Poisson Conditional Expectation ( searching best estimator for h(λ) ) Suppose $X_1$,$X_2$,$X_3$,.....,$X_n$ are i.i.d. random variables with a common density poisson(λ)
(I is an indicator function) (t = a value)
E  $[$$X_2$ - I{$x_1$=1}|$\sum_{i=1}^n X_i=t$$]$ 
=E  $[$$X_2$ |$\sum_{i=1}^n X_i=t$] - E  $[$  I{$x_1$=1}|$\sum_{i=1}^n X_i=t$$]$ 
=E  $[$$X_2$ $]$ -E  $[$  I{$x_1$=1}|$\sum_{i=1}^n X_i$$]$ 
= λ  -E  $[$  I{$x_1$=1}|$\sum_{i=1}^n X_i=t$$]$ 
So my main question is are my calculations right up to this point?
If it is wrong , how should I calculate E  $[$$X_2$ |$\sum_{i=1}^n X_i=t$]
 A: There is a very simple way to find this conditional expectation: $E(X_j|\sum\limits_{i=1}^{n} X_i=t)$ is independent of $j$ because $(X_i)$ is i.i.d.. If you call this $f(t)$ and add the equations $f(t)=E(X_j|\sum\limits_{i=1}^{n} X_i=t)$ you get $nf(t)=E(\sum\limits_{j=1}^{n}X_j|\sum\limits_{i=1}^{n} X_i=t)=t$. Hence the answer is $\frac t n$. 
A: Here is a correct proof, after all.
Express the conditional probability as a ratio of joint and marginal:
$P_k = P(X_i = k|\sum_{j=1}^n X_j=t) = P( X_i = k, \sum_{j=1}^n X_j=t ) /
P(\sum_{j=1}^n X_j=t ) = P( X_i = k) P( \sum_{j<>i}^n X_j=t-k ) /
P(\sum_{j=1}^n X_j=t )
$
Given that all $X_i$ are i.i.d Poisson with the same parameter $\lambda$, the sum of $N$ such random variables is also Poisson with parameter $N\lambda$.
Then, 
$P( X_i = k) = \lambda^k e^{-\lambda}/k! $,
$ P( \sum_{j<>i}^n X_j=t-k ) = [(n-1)\lambda]^{(t-k)} e^{-(n-1)\lambda}/(t-k)!$
$P(\sum_{j=1}^n X_j=t ) =  [n\lambda]^t e^{-n\lambda}/t!$
Little algebra simplifies to
$P_k = $
$t \choose k $ $(1-1/n)^{t-k} (1/n)^k$, which is a p.m.f of a Binomial r.v. with parameters $t$ and $1/n$. Its expectation is $t/n$.
This approach would also work in a more general case when $X_i$ are i.i.d. from Poisson($\lambda_i$). 
Then, $P_k = $
$t \choose k $ $(1-p)^{t-k} p^k$, where $p = \lambda_i/\sum_j^n \lambda_j$ with the expectation $tp$.
