Ratio Distribution of $\frac{Y}{1-X}$ Find the distribution of $\frac{Y}{1-X}$ where the joint distribution of the random variables $X$ and $Y$ is given by:
$$f_{XY}(x,y) = \frac{\Gamma(p_1+p_2+p_3)}{\Gamma(p_1)\Gamma(p_2)\Gamma(p_3)} x^{p_1-1}y^{p_2-1}(1-x-y)^{p_3-1}$$
where, $x\geq0, y\geq0, x+y\leq1 \text{ and } p_1,p_2,p3 > 0$.
I have already shown that
$$f_X(x) = \frac{\Gamma(p_1+p_2+p_3)}{\Gamma(p_1)\Gamma(p_2+p_3)} x^{p_1-1}(1-x)^{p_2+p_3-1}=\text{Beta}(p_1, p_2+p_3)$$
$$f_Y(y) = \frac{\Gamma(p_1+p_2+p_3)}{\Gamma(p_2)\Gamma(p_1+p_3)} y^{p_2-1}(1-y)^{p_1+p_3-1}=\text{Beta}(p_2, p_1+p_3)$$
Can somebody help me how to proceed to calculate the distribution of $\frac{Y}{1-X}$?
 A: This question is an example of ratio distribution and the general procedure is as follow
For an RV, $Z = \frac{X}{Y}$
$$f_Z(z) = \int_{-\infty}^{\infty}|y|f_{XY}(y, zy).dy$$
Now back to the original question.
$$Z = \frac{Y}{1-X}$$
Now to calculate the density of Z at a point z,we can proceed in this way
$$P(Z=z) = \int_{0}^{1}(1-x)f_{XY}(x, z(1-x))dx$$
$$P(Z=z) = k\int_{0}^{1}(1-x)x^{p_1-1}(z(1-x))^{p_2-1}(1-x-z(1-x))^{p_3-1}dx$$
here $k = \frac{\Gamma(p_1+p_2+p_3)}{\Gamma(p_1)\Gamma(p_2)\Gamma(p_3)}$
$$P(Z=z) = kz^{p_2-1}(1-z)^{p_3-1}\int_{0}^{1}x^{p_1-1}(1-x)^{p_2+p_3-1}dx$$
$$P(Z=z) = kz^{p_2-1}(1-z)^{p_3-1}\text{Beta}(p_1, p_2+p_3)$$
Substitute value of k.
$$P(Z=z) = \frac{\Gamma(p_1+p_2+p_3)}{\Gamma(p_1)\Gamma(p_2)\Gamma(p_3)} \ . \ z^{p_2-1}(1-z)^{p_3-1}\ . \frac{\Gamma(p_1), \Gamma(p_2+p_3)}{\Gamma(p_1+p_2+p_3)}$$
$$P(Z=z) = \frac{\Gamma(p_2+p_3)}{\Gamma(p_2)\Gamma(p_3)}z^{p_2-1}(1-z)^{p_3-1}$$
$$Z \sim \text{Beta}(p_2, p_3)$$
A: Here's a solution:
Begin with $U = \frac{Y}{1-X}$ and $V = 1 - X$. Then we'll make a transform to the joint probability itself.
For the time being, let $k = \frac{\Gamma(p_1 + p_2 + p_3)}{\Gamma(p_1)\Gamma(p_2)\Gamma(p_3)}$
then, we first find the inverse transform function h(u,v): $X = h_1(u,v) = 1 - V$ and $Y = 
 h_2(u,v) = UV$
The Jacobian of this inverse function = $-v$
Then the transform formula: $f_{U,V}(u,v) = f_{X,Y}(h_1,h_2) * |Jacobian|$
$$f_{U,V}(u,v) = k* (1-v)^{p_1-1}*(uv)^{p_2-1}*(v-uv)^{p_3-1}*v$$
$$f_{U,V}(u,v) = k* [v^{p_2+p_3-1}(1-v)^{p_1-1}]*[(u)^{p_2-1}(1-u)^{p_3-1}]$$
To find $f_U(u)$ which is the aim, integrate with respect to v from 0 to 1. Observe the integrated is the kernel of the $Beta(p_2+p_3,p_1)$ function.
Thus integrating gives us:
$$f_U(u) = [(u)^{p_2-1}*(1-u)^{p_3-1}] \frac{\Gamma(p_1 + p_2 + p_3)}{\Gamma(p_1)\Gamma(p_2)\Gamma(p_3)}*\frac{\Gamma(p_1)\Gamma(p_2+p_3)}{\Gamma(p_1+p_2+p_3)}$$
$$f_U(u) = \frac{\Gamma(p_2+p_3)}{\Gamma(p_2)\Gamma(p_3)}*[(u)^{p_2-1}*(1-u)^{p_3-1}]$$
Using the usual form of pdf of beta function, We have:
$$f_U(u) = Beta(p_2,p_3)$$
