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1-2. Let $X$ and $Y$ have the joint probability density function $$f_{X,Y}(x,y) = \frac{1}{x},\quad x>1,0<y<\frac{1}{x}, 0\text{ elsewhere.}$$

  1. a) Let $U = XY$. Find the p.d.f. of $U, f_U(u)$.

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Here is the question - I am extremely confused on how to start since $f(x,y)$ is already a joint distribution, how can I find $U = XY$? Any start would be appreciated.

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    $\begingroup$ Formatting tips here. $\endgroup$ – Em. Feb 19 '16 at 2:51
  • $\begingroup$ The difficult part is identifying the "cases". Also, I'm unsure whether im supposed to use convolution or not here? $\endgroup$ – user3874530 Feb 19 '16 at 2:52
  • $\begingroup$ Never mind I have figured it out! $\endgroup$ – user3874530 Feb 19 '16 at 3:26
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From first principles, since the support of $(X,Y)$ is $(1;\infty)\times(0;1/X)$, the support of $XY$ is $(0;1)$

$$\begin{align} f_{XY}(u) = & ~ \frac{\mathrm d}{\mathrm du} \mathsf P(XY\leq u) ~~\mathbf 1_{u\in(0;1)} \\[1ex] = & ~ \frac{\mathrm d}{\mathrm du} ~\int_1^\infty \int_0^{\min(1/x, u/x)} x^{-1}\operatorname d y\operatorname d x ~~\mathbf 1_{u\in(0;1)} \\[1ex] = & ~ \int_1^\infty x^{-2}\operatorname d x ~~ \mathbf 1_{u\in(0;1)} \\[1ex] = & ~ -x^{-1}\Big\vert_{x=1}^{x\to\infty} ~~\mathbf 1_{u\in(0;1)} \\[0ex]~ \\[0ex] \hline \\[0ex] f_{U}(u) = & ~ \mathbf 1_{u\in(0;1)} \end{align}$$

Thus $U\sim\mathcal U(0;1)$

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$$ f_U(u) = \int_{-\infty}^{+\infty} f_{X,Y}(x, \frac{u}{x})~\frac{1}{|x|}~dx $$

where $f_{X,Y}(x, y)$ is joint PDF

Reference: Rohatgi, V.K., 1976. An Introduction to Probability Theory Mathematical Statistics. Wiley, New York.

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