Mean and Variance of the t-Distribution
Formulas
$X \sim t (\nu)$ if $$ E(X) = 0 \qquad , \nu >1 \\ \operatorname{Var}(X) = {{ \nu } \over { \nu - 2 }} \qquad , \nu > 2 $$
Derivation
Strategy: Similar to the chi-squared distribution, the t-distribution also has known moment-generating functions, which we utilize.
Moment of the t-distribution: Assume two random variables $W,V$ are independent and $W \sim N(0,1)$, $V \sim \chi^{2} (r)$. If $k < r$, then $\displaystyle T := { {W} \over {\sqrt{V/r} } }$ has the $k$th moment existing $$ E T^{k} = E W^{k} {{ 2^{-k/2} \Gamma \left( {{ r } \over { 2 }} - {{ k } \over { 2 }} \right) } \over { \Gamma \left( {{ r } \over { 2 }} \right) r^{-k/2} }} $$
- $N \left( \mu , \sigma^{2} \right)$ is a normal distribution with mean $\mu$ and variance $\sigma^{2}$.
- $\chi^{2} \left( r \right)$ is a chi-squared distribution with degrees of freedom $r$.
- $\Gamma$ is the gamma function.
Mean
If we consider $r = \nu$, since $1 = k < r = \nu$, $ET^{1}$ exists, and $W$ follows a standard normal distribution $N(0,1)$, thus $EW^{1} = 0$. Therefore, $ET^{1} = 0$.
■
Variance
If $k=2$ and $W$ follows a standard normal distribution, then $EW^{2} = 1 + 0^{2}$ so $$ \begin{align*} ET^{2} =& EW^{2} {{ 2^{-2/2} \Gamma \left( {{ \nu } \over { 2 }} - {{ 2 } \over { 2 }} \right) } \over { \Gamma \left( {{ \nu } \over { 2 }} \right) \nu^{-2/2} }} \\ =& 1 {{ \nu } \over { 2 }} {{ \Gamma \left( {{ \nu - 1 } \over { 2 }} \right) } \over { \Gamma \left( {{ \nu } \over { 2 }} \right) }} \\ =& {{ \nu } \over { 2 }} {{ 1 } \over { {{ \nu } \over { 2 }} - 1 }} \\ =& {{ \nu } \over { \nu - 2 }} \end{align*} $$
■