Let $_{(1)},\ldots,_{()}$ be the order statistics of a set of $$ independent uniform [0,1] random variables.
- Find the conditional expectation of $_{(1)}$ given that $_{(2)}=x_2~$, i.e. $~\mathbb{E}(_{(1)}|_{(2)}=x_2)$
- Find the conditional expectation of $_{(2)}$ given that $_{(1)}=x_1~$, i.e. $~\mathbb{E}(_{(2)}|_{(1)}=x_1)$
That's what I got for now, but I'm not sure is my result correct.
I just found the conditional pfds $f_{X_1|X_2}$ and $f_{X_2| X_1}$ using the join distribution $f_{X_1,X_2}$
$$f_{X_1,X_2}(x_1,x_2)=n!\frac{x_1^{1-1}}{(1-1)!}\frac{(x_2-x_1)^{2-1-1}}{(2-1-1)!}\frac{(1-x_2)^{n-2}}{(n-2)!} = n(n-1)(1-x_2)^{n-2}$$
and pdfs of order statistics $f_{X_1}$ and $f_{X_2}$ $$f_{X_1}(x_1) = \frac{n!}{(1-1)!(n-1)!}x_1^{1-1}(1-x_1)^{n-1} = n(1-x_1)^{n-1}$$ $$f_{X_2}(x_2) = \frac{n!}{(2-1)!(n-2)!}x_2^{2-1}(1-x_2)^{n-2} = n(n-1)x_2(1-x_2)^{n-2}$$
Hence, $$f_{X_1|X_2}(x_1,x_2) = \frac{f_{X_1,X_2}(x_1,x_2)}{f_{X_2}(x_2)} = \frac{n(n-1)(1-x_2)^{n-2}}{n(n-1)x_2(1-x_2)^{n-2}} = \frac{1}{x_2}$$ $$f_{X_2|X_1}(x_1,x_2) = \frac{f_{X_1,X_2}(x_1,x_2)}{f_{X_1}(x_1)} = \frac{n(n-1)(1-x_2)^{n-2}}{n(1-x_1)^{n-1}} = (n-1)\frac{(1-x_2)^{n-2}}{(1-x_1)^{n-1}}$$
And used them to find conditional expectations.
$$\mathbb{E}(X_1|x_2) = \int\limits_0^{x_2} x_1 f_{X_1|X_2}(x_1,x_2)dx_1 = \int\limits_0^{x_2} \frac{x_1}{x_2}dx_1 = \frac{x_2}{2}$$ $$\mathbb{E}(X_2|x_1) = \int\limits_{x_1}^1 x_2 f_{X_2|X_1}(x_1,x_2)dx_2 = \int\limits_{x_1}^1 x_2 (n-1)\frac{(1-x_2)^{n-2}}{(1-x_1)^{n-1}}dx_2 = \frac{1+(n-1)x_1}{n}$$
Any help is highly appreciated.