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¶
“Student’s t-distribution”, Wikipedia, https://en.wikipedia.org/wiki/Student%27s_t-distribution
Implementation: scipy.stats.t