## You are here

Homegeometric distribution

## Primary tabs

# 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

$\sigma^{2}=E[X^{2}]-E[X]^{2}=G^{{\prime\prime}}(0)-G^{{\prime}}(0)^{2}=\frac{q% }{p^{2}}.$ |

## Mathematics Subject Classification

60E05*no label found*

- Forums
- Planetary Bugs
- HS/Secondary
- University/Tertiary
- Graduate/Advanced
- Industry/Practice
- Research Topics
- LaTeX help
- Math Comptetitions
- Math History
- Math Humor
- PlanetMath Comments
- PlanetMath System Updates and News
- PlanetMath help
- PlanetMath.ORG
- Strategic Communications Development
- The Math Pub
- Testing messages (ignore)

- Other useful stuff
- Corrections