The MSE-optimal scaling of the sample max

Question

For X1, \dots, Xn i.i.d. Uniform(0, 	heta), let M = X{(n)} be the sample maximum. The unbiased estimator is \frac{n+1}{n}M. Among all estimators of the form c\,M, which constant c minimizes the mean squared error, and does it beat the unbiased choice?

Accepted Answer

Two moments of the maximum do all the work: \mathbb{E}[M] = \frac{n}{n+1}	heta, \qquad \mathbb{E}[M^2] = \frac{n}{n+2}	heta^2. Write the mean squared error of cM and minimize over c: 	ext{MSE}(cM) = \mathbb{E}[(cM - 	heta)^2] = c^2\,\mathbb{E}[M^2] - 2c	heta\,\mathbb{E}[M] + 	heta^2. Setting the derivative in c to zero gives c^{} = \dfrac{	heta\,\mathbb{E}[M]}{\mathbb{E}[M^2]}: c^{} = \frac{	heta \cdot \frac{n}{n+1}	heta}{\frac{n}{n+2}	heta^2} = \frac{n+2}{n+1}. Compare. The unbiased estimator uses \frac{n+1}{n}, but the MSE-optimal one uses \frac{n+2}{n+1}, which is slightly smalle…

The MSE-optimal scaling of the sample max

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