How to inverse transform sample with a PDF that maps $\mathbb R^2$ to $\mathbb R$?

Question

Given a PDF: $$ p(x,y) = \frac{\pi^2\sin(\pi x)\sin(\pi y)}{4}. $$ I obtain a CDF using double integral: $$ P(x,y) = \frac{(1-\cos(\pi x))(1-\cos(\pi y))}{4}. $$ And then I'm stuck, normally one would proceed to invert the CDF, but I can't really invert a multivariate function, can I?

Answer 1

Since your distribution has $X$ and $Y$ independent of each other, you can calculate the cdf's of the marginals and invert them separately. The marginal for $x$ is $$p(x)=\frac{\pi}{2}sin(\pi x)$$ with cdf $$P(x)=\frac{1}{2} -\frac{1}{2} cos(\pi x)$$ which inverted gives $$x = \frac{1}{\pi}cos^{-1} (1 - 2P(x))$$