How can I calculate $E\left(\prod_{i=1}^{n}\frac{X_i}{X_{(n)}}\right)$ where $X_1,\ldots,X_n$ are i.i.d $U(0,\theta)$? Suppose $X_1,X_2,\ldots,X_n$ be a random sample rom $U(0,\theta)$. 
How can I calculate $E\left(\prod\limits_{i=1}^{n}\frac{X_i}{X_{(n)}}\right)$ where $X_{(n)}=\max_{1\leq i \leq n}X_i$?
 A: Short answer: it is the same as $\text{E}[\prod_{i=1}^{n-1} Y_i]$ with the factors $Y_i\sim U(0,1)$. If they are all independent (i.e. if $X_1,\ldots,X_n$ are sampled independently), this becomes $2^{1-n}$.
Longer answer: this is really a question about uniform order statistics. If the $X_i$ are independent and uniformly distributed between $0$ and $\theta$, then their joint density is the product of the marginal densities
$$
  f_{X_1,\ldots,X_n} (x_1,\ldots,x_n) = \prod_{i=1}^n f_{X_i} (x_i) = \frac{1}{\theta^n} 1(0\leqslant x_i\leqslant \theta, \forall i)\,,
$$
where $1(A)$ is 1 if $A$ is true and 0 otherwise. So all configurations within the constraint $0\leqslant x_i\leqslant 1, \forall i$, are equally likely. Suppose now we condition this distribution on the fact that $X_{(n)}=a$, i.e. the largest of the values $X_i$ is some $a<\theta$. Because of uniformity, the other $n-1$ values are again uniform within the imposed constraint, namely that they must all be smaller than $a$. In other words, given the largest value $X_{(n)}$, the other values are uniformly distributed over the $(n-1)$-dimensional region where these other values are smaller than $X_{(n)}$. So we can scale and rename the other values, for example assuming $X_{(n)}=X_j$, as
$$
  Y_i = \begin{cases} X_i/X_j \,, & i=1,\ldots,j-1\\
                      X_{i+1}/X_j\,, & i=j=,\ldots,n-1\,, \end{cases}
$$
then these $Y_i$, $i=1,\ldots,n-1$ are uniform in $[0,1]^{n-1}$. Hence:
\begin{align*}
 \text{E}[ \prod_{i=1}^n \frac{X_i}{X_{(n)}}] 
   &  = \sum_{j=1}^n\text{E}[ \prod_{i=1}^n \frac{X_i}{X_{(n)}} \vert X_{(n)}=X_j] \text{Prob}[X_{(n)}=X_j] \\
   &  = \sum_{j=1}^n\text{E}[ Y_1\cdot \ldots\cdot Y_{j-1}\cdot 1 \cdot Y_j \cdot\ldots\cdot Y_{n-1} ] \frac{1}{n}\\
   & = \sum_{j=1}^n \prod_{i=1}^{n-1} \text{E}[Y_i] \frac{1}{n}
     = \sum_{j=1}^n \prod_{i=1}^{n-1} \frac{1}{2} \frac{1}{n}
     = \sum_{j=1}^n 2^{1-n} \frac{1}{n} = 2^{1-n}
\end{align*}
I have a more direct proof lying on my desk here as well, if anyone is interested.
