Finding the non-zero region of a marginal Basic question but: what happens to the region on which a pdf is non-zero when a bivariate is integrated to get a marginal? The example I'm working on (course problem booklet for a mathematics BSc second-year module in statistics, unpublished) is
$$
f_{X,Y}(x,y) = \begin{cases}\frac{1}{4}y^2e^{-x}&&\text{for }0<|y|<x<\infty,\\0&&\text{otherwise}.\end{cases}
$$
I get $f_X(x)=\frac{1}{6}x^3e^{-x}$ for the marginal, but I can't picture what happens to the non-zero region. Presumably it shouldn't depend on $y$ for a function of only $x$. Could someone give me a nudge?

Roughly stated, the support of the marginal is made of all the $x$'s for which the joint in $(x,y)$ is positive for some $y$'s (plus some caution about positive measure of the set of such $y$'s).
More precisely,
$$\left\{x;\ f_X(x)>0 \right\}\subset \bigcup_y \left\{x;\ f_{X,Y}(x,y)>0 \right\}$$
when removing sets of measure zero (under the joint measure).
In the example from the question, for a given $y$,
$$\left\{x;\ f_{X,Y}(x,y)>0 \right\} = (|y|,\infty)$$
and
$$\bigcup_y \left\{x;\ f_{X,Y}(x,y)>0 \right\} = (0,\infty)$$
which is the support of $f_X$, since
$$f_X(x) = \int_0^\infty f_{X,Y}(x,y)\,\text dy\\=\int_{-x}^x \frac{1}{4} y^2 e^{-x} \mathbb I_{x>0}\,\text dy= \dfrac{[y^3]_{-x}^{x}}{12}e^{-x} \mathbb I_{x>0}=\dfrac{x^3}{6}e^{-x} \mathbb I_{x>0}$$
