Show $X = \frac{d_2Y}{d_1(1-Y)}$ has $F$ distribution if $Y$ has $B\left(\frac{d_1}{2} ,\frac{d_2}{2}\right)$ distribution

Question

I search internet alot but I didn't find any prove for this theory.

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If $Y$ has $B\left(\frac{d_1}{2}, \frac{d_2}{2}\right)$ distribution, then show that $X$ with given formula has $F(d_1,d_2)$ distribution.
$$X = \frac{d_2Y}{d_1(1-Y)}$$

I do this but I couldn't do anything because of $\Gamma{}$ integral:
$$Y = \frac{\Gamma({\frac{d_1 + d_2}{2}})}{\Gamma(\frac{d_1}{2})\Gamma(\frac{d_2}{2})}x^{\frac{d_1}{2} - 1}(1-x)^{\frac{d_2}{2}-1}$$

so:
$$X = \frac{d_2Y}{d_1(1-Y)}$$ $$X = \frac{d_2\frac{\Gamma({\frac{d_1 + d_2}{2}})}{\Gamma(\frac{d_1}{2})\Gamma(\frac{d_2}{2})}x^{\frac{d_1}{2} - 1}(1-x)^{\frac{d_2}{2}-1}}{d_1(1-\frac{\Gamma({\frac{d_1 + d_2}{2}})}{\Gamma(\frac{d_1}{2})\Gamma(\frac{d_2}{2})}x^{\frac{d_1}{2} - 1}(1-x)^{\frac{d_2}{2}-1})}$$

$$X = x^{\frac{d_1}{2} - 1}\frac{\Gamma({\frac{d_1 + d_2}{2}})}{\Gamma(\frac{d_1}{2})\Gamma(\frac{d_2}{2})}(\frac{d_2\Gamma(\frac{d_1}{2})\Gamma(\frac{d_2}{2})(1-x)^{\frac{d_2}{2}-1}}{d_1\Gamma(\frac{d_1}{2})\Gamma(\frac{d_2}{2}) - \Gamma({\frac{d_1 + d_2}{2}})x^{\frac{d_1}{2} - 1}(1-x)^{\frac{d_2}{2}-1}})$$

compare it to $F(d_1,d_2)$:

$$F(d_1,d_2)=x^{\frac{d_1}{2} - 1}\frac{\Gamma({\frac{d_1 + d_2}{2}})}{\Gamma(\frac{d_1}{2})\Gamma(\frac{d_2}{2})}(\frac{d_1}{d_2})^{\frac{d_1}{2}-1}\frac{1}{(1+\frac{d_1}{d_2}x)^{\frac{d_1+d_2}{2}}}$$

so there must:

$$(\frac{d_1}{d_2})^{\frac{d_1}{2}-1}\frac{1}{(1+\frac{d_1}{d_2}x)^{\frac{d_1+d_2}{2}}} = (\frac{d_2\Gamma(\frac{d_1}{2})\Gamma(\frac{d_2}{2})(1-x)^{\frac{d_2}{2}-1}}{d_1\Gamma(\frac{d_1}{2})\Gamma(\frac{d_2}{2}) - \Gamma({\frac{d_1 + d_2}{2}})x^{\frac{d_1}{2} - 1}(1-x)^{\frac{d_2}{2}-1}})$$

but I can't prove anymore because of $\Gamma$ integral!

If is it possible, give me a hint for prove this. Thanks.

Answer 1

You should work step by step. And no need to break the Beta function into Gamma functions.

Write down the pdf of the given distribution:

Pdf of $Y$ is $$f_Y(y)=\frac{ y^{d_1/2-1}(1-y)^{d_2/2-1}}{B\left(\frac{d_1}{2},\frac{d_2}{2}\right)}\mathbf1_{0<y<1}$$

You are transforming $Y\to X$ such that$$X=\frac{d_2Y}{d_1(1-Y)}\implies y=\frac{d_1x}{d_1x+d_2}$$

Find the support of the transformed variable:

Since $d_1,d_2>0$, $$0<y<1\implies x>0$$

Find the jacobian of transformation:

$$\frac{\mathrm dy}{\mathrm dx}=\frac{d_1d_2}{(d_1x+d_2)^2}$$

Finally write down the pdf of the transformed variable using the change of variables formula:

So the pdf of $X$ is

\begin{align} f_X(x)&=f_Y\left(\frac{d_1x}{d_1x+d_2}\right)\left|\frac{\mathrm dy}{\mathrm dx}\right|\mathbf1_{x>0} \\\\&= \frac{(d_1x)^{d_1/2-1}\left(1-\frac{d_1x}{d_1x+d_2}\right)^{d_2/2-1}}{B\left(\frac{d_1}{2},\frac{d_2}{2}\right)(d_1x+d_2)^{d_1/2-1}}\cdot\frac{d_1d_2}{(d_1x+d_2)^2}\mathbf1_{x>0} \\\\&= \frac{d_1^{d_1/2}d_2^{d_2/2}}{B\left(\frac{d_1}{2},\frac{d_2}{2}\right)(d_1x+d_2)^{d_1/2+d_2/2}}x^{d_1/2-1}\mathbf1_{x>0} \\\\&=\frac{\left(\frac{d_1}{d_2}\right)^{d_1/2}x^{d_1/2-1}}{B\left(\frac{d_1}{2},\frac{d_2}{2}\right)\left(1+\frac{d_1}{d_2}x\right)^{(d_1+d_2)/2}}\mathbf1_{x>0} \end{align}

The above is the density of an $F$ distribution with $(d_1,d_2)$ degrees of freedom.