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I have a question regarding question 2 of chapter 6 of "All of Statistics" book by Larry Wasserman.

let: $$X_1, ... , X_n \sim \operatorname{Uniform}(0, \theta )$$

and let:

$$\hat{\theta} = \max(X_1, ..., X_n)$$

be the estimator of the parameter $\theta$.

My question is that how one can calculate the Expected value of $\hat{\theta}$. My idea is that since the maximum of these random variable is itself a random variable with the uniform distribution mentioned above, the expected value of it will be:

$$E[\hat{\theta}] = E[X_{\max}] = \theta/2$$

However, I was advised that this answer is wrong. Apparently, the correct approach is trying to find the probability distribution function of $\hat{\theta}$ and after doing the algebra, the answer will be:

$$E[\hat{\theta}] = \frac{n}{n+1}.\theta $$

The problem is that I cannot understand the reason that makes my approach wrong. Can anybody illustrate me?

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    $\begingroup$ The maximum of $X_1, \ldots, X_n$ does not have the same distribution as the individual random variables. $\endgroup$
    – angryavian
    Commented Mar 20 at 0:04
  • $\begingroup$ @angryavian That is actually correct. Thanks. $\endgroup$ Commented Mar 23 at 19:14

1 Answer 1

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The maximum of the $X$s is less than 1/2 only if all $n$ of the observations are less than 1/2, which happens $1/2^n$ of the time.

In general, $$P(\text{all } x\leq t)=t^n$$ for any $t\in[0,1]$, so the CDF of $\max_n X$ is $F(t)=t^n$ for $t\in[0,1]$. From that you can work out the mean.

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