# Conditional expectation and sampling without replacement

Suppose I have $N$ numbers: $x_1,\cdots,x_N$ such that $N$ is even and $\sum_{i=1}^N x_i = 0$. Let $X_j$ be the $j$-th number I get when I sample from $\left\{x_1,\cdots,x_N\right\}$ without replacement. How do I find $$\mathbb{E}\left(X_{k+1} + \cdots + X_M \mid X_k, X_{k-1},\cdots,X_1\right)\,?$$

I can find $\mathbb{E}\left(X_{k+1} \mid X_k,\cdots,X_1\right)$, but I don't know how to find the rest conditional expectations. For example, given $X_k,\cdots,X_1$, how do I know the distribution $X_{k+2}$? I feel like it would have something to do with $X_{k+1}$...

The marginal distributions, and thus the expectations, for $X_{k+1}$ through $X_M$ conditional on $X_1$ through $X_k$ are all the same. Thus, the expectation you want is just $M-k$ times the expectation $\mathbb{E}\left(X_{k+1} \mid X_k,\cdots,X_1\right)$ that you already know how to find.
• PS: ... Because you do know exactly what $\sum\limits_{j=k+1}^N X_j$ will be when you know the first $k$ samples , you may use this method to find to find $\mathsf E\left(X_{k+1}~\middle|~ {(X_i)}_{i=1}^k\right)$ Aug 1, 2018 at 4:51