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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 $$X=\frac{WZ}{1-Z};Y=W$$

$$\frac{\partial(z,w)}{\partial(x,y)}=\frac{1}{(1-z)^2}$$

$$f_{z,w}(z,w)=f_x(\frac{wz}{1-z})f_y(w)(1-z)^2$$ $$f_{z,w}(z,w)=\frac{1}{\Gamma{(a)}}(\frac{wz}{1-z})^{-\alpha-1}e^{-\frac{(1-z)}{wz}}\frac{1}{\Gamma{(\beta)}}w^{-\beta-1}e^{-\frac{1}{w}}(1-z)^2$$

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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    $\begingroup$ Your result is a joint density for $z$ and $w$, but you need marginal density for $z$ only. So if you integrate your $f_{z,w}(z,w)$ by $w$ from 0 to $\infty$ and use property of beta function that $B(\alpha, \beta) = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}$ you'll obtain exactly the same Beta distribution as you want. Note that you don't have to calculate the integral from scratch because expression under it can be seen as inverse gamma distribution. $\endgroup$
    – Alexander Rodin
    Jun 1, 2016 at 18:01
  • $\begingroup$ And yes, $(1 - z)^2$ should be $(1 - z)^{-2}$ (you could see it yourself because otherwise $f_{z,w}(z,w)$ has wrong dimension. $\endgroup$
    – Alexander Rodin
    Jun 1, 2016 at 18:11
  • $\begingroup$ Looks like that should be $\frac{X}{X+Y}\sim Beta(\beta,\alpha)$ $\endgroup$
    – wolfies
    Jun 1, 2016 at 19:45

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Here is my solution from first principles:

$$P\left(z' \leq z\ \left|\ z' = \frac{x}{x+y},\ x \sim X,\ y \sim Y\right. \right) = P\left(\frac{x}{x+y} \leq z\right) = $$ $$ \int dP_{X}(x) \int dP_{Y}(y)\ I\left[ \frac{x}{x+y} \leq z\right] = \iint dx\ dy \ f_{X}(x)\ f_{Y}(y)\ \theta\left(y \frac{z} {1-z} - x\right),$$ where $\theta(x)$ is a Heaviside Step function. Differentiating $P(z)$ by $z$ one can obtain that $$f_{Z}(z) = \frac{d}{dz} P(z' < z) = \int dx\ f_{X}(x) \int dy\ f_{Y}(y)\ \delta\left(y \frac{z} {1-z} - x\right) \frac{y}{(1-z)^2},$$ where $\delta(x)$ is a Dirac Delta function. So integrating over $x$ one could see that

$$f_{Z}(z) = \int dy \ f_{Y}(y)\ f_X\left(y \frac{z}{1-z}\right) \frac{y}{(1-z)^2} =$$ $$\frac{1}{\Gamma(\alpha)\Gamma(\beta)} (1-z)^{-2} \int dy\ y^{-\beta -1}e^{-\frac{1}{y}}\cdot y^{-\alpha - 1} z^{-\alpha -1} (1-z)^{\alpha+1} e^{-\frac{1-z}{yz}},$$ and after simplification and integration by $y$ $$f_{Z}(z) = \frac{\Gamma(\alpha + \beta)}{\Gamma(\alpha)\Gamma(\beta)} (1-z)^{\alpha-1} z^{\beta - 1}$$ which is probability density of beta distribution $Beta(\beta, \alpha).$

P.S. You can avoid using of Step/Delta functions by specifying precise limits of the inner integral, but I find that way less handy.

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