Computing a conditional expectation for uniform RVs

Question

Suppose $X_1, ..., X_n \sim U[0, 1]$ are iid uniform RVs. How would I go about computing $E[X_n | X_{(n)}]$ where $X_{(n)}$ is the nth order statistic, i.e. $\max\{X_1, ..., X_n\}$ ?

I'm stuck without any ideas on this.

Answer 1

A joint independent uniform distribution remains "flat" even after conditioning on what region the values lie in. Thus, your conditional expectation is $c/n + ((n-1)/n)E(Y_c)$ where $c$ is the observed maximum value (which could have been $X_n$), and $Y_c$ is the uniform random variable restricted to $[0,c]$. Thus your conditional expectation is $c/n + ((n-1)/n)c/2$, where $c$ is the given maximum of the $n$ samples.

Answer 2

Assume $Y = X_{(n)}.$ The pdf of $Y$ is $f(y) = ny^{n-1}$. The joint pdf of $Y$ and $X_n$ is $$ f(y,x_n)= \begin{cases} x_n^{n-1}\delta(y-x_n) & x_n = y\\ (n-1)y^{n-2} & x_n < y \end{cases} $$ The conditional pdf of $X_n$ given $Y$ is $$ f(x_n/y) = \begin{cases} \frac{x_n^{n-1}}{ny^{n-1}}\delta(y-x_n) & x_n = y\\ \frac{(n-1)}{ny} & x_n < y \end{cases} $$ So we have $$ E[X_n/Y] = \frac{n+1}{2n}Y $$