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 $X_{(n)}=\max_{1\leq i \leq n}X_i$?

  • $\begingroup$ This can be answered using Basu's theorem: $\prod\limits_{i=1}^{n}\frac{X_i}{X_{(n)}}=\frac{\prod_{i=1}^n X_i}{X_{(n)}^n}$ is an ancillary statistic independent of the complete sufficient statistic $X_{(n)}^n$. $\endgroup$ Commented May 27, 2021 at 14:42

1 Answer 1


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.

  • $\begingroup$ This is a great answer, but it could benefit by explaining its assertions or providing sources. The most important one is that the $X_i/X_{(n)}$ have the same $n-1$-variate distribution as the $Y_i$. $\endgroup$
    – whuber
    Commented May 25, 2016 at 13:46
  • 1
    $\begingroup$ Thank you for the elaboration--it goes well beyond what I expected in terms of detail and explanation! $\endgroup$
    – whuber
    Commented May 26, 2016 at 15:24

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