# What is the bias of uniform distribution parameter estimator?

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?

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

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.