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The random variables $X$ and $Y$ have a joint density function given by; $$f(x,y)= \begin{cases} Kxy,\quad &0\leq x\leq 1,\; 0\leq y\leq 1,\\ 0,&\text{otherwise}. \end{cases} $$

Determine $K$ and find the density functions of the random variables $Z = \max(X,Y)$ and $T = \min(X, Y )$.

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The time might have come to recall that, in order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. Also, many find the use of imperative ("Prove", "Solve", etc.) to be rude when asking for help; please consider rewriting your post (and several very recent similar ones you posted in rapid succession on the site). –  Did Dec 15 '12 at 15:09
This and your other recent questions do not fall under the category probability-theory. –  Stefan Hansen Dec 15 '12 at 15:22
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1 Answer 1

Recall that for some $f(x, y)$ to be a joint density function, two things must be satisfied:

  1. $f(x, y)$ must be non-negative for all $x, y$.
  2. The area under $f(x, y)$ must be 1.

The first part is easily satisfied by noting that $K$ must be non-negative.

The second part can be satisfied by setting the integral of $f(x, y)$ to 1:

$$\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty}f(x, y)\,\mathrm{d}x\mathrm{d}y = 1$$

Which can be simplified to:

$$\int\limits_{0}^{1}\int\limits_{0}^{1}Kxy\,\mathrm{d}x\mathrm{d}y = 1$$

At this point it shouldn't be difficult to solve for $K$.

For the $\max$ and $\min$ questions, you could observe that

$$\max(a, b) = \begin{cases}a\quad a\ge b,\\b\quad \mathrm{otherwise}.\end{cases}$$ $$\min(a, b) = \begin{cases}a\quad a\le b,\\b\quad \mathrm{otherwise}.\end{cases}$$

which ought to help.

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