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

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    $\begingroup$ 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 $j\ne 3$? $\endgroup$ – whuber Apr 29 '19 at 19:12
  • $\begingroup$ is it equal to 1/n ? $\endgroup$ – GAGA Apr 29 '19 at 20:11
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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.

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