# Difference between revisions of "AP Statistics Curriculum 2007 Gamma"

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<br />'''Probability density function''': The waiting time until the hth Poisson event with a rate of change <math>\lambda</math> is | <br />'''Probability density function''': The waiting time until the hth Poisson event with a rate of change <math>\lambda</math> is | ||

− | + | <center><math>P(x)=\frac{\lambda(\lambda x)^{h-1}}{(h-1)!}{e^{-\lambda x}}</math></center> | |

For X~Gamma(k,<math>\theta</math>), where <math>k=h</math> and <math>\theta=1/\lambda</math>, the gamma probability density function is given by | For X~Gamma(k,<math>\theta</math>), where <math>k=h</math> and <math>\theta=1/\lambda</math>, the gamma probability density function is given by | ||

− | + | <center><math>\frac{x^{k-1}e^{-x/\theta}}{\Gamma(k)\theta^k}</math></center> | |

where | where | ||

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<br />'''Cumulative density function''': The gamma cumulative distribution function is given by | <br />'''Cumulative density function''': The gamma cumulative distribution function is given by | ||

− | + | <center><math>\frac{\gamma(k,x/\theta)}{\Gamma(k)}</math></center> | |

where | where | ||

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<br />'''Moment generating function''': The gamma moment-generating function is | <br />'''Moment generating function''': The gamma moment-generating function is | ||

− | + | <center><math>M(t)=(1-\theta t)^{-k}\!</math></center> | |

<br />'''Expectation''': The expected value of a gamma distributed random variable x is | <br />'''Expectation''': The expected value of a gamma distributed random variable x is | ||

− | + | <center><math>E(X)=k\theta\!</math></center> | |

<br />'''Variance''': The gamma variance is | <br />'''Variance''': The gamma variance is | ||

− | + | <center><math>Var(X)=k\theta^2\!</math></center> | |

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===Applications=== | ===Applications=== |

## Revision as of 14:37, 11 July 2011

### Gamma Distribution

**Definition**: Gamma distribution is a distribution that arises naturally in processes for which the waiting times between events are relevant. It can be thought of as a waiting time between Poisson distributed events.

**Probability density function**: The waiting time until the hth Poisson event with a rate of change \(\lambda\) is

For X~Gamma(k,\(\theta\)), where \(k=h\) and \(\theta=1/\lambda\), the gamma probability density function is given by

where

- e is the natural number (e = 2.71828…)
- k is the number of occurrences of an event
- if k is a positive integer, then \(\Gamma(k)=(k-1)!\) is the gamma function
- \(\theta=1/\lambda\) is the mean number of events per time unit, where \(\lambda\) is the mean time between events. For example, if the mean time between phone calls is 2 hours, then you would use a gamma distribution with \(\theta\)=1/2=0.5. If we want to find the mean number of calls in 5 hours, it would be 5 \(\times\) 1/2=2.5.
- x is a random variable

**Cumulative density function**: The gamma cumulative distribution function is given by

where

- if k is a positive integer, then \(\Gamma(k)=(k-1)!\) is the gamma function
- \(\gamma(k,x/\theta)=\int_0^{x/\theta}t^{k-1}e^{-t}dt\)

**Moment generating function**: The gamma moment-generating function is

**Expectation**: The expected value of a gamma distributed random variable x is

**Variance**: The gamma variance is

### Applications

The gamma distribution can be used a range of disciplines including queuing models, climatology, and financial services. Examples of events that may be modeled by gamma distribution include:

- The amount of rainfall accumulated in a reservoir
- The size of loan defaults or aggregate insurance claims
- The flow of items through manufacturing and distribution processes
- The load on web servers
- The many and varied forms of telecom exchange

The gamma distribution is also used to model errors in a multi-level Poisson regression model because the combination of a Poisson distribution and a gamma distribution is a negative binomial distribution.

### Example

Suppose you are fishing and you expect to get a fish once every 1/2 hour. Compute the probability that you will have to wait between 2 to 4 hours before you catch 4 fish.

One fish every 1/2 hour means we would expect to get \(\theta=1/0.5=2\) fish every hour on average. Using \(\theta=2\) and \(k=4\), we can compute this as follows: \[P(2\le X\le 4)=\sum_{x=2}^4\frac{x^{4-1}e^{-x/2}}{\Gamma(4)2^4}=0.12388\]

The figure below shows this result using SOCR distributions