The best scaling of the sample maximum

Question

Values are uniform between 0 and an unknown ceiling $\theta$ . You estimate $\theta$ by $c \cdot M$ , where $M$ is the maximum of $n = 3$ draws and $c$ is a constant you choose.

What value of $c$ minimizes the mean squared error?

Accepted Answer

The idea. The raw max (c = 1) is biased low; the unbiased choice is c = \frac{n+1}{n}. But minimizing MSE, which weighs variance too, lands somewhere else. We scan over all c. Write the MSE. Let a = \mathbb{E}[M]/	heta = \frac{n}{n+1} and V = 	ext{Var}(M)/	heta^2 = \frac{n}{(n+1)^2(n+2)}. Then 	ext{MSE}(c)/	heta^2 = c^2 V + (c a - 1)^2. Minimize. Differentiating and setting to zero gives c(V + a^2) = a, so c^\star = \frac{a}{V + a^2}. Plugging in a and V and simplifying (the denominator collapses to \frac{n}{n+2}): c^\star = \frac{n/(n+1)}{n/(n+2)} = \frac{n+2}{n+1}. For n = 3: c^\star =…

The best scaling of the sample maximum

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