For n ≥ 2, let X₁, …, Xₙ be a random sample with density f(x|θ) = θx^{θ−1} on 0 0 unknown. Which of the following is the UMVUE for 1/θ?

A. −ln Xᵢ ~ Exp(θ) with mean 1/θ, and Σ ln Xᵢ is complete sufficient; the sample mean of −ln Xᵢ is unbiased for 1/θ.

Part BCSIR NET December 2023

Sufficiency, completeness, UMVUE, Cramér–Rao

For n ≥ 2, let $X_{1}, \dots, X_{n}$ be a random sample with density f(x| $\theta) =$ $\theta{}x^{\theta-1}$ on $0 < x < 1, \theta > 0$ unknown. Which of the following is the UMVUE for $1/\theta$ ?

A. $-(1/n) \sum \ln X_{i}$ ✓ Correct
B. $-n/\sum \ln X_{i}$
C. $-(n - 1)/\sum \ln X_{i}$
D. $-(2/n) \sum \ln X_{i}$

The idea, in plain words

Take the negative logarithm of each observation and this awkward family turns into the standard waiting-time family, whose average is one over the parameter. From there everything is a familiar computation.

Solution

$-\ln X_{i}$ ~ Exp $(\theta)$ with mean $1/\theta$ , and $\sum \ln X_{i}$ is complete sufficient; the sample mean of $-\ln X_{i}$ is unbiased for $1/\theta$ .

Why each option is right or wrong

Checked against 3 symbolic computations

A.Correct
True.

Transform: P(−ln X > t) = P(X < $e^{-t})$ = $e^{-\theta{}t}$

which SymPy confirms at $\theta = 3$

so −ln X ~ Exp $(\theta)$ with mean $1/\theta ($ verified as 1/2 at $\theta = 2$ and 1/4 at $\theta = 4)$ .

The sample mean of the $-\ln X_{i}$ is therefore unbiased for $1/\theta$ , and $\sum \ln X_{i}$ is complete sufficient

so Lehmann–Scheffé makes $-(1/n)\sum \ln X_{i}$ the UMVUE.

The trigger: a density of the form $\theta{}x^{\theta-1}$ is an exponential family in −ln x — transform first.
B.WrongConverse assumed
False — this estimates $\theta$ , not $1/\theta$ .

$-n/\sum \ln X_{i}$ is the reciprocal of the correct estimator and is in fact the MLE of $\theta$ .

Reciprocals do not commute with expectation

so an unbiased estimator of $1/\theta$ cannot simply be inverted to give one for $\theta$ .
C.WrongExecution slip
False — this is the UMVUE of $\theta$ itself, with the (n − 1) correcting the bias of the MLE.

It is the right estimator for the wrong quantity.

Read which of $\theta$ and $1/\theta$ the question asks for before choosing.
D.WrongExecution slip
False.

This is twice the correct estimator

so its expectation is $2/\theta$ rather than $1/\theta$ — it is biased by a factor of 2 and cannot be a UMVUE.

3 of these 3 wrong options correspond to a named reasoning error — the same errors recur across subjects.

Related counterexample: A sufficient statistic is complete

Sufficiency, completeness, UMVUE, Cramér–Rao

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