0

Suppose $X_1 ,..., X_n$ is a random sample from u{0,$\theta$} consider the estimator $\hat{\theta} = X_n$, is it unbiased?

So far I have E($\hat{\theta}$) = E($X_n$) = ${\theta}$($\frac{n}{n+1}$) therefore $X_n$ is a biased estimator but I have no clue if I'm even on the right track.

V. Vancak
  • 16,444
J.Doe
  • 9
  • 2
  • The maximum in every case must be less than or equal to $\theta,$ so intuitively the maximum must be biased (downward) as the Answer by @V.V. shows. However, by my multiplying by the appropriate constant (depending on the sample size $n$) the max can be made unbiased. Notice that double the sample mean and double the sample median are also unbiased estimates of $\theta,$ bit the unbiased max is the smallest variance of the three. – BruceET Feb 06 '17 at 07:33

2 Answers2

0

We have: $$ \def\E{\mathbf E_\theta}\E[\hat\theta] = \E[X_n] = \frac\theta 2 \ne \theta $$ hence, it's a biased estimator.

martini
  • 84,101
0

I guess you meant $X_{(n)} = \max\{X_1,...,X_n\}$, then your calculations are right. Recall that $$ \text{bias}{ (\hat{\theta})} = E(\hat\theta - \theta)= \frac{n\theta}{n+1} - \theta=-\frac{\theta}{n}. $$ Thus the estimator is biased downward for every $n\in\mathbb{N}$.

V. Vancak
  • 16,444