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.. _continuous-t:
Student t Distribution
======================
There is one shape parameter :math:`\nu>0` and the support is :math:`x\in\mathbb{R}`.
.. math::
:nowrap:
\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*}
.. math::
:nowrap:
\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 :math:`I\left(x; a,b\right)` is the incomplete beta integral and :math:`I^{-1}\left(I\left(x; a,b\right); a,b\right)=x`.
As :math:`\nu\rightarrow\infty,` this distribution approaches the standard normal distribution.
.. math::
h\left[X\right]=\frac{1}{4}\log\left(\frac{\pi c\Gamma^{2}\left(\frac{c}{2}\right)}{\Gamma^{2}\left(\frac{c+1}{2}\right)}\right)-\frac{\left(c+1\right)}{4}\left[\Psi\left(\frac{c}{2}\right)-cZ\left(c\right)+\pi\tan\left(\frac{\pi c}{2}\right)+\gamma+2\log2\right]
where
.. math::
Z\left(c\right)=\,_{3}F_{2}\left(1,1,1+\frac{c}{2};\frac{3}{2},2;1\right)=\sum_{k=0}^{\infty}\frac{k!}{k+1}\frac{\Gamma\left(\frac{c}{2}+1+k\right)}{\Gamma\left(\frac{c}{2}+1\right)}\frac{\Gamma\left(\frac{3}{2}\right)}{\Gamma\left(\frac{3}{2}+k\right)}
Implementation: `scipy.stats.t`
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