# Finding distribution of $\frac{X^2}{Y}$ given joint pdf of $X$ and $Y$

So I have the joint distribution of $$X$$ and $$Y$$: \begin{align} f(x,y) = cx, \ 0 < x^2 < y < \sqrt{x} < 1 \end{align} and I want to find the distribution of $$X^2/Y$$. So I set $$U = X^2/Y$$ and introduced the auxiliary variable $$V = Y$$. The inverse yields

\begin{align} &U = \frac{X^2}{Y} \implies X = \sqrt{UV} \\ &V = Y \implies Y = V \end{align}

It is obvious that $$0 < u < 1$$, but I am struggling to specify the domain of $$V$$. So my question is, how can I specify the limits of $$V$$ here?

• Hint: Draw a sketch of the unit square and indicate on it which points $(x,y)$ in the interior satisfy $0 < x^2 < y < \sqrt{x}$. Pick one point in this region (say $(x.y) = (0.25, 0.4)$) and figure out the answer for this point. Repeat for other points till you see a pattern emerge upon looking at all the calculations. – Dilip Sarwate Nov 22 '19 at 14:57
• Sketching the support of $f$ is simplified by recognizing it is the intersection of $0\lt x^2 \lt y \lt 1$ and $0 \lt y^2 \lt x \lt 1,$ which are symmetrically situated about the diagonal. Thus, all you need to do is plot the graph of $y=x^2,$ reflect that about the diagonal $x=y,$ and shade in the region bounded by those two graphs. – whuber Nov 22 '19 at 15:04
• Thank you for answer! I don't really see how this will help me defining the limits of V in terms of U. Kindly provide more hints. – FAHRB Nov 24 '19 at 11:28