doc/source/tutorial/stats/continuous_t.rst
.. _continuous-t:
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{\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 :math:\psi(x) is the digamma function and :math:B(x, y) is the
beta function.
Implementation: scipy.stats.t