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Given $X_1,...,X_n \sim f(x)$ How do I find $E(X_{(1)} | X_{(2)})$? Would I have to find the conditional pdf and integrate wrt x?

I get the conditional distribution to be $f_{X|Y}(x|y) =\frac{\frac{n!}{(n-2)!}[1-F_X(y)]^{n-2}f_X(x)f_X(y)}{\frac{n!}{(n-2)!}[F_X(x)][1-F_X(x)]^{n-2}f_X(x)} = \frac{f_X(y)}{F_X(x)}$ but I'm not sure how to integrate.

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  • $\begingroup$ Please provide more details on the problem and on why you cannot solve it. As is, it is unclear. Also add self-study as a tag. $\endgroup$ – Xi'an Apr 26 '15 at 10:42
  • $\begingroup$ Check your calculations. Your conditional density is a function of $x$ with $y$ as a parameter, but what you have is a function that is proportional to $\frac{1}{F_X(x)}$. $\endgroup$ – Dilip Sarwate Apr 26 '15 at 13:53

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