Poisson Conditional Expectation ( searching best estimator for h(λ) )

Question

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$ |$\sum_{i=1}^n X_i=t$]

Answer 1

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 $\frac t n$.

Answer 2

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 $tp$.