If $Y_1, \ldots, Y_n \sim Poisson(\lambda)$ are iid, how to show that $E(Y_1 | T = \sum_{i=1}^{n}Y_i) = \frac{T}{n}$? Suppose I have that $Y_1, \ldots, Y_n \sim Poisson(\lambda)$ are iid. I saw a line in a book that said that if $T = \sum_{i=1}^{n}Y_i$, then:
$$
E(Y_1 | T) = \frac{T}{n}
$$
I am lost as to how they obtain this. One approach I did was:
$$
E(Y_1 | T) = E\left(Y_1 | \sum_{i=1}^{n}Y_i = t\right)  = E\left(Y_1 | Y_1 = t-\sum_{i=2}^{n}Y_i \right) = E\left(T-\sum_{i=2}^{n}Y_i\right) = T- (n-1)\lambda
$$
However, I know this is wrong but dont know why. 
 A: By linearity of expectation (since $T=\sum\limits_{i=1}^n Y_i$):
$$\sum\limits_{i=1}^n\mathsf E(Y_i\mid T) = \mathsf E(T\mid T)$$
Then by symmetry (since they are identical and independent): $$n~\mathsf E(Y_1\mid T)=T$$
That is all.

Where you went wrong is that the condition that $T=t$ affects all $Y_i$ and must "remain in play" when you replace $Y_1$ in step $\star$.
$$\begin{align}
\mathsf E\left(Y_1 ~\middle\vert~ T=t\right) = & ~ \mathsf E\left(Y_1 ~\middle\vert~ \sum_{i=1}^{n}Y_i = t\right) \\ = & \mathsf E\left(Y_1 ~\middle\vert~ Y_1 = t-\sum_{i=2}^{n}Y_i, \color{red}{T=t} \right) \\ = & \mathsf E\left(T-\sum_{i=2}^{n}Y_i ~\middle\vert~ \color{red}{T=t}\right) & \star
\\ = & ~ \mathsf E(T\mid T=t) - \color{red}{\sum_{i=2}^n \mathsf E\left(Y_i~\middle\vert~T=t\right)}
\end{align}$$
A: If $X_1,X_2,\ldots,X_n$ are independent Poisson random variables with parameters $\lambda_1,\lambda_2,\ldots,\lambda_n$, the distribution of $X_i$ given $T=\sum_{i=1}^nX_i$ is the binomial distribution with parameters $T$ and $\lambda_i/\sum_{j=1}^n\lambda_j$, that is,
$$
X_i\sim\mathrm{Binom}\biggl(T,\frac{\lambda_i}{\sum_{j=1}^n\lambda_j}\biggr)
$$
(see, for example, here). 
In our case, $\lambda_1=\lambda_2=\ldots=\lambda_n=\lambda$. Hence,
$$
\operatorname E[X_1\mid T]=T\cdot\frac{\lambda_1}{\sum_{j=1}^n\lambda_j}=T\cdot\frac{\lambda}{n\lambda}=\frac Tn.
$$
I am not sure if there is a simpler way to evaluate $\operatorname E[X_1\mid T]$. I hope this is useful.
