Find $E[X_1\mid\max\{X_1,X_2,...,X_n\}]$ for $(X_k)$ i.i.d uniform on $[0,1]$ Let $X_1,X_2,...,X_n$ be i.i.d uniform random variables on $[0,1]$. Define $Y=max${$X_1,X_2,...,X_n$}. Find $E[X_1|Y]$.   
My answer:
$P(X_i|Y)=P(X_j|Y)$ for all $i,j=1,2,...,n$.
Since $\sum_{i=n}^{n}P(X_i|Y)=1$, I get $P(X_i|Y)=1/n$.
Then $$E[X_1|Y]=\int_{0}^{1}x_1\times{1\over n}dx_1={1\over 2n}$$
Is it correct for $P(X_i|Y)=P(X_j|Y)$ and $\sum_{i=n}^{n}P(X_i|Y)=1$? 
 A: The equality $\sum_i \mathbb P[X_i|Y]=1$ is false, you don't get the right result for this reason.
Observe that $\mathbb P[Y\leq y] = \mathbb P[X_1\leq y\land X_2\leq y\land\dots\land X_n\leq y]=y^n$ hence $f_Y(y)=n y^{n-1}$. Similarly $\mathbb P[X_1\leq x,Y\leq y]=\mathbb P[X_1\leq \min(x,y)\land X_2\leq y\land\dots\land X_n\leq y]=\min(x,y) y^{n-1}$, hence for $x\leq y$, $f_{X_1,Y}(x,y)=(n-1) y^{n-2}$.
Using that, you can obtain the conditional distribution of $X_1$ given that $Y$, for $x<y$
\begin{align*}
f_{X_1|Y}(x|y) &= \frac{f_{X_1,Y}(x,y)}{f_Y(y)}\\
&=\frac{(n-1)y^{n-2}}{n y^{n-1}}\\
&=\frac{n-1}{ny}
\end{align*}
For $x=y$ a discontinuity happens and $\mathbb P[X_1=y|Y=y]=\frac{1}{n}$ by symmetry. Hence for any $x\leq y$,
\begin{align*}
f_{X_1|Y}(x|y) =\frac{n-1}{ny}+\frac{1}{n}\delta(x-y)
\end{align*}
finally the conditional expectation is given by
\begin{align*}
\mathbb E[X_1|Y=y] &= \int_0^{y} x \cdot f_{X_1|Y}(x|y) dx\\
&=\frac{n-1}{ny} \cdot \frac{y^2}{2}+\frac{y}{n}\\
&=\frac{y(n+1)}{2n}
\end{align*}
And hence $\mathbb E[X_1|Y]=\frac{Y(n+1)}{2n}$
