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Let IG denote Inverse-Gamma distribution Inverse-Gamma. If $X\sim IG(\alpha,1)$ and $Y\sim IG(\beta,1)$. Show that $\frac{X}{X+Y}\sim Beta(\alpha,\beta)$

I tried with jacobian transformation taking $Z=\frac{X}{X+Y}$ and $W=Y$ then \begin{align}X&=\frac{WZ}{1-Z};\quad Y=W \\[0.2cm]\frac{\partial(z,w)}{\partial(x,y)}&=\frac{1}{(1-z)^2}\\[0.3cm]f_{Z,W}(z,w)&=f_X\left(\frac{wz}{1-z}\right)f_Y(w)(1-z)^2\\[0.3cm]f_{Z,W}(z,w)&=\frac{1}{\Gamma{(\alpha)}}\left(\frac{wz}{1-z}\right)^{-\alpha-1}e^{-\frac{(1-z)}{wz}}\frac{1}{\Gamma{(\beta)}}w^{-\beta-1}e^{-\frac{1}{w}}(1-z)^2\end{align}

but I'm stuck, in some place I read that $\frac{X}{X+Y}$ is a type 3 Beta distribution, but I can't show that.

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  • $\begingroup$ I find $$\frac{\partial(z,w)}{\partial(x,y)}=\frac{1}{(1-Z)^2}$$ Without the $W$ in any case $\endgroup$ – Jimmy R. Jun 1 '16 at 14:55
  • $\begingroup$ @JimmyR. That is right, I made a mistake $\endgroup$ – Roland Jun 1 '16 at 15:11
  • $\begingroup$ I think this should be $Beta(\beta+1,\alpha)$ and not $Beta(\alpha,\beta)$ $\endgroup$ – Jimmy R. Jun 1 '16 at 16:53
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Assuming that $X,Y$ are independent, you have that \begin{align}f_{Z,W}(z,w)&=f_{X,Y}\left(\frac{wz}{1-z},w\right)\left|\frac{\partial(x,y)}{\partial(z,w)}\right|=f_{X}\left(\frac{wz}{1-z}\right)f_{Y}\left(w\right)\left|\frac{1}{(1-z)^2}\right|\\[0.2cm]&=\frac{1}{\Gamma(\alpha)\Gamma(\beta)}e^{-\frac{1-z}{wz}}\left(\frac{wz}{1-z}\right)^{-\alpha-1}e^{-\frac1w}w^{-\beta-1}\frac{1}{(1-z)^2}\\[0.2cm]&=\frac{1}{\Gamma(\alpha)\Gamma(\beta)}e^{-\frac1{wz}}z^{-\alpha-1}\left(1-z\right)^{\alpha-1}w^{-\alpha-\beta-2}\end{align} with $0<z<1$ and $0<w$. Hence \begin{align}f_Z(z)&=\int_0^{+\infty}f_{Z,W}(z,w)dw\\[0.3cm]&=\frac{1}{\Gamma(\alpha)\Gamma(\beta)}z^{-\alpha-1}\left(1-z\right)^{\alpha-1}\int_{0}^{+\infty}w^{-\alpha-\beta-2}e^{-\frac{1}{zw}}dw\end{align} Now you can calculate the integral if you observe that you can construct an inverse gamma distribution with parameters $\alpha+\beta+1$ and $1/z$. Indeed \begin{align}\int_{0}^{+\infty}w^{-\alpha-\beta-2}e^{-\frac1{wz}}dw&=\int_{0}^{+\infty}w^{-(\alpha+\beta+1)-1}e^{-\frac{1/z}{w}}dw\\[0.2cm]&=\frac{\Gamma(\alpha+\beta+1)}{(1/z)^{\alpha+\beta+1}}\cdot\int_{0}^{+\infty}\frac{(1/z)^{\alpha+\beta+1}}{\Gamma(\alpha+\beta+1)}w^{-(\alpha+\beta+1)-1}e^{-\frac{1/z}{w}}dw\\[0.2cm]&=\frac{\Gamma(\alpha+\beta+1)}{(1/z)^{\alpha+\beta+1}}\cdot1\end{align} Hence \begin{align}f_Z(z)&=\frac{1}{\Gamma(\alpha)\Gamma(\beta)}\left(1-z\right)^{\alpha-1}z^{-\alpha-1}\frac{\Gamma(\alpha+\beta+1)}{(1/z)^{\alpha+\beta+1}}=\frac{\Gamma(\alpha+\beta+1)}{\Gamma(\alpha)\Gamma(\beta)}z^{\left(\beta+1\right)-1}(1-z)^{\alpha-1}\\[0.2cm]&=\beta\frac{\Gamma(\alpha+\beta+1)}{\Gamma(\alpha)\Gamma(\beta+1)}z^{\left(\beta+1\right)-1}(1-z)^{\alpha-1}\end{align} which shows that $Z\sim Beta(\beta+1,\alpha)$. This all would be (probably) correct, if I hadn't this $\beta$ in the last formula. However I cannot get rid of it.

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If $X\sim\mathrm{InvGamma}(\alpha,1)$, then $\frac{1}{X} \sim \mathrm{Gamma}(\alpha,1)$. Since $\frac{X}{X+Y}= \frac{1/Y}{1/X+1/Y}$, it is enough to show that if $X'\sim \mathrm{Gamma}(\alpha,1)$ and $Y'\sim \mathrm{Gamma}(\beta,1)$ then $\frac{Y'}{X'+Y'}\sim \mathrm{Beta}(\alpha,\beta)$. Then see for example this answer.

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