Choosing between unbiased estimators of a uniform endpoint
A reliability laboratory observes n=10 independent lifetimes, in hundreds of hours, from a uniform model on (0,θ), with θ>0 unknown. The sample mean is 4.2 and the sample maximum is 9.1. Two unbiased estimators are under consideration: T1=2X̄ and T2=((n+1)/n)X_(n), where X_(n) is the sample maximum. A junior analyst prefers T1 because it uses all observations and is a moment estimator; a senior analyst prefers T2 because it is a function of the maximum. The laboratory must choose one estimator for n=10 using mean squared error at the same θ, and must also report whether each estimator is consistent as n→∞. A third candidate, the maximum-likelihood estimator M=X_(n), is tabled for comparison; its bias and MSE must be computed, not merely asserted. All derivations should use the distribution of X_(n) and standard variance formulas for the sample mean.
1. Compute the numerical estimates produced by T1, T2, and M.
Read the worked answer
T1=2X̄=2(4.2)=8.4. T2=((n+1)/n)X_(n)=(11/10)(9.1)=10.01. M=X_(n)=9.1. These are estimates of θ; the units are hundreds of hours. The three values differ substantially, which already shows that the choice of estimator matters at n=10.
2. Verify the bias of each estimator and compare their mean squared errors at n=10.
Read the worked answer
X_(n) has CDF (x/θ)^n on (0,θ), so E[X_(n)]=nθ/(n+1) and Var(X_(n))=nθ²/((n+1)²(n+2)). Thus T2 is unbiased, with Var(T2)=((n+1)/n)²Var(X_(n))=θ²/(n(n+2))=θ²/120≈0.00833θ². T1 is unbiased because E[X]=θ/2; Var(X)=θ²/12, so Var(T1)=4Var(X)/n=θ²/(3n)=θ²/30≈0.0333θ². For M, bias=-θ/(n+1)=-θ/11, Var=10θ²/(121·12)=5θ²/726≈0.00689θ², and MSE=Var+bias²≈0.00689θ²+0.00826θ²=θ²/66≈0.01515θ². Ranking at n=10: T2 best, then M, then T1.
3. Discuss consistency and sufficiency for these estimators. Does sufficiency alone settle the MSE choice?
Read the worked answer
All three are consistent. T1→θ by the weak law of large numbers, since E[X]=θ/2. T2→θ because X_(n)→θ in probability: P(θ-X_(n)>ε)=(1-ε/θ)^n→0, and its variance also tends to 0. M→θ for the same reason, despite its bias, because the bias tends to 0. For sufficiency, the joint density is θ^{-n}1{0<x_i<θ}; the factorization criterion shows that X_(n) is a minimal sufficient statistic. T2 and M are functions of X_(n), while T1 is not sufficient because it discards the maximum information. Sufficiency does not settle the MSE comparison: M is sufficient but has larger MSE than T2 at n=10.