Conditional Expectation of Poisson Suppose $X_1$,$X_2$,$X_3$,.....,$X_n$ are i.i.d. random variables with a common pmf poisson(λ)
(t = a value)
How would you calculate the below without using intuition 
(I would appreciate if you could show me with steps )
E  $[$$X_3$ |$\sum_{i=1}^n X_i=t$]
 A: IID (independent and identically distributed) implies exchangeability: this means that no reordering of the indices in the vector $(X_1, X_2, \ldots, X_n)$ changes its distribution.
Writing $S=X_1+X_2+\cdots+X_n$, notice that $S$ is unchanged whenever the indices are reordered.  Therefore, because they can be reordered to place any one of the components first in the vector,
$$E[X_1\mid S=t] = E[X_2\mid S=t] = \cdots = E[X_n\mid S=t].$$
Add these $n$ values--let's temporarily call their common value $\mu$--and use linearity of expectation to express that as
$$\eqalign{
n\mu &= \mu + \mu + \cdots + \mu \\
&= E[X_1\mid S=t] + E[X_2\mid S=t] + \cdots + E[X_n\mid S=t] \\
&= E[X_1+X_2+\cdots+X_n\mid S=t] \\
&= E[S\mid S=t] \\
&= t.
}$$
The last step is called "taking out what is known" in conditional expectations.
Now you can solve for $\mu,$ producing (among other things) the result
$$E[X_3 \mid S=t] = \frac{t}{n}.$$
Intuitively, when the $X_i$ are exchangeable they must contribute equally to the sum, so once you assume the sum equals $t,$ each $X_i$ must contribute $1/n$ of that total on average.
