Conditional expectation to de maximum $E(X_1\mid X_{(n)})$ Let $X_1, \ldots, X_n$ a random sample of a Uniform(0,1):
Which is $E(X_1\mid X_{(n)})$ ?
where $X_{(n)}=\max\{X_1,\ldots,X_n\}$
 A: Let $Y$ be the index of the maximum value: $X_Y=X_{(n)}$ (the event of ties is unimportant, it has zero probability). Let $Z=X_{(n)}$
Then 
$$E(X_1\mid Z ,Y) = 
  \begin{cases} 
   Z & \text{ if } Y=1 \\
   \frac{1}{2}Z & \text{ elsewhere} \\
  \end{cases}
$$
Then, applying the iterated expectations property, and because  $P(Y=k)=1/n$:
$$E(X_1\mid Z)=E( E(X_1\mid Z , Y))=\frac{1}{n}Z+ \frac{n-1}{n}\frac{Z}{2} = \frac{n+1}{n}\frac{Z}{2}  $$
Edit: this is not need to answer the question, but (from a comment) the full probability can be obtained (some abuse of notation follows) thus:
$$P(X_1\mid Z ) =\sum_{Y=1}^n P(X_1,Y\mid Z )\\
=\sum_{Y=1}^n P(X_1\mid Y,Z ) P(Y \mid Z)\\
=\delta(X_1 -Z)\frac{1}{n}+\sum_{Y=2}^n P(X_1\mid X_1<Z ) \frac{1}{n}\\
=\frac{1}{n}\delta(X_1 -Z)+ \frac{n-1}{n}U[0,Z]
$$
So, yes $P(X_1\mid X_{(n)} )$ can be seen a mixing of a delta with a (truncated) uniform.
A: You're looking for $E(X_1\mid \max)$.  The probability that $X_1=\max$ is $1/n$. So
$$
E(X_1\mid \max) = \frac 1 n E(X_1\mid X_1=\max, \max) + \frac{n-1} n E(X_1\mid X_1\ne \max, \max).
$$
The first term is easy to find: $E(X_1\mid X_1=\max,\max) = \max$, so we have
$$
E(X_1\mid \max) = \frac 1 n \max + \frac{n-1} n E(X_1\mid X_1\ne \max, \max).
$$
Then you need to show that the conditional distribution of $X_1$ given $\max$ and given the event that $X_1\ne\max$ is uniform on the interval from $0$ to $\max$.
