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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?

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$-x \leq y \leq x$ 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 $\infty$.

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