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Suppose I have some random variable $X$ which only takes on values over some finite region of the real line, and I want to estimate the maximum value of this random variable. Obviously one crude method is to take many measurements, lets say $X_1$, $X_2$, $\ldots, X_n$ (which we'll say are all iid) and to use $$X_{max} = \text{max}(X_1, \ldots X_n)$$ as my guess, and as long as $n$ is large enough this should be good enough. However, $X_{max}$ is always less than the actual maximum, and I'm wondering if there's any way to modify $X_{max}$ so it gives a guess (still with some uncertainty) which is centred around the actual maximum value, rather than always a little less than it.

Thanks

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Got something from an answer below? –  Did Jun 30 '12 at 12:17
    
You may try to show that maximum of the samples of $X_n$ is the maximum likelihood estimator of the maximum of $X$. –  user103255 Oct 25 '13 at 19:59

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up vote 3 down vote accepted

When the random variables are uniform on $(0,M)$ for an unknown $M\gt0$, one can check that the maximum $X_n^*$ of an i.i.d. sample of size $n$ is such that $\mathrm P(X_n^*\leqslant x)=(x/M)^n$ for every $x$ in $(0,M)$ hence $\mathrm E(X_n^*)=\frac{n}{n+1}M$. This means that an unbiased estimate of $M$ is $\widehat M_n=\frac{n+1}nX_n^*$.

For other distributions, the result is different. For example, starting from the distribution with density $ax^{a-1}/M^a$ on $(0,M)$ for some known $a\gt0$, one gets $\mathrm P(X_n^*\leqslant x)=(x/M)^{an}$ for every $x$ in $(0,M)$ hence $\mathrm E(X_n^*)=\frac{an}{an+1}M$ and an unbiased estimate of $M$ is $\widehat M_n=\frac{an+1}{an}X_n^*$.

Likewise, if the density is $a(1-x)^{a-1}/M^a$ on $(0,M)$ for some known $a\gt0$, one gets $\mathrm E(X_n^*)=M(1-c_n)$ with $c_n\sim\frac{\Gamma(1+1/a)}{n^{1/a}}$ and an unbiased estimate of $M$ is $\widehat M_n=\frac1{1-c_n}X_n^*$.

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You might ask whether or not an unbiased estimate is desirable (i.e. more accurate than a biased one in terms of mean square error). The unbiased estimators that "did" suggested require knowledge of the form fo the distribution and as he notes an estimate that is unbiased in one case may not be in another. While the sample maximum is slightly biased on the low side the unbiased estimate can be above the actual maximum as well as below. –  Michael Chernick Jun 17 '12 at 12:37

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