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Suppose I have $\{X_i\}$ for $i$ from $1$ to $n$; I know that the $X_i$ are all independent and identically distributed, and also $P(X_i=1)=p$,$P(X_i=-1)=q$. I want to calculate the conditional expectation $E[X_1\mid X_1+\cdots+X_n]$; how can I do that?

joriki
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VadimStacheff
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  • Hello Vadim. What do you know about the conditional expectation ? – Hamdiken Feb 14 '23 at 22:32
  • First, lets recall definition E[X|A]=$\sum_{p \in \Omega} p*P(X=p|A)$, this is when we have X - discrete,so in my case. So i must do something like that $EX_1|X_1+...+X_n=x = -P(X_1=-1|X_1+...+X_n=x)+P(X_1=1|X_1+...+X_n=x)$ – VadimStacheff Feb 14 '23 at 22:46
  • $E[X_1|X_1+...+X_n]=\frac 1 n (X_1+...+X_n)$ whenvever $(X_i)$ is i.i.d. with finite mean. – geetha290krm Feb 14 '23 at 23:14
  • @geetha290krm, can you please explain why is that so? – VadimStacheff Feb 15 '23 at 04:09
  • The reason is the joint distrbution of $(X_i)$ remains invariant under permutations. This result has appeared many times on this site. See, for example, https://math.stackexchange.com/questions/78546/conditional-expectation-for-a-sum-of-iid-random-variables-e-xi-mid-xi-eta-e – geetha290krm Feb 15 '23 at 05:05

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