19 Geometric Distribution (0)

The Geometric Distribution random variable may be considered as the number of failures that occur before the first success. The support of the distribution is the non-negative integers.

19.1 Probability Mass Function

A random variable $$X$$ is said to have a Geometric Distribution with parameter $$p$$ if its probability mass function is

$\Pr(X = x) = \left\{ \begin{array}{ll} p(1-p)^{x}, & x= 0,1,2,3,\ldots\\ 0 & otherwise \end{array} \right.$

where $$p$$ is the probability of a success on any given trial and $$x$$ is the number of trials until the first success.

19.2 Cumulative Distribution Function

The cumulative probability of $$x$$ is computed as the $$x^{th}$$ partial sum of the Geometric Series See 21.1.

\begin{aligned} \Pr(X \leq x) &= \sum\limits_{i=0}^{x}p(1-p)^{x+1} \\ &= \frac{p-p(1-p)^{x+1}}{1-(1-p)} \\ &= \frac{p[1-(1-p)^{x + 1}]}{p} \\ &= 1-(1-p)^{x + 1} \end{aligned}

So

$\Pr(x) = \left\{ \begin{array}{ll} 1-(1-p)^{x + 1},& x=0, 1,2,3,\ldots\\ 0 & otherwise \end{array} \right.$

A recursive form of the cdf can be derived and has some usefulness in computer applications. With it, one need only initiate the first value and additional cumulative probabilities can be calculated. It is derived as follows:

\begin{aligned} P(X=x+1) &= p \cdot (1-p)^{x+1 + 1} \\ &= p \cdot (1 - p) ^ {x + 2} \\ &= (1-p) \cdot p \cdot (1-p)^{x+1} \\ &= (1-p) \cdot \Pr(X = x) \end{aligned}

19.3 Expected Values

\begin{aligned} {\mathrm{E}}[X] &= \sum\limits_{i = 0}^{\infty} x \cdot p \cdot (1 - p)^x \\ &= p \cdot \sum\limits_{i = 0}^{\infty} x \cdot (1 - p) ^ x \\ ^{[1]} &= p \cdot \sum\limits_{i = 0}^{\infty} x \cdot q ^ x \\ &= p \cdot q \cdot \sum\limits_{i = 0}^{\infty} x \cdot q ^ {x - 1} \\ ^{[2]} &= p \cdot q \frac{d}{dq} \sum\limits_{i = 0}{\infty} q ^ x \\ &= p \cdot q \frac{d}{dq} \sum\limits_{i = 0}{\infty} q \cdot q^{x - 1} \\ ^{[3]} &= p \cdot q \frac{d}{dq} \frac{q}{1 - q} \\ ^{[4]} &= p \cdot q \frac{(1 - q) \cdot 1 - q \cdot (-1)}{(1 - q)^2} \\ &= p \cdot q \frac{(1 - q) + q}{(1 - q)^2} \\ &= p \cdot q \frac{1}{(1 - q)^2} \\ &= \frac{p \cdot q}{(1 - q)^2} \\ ^{[5]} &= \frac{p \cdot (1 - p)}{\left(1 - (1 - p)\right)^2} \\ &= \frac{p \cdot (1 - p)}{p^2} \\ &= \frac{1 - p}{p} \end{aligned}

1. Let $$q = 1 - p$$
2. $$a\cdot x^{a-1}=\frac{d}{dx}(x^a)$$
3. $$\sum\limits_{i=0}^{\infty}ar^{i}=\frac{a}{1-r}$$ (Geometric Series 21.1).
4. Product Rule for Differentiation:
$$\frac{d}{dx}(\frac{f(x)}{g(x)}) = \frac{g\prime(x)\cdot f(x)-f\prime(x)\cdot g(x)}{g^2(x)}$$
5. $$q = 1- p \ \ \Rightarrow p=1-q$$