# conditional distribution of sum of order statistic

When I read Ross's book "Statistic Process" , I find the lemma, but I cannot prove it. The lemma states that Let $$Y_1, \cdots, Y_n$$ be iid nonnegative random variables then $$E[Y_1+\cdots + Y_k| Y_1+\cdots+Y_n = y] =\frac{ky}{n}$$, $$k=1, .., n$$.

I tried to calculate the expectation directly, Let $$S_k = Y_1+\cdots + Y_k, k=1, 2, ..., n$$, then $$E[S_k| S_n = y] = \int_{0}^{y} t P\{S_k= t | S_n = y\}dt = \int_{0}^{y}P\{S_k \in [t, t+\delta], S_{k-1}, .., S_{1} \in [0, t)\}t dt= \int_{0}^{y}n\frac{1}{y} P\{S_{k-1}, .., S_{1} \in [0, t)\} dt= \int_{0}^{y}\frac{n}{y}C_{n-1}^{k-1} (\frac{t}{y})^{k-1}(1-\frac{t}{y})^{n-k}dt$$.

I can not calculdate this integration. So I want to look for some other explaination about the result or anyone who can help me calculate the expectation.

We can use the following: for each permutation $$\sigma$$ from $$\{1,\dots,n\}$$ to itself, $$\tag{*} \mathbb E\left[Y_1+\dots+Y_k\mid Y_1+\dots+Y_n\right]=\mathbb E\left[Y_{\sigma(1)}+\dots+Y_{\sigma(k)}\mid Y_1+\dots+Y_n\right],$$ which is due to the fact that the vectors $$\left(Y_1+\dots+Y_k,Y_1+\dots+Y_n\right)$$ and $$\left(Y_{\sigma(1)}+\dots+Y_{\sigma(k)}, Y_1+\dots+Y_n\right)$$ have the same distribution.
Then sum (*) over all the permutations and rearrange $$\sum_{\sigma\in\mathcal S_n}\left(Y_{\sigma(1)}+\dots+Y_{\sigma(k)}\right)$$.
• Thanks for your answer. But I am still confused about the probability $P\{Y_{\sigma(1)} + \cdots + Y_{\sigma(k)}| \sum_{i=0}^{n}Y_i = y\}$. Could you explain more? Actually, I am not familar with order statistic. – Icy Oct 3 '19 at 16:53
• Is this the probability? $P\{ Y_{\sigma(1)} + \cdots + Y_{\sigma(k)}=t | \sum_{i=0}^{n} Y_i = y \} = (\frac{t}{y})^k(\frac{y-t}{y})^{n-k}$ – Icy Oct 3 '19 at 16:57
• Actually there are no order statistic here. It is just the conditional expectation of $Y_{\sigma(1)}+\dots+Y_{\sigma(k)}$ with respect the $\sigma$-algebra generated by $S_n$. – Davide Giraudo Oct 3 '19 at 17:22