Deriving Marginal Distribution from Two Independent Beta Distributions

Question

I have two independent random variables:

1. $X$ is Beta($a, b$), $x \in [0, 1]$
2. $Y$ is Beta($a+b, \gamma$), $y \in [0, 1]$

$$f(x,y) = \frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)} x^{a-1}(1-x)^{b-1}\frac{\Gamma(a+b+\gamma)}{\Gamma(a+b)\Gamma(\gamma)}y^{a+b-1} (1-y)^{\gamma-1}$$

I want to perform the transformations for $U = XY$, $V=Y$. $$U=XY \rightarrow U=XY \rightarrow X=U/V, \qquad u \in [0,1]$$ $$V=Y \rightarrow Y=V, \qquad v\in[0,1]$$

$$\det{J}= \frac{1}{v}*1-0=\frac{1}{v}$$ So I have,

$$f(u, v) = \frac{\Gamma(a+b+\gamma)}{\Gamma(a+b)\Gamma(\gamma)} (u/v)^{a-1}(1-(u/v))^{b-1}v^{a+b-1} (1-v)^{\gamma-1} * \frac{1}{v}, \qquad 0 \le u \le v \le 1$$

I want to find the marginal distribution of $u$.

$$f_{U}(u) = \frac{\Gamma(a+b+\gamma)}{\Gamma(a)\Gamma(b)\Gamma(\gamma)} u^{a-1} \int^1_{u} (1-(u/v))^{b-1}v^{a+b-1} (1-v)^{\gamma-1} \frac{1}{v} dv$$

QUESTION: However, I am unsure how to go about this integral. How can I proceed with this integration find the marginal distribution for $U$?

Answer 1

Suppose $X\sim \text{Beta}(a,b)$ and $Y\sim \text{Beta}(c,d)$ are independently distributed with $c=a+b$.

Then $(X,Y)$ has density $$f_{X,Y}(x,y)\propto x^{a-1}(1-x)^{b-1}y^{c-1}(1-y)^{d-1}\mathbf1_{0<x,y<1}\,,$$ where $a,b,c,d>0$.

Transforming $(X,Y)\mapsto (U,V)=(XY,Y)$, density of $(U,V)$ is

\begin{align} f_{U,V}(u,v)&=f_{X,Y}\left(\frac{u}{v},v\right)\cdot\frac1v \\&\propto u^{a-1}(v-u)^{b-1}(1-v)^{d-1}\mathbf1_{0<u<v<1} \end{align}

Pdf of $U$ is therefore

$$f_U(u)\propto u^{a-1}\int_u^1 (v-u)^{b-1}(1-v)^{d-1}\,dv\,\mathbf1_{0<u<1}$$

Now you can turn this integral into a Beta function if you change variables $v\mapsto t$ such that $t=0$ when $v=u$ and $t=1$ when $v=1$. This yields the relation $v-u=t(1-u)$.

On simplification you have another Beta distribution:

$$f_U(u)\propto u^{a-1}(1-u)^{b+d-1}\,\mathbf1_{0<u<1}$$

I have ignored the normalizing constants here since these are all guaranteed to be valid densities. You can of course verify that you end up with the correct constant.

Answer 2

The joint density results to me

$$f(u,v-u,1-v)=\frac{\Gamma(a+b+\gamma)}{\Gamma(a)\Gamma(b)\Gamma(\gamma)}u^{a-1}(v-u)^{b-1}(1-v)^{\gamma-1}$$

This is a Dirichlet distribution with parameters $(a,b,\gamma)$ thus it is immediate to derive the 3 marginal laws:

$$ \bbox[5px,border:2px solid black] { U\sim Beta(a; b +\gamma) \qquad (1) } $$

$$ \bbox[5px,border:2px solid red] { (V-U)\sim Beta(b; a +\gamma) \qquad (2) } $$

$$ \bbox[5px,border:2px solid blue] { (1-V)\sim Beta(\gamma; a+b) \qquad (3) } $$

(3) is an obvious result, knowing that $V\sim Beta(a+b;\gamma)$

Answer 3

See Statistical Inference by Casella & Berger Example 4.3.3; they do this exact example. You need to do a change of variable $y=(u/v-u)/(1-u)$ (non-obvious substitution), so that $dy=-u[v^2(1-u)]dv$. The main reason for this is to change the bounds of the integral where the lower bound is $0$ and the upper bound is $1$ so you can use the Beta pdf to compute the new integral.

Answer:

If $X\sim\text{Beta}(\alpha,\beta)$ and $Y\sim\text{Beta}(\alpha+\beta,\gamma)$ are independent, then $U=XY$ and $V=X$ yields $U\sim\text{Beta}(\alpha,\beta+\gamma)$.