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I've been doing some multivariate pdf problems such as this: we have $f(x,y,z)=xyz$ and we need to find $f(z)$ in which case we "integrate out" $x$ and $y$ over their respective ranges. I concretely understand this concept, however I do not understand how to generalize the following. If we have $f(x,y)=g(x)h(y)$ then the marginal pdf $f(y)=\frac{1}{k}h(y)$ where $k=\int_{-\infty}^\infty h(y) dy$. Is this the same concept as the example above?

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Yes, it is.

Since $f(x,y)=g(x)h(y)$ must a proper pdf, it must integrate to one. However, the single $g$ and $h$ need not necessarily integrate to one (just their product; thanks Dilip for pointing that out). To obtain the marginal w.r.t. $y$ you can integrate out $x$ as before (which gets you rid of $f$) leaving you with some constant times $h$, i.e. $\int f(x,y) = c h(y)$. $h(y)$ is also not necessarily a pdf and must be normalized to be a proper marginal. This is done via $$\frac{h(y)c}{\int h(y) c dy} = \frac{h(y)}{\int h(y) dy}.$$

Therefore, the constant you got from $g$ is not important since it cancels.

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"If $f(x,y)=g(x)h(y)$ is a proper pdf, then the marginal w.r.t. $y$ is just $h(y)$" is an ambiguous statement since we could instead write $f(x,y) = (2g(x))(0.5h(y)) = \hat{g}(x)\hat{h}(y)$ and conclude that the marginal pdf of $Y$ is $\hat{h}(y) = 0.5h(y)$. All that we can conclude is that the marginal pdf is proportional to $h(y)$ and that the marginal pdfs are $c\cdot g(x)$ and $c^{-1}h(y)$ where $$c = \int_{-\infty}^\infty h(y) \mathrm dy$$ which is exactly what the OP has. If you begin with $f(x,y)$ and "integrate out" w.r.t. $x$, you will be computing $\int g(x) dx$ which is $c^{-1}$. – Dilip Sarwate Mar 29 '12 at 20:45
Oops, of course. Thanks for pointing it out. I changed the answer. – fabee Mar 30 '12 at 6:37

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