Question

Show that if Q ∼ beta(1/2α, 1/2β) then βQ/α(1−Q) ∼ Fα,β.

Answer 1

We are given that $\small Q \sim Beta \left( \frac{\alpha}{2}, \frac{\beta}{2} \right )$

Thus, the pdf of Q is given by:

$\small f_Q(Q) = \frac{1}{B \left( \frac{\alpha}{2}, \frac{\beta}{2} \right )} Q^{\frac{\alpha}{2}-1} (1-Q)^{\frac{\beta}{2}-1} \ \ \ \ \ ; 0 \le Q \le 1$

where B(.,.) denotes the beta funtion.

Now, let

$\small \begin{align*} \\ F &= \frac{\beta Q}{\alpha(1-Q)} \\ \Rightarrow \alpha F - \alpha F Q &= \beta Q \\ \Rightarrow (\beta + \alpha F)Q &= \alpha F \\ \Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ Q &= \frac{\alpha F}{\beta + \alpha F} \end{align*}$

Thus, we get:

$\small \begin{align*} \\ \frac{\partial Q}{\partial F} &= \frac{\partial }{\partial F}\left( \frac{\alpha F} {\beta + \alpha F} \right ) \\ &= \frac{(\alpha)(\beta + \alpha F) - (\alpha F )(\alpha)}{(\beta + \alpha F)^2} \\ &= \frac{\alpha \beta}{(\beta + \alpha F)^2} \end{align*}$

Now, the pdf of F is given by:

$\small \begin{align*} f_F(F) &= f_Q \left( \frac{\alpha F}{\beta + \alpha F} \right ) *\left | \frac{\partial Q}{\partial F} \right | \\ &= \frac{1}{B\left(\frac{\alpha}{2},\frac{\beta}{2} \right )}* \left( \frac{\alpha F}{\beta + \alpha F} \right )^{\frac{\alpha}{2}-1}* \left( 1- \frac{\alpha F}{\beta + \alpha F} \right )^{\frac{\beta}{2}-1} \frac{\alpha \beta}{(\beta+\alpha F)^2} \\ &= \frac{1}{B\left(\frac{\alpha}{2},\frac{\beta}{2} \right )}* \left( \frac{\alpha F}{\beta + \alpha F} \right )^{\frac{\alpha}{2}-1}* \left( \frac{\beta}{\beta + \alpha F} \right )^{\frac{\beta}{2}-1} \frac{\alpha \beta}{(\beta+\alpha F)^2} \\ &= \frac{\alpha^{\frac{\alpha}{2}-1+1}\beta^{\frac{\beta}{2}-1+1}}{B\left(\frac{\alpha}{2},\frac{\beta}{2} \right )}* \frac{F^{\frac{\alpha}{2}-1}}{(\beta+\alpha F)^{\frac{\alpha}{2}-1+\frac{\beta}{2}-1+2}} \end{align*}$

$\small \begin{align*} f_F(F) &= \frac{\alpha^{\frac{\alpha}{2}}\beta^{\frac{\beta}{2}}}{B\left(\frac{\alpha}{2},\frac{\beta}{2} \right )}* \frac{F^{\frac{\alpha}{2}-1}}{(\beta+\alpha F)^{\frac{\alpha+\beta}{2}}} \end{align*}$

which is the pdf of $\small F_{\alpha,\beta}$ distribution.

Thus, $\small F = \frac{\beta Q}{\alpha (1-Q)} \sim F_{\alpha,\beta}$

Hence Proved

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