geometric distribution

Suppose that a random experiment has two possible outcomes, success with probability $p$ and failure with probability $q=1-p$. The experiment is repeated until a success happens. The number of trials before the success is a random variable $X$ with density function

 $f(x)=q^{(x-1)}p.$

The distribution function determined by $f(x)$ is called a geometric distribution with parameter $p$ and it is given by

 $F(x)=\sum_{k\leq x}q^{(k-1)}p.$

The picture shows the graph for $f(x)$ with $p=1/4$. Notice the quick decreasing. An interpretation is that a long run of failures is very unlikely.

We can use the moment generating function method in order to get the mean and variance. This function is

 $G(t)=\sum_{k=1}^{\infty}e^{tk}q^{(k-1)}p=pe^{t}\sum_{k=0}^{\infty}(e^{t}q)^{k}.$

The last expression can be simplified as

 $G(t)=\frac{pe^{t}}{1-e^{t}q}.$

The first derivative is

 $G^{\prime}(t)=\frac{e^{t}p}{(1-e^{t}q)^{2}}$

so the mean is

 $\mu=E[X]=G^{\prime}(0)=\frac{1}{p}.$

In order to find the variance, we use the second derivative and thus

 $E[X^{2}]=G^{\prime\prime}(0)=\frac{2-p}{p^{2}}$

and therefore the variance is

 $\sigma^{2}=E[X^{2}]-E[X]^{2}=G^{\prime\prime}(0)-G^{\prime}(0)^{2}=\frac{q}{p^% {2}}.$
 Title geometric distribution Canonical name GeometricDistribution Date of creation 2013-03-22 13:03:07 Last modified on 2013-03-22 13:03:07 Owner Mathprof (13753) Last modified by Mathprof (13753) Numerical id 14 Author Mathprof (13753) Entry type Definition Classification msc 60E05 Synonym geometric random variable Related topic RandomVariable Related topic DensityFunction Related topic DistributionFunction Related topic Mean Related topic Variance Related topic BernoulliDistribution Related topic ArithmeticMean