# Ratio of 2 Gamma distributions

1. Mar 4, 2007

### jimmy1

If X and Y are gamma distributed random variables, then the ratio X/Y, I was told follows a beta distribution, but all I can find so for is that the ratio X/(X+Y) follows a beta distrinbution.
So is it true that X/Y follows a beta distribution?

2. Mar 4, 2007

### jimmy1

Ok, I found the answer (just had a bit of a brain freeze!!). It is X/(X+Y), and not X/Y

3. Mar 4, 2007

### ZioX

X/Y does follow a beta distribution! (Assuming they have the same second parameter. This is very important). It's called the beta distribution of the second kind with parameters alpha_x and alpha_y. The F distribution is simply b*X/Y where b>0.

I'll show you why X/Y is called the beta distribution of the second kind.

Suppose X~$$\Gamma(\alpha_1,\beta)$$ and Y~$$\Gamma(\alpha_2,\beta)$$, X and Y independent. What is the distribution of $$U_1=\frac{X}{Y}$$?

Now this is a multivariate transformation, (http://www.ma.ic.ac.uk/~ayoung/m2s1/Multivariatetransformations.PDF see here if you don't know how to do these), so we will use $$U_2=Y$$ as an auxillary equation.

So, $$g_1(x,y)=x/y$$ and $$g_2(x,y)=y$$ where x,y are positive reals (because they come from a gamma distribution) now it should be clear to see that $$g_1^{-1}(x,y)=xy$$ and $$g_2^{-1}(x,y)=y$$. Note how g_1 and g_2 have range (0,+infty).

Therefore, $$f_{(U_1,U_2)}(u_1,u_2)=f_{(x,y)}(g_1^{-1}(u_1,u_2),g_2^{-1}(u_1,u_2))|J|$$. As an exercise you can show that $$|J|=u_2$$

Since X and Y are independent $$f_{(X,Y)}=f_X f_Y$$

Now $$f_{(U_1,U_2)}(u_1,u_2)=f_x(u_1u_2)f_y(u_2)u_2=\frac{e^{-\frac{1}{\beta}(1+u_1)u_2}u_1^{\alpha_1-1}u_2^{\alpha_1+\alpha_2-1}}{\beta^{\alpha_1+\alpha_2}\Gamma(\alpha_1)\Gamma(\alpha_2)}$$

(I have done some simplifying)

Now, we don't want the pdf of (U_1,U_2) we want the pdf of U_1, so we integrate over the joint to get the marginal distribution of U_1.

$$f_{U_1}(u_1)=\frac{u_1^{\alpha_1-1}}{\beta^{\alpha_1+\alpha_2}\Gamma(\alpha_1)\Gamma(\alpha_2)}\int_0^{+\infty}u_2^{\alpha_1+\alpha_2-1}e^{-\frac{1}{\beta}(1+u_1)u_2}du_2$$

But the integral is just a gamma function (after we change variables). So this means that $$\int_0^{+\infty}u_2^{\alpha_1+\alpha_2-1}e^{-\frac{1}{\beta}(1+u_1)u_2}du_2=\frac{\Gamma(\alpha_1+\alpha_2)\beta^{\alpha_1+\alpha_2}}{(1+u_1)^{\alpha_1+\alpha_2}}$$.

Plugging this in we get $$f_{U_1}(u_1)=\frac{\Gamma(\alpha_1+\alpha_2)u_1^{\alpha_1-1}}{\Gamma(\alpha_1)\Gamma(\alpha_2)(1+u_1)^{\alpha_1+\alpha_2}}=\frac{u_1^{\alpha_1-1}}{\beta(\alpha_1,\alpha_2)(1+u_1)^{\alpha_1+\alpha_2}}$$

So there we go! U_1=X/Y is distributed as that. Now why is this called a beta distribution of the second kind? If you do some transformations you should see that $$\beta(\alpha_1,\alpha_2)=\int_0^1x^{\alpha_1}(1-x)^{\alpha_2-1}dx=\int_0^{+\infty}\frac{x^{\alpha_1-1}}{(1+x)^{\alpha_1+\alpha_2}}dx$$

I hope someone finds this interesting ;0

Last edited by a moderator: Apr 22, 2017
4. Jan 25, 2010

### alexis_k

Hello!
I desperately need a proof of the fact that x/(x+y) has a beta distribution.

5. Jan 25, 2010

### SW VandeCarr

The mean of the beta distribution is $$\mu=\frac{\alpha}{\alpha+\beta}$$. Does this help you?

Edit: Look up the PDF and the MGF of the beta distribution. I assume you know the relationship between the gamma and beta functions. By the way, just saying x/(x+y) doesn't mean much by itself. I'm assuming it's relevant to the ratio of two independent gamma distributions.

Last edited: Jan 26, 2010