Check if the estimator is unbiased

Question

For $X_i\sim U[0,a]$ where $i=1,2,\dots,n$ so, $E(X_i)=\dfrac a2$.

Is $a'=\max\{X_1,X_2,\dots,X_n\}$ an unbiased estimator of $a$?

This is what I thought.

Since $a'=\max\{X_1,X_2,\dots,X_n\}=X_k$ such that $X_k\ge X_h$ for any $h$,

$E(a')=E(X_k)=\dfrac a2$.

Therefore $a'$ is not an unbiased estimator.

However, I don't think it is properly solved...

Answer 1

The $\text{cdf}$ of the maximum of $n$ iid random variables is the $n^{th}$ power of the single $\text{cdf}$. In this case, $$\text{cdf}(X)=\left(\frac xa\right)^n.$$

The expectation is

$$\int_0^a x\frac na\left(\frac xa\right)^{n-1}dx=\frac{na^{n+1}}{(n+1)a^n},$$ which is smaller than $a$ (but asymptotically unbiaised for large $n$).

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Your reasoning is wrong in that the distribution of $\max\{X_k\}$ is not the same as that of a single $X_k$ (obviously the $\max$ favors the larger values), and the expectations do not match.

Answer 2

By definition, $B_a(a')$ is the bias of the estimator $a'$ for $a$. That is, $$B_a(a')=\mathbb E \left [a'-a \right]$$ Always, $a' < a$, because $P(X=a)=0$. Therefore $B_a(a')<0$ and (more important) $B_a(a') \ne 0$. It follows that $a'$ is biased.

By the way: $a'$ is the MLE of $a$. Asymptotically (or approximately for large samples) it is unbiased.