# expected value of conditional order statistics

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.

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.