# 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

$$F(x)=\sum _{k\le x}{q}^{(k-1)}p.$$ |

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

$$G(t)=\sum _{k=1}^{\mathrm{\infty}}{e}^{tk}{q}^{(k-1)}p=p{e}^{t}\sum _{k=0}^{\mathrm{\infty}}{({e}^{t}q)}^{k}.$$ |

The last expression can be simplified as

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

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 |