Conditional Expectation of Poisson Random Variables Suppose $X_1,X_2,\ldots,X_n$ is a random sample from a Poisson Distribution with mean $\theta$. How can I find the conditional expectation $E \left( X_1+X_2+3X_3 |\sum_{i=1}^n X_i \right)$?
I know that $\sum X_i $ has a $poisson (n\theta$) distribution. Similarly the random variable $X_1+X_2+X_3$ has a $poisson (3\theta)$ distribution. I get confused with the required summations afterwards though.
Thank you.
 A: Define $S_n=\sum_{i=1}^n X_i$. By symmetry,
$$
  \mathrm{E}\left[ X_1 \mid S_n \right] = \mathrm{E}\left[ X_2 \mid S_n \right] = \dots = \mathrm{E}\left[ X_n \mid S_n \right] \quad \textrm{a.s.} \quad (*)
$$
Hence, using $(*)$ and the linearity of the conditional expectation, we have
$$
  \mathrm{E}\left[ X_1 \mid S_n \right] = \frac{1}{n} \mathrm{E}\left[ X_1+\dots+X_n \mid S_n \right] = \frac{1}{n} \mathrm{E}\left[ S_n \mid S_n \right] = \frac{S_n}{n} \quad \textrm{a.s.}
$$
The same reasoning leads to
$$
  \mathrm{E}\left[ X_1 +X_2 +3X_3\mid S_n \right] = 5\,\mathrm{E}\left[ X_1 \mid S_n \right] = \frac{5\,S_n}{n} \quad \textrm{a.s.}
$$
Now, remember that $S_n\sim \mathrm{Poisson}(n\theta)$, and find the pmf of $5\,S_n/n$ (consider its support).
A: The OP has apparently found the way, so I am posting an answer.
I will denote $Z \equiv \sum_{i=1}^n X_i$. By linearity of the expected value we have
$$E \left( X_1+X_2+3X_3 |Z \right)= E \left( X_1 |Z \right)+E \left( X_2 |Z \right)+3E \left(X_3 |Z \right)$$
Since the variables are i.i.d. they are also exchangeable,at least with respect to $Z$ (to which they have a symmetric relationship), so the three conditional expected values will be equal:
$$E \left( X_1+X_2+3X_3 |Z \right)= 5E \left( X_1 |Z \right)$$
Moreover, it is a known result that the conditional distribution of $X_1$ conditional on $Z=k$ is a Binomial,
$$X_1 | Z=k \sim Bin\left(k, \frac {E(X_1)}{E(Z)}\right) = Bin\left(k, 1/n\right)$$
and so 
$$5E \left( X_1 |Z=k \right) = 5\frac kn$$
Our conditional expectation is viewed as a function of $Z$, is not conditioned just on $Z$ acquiring a specific value. Generalizing the last equation we obtain
$$5E \left( X_1 |Z \right) = \frac 5n Z= 5 \frac 1n \sum_{i=1}^n X_i$$
Note that 
$$E \left( X_1 |Z \right) \rightarrow_p E(X_1) \;\;\text {as}\;\; n\rightarrow \infty$$
which should be intuitive.
