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 The example I'm working on (course problem booklet for a mathematics BSc second-year module in statistics, unpublished) is $$ f_{X,Y}(x,y) = \begin{cases}\frac{1}{4}y^2e^{-x}&&\text{for }0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?
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1$\begingroup$ What do you mean by "what happens"? $\endgroup$– J. DelaneyCommented May 22, 2022 at 18:03
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$\begingroup$ @J.Delaney I mean "On what region is the marginal pdf non-zero?", or "How should the region on which the bivariate pdf is non-zero be transformed to reflect the transformation (by integration) of the bivariate pdf into the marginal?". $\endgroup$– mjcCommented May 22, 2022 at 18:56
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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 $$f_X(x) = \int_0^\infty f_{X,Y}(x,y)\,\text dy\\=\int_{-x}^x \frac{1}{4} y^2 e^{-x} \mathbb I_{x>0}\,\text dy= \dfrac{[y^3]_{-x}^{x}}{12}e^{-x} \mathbb I_{x>0}=\dfrac{x^3}{6}e^{-x} \mathbb I_{x>0}$$
