Statistics/Distributions/Geometric

Geometric Distribution

Template:Probability distribution 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:

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

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 ${(1-p)}^k$.

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?

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

Unshifted

The probability mass function is defined as:

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

Mean

$\mathrm{E}[X]=\sum _{i}f(x_i)x_i=\sum _0^{\mathrm{\infty }}p(1-p{)}^{x}x$

Let q=1-p

$\mathrm{E}[X]=\sum _0^{\mathrm{\infty }}(1-q)q^{x}x$

$\mathrm{E}[X]=\sum _0^{\mathrm{\infty }}(1-q)q{q}^{x-1}x$

$\mathrm{E}[X]=(1-q)q\sum _0^{\mathrm{\infty }}{q}^{x-1}x$

$\mathrm{E}[X]=(1-q)q\sum _0^{\mathrm{\infty }}\frac{d}{dq}{q}^x$

We can now interchange the derivative and the sum.

$\mathrm{E}[X]=(1-q)q\frac{d}{dq}\sum _0^{\mathrm{\infty }}{q}^x$

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

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

$\mathrm{E}[X]=q\frac{1}{(1-q)}$

$\mathrm{E}[X]=\frac{(1-p)}{p}$

Variance

We derive the variance using the following formula:

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

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

$\mathrm{E}[X^2]=\sum _{i}f(x_i)\cdot x^2$

$\mathrm{E}[X^2]=\sum _0^{\mathrm{\infty }}p(1-p{)}^x{x}^2$

Let q=1-p

$\mathrm{E}[X^2]=\sum _0^{\mathrm{\infty }}(1-q)q^x{x}^2$

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

$\mathrm{E}[X^2]=(1-q)\sum _0^{\mathrm{\infty }}{q}^x[(x^2-x)+x]$

$\mathrm{E}[X^2]=(1-q)[\sum _0^{\mathrm{\infty }}{q}^x(x^2-x)+\sum _0^{\mathrm{\infty }}{q}^{x}x]$

$\mathrm{E}[X^2]=(1-q)[q^2\sum _0^{\mathrm{\infty }}{q}^{x-2}x(x-1)+q\sum _0^{\mathrm{\infty }}{q}^{x-1}x]$

$\mathrm{E}[X^2]=(1-q)q[q\sum _0^{\mathrm{\infty }}\frac{d^2}{(dq{)}^2}{q}^x+\sum _0^{\mathrm{\infty }}\frac{d}{dq}{q}^x]$

$\mathrm{E}[X^2]=(1-q)q[q\frac{d^2}{(dq{)}^2}\sum _0^{\mathrm{\infty }}{q}^x+\frac{d}{dq}\sum _0^{\mathrm{\infty }}{q}^x]$

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

$\mathrm{E}[X^2]=(1-q)q[q\frac{2}{(1-q{)}^3}+\frac{1}{(1-q{)}^2}]$

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

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

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

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

We then return to the variance formula

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

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

External links

• Interactive Geometric Distribution Web Applet (Java)