# Student t Distribution#

There is one shape parameter $$\nu>0$$ and the support is $$x\in\mathbb{R}$$.

\begin{eqnarray*} f\left(x;\nu\right) & = & \frac{\Gamma\left(\frac{\nu+1}{2}\right)}{\sqrt{\pi\nu}\Gamma\left(\frac{\nu}{2}\right)\left[1+\frac{x^{2}}{\nu}\right]^{\frac{\nu+1}{2}}}\\ F\left(x;\nu\right) & = & \left\{ \begin{array}{ccc} \frac{1}{2}I\left(\frac{\nu}{\nu+x^{2}}; \frac{\nu}{2},\frac{1}{2}\right) & & x\leq0\\ 1-\frac{1}{2}I\left(\frac{\nu}{\nu+x^{2}}; \frac{\nu}{2},\frac{1}{2}\right) & & x\geq0 \end{array} \right.\\ G\left(q;\nu\right) & = & \left\{ \begin{array}{ccc} -\sqrt{\frac{\nu}{I^{-1}\left(2q; \frac{\nu}{2},\frac{1}{2}\right)}-\nu} & & q\leq\frac{1}{2}\\ \sqrt{\frac{\nu}{I^{-1}\left(2-2q; \frac{\nu}{2},\frac{1}{2}\right)}-\nu} & & q\geq\frac{1}{2} \end{array} \right. \end{eqnarray*}
\begin{eqnarray*} m_{n}=m_{d}=\mu & = & 0\\ \mu_{2} & = & \frac{\nu}{\nu-2}\quad\nu>2\\ \gamma_{1} & = & 0\quad\nu>3\\ \gamma_{2} & = & \frac{6}{\nu-4}\quad\nu>4\end{eqnarray*}

where $$I\left(x; a,b\right)$$ is the incomplete beta integral and $$I^{-1}\left(I\left(x; a,b\right); a,b\right)=x$$. As $$\nu\rightarrow\infty,$$ this distribution approaches the standard normal distribution.

$h\left[X\right]=\frac{\nu+1}{2} \left[\psi \left(\frac{1+\nu}{2} \right) -\psi \left(\frac{\nu}{2} \right) \right] + \ln \left[ \sqrt{\nu} B \left( \frac{\nu}{2}, \frac{1}{2} \right) \right]$

where $$\psi(x)$$ is the digamma function and $$B(x, y)$$ is the beta function.

## References#

Implementation: scipy.stats.t