Expectation of Chi Distribution

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Theorem

Let $n$ be a strictly positive integer.

Let $X \sim \chi_n$ where $\chi_n$ is the chi distribution with $n$ degrees of freedom.


Then the expectation of $X$ is given by:

$\expect X = \sqrt 2 \dfrac {\map \Gamma {\paren {n + 1} / 2} } {\map \Gamma {n / 2} }$

where $\Gamma$ is the gamma function.


Proof

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

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

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

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

So:

\(\ds \expect X\) \(=\) \(\ds \frac 1 {2^{\paren {n / 2} - 1} \map \Gamma {n / 2} } \int_0^\infty x^n e^{- x^2 / 2} \rd x\)
\(\ds \) \(=\) \(\ds \frac 1 {2^{\paren {n / 2} - 1 + \paren {1 / 2} } \map \Gamma {n / 2} } \int_0^\infty \frac 1 {\sqrt t} \paren {\sqrt {2 t} }^n e^{- t} \rd t\) substituting $x = \sqrt {2 t}$
\(\ds \) \(=\) \(\ds \frac {2^{\paren {n / 2} } } {2^{\paren {n / 2} - \paren {1 / 2} } \map \Gamma {n / 2} } \int_0^\infty t^{\paren {n - 1} / 2} e^{- t} \rd t\)
\(\ds \) \(=\) \(\ds \sqrt 2 \dfrac {\map \Gamma {\paren {n + 1} / 2} } {\map \Gamma {n / 2} }\) Definition of Gamma Function

$\blacksquare$