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What distribution does the following r.v follow:

$$X/(X+Y)$$

$$X \sim Gamma(a,1)$$ $$ Y \sim Gamma(b,1)$$

and the variables are independent.

Further, how to prove that the random variable is independent of:

$X+Y \sim Gamma(a+b,1)$?

I am sure there is some kind of a hack to get the result without using the convolution technique, and only relying on the moment generating functions. But I can't come up with it.

1 Answers 1

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Here is a sketch of the answer:

Define $V = X + Y =g_1(X,Y)$ and $U = \frac{X}{X+Y}=g_2(X,Y)$, then compute the joint density $$ f_{U,V}(u,v) = f_{X}(g_2^{-1}(u,v))f_Y(g_2^{-1}(u,v))|J|=\mathcal{G}amma(a+b,1)\mathcal{B}eta(a,b), $$ where $J$ is the Jacobian of the change of variables $$ J= \begin{vmatrix} v & u \\ -v & 1-u \end{vmatrix} = v . $$ Now, you can get that $$ f_{U}(u) = \int_{\mathbb{R}^+}f_{U,V}dv = \mathcal{B}eta(a,b). $$

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    This neglects the (crucial) jacobian factor of the change of variables. – 2017-02-06
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    @V.V Thank you, this is the way I was looking for indeed. What is this theorem called? Why does integrating over the joint distribution equal the distribution? – 2017-02-06
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    1. Just change of variables (from multivariable calculus) - I don't think that it has any special name besides the use of Jacobian. 2. Check out "marginal distribution": https://en.wikipedia.org/wiki/Marginal_distribution – 2017-02-06