For $Y_1, \ldots, Y_n$ i.i.d. with Poisson distribution, and $S_n=\sum\limits_{i=1}^nY_i$, how to show $E(Y_1 \mid S_n) = \frac1nS_n$?

Question

Suppose that $Y_1, \ldots, Y_n \sim \text{Pois}(\lambda)$ are iid observations from a Poisson distribution. I am trying to show why:

$$\mathbb E\left(Y_1 \;\;\middle| \;\;\sum_{i=1}^nY_i\right) = \frac1n\sum_{i=1}^nY_i$$

I know that one way is to directly go to the infinite summation, but was wondering if there was an easier way to get this. Thanks.

Answer 1

Hint: By symmetry, the conditional expectation has the same value if you replace $Y_1$ with $Y_i$ on the left-hand side. Use this and the equation $E[X \mid X]=X$ almost surely.

Note that this approach does need the fact that the random variables are independent and identically distributed (or more generally, exchangeable) for $E[Y_1 \mid \sum Y_i] = E[Y_j \mid \sum Y_i]$ to hold. The fact that the distribution is Poisson does not play a significant role - the proof works for any other distribution with finite first moment.