Given the joint probability density function :
$$f_{x,y} (x,y) =
\left\{
\begin{array}{ll}
2 \qquad ,\, 0\le x \le 1 \, ,0\le y \le x \\
0 \qquad ,\, otherwise
\end{array}
\right.$$
and Given : $$ Z=XY^3$$
we need to calculate the cumulative distribution function of Z at 0.3
in other words we want to calculate : $$F_z(0.3)$$
which means we have to calculate : $$P(Z<z) = \iint\limits_D f_{x,y}(x,y)\, \mathrm{d} x\,\mathrm{d} y \quad$$
$$D : yx^3\le z $$ (could not write the regin inside the integral)
I'm trying to find the integral limits but not succeeding could you give some guidance ?
Best Answer
$F_Z(z)=\int_0^{1} \int_0^{g(x)} 2dydx$ where $g(x)=\min \{x,\frac z {x^{3}}\}=2\int_0^{x} g(x)dx$. Split the integral into integral from $0$ to $z^{1/4}$ and the integral from $z^{1/4}$ to $1$. Note that $g(x)=x$ for the first integral and $g(x)=\frac z {x^{3}}$ for the second integral. I will leave the rest to you.