Statistics/Distributions/Gamma

Gamma Distribution

Gamma

Probability density function
Cumulative distribution function
Parameters	$\scriptstyle k\;>\;0$ shape $\scriptstyle \theta \;>\;0\,$ scale
Support	$\scriptstyle x\;\in \;(0,\,\infty )\!$
PDF	$\scriptstyle {\frac {1}{\Gamma (k)\theta ^{k}}}x^{k\,-\,1}e^{-{\frac {x}{\theta }}}\,\!$
CDF	$\scriptstyle {\frac {1}{\Gamma (k)}}\gamma \left(k,\,{\frac {x}{\theta }}\right)\!$
Mean	$\scriptstyle \operatorname {E} [X]=k\theta \!$ $\scriptstyle \operatorname {E} [\ln X]=\psi (k)+\ln(\theta )\!$ (see digamma function)
Median	No simple closed form
Mode	$\scriptstyle (k\,-\,1)\theta {\text{ for }}k\;>\;1\,\!$
Variance	$\scriptstyle \operatorname {Var} [X]=k\theta ^{2}\,\!$ $\scriptstyle \operatorname {Var} [\ln X]=\psi _{1}(k)\!$ (see trigamma function )
Skewness	$\scriptstyle {\frac {2}{\sqrt {k}}}\,\!$
Ex. kurtosis	$\scriptstyle {\frac {6}{k}}\,\!$
Entropy	$\scriptstyle {\begin{aligned}\scriptstyle k&\scriptstyle \,+\,\ln \theta \,+\,\ln[\Gamma (k)]\\\scriptstyle &\scriptstyle \,+\,(1\,-\,k)\psi (k)\end{aligned}}$

The Gamma distribution is very important for technical reasons, since it is the parent of the exponential distribution and can explain many other distributions.

The probability distribution function is:

$f_{x}(x)={\begin{cases}{\frac {1}{a^{p}\Gamma (p)}}x^{p-1}e^{-x/a},&{\mbox{if }}x\geq 0\\0,&{\mbox{if }}x<0\end{cases}}\quad a,p>0$

Where $\Gamma (p)=\int _{0}^{\infty }t^{p-1}e^{-t}\,dt\,$ is the Gamma function. The cumulative distribution function cannot be found unless p=1, in which case the Gamma distribution becomes the exponential distribution. The Gamma distribution of the stochastic variable X is denoted as $X\in \Gamma (p,a)$.

Alternatively, the gamma distribution can be parameterized in terms of a shape parameter $\alpha =k$ and an inverse scale parameter $\beta =1/\theta$, called a rate parameter:

$g(x;\alpha ,\beta )=Kx^{\alpha -1}e^{-\beta \,x}\ \mathrm {for} \ x>0\,\!.$

where the $K$ constant can be calculated setting the integral of the density function as 1:

$\int _{-\infty }^{+\infty }g(x;\alpha ,\beta )\mathrm {d} t\,=\int _{0}^{+\infty }Kx^{\alpha -1}e^{-\beta \,x}\mathrm {d} x\,=1$

following:

$K\int _{0}^{+\infty }x^{\alpha -1}e^{-\beta \,x}\mathrm {d} x\,=1$

$K={\frac {1}{\int _{0}^{+\infty }x^{\alpha -1}e^{-\beta \,x}\mathrm {d} x}}$

and, with change of variable $y=\beta x$ :

${\begin{aligned}K&={\frac {1}{\int _{0}^{+\infty }{\frac {y^{\alpha -1}}{\beta ^{\alpha -1}}}e^{-y}{\frac {\mathrm {d} y}{\beta }}}}\\&={\frac {1}{{\frac {1}{\beta ^{\alpha }}}\int _{0}^{+\infty }y^{\alpha -1}e^{-y}\mathrm {d} y}}\\&={\frac {\beta ^{\alpha }}{\int _{0}^{+\infty }y^{\alpha -1}e^{-y}\mathrm {d} y}}\\&={\frac {\beta ^{\alpha }}{\Gamma (\alpha )}}\end{aligned}}$

following:

$g(x;\alpha ,\beta )=x^{\alpha -1}{\frac {\beta ^{-\alpha }\,e^{-\beta \,x}}{\Gamma (\alpha )}}\ \mathrm {for} \ x>0\,\!.$

Probability Density Function

We first check that the total integral of the probability density function is 1.

$\int _{-\infty }^{\infty }{\frac {1}{a^{p}\Gamma (p)}}x^{p-1}e^{-x/a}dx$

Now we let y=x/a which means that dy=dx/a

${\frac {1}{\Gamma (p)}}\int _{0}^{\infty }y^{p-1}e^{-y}dy$

${\frac {1}{\Gamma (p)}}\Gamma (p)=1$

Mean

$\operatorname {E} [X]=\int _{-\infty }^{\infty }x\cdot {\frac {1}{a^{p}\Gamma (p)}}x^{p-1}e^{-x/a}dx$

Now we let y=x/a which means that dy=dx/a.

$\operatorname {E} [X]=\int _{0}^{\infty }ay\cdot {\frac {1}{\Gamma (p)}}y^{p-1}e^{-y}dy$

$\operatorname {E} [X]={\frac {a}{\Gamma (p)}}\int _{0}^{\infty }y^{p}e^{-y}dy$

$\operatorname {E} [X]={\frac {a}{\Gamma (p)}}\Gamma (p+1)$

We now use the fact that $\Gamma (z+1)=z\Gamma (z)$

$\operatorname {E} [X]={\frac {a}{\Gamma (p)}}p\Gamma (p)=ap$

Variance

We first calculate E[X^2]

$\operatorname {E} [X^{2}]=\int _{-\infty }^{\infty }x^{2}\cdot {\frac {1}{a^{p}\Gamma (p)}}x^{p-1}e^{-x/a}dx$

Now we let y=x/a which means that dy=dx/a.

$\operatorname {E} [X^{2}]=\int _{0}^{\infty }a^{2}y^{2}\cdot {\frac {1}{a\Gamma (p)}}y^{p-1}e^{-y}ady$

$\operatorname {E} [X^{2}]={\frac {a^{2}}{\Gamma (p)}}\int _{0}^{\infty }y^{p+1}e^{-y}dy$

$\operatorname {E} [X^{2}]={\frac {a^{2}}{\Gamma (p)}}\Gamma (p+2)=pa^{2}(p+1)$

Now we use calculate the variance

$\operatorname {Var} (X)=\operatorname {E} [X^{2}]-(\operatorname {E} [X])^{2}$

$\operatorname {Var} (X)=pa^{2}(p+1)-(ap)^{2}=pa^{2}$

External links

• Interactive Gamma Distribution Web Applet (Java)