# Transformation of random variable bivariate

If the random variables $X$ and $Y$ have the joint density

$$f(x,y) = \left\{ \begin{array}{ll} \frac{6}{7}x & \mbox{for 1 \leq x+y \leq 2,x \geq 0,y \geq 0};\\ 0 & \mbox{otherwise}.\end{array} \right.$$

what is the density of $\frac{X}{Y}$?

I already know that the answer to this question is

$$g_1(u) = \left\{ \begin{array}{ll} \frac{2u}{(1+u)^3} & \mbox{if 0 \leq u \leq \infty};\\ 0 & \mbox{otherwise}.\end{array} \right.$$

How can we arrive to that answer? Any help will be appreciated.

• Could you clarify the domain of support for $f(x,y)$? – PiE May 13 '17 at 12:09
• $1 \leq x+y \leq 2$ – geniwebb May 13 '17 at 12:22
• You should probably go through the transformation of random variables chapter in Casella and Berger, basically you just find the inverse function of the transform given by $U = \frac{X}{Y}$ and $V=Y$ and then integrate out over V – user2879934 May 13 '17 at 12:35
• I got $g(u,v) = \frac{6}{7}uv^2$. When I tried to get $g_1(u)$ I generated $g_1(u) = \frac{4}{7}u (1+u)^3$ instead. – geniwebb May 13 '17 at 12:55