9f4beb)
![{\displaystyle \operatorname {E} [X]=\sum _{0}^{\infty }(1-q)q^{x}x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f9ad822d6cb59a023236e1c2dba4ea8ce8fe0bb0)
![{\displaystyle \operatorname {E} [X]=\sum _{0}^{\infty }(1-q)qq^{x-1}x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b173d51fc1397b9d8290c8e8a58dbac31d80e0cd)
![{\displaystyle \operatorname {E} [X]=(1-q)q\sum _{0}^{\infty }q^{x-1}x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3694fabe53f333614812d1e6b7a75577a8b00d2c)
![{\displaystyle \operatorname {E} [X]=(1-q)q\sum _{0}^{\infty }{\frac {d}{dq}}q^{x}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/00ecb1f87eadc5d1bec6706a38d2e526254149b0)
![{\displaystyle \operatorname {E} [X]=(1-q)q{\frac {d}{dq}}\sum _{0}^{\infty }q^{x}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/29126ac9a1e1420eb4b6f0b292b11542191314ac)
![{\displaystyle \operatorname {E} [X]=(1-q)q{\frac {d}{dq}}{1 \over 1-q}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c28cabe4165678e022964b678cb3bed82f7d0a7f)
![{\displaystyle \operatorname {E} [X]=(1-q)q{1 \over (1-q)^{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0088bfe78733dbdd446cdcee8839d8b06334e2b5)
![{\displaystyle \operatorname {E} [X]=q{1 \over (1-q)}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d22bd1347ab70303be5bafa8d54e6b05a41fd008)
![{\displaystyle \operatorname {E} [X]={(1-p) \over p}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/005e463e9bbf3557dcc3ad6930b3321e09013809)
![{\displaystyle \operatorname {Var} [X]=\operatorname {E} [X^{2}]-(\operatorname {E} [X])^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d3a927c2af9ce191701ce464bf66fe7a6fe2190c)
![{\displaystyle \operatorname {E} [X^{2}]=\sum _{i}f(x_{i})\cdot x^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9b457462143b3c9a9d2d1c9de47dd57fae3dbec2)
![{\displaystyle \operatorname {E} [X^{2}]=\sum _{0}^{\infty }p(1-p)^{x}x^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5683649108ad3852d33b3f0f8ca642e237678fd5)
![{\displaystyle \operatorname {E} [X^{2}]=\sum _{0}^{\infty }(1-q)q^{x}x^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3d5c3ee343acd07e0bb8c579be8dc55ca22fc589)
![{\displaystyle \operatorname {E} [X^{2}]=(1-q)\sum _{0}^{\infty }q^{x}[(x^{2}-x)+x]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/75d3eafe380f83410881e368fe533b827ca74f92)
![{\displaystyle \operatorname {E} [X^{2}]=(1-q)\left[\sum _{0}^{\infty }q^{x}(x^{2}-x)+\sum _{0}^{\infty }q^{x}x\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/24f93fd82a11d90cda7b6d70be13b8244e3a7d05)
![{\displaystyle \operatorname {E} [X^{2}]=(1-q)\left[q^{2}\sum _{0}^{\infty }q^{x-2}x(x-1)+q\sum _{0}^{\infty }q^{x-1}x\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e6b1637a45a96e3347eac070309b680a9d4ff66c)
![{\displaystyle \operatorname {E} [X^{2}]=(1-q)q\left[q\sum _{0}^{\infty }{\frac {d^{2}}{(dq)^{2}}}q^{x}+\sum _{0}^{\infty }{\frac {d}{dq}}q^{x}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a670c354fefd15e8e9c92ae0e84f11f2aaa924ea)
![{\displaystyle \operatorname {E} [X^{2}]=(1-q)q\left[q{\frac {d^{2}}{(dq)^{2}}}\sum _{0}^{\infty }q^{x}+{\frac {d}{dq}}\sum _{0}^{\infty }q^{x}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eb989d8fec7d63001d1753487542e9c66ee35688)
![{\displaystyle \operatorname {E} [X^{2}]=(1-q)q\left[q{\frac {d^{2}}{(dq)^{2}}}{1 \over 1-q}+{\frac {d}{dq}}{1 \over 1-q}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dc28f609f6a73d90dd592e92f471e1f9c892f962)
![{\displaystyle \operatorname {E} [X^{2}]=(1-q)q\left[q{2 \over (1-q)^{3}}+{1 \over (1-q)^{2}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/569b086d793e05931e2f107f0faa3669e3a2e38e)
![{\displaystyle \operatorname {E} [X^{2}]={2q^{2} \over (1-q)^{2}}+{q \over (1-q)}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f1d4242ae0316880299a80f3b86026827724baa2)
![{\displaystyle \operatorname {E} [X^{2}]={2q^{2}+q(1-q) \over (1-q)^{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5fa45c19cb12088654ff064a34fe60e5d01145e3)
![{\displaystyle \operatorname {E} [X^{2}]={q(q+1) \over (1-q)^{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0439f27f91edf1d341788b23bebb1f52f7aed08e)
![{\displaystyle \operatorname {E} [X^{2}]={(1-p)(2-p) \over p^{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a65e09f63d237df831cf41227667ef09b8663043)
![{\displaystyle \operatorname {Var} [X]=\left[{(1-p)(2-p) \over p^{2}}\right]-\left({1-p \over p}\right)^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5083e58db627dd9b8b3c9538630be47a30f52318)
![{\displaystyle \operatorname {Var} [X]={(1-p) \over p^{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/425211a2e0d5b31cb138f0691612a7009fa16060)