geometric distributionGeometricDistribution22002-09-15 00:03:382007-06-22 17:24:15Definition\usepackage{graphicx}
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\newcommand{\figuraex}[2]{\begin{center}\includegraphics[#2]{#1}\end{center}}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 \emph{geometric distribution} with parameter $p$ and it is given by
$$F(x) = \sum_{k\leq x}q^{(k-1)}p.$$
\figura{geomet}
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
$$G(t)=\sum_{k=1}^\infty e^{tk}q^{(k-1)}p=p\sum_{k=0}^\infty (e^tq)^k.$$
The last expression can be simplified as
$$G(t)=\frac{p}{1-e^tq}.$$
The first and second derivatives are
$$G'(t)=\frac{e^tpq}{(1-e^tq)^2}$$
so the mean is
$$\mu=E[X]=G'(0)=\frac{q}{p}.$$
In order to find the variance, we get the second derivative and thus
$$E[X^2]=G''(0)=\frac{2q^2}{p^2}+\frac{q}{p}$$
and therefore the variance is
$$\sigma^2=E[X^2]-E[X]^2 = G''(0) - G'(0)^2 = \frac{q}{p^2}.$$