I was working on a problem in which $X$ and $Y$ are continuous random variables which both have the uniform distribution $U[0, 1]$. The question was: what is the probability that $X^2 > Y^3$.
This solution is easy enough, it is the area between $0$ and $1$ under the curve $y = x ^ {\frac{2}{3}}$.
My question is, if this were anything other than a uniform density, how is that solved? Is it to integrate $x ^ {\frac{2}{3}} \times f(x,y)$ over the stated interval?
Thanks!
In general, calculating the probability of an event corresponds to (more specifically, this is the definiton) integrating the indicator of that event, i.e. $P[A] = E[1_{A}]$. In the case that the event $A$ can be described in terms of one (or more) random variables this can be simplified.
Suppose that $X$ and $Y$ have the joint density $f_{X,Y}(x,y)$. Then, in your example, taking $A = \{X^2 > Y^3 \}$, we have $$ P[A] = E[1_{A}] = \iint_{\mathbb{R}^2} 1_{X^2 > Y^3}(x,y)f_{X,Y}(x,y)dydx. $$
Now in a real example we would hope that this expression may be simplified even more. In the case that $X$ and $Y$ are uniform on $[0,1]$ and independent, we know $f_{X,Y}(x,y)=1$ inside $[0,1]\times [0,1]$ and it is zero otherwise. We say that $[0,1]\times[0,1]$ is the support of the density.
In this case $$ \iint_{\mathbb{R}^2} 1_{X^2 > Y^3}(x,y)f_{X,Y}(x,y)dydx= \iint_{[0,1]\times[0,1]} 1_{X^2 > Y^3}(x,y)\cdot 1dydx. $$ Now the region where $X^2 > Y^3$ on $[0,1]\times[0,1]$ can be simplified even further to $\{0 \leq X \leq 1, 0 \leq Y \leq X^{2/3}\}$ and now the double integral may be expressed as $$ \int_0^1\int_0^{x^{2/3}}dydx = \int_0^1x^{2/3}dx $$ and you know how to proceed from here.
Now suppose that the density was not uniform. Do you see how my argument would change? Firstly, the support of the density might be different, but that's okay, we can still express the region in a similar manner, $\{-\infty \leq X \leq \infty, -\infty \leq Y \leq X^{2/3}\}$ (and it may be simplified further if the support is not all of $\mathbb{R}^2$), so the double integral becomes $$ \iint_{\mathbb{R}^2} 1_{X^2 > Y^3}(x,y)f_{X,Y}(x,y)dydx = \int_{-\infty}^\infty\int_{-\infty}^{x^{2/3}} f_{X,Y}(x,y)dydx. $$ Unless we know more about $f_{X,Y}$ we can't simplify any further. If $X$ and $Y$ are independent, so that $f_{X,Y}(x,y) = f_X(x)f_Y(y)$ then we can say $$ \int_{-\infty}^\infty\int_{-\infty}^{x^{2/3}} f_X(x)f_Y(y)dydx= \int_{-\infty}^\infty\left(\int_{-\infty}^{x^{2/3}}f_Y(y)dy\right) f_X(x)dx = \int_{-\infty}^\infty F_Y(x^{2/3}) f_X(x)dx $$ where here $F_Y(y)$ denotes the CDF (cumulative distribution function) of $Y$.
The point to take away here is that, in general, one should start with the definition $P[A] = E[1_A]$ and make simplifications when possible.