calculate probability using joint density function I'm stuck with this question:
X,Y are random variables and thier joint density function is:
$$f_X,_Y(x,y)=2 ,0<=x<=1, 0<=y<=x$$
Now we define new random variable Z: $$Z=XY^3$$
I need to calculate the value of $$F_Z(0.3)$$
and i'm not so sure which bounds i should integrate with the joint function of X and Y.
 A: Let $\mathcal A_t= \left \{0 \leq x \leq 1, 0 \leq y \leq x : xy^3 \leq t \right\}$
The probability $\mathbb P(XY^3 \leq t)$ can be seen as a double integral over $\mathcal A_t$:
$$
\mathbb P(XY^3 \leq t) = 2 \int_{A_t} dxdy
$$
The condition $xy^3 \leq t$ imply that $y \leq \left( \frac{t}{x} \right)^{\frac{1}{3}}$
and $\left( \frac{t}{x} \right)^3 \leq x \Rightarrow x \geq t^{\frac{1}{4}}$.
So in the above integral, for a fixed $x$, if $x \geq t^{\frac{1}{4}}$ then integration over $y$ ranges from $0$ to $\left( \frac{t}{x} \right)^{\frac{1}{3}}$ (it is $0$ otherwise) and if $x \leq t^{\frac{1}{4}}$ it ranges from $0$ to $x$.
Hence we have:
\begin{align*}
\int_{A_t} dxdy  &= \int_0^{t^{\frac{1}{4}}} x dx + \int_{t^{\frac{1}{4}}}^1\left( \frac{t}{x} \right)^{\frac{1}{3}} dx \\
&= \left[ \frac{x^2}{2} \right]_0^{t^{\frac{1}{4}}} + t^{\frac{1}{3}} \left[ \frac{3}{2} x^{\frac{2}{3}}\right ]_{t^{\frac{1}{4}}}^1 \\
&= \frac{\sqrt{t}}{2} + \frac{3}{2}t^{\frac{1}{3}}\left(1-t^{\frac{1}{6}} \right) \\
&= \frac{\sqrt{t}}{2} + \frac{3}{2}t^{\frac{1}{3}}-\frac{3\sqrt{t}}{2} \\
&=\frac{3}{2}t^{\frac{1}{3}} - \sqrt{t}
\end{align*}
Multiplying this by $2$ yields the desired probability:
\begin{align*}
\mathbb P(Z \leq t) = 3t^{\frac{1}{3}} - 2 \sqrt{t}
\end{align*}
For $t=0.3$ this gives $\mathbb P(Z \leq t) \approx 0.913$.
