Let $(X_1, \ldots, X_n) \sim \operatorname{Unif}(0,b), b>0$. Find $E\left[\sum \frac{X_i }{X_{(n)}}\right]$

Question

Let $(X_1, \ldots, X_n) \sim \operatorname{Unif}(0,b), b>0$. Find $E\left[\sum \frac{X_i }{X_{(n)}}\right]$ where $X_{(n)} = \max_i X_i$.

It was suggested to use Basu's Theorem which I am unfamiliar with.

There are finitely many terms so we can rearrange using order statistics and write it as:

\begin{align} E\left[\sum_{i = 1}^n \frac{X_i }{X_{(n)}}\right] & = E\left[\frac{X_{(1)}}{X_{(n)}}\right] + E\left[\frac{X_{(2)}}{X_{(n)}}\right]+ \cdots +E[1] \\[8pt] & = (n-1) E\left[\frac{X_i}{X_{(n)}}\right] + 1 \end{align}

If this is correct then I will need to calculate a conditional expectation to calculate this so I wanted to see if this is even correct before moving forward. Or if someone familiar with Basu's theorem can explain how I apply that here.

Answer 1

Assuming $X_1,\ldots,X_n$ are independent and identically distributed,

$$E\left[\sum_{i=1}^n \frac{X_i}{X_{(n)}}\right]=\sum_{i=1}^n E\left[\frac{X_{i}}{X_{(n)}}\right]=nE\left[\frac{X_1}{X_{(n)}}\right]$$

This can be evaluated by showing that $X_1/X_{(n)}$ is independent of $X_{(n)}$, so that

$$E\left[X_1\right]=E\left[\frac{X_1}{X_{(n)}}\cdot X_{(n)}\right]=E\left[\frac{X_1}{X_{(n)}}\right]\cdot E\left[X_{(n)}\right]$$

Thus giving $$E\left[\frac{X_1}{X_{(n)}}\right]=\frac{E\left[X_1\right]}{E\left[X_{(n)}\right]}$$

The independence can be argued using Basu's theorem since the distribution of $\frac{X_1}{X_{(n)}}=\frac{X_1/b}{X_{(n)}/b}$ is free of $b$ (making $\frac{X_1}{X_{(n)}}$ an ancillary statistic) and $X_{(n)}$ is a complete sufficient statistic for $b$.

Answer 2

Basu's theorem says that a complete sufficient statistic is independent of an ancillary statistic.

To say that $X_{(n)} = \max\{X_1,\ldots,X_n\}$ is sufficient for this family of distributions (parametrized by $b$) means that the conditional distribution of $X_1,\ldots,X_n$ given $X_{(n)}$ does not depend on the value of $b.$

To say that $X_{(n)}$ is complete means there is no function $g$ (not depending on $b$) for which $\operatorname E(g(X_{(n)}))$ remains equal to $0$ as $b$ changes.

To say that $\dfrac{X_i}{X_{(n)}}$ is an ancillary statistic means that its probability distribution remains the same as $b$ changes, even though $\dfrac{X_i}{X_{(n)}}$ depends on $(X_1,\ldots,X_n,b)$ only through $(X_1,\ldots,X_n).$

If all of that is shown, then what Basu's theorem tells you is that $\dfrac{X_i}{X_{(n)}}$ and $X_{(n)}$ are independent.

That is what makes it possible to conclude that the expected value of their product is the product of their expected values, and then the rest is routine.