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Let X1, X2, . . . , Xn be i.i.d. random variables and let X(1),X(2), . . . ,X(n) be the order variables.

Show: E(X1 | X(1),X(2), . . . ,X(n)) = $\sum_{k=1}^n \frac{Xk}{n}$

This is a question from a textbook with no explained answer. I understand this question could be done from first principles that is finding the conditional distribution of f X1 | X(1),X(2), . . . ,X(n) and then applying the expected value formula. But, I believe there is a shorter way based on the relationship between X1 and X(1),X(2), . . . ,X(n). Would someone be able to provide an intuition on a shorter way to calculate this problem?

Something's I understand, but not sure if they apply to the problem:

f(x) = $\sum_{k=1}^n \frac{fk(x)}{n}$ where fk(x) is for the kth order statistic.

E[X] = $\sum_{k=1}^n \frac{Ek[x]}{n}$ where Ek(x) is for the kth order statistic.

Thank you in advance!

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Since the variables are i.i.d. the conditional expectation does not change if you permute the random variables. Hence the left had side remains the same if you change $X_1$ to any $X_i$. It follows that it does not change if you replace $X_1$ by the average of teh $X_i$'s which is the RHS. However RHS is already measurable with respect to $\sigma \{X_{(1)},X_{(2)},...,X_{(n)}\}$ so conditioning goes away and you get LHS=RHS.

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