Let $U_1, ... U_n$ be iid standard uniform variables. Let $X = max(U_i)$ and $Y = min(U_i)$. The goal is to compute the joint PDF of $X, Y$!

I have already computed the PDFs of $X$ and $Y$ separately. But I am not sure if that is so useful because $X, Y$ are not independent (by definition $X \geq Y$) so it is not correct to just multiply the marginal PDFs of $X$ and $Y$.

I thought about starting from the CDF and taking advantage of the fact that all of the $U_i$'s are independent. So something like: $P(X \leq x, Y \leq y) = P(U_1, ...U_n \leq x, \text{at least one $U_i$ is less than }y)$ and you can already see the problem here -- I don't know how to express the relation for $Y$ in terms of all the $U_i$'s like that. when computing $Y$'s CDF, I did $1 - P(U_1 \geq y, ... U_n \geq y)$ and was able to take advantage of the independence of all the $U_i$'s there. But I'm not sure how to translate that to the joint PDF of $X, Y$.


First, on $A=\{0\le y<x\le 1\}$, \begin{align} \mathsf{P}(X\le x,Y\le y)&=\mathsf{P}(X\le x)-\mathsf{P}(Y>y,X\le x) \\ &=\mathsf{P}\left(\bigcap_{1\le i\le n}(U_i\le x)\right)-\mathsf{P}\left(\bigcap_{1\le i\le n}(y<U_i\le x)\right) \\ &=x^n-(x-y)^n. \end{align} Therefore, on $A$, $$ f_{X,Y}(x,y)=\frac{d^2}{dxdy}(x^n-(x-y)^n)=n(n-1)(x-y)^{n-2}. $$

  • $\begingroup$ Thank you so much for your answer. I have a question, though -- would you mind explaining the first and third equalities a little more? I am really not seeing how you arrived at the first one in particular (I'm not questioning if it's correct, I just do not follow!). $\endgroup$ – 0k33 Dec 13 '18 at 4:11
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    $\begingroup$ The event $\{X \le x\}$ can be partitioned into two disjoint events, based on whether $Y \le y$ or $Y > y$. The probability of $\{X \le x\}$ is therefore the sum of probabilities of these two smaller events. $\endgroup$ – angryavian Dec 13 '18 at 4:15
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    $\begingroup$ @0k33 The third equality follows from independence and the fact that $\mathsf{P}(y<U_i\le x)=x-y$. $\endgroup$ – d.k.o. Dec 13 '18 at 4:36
  • $\begingroup$ Thank you both so much! It's clear now! $\endgroup$ – 0k33 Dec 13 '18 at 5:29

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