Let $S_1,S_2...$ be the arrival times of a Poisson process with parameter $\lambda$. Given the time of the $n$:th arrival, find $E(S_1|S_n)$.
I know that $S_n=X_1+X_2+...+X_n$ where $X_i$ are the arrival times. Conditioning on $S_n$ I get
$$E(S_1|S_n=t)=E(S_1|S_{n-1}+X_n=t),$$
I'm not sure how to rewrite this. First I thought that
$$E(X_1 \ | \ X_1+...+X_n=t)$$
would be a good idea since all the $X_i$ are independent of eachother. But ofcourse $X_1$ can't be independent of itself.
$$S_n = \sum_{i=1}^n X_i = X_1 + \sum_{i=2}^n X_i = S_1 + \sum_{i=2}^n X_i$$.
Thus, $$\mathbb{E}[S_n|S_n] = \mathbb{E}[S_1|S_n] + \sum_{i=2}^n \mathbb{E}[X_i|S_n]$$, implying
$$\mathbb{E}[S_1|S_n] = S_n - \sum_{i=2}^n \mathbb{E}[X_i|S_n]$$.
As @Did pointed out, by symmetry $\mathbb{E}[X_i|S_n]$ is independent of $i$ for $1 \leq i \leq n$, implying $n\mathbb{E}[X_i|S_n] = \mathbb{E}[S_n|S_n] = S_n$, i.e. $\mathbb{E}[X_i|S_n] = {S_n \over n}$.
Substituting back, we get: $$\mathbb{E}[S_1|S_n] = S_n - (n-1){S_n \over n} = {S_n \over n}$$