Given two random uniform variables, $U$ and $V$, that are uniformly distributed over [0,1], how do you calculate the joint pdf of $X$, $Y$ where $X = F(U,V)$ and $Y = G(U,V)$ and where is the joint pdf zero.

Specifically, where $$X = + \sqrt(U) \\ Y = UV$$

I understand this is similar to Joint PDF of two random variables and their sum but I am not interested in the marginal distribution and looking for more of an explanation.

  • $\begingroup$ Maybe you want $Y = G(U, V)$ where $G$ is another function that can be different from $F$. $\endgroup$ – Tunococ Sep 4 '13 at 23:15
  • $\begingroup$ @Tunococ, yes and I've updated the question. I tried to pose the question as general as possible and then added the specific HW question once the general form is explained. $\endgroup$ – Elpezmuerto Sep 4 '13 at 23:21

For any $x, y$ in $[0, 1]$, \begin{align} \Pr(X \le x, Y \le y) & = \Pr(\sqrt U \le x, UV \le y) \\ & = \Pr\left(\sqrt U \le x, U \le \frac yV\right) \\ & = \int_0^1 \Pr\left(U \le x^2, U \le \frac yV \mid V = v\right) dv \\ & = \int_0^1 \min\left\{x^2, \frac yv\right\} dv \\ & = \begin{cases} x^2 & ; x^2 \le y \\ \int_0^{y/x^2} x^2 dv + \int_{y/x^2}^1 \frac yv dv & ; x^2 \ge y \\ \end{cases}\\ & = \begin{cases} x^2 & ; x^2 \le y \\ y - y \log y + 2y\log x & ; x^2 \ge y \\ \end{cases} \end{align} To obtain the probability density function, you take the derivative with respect to $x$ and $y$: $$ f_{X,Y}(x,y) = \begin{cases} 0 & ; x^2 \le y \\ 2/x & ; x^2 \ge y \\ \end{cases} $$

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  • $\begingroup$ Thanks, this really helps. Just a few questions to clarify. First, what does PR stand for? Second, why have the conditional probabilities with respect to $y$? Third, the derivative of $x^2$ is $2x$, not $0$? $\endgroup$ – Elpezmuerto Sep 5 '13 at 12:36
  • $\begingroup$ First, $\Pr$ stands for "probability". Second, you could condition on $X$ or $Y$. I chose $Y$ because I thought the calculation would be easier with this choice. Third, you are right, but then you have to take the derivative of $2x$ with respect to $y$. $\endgroup$ – Tunococ Sep 6 '13 at 9:39

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