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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?

kjetil b halvorsen
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S.Rana
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    See https://math.stackexchange.com/questions/60497/unbiased-estimator-for-a-uniform-variable-support. – StubbornAtom Oct 22 '18 at 12:59
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    Also possibly answered [here](https://stats.stackexchange.com/questions/252129/is-there-an-example-where-mle-produces-a-biased-estimate-of-the-mean). – StubbornAtom Oct 22 '18 at 13:02

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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 $\frac{n}{n+1}\theta$.

kjetil b halvorsen
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