Let $X$ and $Y$ be random variables having the following joint probability density function
$f(x,y)=\begin{cases}
\frac{3}{8}xy & x\geq0,\,y\geq0,\:x+y\leq2,\\
0 & \mbox{otherwise}.
\end{cases}$
Find the joint probability density function of $X^2$ and $Y^2$.
This is my solution:
$z=x^2, \Longrightarrow x=\sqrt z$, because $x\geq0$
$w=y^2, \Longrightarrow y=\sqrt w$, because $y\geq0$.
The Jacobian $J$ of the inverse transformation would then equal:
$J={ \left|\begin{array}{cc}
\frac{1}{2\sqrt{z}} & 0\\
0 & \frac{1}{2\sqrt{w}}
\end{array}\right|}=\frac{1}{4\sqrt{wz}}$
so, the joint probability density function
\begin{align}
g(z,w)&=\begin{cases}
\frac{3}{8}\sqrt z \sqrt w \frac{1}{4\sqrt wz}& w\geq0,\,z\geq0,\:\sqrt w+\sqrt z\leq2\\
0 & \mbox{otherwise}
\end{cases} \\
&=\begin{cases}
\frac{3}{32}\,\,\,\,& w\geq0,\,z\geq0,\:\sqrt w+\sqrt z\leq2\\
0 & \mbox{otherwise}
\end{cases}
\end{align}
but I have a problem, $\int_{0}^{4}\int_{0}^{w-4\sqrt{w}+4}\frac{3}{32}\,dz\,dw=\frac{1}{4}\neq1$.
Please help me at my mistake
Best Answer
I think the issue is the original pdf: $$ \int_{\mathbb{R}^2}f(x,y)\;dydx=\frac{3}{8}\int_0^2\int_{0}^{2-x}xy\;dydx=\frac{3}{8}\int_0^2\frac{x(2-x)^2}{2}\;dx$$ $$ =\frac{3}{8}\int_0^2\frac{(2-x)x^2}{2}\;dx=\frac{3}{8}\cdot\frac{2}{3}=\frac{1}{4}$$