Ratio Distribution of $\frac{Y}{1-X}$

Question

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 $\frac{Y}{1-X}$?

Answer 1

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. $$P(Z=z) = \frac{\Gamma(p_1+p_2+p_3)}{\Gamma(p_1)\Gamma(p_2)\Gamma(p_3)} \ . \ z^{p_2-1}(1-z)^{p_3-1}\ . \frac{\Gamma(p_1), \Gamma(p_2+p_3)}{\Gamma(p_1+p_2+p_3)}$$ $$P(Z=z) = \frac{\Gamma(p_2+p_3)}{\Gamma(p_2)\Gamma(p_3)}z^{p_2-1}(1-z)^{p_3-1}$$ $$Z \sim \text{Beta}(p_2, p_3)$$

Answer 2

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:

$$f_U(u) = Beta(p_2,p_3)$$