0
$\begingroup$

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?

1 Answers 1

1

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.

  • 0
    Oops, of course. Thanks for pointing it out. I changed the answer.2012-03-30