X_1 through X_n are iid Uniform(0, theta). With n = 20 and a sample maximum of 9.5, give the maximum likelihood estimate of theta, its expectation, the unbiased correction, and how its mean squared error compares with the method of moments estimator 2*xbar.

X_1 through X_n are iid Uniform(0, theta). With n = 20 and a sample maximum of 9.5, give the maximum likelihood estimate of theta, its expectation, the unbiased correction, and how its mean squared error compares with the method of moments estimator 2*xbar.

Approach: The likelihood has a support constraint, so maximise it by inspection rather than by differentiating, then use the distribution of the sample maximum for its moments.

9.5. The likelihood is theta^{-n} for theta >= max(x_i) and zero below, so it is decreasing wherever it is positive and is maximised at the sample maximum M = 9.5. Since P(M <= m) = (m/theta)^n, the density is n*m^{n-1}/theta^n and E[M] = n*theta/(n + 1) = (20/21)*theta, a downward bias, so (n + 1)/n * M = (21/20)*9.5 = 9.975 is unbiased. The mean squared error of M is 2*theta^2/((n + 1)*(n + 2)) = theta^2/231, while 2*xbar is unbiased with variance 4*(theta^2/12)/n = theta^2/60. The maximum wins by a factor of 3.85 because it uses the boundary information that averaging discards, and its error shrinks like 1/n rather than 1/sqrt(n). The method of moments estimator can also fall below the observed maximum, which is an impossible value for theta.

Follow-up: Which estimator of the form c*M minimises mean squared error, and why is c larger than the unbiasing constant?

Key concepts: maximum likelihood, bias correction, method of moments, mean squared error.