Geometric Distribution¶

A geometric distribution represents the probability distribution for the number of failures in Bernoulli trials till the first success.

\[ \begin{aligned} P(X = k) &= (1-p)^{k}p \quad k = 0, 1, 2, \ldots \end{aligned} \]

Another definition is to consider \(k\) as the number of trials before the first success. The distribution will then be defined on \(k = 1, 2, \ldots\) and is often called the shifted distribution.

Moment Generating Function¶

To calculate mean and variance, we first calculate the moment generating function

\[ \begin{aligned} E[e^{tX}] &= \sum_{k=0}^{\infty} (1-p)^{k}p e^{tk}\newline \phi(k) &= p \sum_{k=0}^{\infty} ((1-p)e^{t})^{k}\newline &= p \frac{1}{1 - (1-p)e^{t}}\newline \diffone{\phi}(t) &= \frac{p(1-p)e^{t}}{(1 - (1-p)e^{t})^{2}}\newline \difftwo{\phi}(t) &= \frac{p(1-p)e^{t}\squarebr{1 + (1-p)e^{t}}}{(1 - (1-p)e^{t})^{3}} \end{aligned} \]

Expected number of failures will be \(1/p - 1 = (1-p)/p\) (since \(1/p\) is the total expected trials and we subtract the last trial which is a success).

Mean and Variance¶

With the moment generating function, mean and variance are easy to calculate

\[ \begin{aligned} E[X] &= \diffone{\phi}(0) = \frac{1-p}{p}\newline E[X^{2}] &= \difftwo{\phi}(0) = \frac{(1-p)(2-p)}{p^{2}}\newline Var(X) &= \frac{(1-p)(2-p)}{p^{2}} - \frac{(1-p)^{2}}{p^{2}} = \frac{1-p}{p^{2}} \end{aligned} \]

If \(X\) denoted the total trials till first success, \(E[X]\) would be \((1-p)/p + 1 = 1/p\) and variance would be same since its only a shifted distribution.