This exercise comes from Rice 3.12:
let $f_{XY}(x,y)=c(x^2-y^2)e^{-x}, 0\leq x <\infty, -x \leq y \leq x$
b) find the marginal densities
I have found thatc $c=\frac18$ and also that $f_X(x)=\frac16 x^3e^{-x}, 0 \leq x < \infty$.
for the marginal $f_Y(y)$ i wasnt sure how to determine the limits of integration. I have found an solution that states:
for a specific value of $y$, $x$ ranges from $|y|$ to $\infty$.
$$ f_Y(y)= \int_{|y|}^\infty cx^2e^{-x}dx - \int_{|y|}^\infty cy^2e^{-x}dx$$
Why are there absolute values of y? How are the limits of integration determined?
$-x \leq y \leq x$ is equivalent to $|y| \leq x$ or $x \geq |y|$. when you integrate w.r.t. $x$ you have to all the given constraints into account. The only constraint on $x$ in this case is $-x \leq y \leq x$ so you integrate from $|y|$ to $\infty$.