Home page for accesible maths 5 Models for discrete random variables 5.4.1 Expectation and variance 5.6 Poisson random variables

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5.5 Geometric random variables

Consider an experiment based on independent Bernoulli trials, each with the probability of a success being ${\theta}$ . Now define the variable of interest, ${R}$ , to be the number of trials up to BUT NOT including the first success. Here the induced sample space is ${{\mathcal{S}}=\{0,1,2,\ldots\}}$ , and is infinite, corresponding to outcomes in the original sample space

\displaystyle\Omega=\{S,FS,FFS,FFFS,\dots\}.

If, for example, the sequence ${FFFFS}$ occurs then random variable ${R(FFFFS)=4}$ .

Such a random variable is called a Geometric random variable, examples of which include:

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the number of heads of a coin toss before the first tail,
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the number of boys born before the first girl,
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the number of black cars passed before a red car,
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the number of years to pass before the Scotland football team qualify for anything.

We say ${R\sim{\rm Geometric}(\theta)}$ .

Exercise 5.14.

Use the independence of the Bernoulli random variables to derive the pmf of the geometric random variable. Hint: $R=4$ corresponds to the sample point $F F F F S$ .

Solution.

$\displaystyle p_{R}(r)$	$\displaystyle=$	$\displaystyle{\rm P}(R=r)={\rm P}(\{\underbrace{FF\dots F}_{r}S\})$
	$\displaystyle=$	$\displaystyle{\rm P}(F){\rm P}(F)\dots{\rm P}(F){\rm P}(S)\color[rgb]{1,1,1}% \quad\mbox{indep}$
	$\displaystyle=$	$\displaystyle(1-\theta)^{r}\theta\mbox{ for }r=0,1,2,\dots$

Exercise 5.15.

The rv ${R\sim{\rm Geometric}(0.3)}$ . Use R to evaluate and plot the pmf of ${R}$ for ${r=0,1,2,\dots,5}$ , with the commands

dgeom(0:5,prob=0.3)
dgeom(0:5,prob=0.4)
            # Note how the probabilities change
barplot( dgeom(0:5,prob=0.4),names.arg=c(0:5) )

Repeat with ${\theta=0.4}$ and plot.

Exercise 5.16.

Verify that ${\sum_{r=0}^{\infty}\;p_{R}(r)=1}$ for the geometric pmf. This requires the mathematical formulae for sums of geometric type series given at the start of this chapter.

Solution.

$\displaystyle\sum_{r=0}^{\infty}\;p_{R}(r)$	$\displaystyle=$	$\displaystyle\sum_{r=0}^{\infty}\,(1-\theta)^{r}\theta$
	$\displaystyle=$	$\displaystyle\theta\sum_{r=0}^{\infty}\,(1-\theta)^{r}$
	$\displaystyle=$	$\displaystyle\theta\frac{1}{1-(1-\theta)}=1.$

Example 5.17.

For a general ${R\sim{\rm Geometric}(\theta)}$ , find ${{\rm P}(R\geq r)}$ .

Solution.

$\displaystyle{\rm P}(R\geq r)$	$\displaystyle=$	$\displaystyle\sum_{s=r}^{\infty}p_{R}(s)$
	$\displaystyle=$	$\displaystyle\sum_{s=r}^{\infty}(1-\theta)^{s}\theta$
	$\displaystyle=$	$\displaystyle(1-\theta)^{r}\theta\sum_{s=r}^{\infty}(1-\theta)^{s-r}$
	$\displaystyle=$	$\displaystyle(1-\theta)^{r}\theta\sum_{s^{\prime}=0}^{\infty}(1-\theta)^{s^{% \prime}}\quad\mbox{setting ${s^{\prime}=s-r}$}$
	$\displaystyle=$	$\displaystyle(1-\theta)^{r}\theta\frac{1}{1-(1-\theta)}$
	$\displaystyle=$	$\displaystyle(1-\theta)^{r}$

Note that this is simply the probability that the first ${r}$ Bernoulli trials are all ${F}$ .

Example 5.18.

Find ${{\rm E}(R)}$ and ${\rm Var}(R)$ for a geometric random variable.

We need to use the basic identities from Section 5.1 on page 5.1.

Solution.

$\displaystyle{\rm E}(R)$	$\displaystyle=$	$\displaystyle\sum_{r=0}^{\infty}r\,p_{R}(r)$
	$\displaystyle=$	$\displaystyle 0+\sum_{r=1}^{\infty}r\,(1-\theta)^{r}\theta$
	$\displaystyle=$	$\displaystyle(1-\theta)\theta\sum_{r=1}^{\infty}r(1-\theta)^{r-1}$
	$\displaystyle=$	$\displaystyle(1-\theta)\theta[(1-(1-\theta))^{-2}]$
	$\displaystyle=$	$\displaystyle\frac{1-\theta}{\theta}.$

Now to find ${\rm Var}(R)$ we begin by calculating ${\rm E}[R(R-1)]$ :

$\displaystyle{\rm E}[R(R-1)]$	$\displaystyle=$	$\displaystyle\sum_{r=0}^{\infty}r(r-1)p_{R}(r)$
	$\displaystyle=$	$\displaystyle\sum_{r=0}^{\infty}r(r-1)(1-\theta)^{r}\theta$
	$\displaystyle=$	$\displaystyle(1-\theta)^{2}\theta\sum_{r=2}^{\infty}r(r-1)(1-\theta)^{r-2}$
	$\displaystyle=$	$\displaystyle(1-\theta)^{2}\theta 2(1-(1-\theta))^{-3}$
	$\displaystyle=$	$\displaystyle\frac{2(1-\theta)^{2}}{\theta^{2}}$

This is then plugged in to get

$\displaystyle{\rm Var}(R)$	$\displaystyle=$	$\displaystyle{\rm E}[R(R-1)]+{\rm E}[R]-({\rm E}[R])^{2}$
	$\displaystyle=$	$\displaystyle\frac{2(1-\theta)^{2}}{\theta^{2}}+\frac{1-\theta}{\theta}-\left(% \frac{1-\theta}{\theta}\right)^{2}$
	$\displaystyle=$	$\displaystyle\frac{(1-\theta)^{2}+(1-\theta)\theta}{\theta^{2}}$
	$\displaystyle=$	$\displaystyle\frac{1-\theta}{\theta^{2}}$

Yuck!

Note that as the Bernoulli probability ${\theta\downarrow 0}$ then the expected number of trials (and the variance) goes to ${\infty}$ . To summarise

For a geometric random variable ${R\sim{\rm Geometric}(\theta)}$ $\displaystyle p_{R}(r)$ $\displaystyle=$ $\displaystyle(1-\theta)^{r}\theta\mbox{ for }r=0,1,2,,\dots$ $\displaystyle p_{R}(r)$ $\displaystyle=$ $\displaystyle 0\quad\mbox{otherwise}$ $\displaystyle{\rm E}(R)$ $\displaystyle=$ $\displaystyle\frac{1-\theta}{\theta}$ $\displaystyle{\rm Var}(R)$ $\displaystyle=$ $\displaystyle\frac{1-\theta}{\theta^{2}}$