# Fratio (or F) DistributionΒΆ

The distribution of $$\left(X_{1}/X_{2}\right)\left(\nu_{2}/\nu_{1}\right)$$ if $$X_{1}$$ is chi-squared with $$v_{1}$$ degrees of freedom and $$X_{2}$$ is chi-squared with $$v_{2}$$ degrees of freedom. The support is $$x\geq0$$.

\begin{eqnarray*} f\left(x;\nu_{1},\nu_{2}\right) & = & \frac{\nu_{2}^{\nu_{2}/2}\nu_{1}^{\nu_{1}/2}x^{\nu_{1}/2-1}}{\left(\nu_{2}+\nu_{1}x\right)^{\left(\nu_{1}+\nu_{2}\right)/2}B\left(\frac{\nu_{1}}{2},\frac{\nu_{2}}{2}\right)}\\ F\left(x;v_{1},v_{2}\right) & = & I\left(\frac{\nu_{1}x}{\nu_{2}+\nu_{1}x}; \frac{\nu_{1}}{2},\frac{\nu_{2}}{2}\right)\\ G\left(q;\nu_{1},\nu_{2}\right) & = & \left(\frac{\nu_{2}} {I^{-1}\left(q; \nu_{1}/2,\nu_{2}/2\right)}-\frac{\nu_{1}}{\nu_{2}}\right)^{-1}\\ \mu & = & \frac{\nu_{2}}{\nu_{2}-2}\quad\textrm{for }\nu_{2}>2\\ \mu_{2} & = & \frac{2\nu_{2}^{2}\left(\nu_{1}+\nu_{2}-2\right)}{\nu_{1}\left(\nu_{2}-2\right)^{2}\left(\nu_{2}-4\right)}\quad\textrm{for } v_{2}>4\\ \gamma_{1} & = & \frac{2\left(2\nu_{1}+\nu_{2}-2\right)}{\nu_{2}-6}\sqrt{\frac{2\left(\nu_{2}-4\right)}{\nu_{1}\left(\nu_{1}+\nu_{2}-2\right)}}\quad\textrm{for }\nu_{2}>6\\ \gamma_{2} & = & \frac{3\left(8+\left(\nu_{2}-6\right)\gamma_{1}^{2}\right)}{2\nu-16}\quad\textrm{for }\nu_{2}>8\end{eqnarray*}

where $$I\left(x;a,b\right)=I_{x}\left(a,b\right)$$ is the regularized incomplete Beta function.

Implementation: scipy.stats.f

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