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Another question from the past test papers!

The joint density function of $X$ and $Y$ is given by $f_{X,Y}(x,y) = \frac{1}{x^2 y^2}$, $x \geq 1, y \geq 1$.
(i) Find the joint density function of $U=XY$ and $V=X/Y$.
(ii) What are the marginal densities of $U$ and $V$?
(iii) Evaluate the expectation of $\frac{1}{UV}$

The first 2 parts are rather straight-forward, so I won't talk about them here. Here's what I did for part (iii): first evaluate the marginal density of $X$ and then compute the required expectation (since $\frac{1}{UV}$ is actually $\frac{1}{X^2}$). I just want to ask if there are alternative ways of solving this problem. I think I'm supposed to calculate the expectation using the results from the previous parts, without explicitly computing $f_X(x)$.
Hope someone could help me clarify this, thanks a lot!

Michael Hardy
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drawar
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  • Pretty much the only other thing you can do is use there joint distribution in (i) to do the integration - is this any easier? – not all wrong Apr 21 '13 at 14:56
  • I changed \frac{1}{{{x^2}{y^2}}} to \frac{1}{x^2 y^2}. At least this time it wasn't \frac{1}{{{{x}^{2}}{{y}^{2}}}}. – Michael Hardy Apr 21 '13 at 15:33
  • @MichaelHardy And you could even have used \frac1{x^2y^2} (three strokes less). – Did Apr 21 '13 at 15:47
  • Cool, can you show me how to do it please? @Sharkos. And is it possible to evaluate the expectation of $\frac{1}{UX}$ or $\frac{Y}{V}$ as well? – drawar Apr 22 '13 at 01:13

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$E[g(U,V)] = \int\int g(u,v)\times f_{U,V} (u,v) du dv$ just like you'd expect. This is the only thing I can think of to do. If the integral is easier this is probably what they were thinking.

For functions $h(U,X)$ either substitute for one of the variables or, if you feel adventurous, do the change of variables $X,Y \to X,U$.

not all wrong
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