# 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 $$[X_3$$ |$$\sum_{i=1}^n X_i=t$$]

• Hints: (1) You don't even need to know the common distribution of the $X_i$: it suffices that they be exchangeable (and have finite expectations). (2) How would $E\left[X_j\mid \sum_{i=1}^n X_i=1\right]$ compare to the answer when $j\ne 3$?
– whuber
Apr 29, 2019 at 19:12
• is it equal to 1/n ?
– GAGA
Apr 29, 2019 at 20:11
• stats.stackexchange.com/q/374989/119261 Apr 27, 2021 at 16:07

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.