I have two independent random variables:
- $X_{1}$ is governed by an Inverse Gamma(a, b).
- $X_{2}$ is governed by an Inverse Gamma(a, b).
Given the transformation $Y = X_{1}/X_{2}$, I want to find the marginal distribution for $Y$, $f_{Y}(y)$; however, I am a little confused how to proceed with the transformation. Here is how I tried thinking of it:
To make things easier, I tried assigning $W = X_{2}.$
$$Y = X_{1}/X_{2} \rightarrow Y = X_{1}/W$$
$$W = X_{2} \rightarrow X_{2} = W$$
I am unsure if this is correct and how to proceed.
To simplify the notation let's set $Z=\frac{X}{Y}$
Let's set
$$\begin{cases} z=\frac{x}{y}\\ u=x, \end{cases}\rightarrow\begin{cases} x=u\\ y=\frac{u}{z}, \end{cases}$$
The jacobian is $|J|=\frac{u}{z^2}$ and thus the joint density $(U,Z)$ is
$$f_{UZ}(u,z)=\frac{z^{a-1}}{\Gamma(a)\Gamma(a)}\cdot \frac{b^{2a}e^{-\frac{b}{u}(z+1)}}{u^{2a+1}}$$
This can be rewritten as
$$\frac{\Gamma(a+a)}{\Gamma(a)\Gamma(a)}z^{a-1}(1+z)^{-a-a}\cdot \frac{b^{2a}e^{-\frac{b}{u}(z+1)}}{\Gamma(2a)}\Bigg(\frac{1+z}{u}\Bigg)^{2a+1}\cdot\frac{1}{1+z}$$
To get your $f_Z(z)$ you must integrate
$$\frac{\Gamma(a+a)}{\Gamma(a)\Gamma(a)}z^{a-1}(1+z)^{-a-a}\cdot\underbrace{\int_0^{\infty} \frac{b^{2a}e^{-\frac{b}{u}(z+1)}}{\Gamma(2a)}\Bigg(\frac{1+z}{u}\Bigg)^{2a+1}d\Bigg(\frac{u}{1+z}\Bigg)}_{=1}$$
Thus your density is a Beta prime as the integral is an integral of an Inverse gamma
$$ \bbox[5px,border:2px solid red] { f_Z(z)=\frac{\Gamma(a+a)}{\Gamma(a)\Gamma(a)}z^{a-1}(1+z)^{-a-a}\cdot\mathbb{1}_{[0;+\infty)}(z) \ } $$
If you do not like to deal with Inverse Gamma rv's you can rewrite you $Z$ in the following way
$$Z=\frac{X}{Y}=\frac{\frac{1}{Y}}{\frac{1}{X}}$$
and thus you can use Gamma distributions if you are familiar with them.
In fact, if $X\sim Gamma$ then $\frac{1}{X}\sim InverseGamma$