1.4 Geometric distribution

Geometric distribution

To obtain a binomial random variable, we observed a sequence of n Bernoulli trials and counted the number of successes. Suppose now that we do not fix the number of Bernoulli trials in advance but instead continue to observe the sequence of Bernoulli trials until a certain number r , of successes occurs. The random variable of interest is the number of trials needed to observe the r th success .

Let first discuss the problem when r =1. That is, consider a sequence of Bernoulli trials with probability p of success. This sequence is observed until the first success occurs. Let X denot the trial number on which the first success occurs.

For example , if F and S represent failure and success, respectively, and the sequence starts with F,F,F,S,…, then X =4. Moreover, because the trials are independent, the probability of such sequence is

$$P\left(X=4\right)=\left(q\right)\left(q\right)\left(q\right)\left(p\right)=q^{3}p={\left(1-p\right)}^{3}p.$$

In general, the p.d.f. $f\left(x\right)=P\left(X=x\right)$ , of X is given by $f\left(x\right)={\left(1-p\right)}^{x-1}p,$ $x=\mathrm{1,2,...}$ , because there must be x -1 failures before the first success that occurs on trail x . We say that X has a geometric distribution .

Thus, $${\displaystyle \sum _{x=1}^{\infty }f\left(x\right)={\displaystyle \sum _{x=1}^{\infty }{\left(1-p\right)}^{k-1}p=\frac{p}{1-\left(1-p\right)}=1}},$$ so that $f\left(x\right)$ does satisfy the properties of a p.d.f..

From the sum of geometric series we also note that, when k is an integer,

$$P\left(X>k\right)={\displaystyle \sum _{x=k+1}^{\infty }}{\left(1-p\right)}^{x-1}p=\frac{{\left(1-p\right)}^{k}p}{1-\left(1-p\right)}={\left(1-p\right)}^k=q^k,$$

and thus the value of the distribution function at a positive integer k is

$$P\left(X\le k\right)={\displaystyle \sum _{x=k+1}^{\infty }}{\left(1-p\right)}^{x-1}p=1-P\left(X>k\right)=1-{\left(1-p\right)}^k=1-q^k.$$

To find a mean and variance for the geometric distribution, let use the following results about the sum and the first and second derivatives of a geometric series. For $-1<r<1$ , let $$g\left(r\right)={\displaystyle \sum _{k=0}^{\infty }a{r}^k}=\frac{a}{1-r}.$$

Then $$g\text{'}\left(r\right)={\displaystyle \sum _{k=1}^{\infty }ak{r}^{k-1}}=\frac{a}{{\left(1-r\right)}^2},$$ and $$g\text{'}\text{'}\left(r\right)={\displaystyle \sum _{k=2}^{\infty }ak\left(k-1\right)r^{k-2}}=\frac{2a}{{\left(1-r\right)}^3}.$$

If X has a geometric distribution and $0<p<1$ , then the mean of X is given by

using the formula for $g\text{'}\left(x\right)$ with $a=p$ and $r=q$ .

To find the variance of X , let first find the second factorial moment $E\left[X\left(X-1\right)\right]$ . We have

$$E\left[X\left(X-1\right)\right]={\displaystyle \sum _{x=1}^{\infty }x\left(x-1\right)q^{x-1}p=}{\displaystyle \sum _{x=1}^{\infty }pqx\left(x-1\right)q^{x-2}=\frac{2pq}{{\left(1-q\right)}^3}=\frac{2q}{p^2}}.$$

Using formula for $g\text{'}\text{'}\left(x\right)$ with $a=pq$ and $r=q$ . Thus the variance of X is

$$\begin{array}{l}Var\left(X\right)=E\left(X^2\right)-{\left[E\left(X\right)\right]}^2=\left\{E\left[X\left(X-1\right)\right]+E\left(X\right)\right\}-{\left[E\left(X\right)\right]}^2=\\ =\frac{2q}{p^2}+\frac{1}{p}-\frac{1}{p^2}=\frac{2q+p-1}{p^2}=\frac{1-p}{p^2}.\end{array}$$

The standard deviation of X is $$\sigma =\sqrt{\left(1-p\right)/p^2}.$$