$X$ and $Y$ are distributed identically, with probability density function:
\begin{equation}
f(x)=
\begin{cases}
2x, & \text{if}\ 0 < x <1 \\
0, & \text{otherwise}
\end{cases}
\end{equation}
And cumulative distribution function:
\begin{equation}
F(x)=
\begin{cases}
0, & \text{if}\ x \leq 0 \\
x^2, & \text{if}\ x \in ]0,1[ \\
1, & \text{if}\ x \geq 1
\end{cases}
\end{equation}
Upon looking at the answer, it is of the form of a double integral:
$$\int_{0}^{1} \int_{0}^{x^2} (2x)(2y) dydx=\frac{1}{3}$$
Performing the integration is not the issue, it's how to get there, to be honest. I am not sure what property of the independent random variables allows us to solve it in this way. Is it something to do with the fact that:
$$F_{X,Y}(x,y) = F_X(x) \cdot F_Y(y)$$
I saw additionally this post, which shows the same kind of formula. I can't find it anywhere in my probability textbook, so assume it is derived from another one of the properties we have been given, but I'm not sure which.
Best Answer
The independence of $X$ and $Y$ gives you the joint density: $$f_{X,Y}(x,y)=f(x)f(y).$$
Once you have the joint density, $$ P(X^2>Y)=\iint_{\{(x,y)\in\mathbb{R}^2|x^2>y\}}f_{X,Y}(x,y)\ dxdy\\ =\iint_{\{(x,y)\in\mathbb{R}^2|x^2>y\}}f(x)f(y)\ dxdy. $$
Now, it is an exercise to show (using the definition of $f$) that $$ \iint_{\{(x,y)\in\mathbb{R}^2|x^2>y\}}f(x)f(y)\ dxdy=\int_{0}^{1} \int_{0}^{x^2} (2x)(2y) dydx. $$