# joint PDF of max and min of $n$ iid standard uniform random variables

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 $$A=\{0\le y, \begin{align} \mathsf{P}(X\le x,Y\le y)&=\mathsf{P}(X\le x)-\mathsf{P}(Y>y,X\le x) \\ &=\mathsf{P}\left(\bigcap_{1\le i\le n}(U_i\le x)\right)-\mathsf{P}\left(\bigcap_{1\le i\le n}(y Therefore, on $$A$$, $$f_{X,Y}(x,y)=\frac{d^2}{dxdy}(x^n-(x-y)^n)=n(n-1)(x-y)^{n-2}.$$
• 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 4:15
• @0k33 The third equality follows from independence and the fact that $\mathsf{P}(y<U_i\le x)=x-y$. – d.k.o. Dec 13 '18 at 4:36