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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 $X, Y$.

0k33
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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 $A$, $$ f_{X,Y}(x,y)=\frac{d^2}{dxdy}(x^n-(x-y)^n)=n(n-1)(x-y)^{n-2}. $$

  • 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!). – 0k33 Dec 13 '18 at 04:11
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    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. – angryavian Dec 13 '18 at 04:15
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    @0k33 The third equality follows from independence and the fact that $\mathsf{P}(y<U_i\le x)=x-y$. –  Dec 13 '18 at 04:36
  • Thank you both so much! It's clear now! – 0k33 Dec 13 '18 at 05:29
  • @angryavian Hello! I apologise for bringing up such an old post, but I am trying to solve a similar problem to the OP's. Do you mind explaining your comment in greater detail? In particular, I do not quite understand how "the event ${X \leq x}$ can be partitioned into two disjoint events, based on whether $Y \leq y$ or $Y > y$", or rather, I am still not clear as to how the first equality is derived. – Ethan Mark Aug 13 '21 at 15:52
  • @EthanMark The event ${X \le x}$ is the disjoint union of the events ${X \le x, Y \le y}$ and ${X \le x, Y > y}$. In general $P(A \cup B) = P(A) + P(B)$ when $A$ and $B$ are disjoint. – angryavian Aug 13 '21 at 16:08
  • @EthanMark The first equality is $P(A) = P(A \cup B) - P(B)$. – angryavian Aug 13 '21 at 16:15
  • @EthanMark The statement "${X \le x}$ is the disjoint union of the events ${X \le x} \cap {Y \le y}$ and ${X \le x} \cap {Y > y}$" holds for any random variables $X$ and $Y$, not just in this problem. Try to show that for any events $E_1$ and $E_2$ we have that $E_1 \cap E_2$ and $E_1 \cap E_2^c$ are disjoint events whose union is $E_1$. – angryavian Aug 13 '21 at 16:23
  • Oh okay I think I see it now... It's simply the law of total probability isn't it. Since $E_2$ and $E^c_2$ form a partition then $(E_1 \cap E_2) \cup (E_1 \cap E^c_2) = E_1$? – Ethan Mark Aug 13 '21 at 16:27