Expectation of Chi Distribution
Jump to navigation
Jump to search
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:
- $\map {f_X} x = \dfrac 1 {2^{\paren {n / 2} - 1} \map \Gamma {n / 2} } x^{n - 1} e^{- x^2 / 2}$
From the definition of the expected value of a continuous random variable:
- $\ds \expect X = \int_0^\infty x \map {f_X} x \rd x$
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 |
$\blacksquare$