How do I show that the maximum likelihood estimator for uniform distribution on $[0, \theta]$ for a random sample of size $n$ is biased? I've calculated the MLE as $\max_i\{X_i\}$.

Intuitively, we can say that estimator is biased because only the maximum of $X_i$ from the sample would be the estimator. How do I derive the distribution of MLE and check for biasedness?


1 Answer 1


So $X_1, \dotsc, X_n$ is iid uniform on $(0, \theta)$ with $\theta > 0$. Then the maximum likelihood estimator (also sufficient statistic) of $\theta$ is $M=\max_i X_i$. Now clearly $M < \theta$ with probability one, so the expected value of $M$ must be smaller than $\theta$, so $M$ is a biased estimator.

We need to find the distribution of $M$. Use that $$ P(M \le m)= P(X_1\le m, X_2\le m, \dotsc, X_n\le m)=\left(m/\theta\right)^n $$ and by differentiation you can find the density $f(m)=n\left(\frac{m}{\theta}\right)^{n-1}\frac1\theta$, Then integration will yield the expected value as $\frac{n}{n+1}\theta$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.