Raw Moment of Chi-Squared Distribution

Theorem
Let $n$ and $m$ be strictly positive integers.

Let $X \sim \chi^2_n$ where $\chi^2_n$ is the chi-squared distribution with $n$ degrees of freedom.

Then the $m$th raw moment $\expect {X^m}$ of $X$ is given by:


 * $\displaystyle \expect {X^m} = \prod_{k \mathop = 0}^{m - 1} \paren {n + 2 k}$

Proof
From the definition of the chi-squared distribution, $X$ has probability density function:


 * $\displaystyle \map {f_X} x = \dfrac 1 {2^{n / 2} \map \Gamma {n / 2} } x^{\paren {n / 2} - 1} e^{- x / 2}$

From the definition of the expected value of a continuous random variable:


 * $\displaystyle \expect {X^m} = \int_0^\infty x^m \map {f_X} x \rd x$

So: