Chi distribution

The probability density function (pdf) of the chi-distribution is

${\displaystyle f(x;k)={\begin{cases}{\dfrac {x^{k-1}e^{-x^{2}/2}}{2^{k/2-1}\Gamma \left({\frac {k}{2}}\right)}},&x\geq 0;\\0,&{\text{otherwise}}.\end{cases}}}$

where ${\displaystyle \Gamma (z)}$ is the gamma function.

The cumulative distribution function is given by:

${\displaystyle F(x;k)=P(k/2,x^{2}/2)\,}$

where ${\displaystyle P(k,x)}$ is the regularized gamma function.

The moment-generating function is given by:

${\displaystyle M(t)=M\left({\frac {k}{2}},{\frac {1}{2}},{\frac {t^{2}}{2}}\right)+t{\sqrt {2}}\,{\frac {\Gamma ((k+1)/2)}{\Gamma (k/2)}}M\left({\frac {k+1}{2}},{\frac {3}{2}},{\frac {t^{2}}{2}}\right),}$

where ${\displaystyle M(a,b,z)}$ is Kummer's confluent hypergeometric function. The characteristic function is given by:

${\displaystyle \varphi (t;k)=M\left({\frac {k}{2}},{\frac {1}{2}},{\frac {-t^{2}}{2}}\right)+it{\sqrt {2}}\,{\frac {\Gamma ((k+1)/2)}{\Gamma (k/2)}}M\left({\frac {k+1}{2}},{\frac {3}{2}},{\frac {-t^{2}}{2}}\right).}$

The raw moments are then given by:

${\displaystyle \mu _{j}=\int _{0}^{\infty }f(x;k)x^{j}\mathrm {d} x=2^{j/2}\ {\frac {\ \Gamma \left({\tfrac {1}{2}}(k+j)\right)\ }{\Gamma \left({\tfrac {1}{2}}k\right)}}}$

where ${\displaystyle \ \Gamma (z)\ }$ is the gamma function. Thus the first few raw moments are:

${\displaystyle \mu _{1}={\sqrt {2\ }}\ {\frac {\ \Gamma \left({\tfrac {1}{2}}(k+1)\right)\ }{\Gamma \left({\tfrac {1}{2}}k\right)}}}$

${\displaystyle \mu _{2}=k\ ,}$

${\displaystyle \mu _{3}=2{\sqrt {2\ }}\ {\frac {\ \Gamma \left({\tfrac {1}{2}}(k+3)\right)\ }{\Gamma \left({\tfrac {1}{2}}k\right)}}=(k+1)\ \mu _{1}\ ,}$

${\displaystyle \mu _{4}=(k)(k+2)\ ,}$

${\displaystyle \mu _{5}=4{\sqrt {2\ }}\ {\frac {\ \Gamma \left({\tfrac {1}{2}}(k\!+\!5)\right)\ }{\Gamma \left({\tfrac {1}{2}}k\right)}}=(k+1)(k+3)\ \mu _{1}\ ,}$

${\displaystyle \mu _{6}=(k)(k+2)(k+4)\ ,}$

where the rightmost expressions are derived using the recurrence relationship for the gamma function:

${\displaystyle \Gamma (x+1)=x\ \Gamma (x)~.}$

From these expressions we may derive the following relationships:

Mean: ${\displaystyle \mu ={\sqrt {2\ }}\ {\frac {\ \Gamma \left({\tfrac {1}{2}}(k+1)\right)\ }{\Gamma \left({\tfrac {1}{2}}k\right)}}\ ,}$ which is close to ${\displaystyle {\sqrt {k-{\tfrac {1}{2}}\ }}\ }$ for large k.

Variance: ${\displaystyle V=k-\mu ^{2}\ ,}$ which approaches ${\displaystyle \ {\tfrac {1}{2}}\ }$ as k increases.

Skewness: ${\displaystyle \gamma _{1}={\frac {\mu }{\ \sigma ^{3}\ }}\left(1-2\sigma ^{2}\right)~.}$

Kurtosis excess: ${\displaystyle \gamma _{2}={\frac {2}{\ \sigma ^{2}\ }}\left(1-\mu \ \sigma \ \gamma _{1}-\sigma ^{2}\right)~.}$

The entropy is given by:

${\displaystyle S=\ln(\Gamma (k/2))+{\frac {1}{2}}(k\!-\!\ln(2)\!-\!(k\!-\!1)\psi ^{0}(k/2))}$

where ${\displaystyle \psi ^{0}(z)}$ is the polygamma function.

We find the large n=k+1 approximation of the mean and variance of chi distribution. This has application e.g. in finding the distribution of standard deviation of a sample of normally distributed population, where n is the sample size.

The mean is then:

${\displaystyle \mu ={\sqrt {2}}\,\,{\frac {\Gamma (n/2)}{\Gamma ((n-1)/2)}}}$

We use the Legendre duplication formula to write:

${\displaystyle 2^{n-2}\,\Gamma ((n-1)/2)\cdot \Gamma (n/2)={\sqrt {\pi }}\Gamma (n-1)}$,

so that:

${\displaystyle \mu ={\sqrt {2/\pi }}\,2^{n-2}\,{\frac {(\Gamma (n/2))^{2}}{\Gamma (n-1)}}}$

Using Stirling's approximation for Gamma function, we get the following expression for the mean:

${\displaystyle \mu ={\sqrt {2/\pi }}\,2^{n-2}\,{\frac {\left({\sqrt {2\pi }}(n/2-1)^{n/2-1+1/2}e^{-(n/2-1)}\cdot [1+{\frac {1}{12(n/2-1)}}+O({\frac {1}{n^{2}}})]\right)^{2}}{{\sqrt {2\pi }}(n-2)^{n-2+1/2}e^{-(n-2)}\cdot [1+{\frac {1}{12(n-2)}}+O({\frac {1}{n^{2}}})]}}}$

${\displaystyle =(n-2)^{1/2}\,\cdot \left[1+{\frac {1}{4n}}+O({\frac {1}{n^{2}}})\right]={\sqrt {n-1}}\,(1-{\frac {1}{n-1}})^{1/2}\cdot \left[1+{\frac {1}{4n}}+O({\frac {1}{n^{2}}})\right]}$

${\displaystyle ={\sqrt {n-1}}\,\cdot \left[1-{\frac {1}{2n}}+O({\frac {1}{n^{2}}})\right]\,\cdot \left[1+{\frac {1}{4n}}+O({\frac {1}{n^{2}}})\right]}$

${\displaystyle ={\sqrt {n-1}}\,\cdot \left[1-{\frac {1}{4n}}+O({\frac {1}{n^{2}}})\right]}$

And thus the variance is:

${\displaystyle V=(n-1)-\mu ^{2}\,=(n-1)\cdot {\frac {1}{2n}}\,\cdot \left[1+O({\frac {1}{n}})\right]}$