Statistics/Distributions/Geometric

Geometric Distribution

Geometric

Probability mass function
Cumulative distribution function
Parameters	$0< p \leq 1$ success probability (real)
Support	$k \in \{1,2,3,\dots\}\!$
PMF	$(1 - p)^{k-1}\,p\!$
CDF	$1-(1 - p)^k\!$
Mean	$\frac{1}{p}\!$
Median	$\left\lceil \frac{-1}{\log_2(1-p)} \right\rceil\!$ (not unique if $-1/\log_2(1-p)$ is an integer)
Mode	$1$
Variance	$\frac{1-p}{p^2}\!$
Skewness	$\frac{2-p}{\sqrt{1-p}}\!$
Ex. kurtosis	$6+\frac{p^2}{1-p}\!$
Entropy	$\tfrac{-(1-p)\log_2 (1-p) - p \log_2 p}{p}\!$
MGF	$\frac{pe^t}{1-(1-p) e^t}\!$, for $t<-\ln(1-p)\!$
CF	$\frac{pe^{it}}{1-(1-p)\,e^{it}}\!$

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.

Shifted

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:

$P\left( X=i \right) =p\left( 1-p\right)^{i-1}$

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 $\left( 1-p\right)^k$.

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?

$P\left( X=4 \right) =\frac{1}{10}\left( 1-\frac{1}{10}\right)^{4-1}=\frac{1}{10}\left( \frac{9}{10}\right)^{3}=0.0729$

Unshifted

The probability mass function is defined as:

$f(x) = p(1-p)^x \,$ for $x \in \{0, 1, 2, \}$

Mean

$\operatorname{E}[X] = \sum_i f(x_i) x_i = \sum_0^{\infin} p(1-p)^x x$

Let q=1-p

$\operatorname{E}[X] = \sum_0^{\infin}(1-q) q^x x$

$\operatorname{E}[X] = \sum_0^{\infin}(1-q)q q^{x-1} x$

$\operatorname{E}[X] = (1-q)q\sum_0^{\infin} q^{x-1} x$

$\operatorname{E}[X] = (1-q)q\sum_0^{\infin} \frac{d}{dq}q^x$

We can now interchange the derivative and the sum.

$\operatorname{E}[X] = (1-q)q\frac{d}{dq}\sum_0^{\infin} q^x$

$\operatorname{E}[X] = (1-q)q\frac{d}{dq}{1 \over 1-q}$

$\operatorname{E}[X] = (1-q)q{1 \over (1-q)^2}$

$\operatorname{E}[X] = q{1 \over (1-q)}$

$\operatorname{E}[X] = {(1-p) \over p}$

Variance

We derive the variance using the following formula:

$\operatorname{Var}[X] = \operatorname{E}[X^2] - (\operatorname{E}[X])^2$

We have already calculated E[X] above, so now we will calculate E[X²] and then return to this variance formula:

$\operatorname{E}[X^2] = \sum_i f(x_i) \cdot x^2$

$\operatorname{E}[X^2] = \sum_0^{\infin} p(1-p)^x x^2$

Let q=1-p

$\operatorname{E}[X^2] = \sum_0^{\infin} (1-q)q^x x^2$

We now manipulate x² so that we get forms that are easy to handle by the technique used when deriving the mean.

$\operatorname{E}[X^2] = (1-q)\sum_0^{\infin} q^x [(x^2-x)+x]$

$\operatorname{E}[X^2] = (1-q)\left[\sum_0^{\infin} q^x (x^2-x)+\sum_0^{\infin}q^x x\right]$

$\operatorname{E}[X^2] = (1-q)\left[q^2\sum_0^{\infin} q^{x-2} x(x-1)+q\sum_0^{\infin}q^{x-1} x\right]$

$\operatorname{E}[X^2] = (1-q)q\left[q\sum_0^{\infin} \frac{d^2}{(dq)^2}q^x+\sum_0^{\infin}\frac{d}{dq}q^x\right]$

$\operatorname{E}[X^2] = (1-q)q\left[q\frac{d^2}{(dq)^2}\sum_0^{\infin} q^x+\frac{d}{dq}\sum_0^{\infin}q^x\right]$

$\operatorname{E}[X^2] = (1-q)q\left[q\frac{d^2}{(dq)^2}{1 \over 1-q}+\frac{d}{dq}{1 \over 1-q}\right]$

$\operatorname{E}[X^2] = (1-q)q\left[q{2 \over (1-q)^3}+{1 \over (1-q)^2}\right]$

$\operatorname{E}[X^2] = {2q^2 \over (1-q)^2}+{q \over (1-q)}$

$\operatorname{E}[X^2] = {2q^2 +q(1-q) \over (1-q)^2}$

$\operatorname{E}[X^2] = {q(q+1) \over (1-q)^2}$

$\operatorname{E}[X^2] = {(1-p)(2-p) \over p^2}$

We then return to the variance formula

$\operatorname{Var}[X] = \left[{(1-p)(2-p) \over p^2}\right] - \left({1-p \over p}\right)^2$

$\operatorname{Var}[X] = {(1-p) \over p^2}$

External links

• Interactive Geometric Distribution Web Applet (Java)