I've seen some questions regarding this problem(order statistics). I am aware of the proof by curling the unit interval into a circle, and finding the expectation as a certain portion of the circumference. I do not want such proof, and just by pure computation. The problem is as following:
Let $U_k \sim \text{Unif}[0, 1]$ for $1\leq k\leq n$, and each $U_k$'s being independent. Let $Y_k$ be the $k$th smallest $U_k$ when ordered by size, so that $Y_1=\min(U_1, U_2, \cdots U_n)$, and $Y_n=\max(U_1, U_2, \cdots U_n)$. Prove that $E(Y_k)=\frac{k}{n+1}$.
I proved the case if $k=1$ or $k=n$, the boundary cases, and that was easy. I want to prove for the case when $1<k<n$. My work on this problem is this:
$$F_{Y_k}(y)=\Pr(Y_k\leq y)=\Pr (\text{k of them are} \leq y \text{, others are > k})=\binom{n}{k}\Pr(U_1\leq y)^k \Pr (U_1>y)^{n-k}$$
where I have used the fact that every $U_i$'s are i.i.d. $F_{U_1}(y)=y$ for $0\leq y\leq 1$, so we get
$$f_{Y_k}(y)=\binom{n}{k}ky^{k-1}(1-y)^{n-k}-\binom{n}{k}(n-k)y^k(1-y)^{n-k-1}=(k-ny)\binom{n}{k}y^{k-1}(1-y)^{n-k-1}$$
I know that $E[Y_k]=\int_{0}^{1}yf_{Y_k}(y)dy$, but I don't know how to integrate this. First of all, is my process up to now correct? If is it so, then how can I proceed from here?
Best Answer
Your computation of the CDF is not correct: for instance, if all of the $U_i$ are $\le y$, then $Y_k \le y$ as well, but you excluded this case.
You seem to be conflating with reasoning behind the derivation for the density of $Y_k$, which is $f_{Y_k}(y) = n\binom{n-1}{k-1} y^{k-1} (1-y)^{n-k}$. (Pick one $U_i$ to be equal to $y$, then from the remaining $n-1$ choose $k-1$ of them to be $< y$, and the rest are $> y$.)
Then $$E[Y_k] = \int_0^1 y n\binom{n-1}{k-1} y^{k-1} (1-y)^{n-k} \, dy = n\binom{n-1}{k-1} \int_0^1 y^k (1-y)^{n-k}.$$
By pattern matching you might guess that $\int_0^1 (n+1)\binom{n}{k} y^k (1-y)^{n-k} \, dy = 1$; indeed this integrand is the density of the $k+1$ order statistic of $n+1$ i.i.d. uniform random variables. Thus, $$E[Y_k] = \frac{n \binom{n-1}{k-1}}{(n+1)\binom{n}{k}} = \frac{k}{n+1}.$$