Point and Interval Estimation Questions and Answers

Answer: A

The true statements are (I), (II) and (III).

(I) is Basu's theorem: a boundedly complete sufficient statistic is independent of every ancillary statistic.

(II) is true because without completeness the independence can fail; when the sufficient statistic is not complete, an ancillary statistic can be a function of it.

(III) is true: the Cramer-Rao bound applies to unbiased estimators under regularity conditions. A biased estimator, such as a constant, has variance 0, below the bound, and when the regularity conditions fail (for example the uniform (0, θ) family) estimators can beat the bound.

(IV) is false: the bound \([\psi'(\theta)]^{2}/I(\theta)\) needs \(\psi\) to be differentiable. That rules out every option containing (IV).

Answer: C

The law that is not a member of the (one-parameter) exponential family is N(θ, θ²).

Writing the density,

\[ \begin{aligned} f(x) &\propto \frac{1}{\theta}\exp(Q), \\ Q &= -\frac{x^{2}}{2\theta^{2}} + \frac{x}{\theta} - \frac{1}{2} \end{aligned} \]

the exponent has two terms, \(x^{2}\) and x, with coefficients \(-1/(2\theta^{2})\) and \(1/\theta\) depending on the single parameter. It cannot be written as \(c(\theta)T(x)\) with one statistic, so it is a curved family, not a one-parameter exponential family.

The others are members: \(\theta(1 - x)^{\theta - 1} = \exp[(\theta - 1)\log(1 - x) + \log\theta]\); the Bernoulli is \(\exp[x\log\frac{\theta}{1 - \theta} + \log(1 - \theta)]\); and for \(N(\theta, \theta)\) the x term reduces to the constant 1, leaving only \(-x^{2}/(2\theta)\).

Answer: C

The consistent estimator is max{X₁, X₂/2, ..., X_n/n}.

Each \(Y_{i} = X_{i}/i\) is uniform on (0, θ), and the \(Y_{i}\) are independent. Their maximum \(M_{n}\) satisfies

\[ \begin{aligned} P(M_{n} \le \theta - \varepsilon) &= \left(1 - \frac{\varepsilon}{\theta}\right)^{n} \\ &\to 0 \end{aligned} \]

so \(M_{n} \to \theta\) in probability: it is consistent.

The first option is wrong because \(\frac{2}{n}\sum Y_{i}\) is twice the mean of i.i.d. U(0, θ) variables, which tends to \(\theta\), so it is consistent. The sample mean has expectation \(\theta(n + 1)/4\), which grows with n. And \(X_{(n)}\) is driven by \(X_{n} \sim U(0, n\theta)\), so it grows without bound.

Answer: B

The asymptotic dispersion matrix is diag(σ², 2σ⁴).

For a normal sample, \(\bar{X}_{n}\) and \(S_{n}^{2}\) are independent, so the off-diagonal terms are 0. Next,

\[ \begin{aligned} Var(\sqrt{n}\,\bar{X}_{n}) &= \sigma^{2} \\ Var(\sqrt{n}\,S_{n}^{2}) &\to 2\sigma^{4} \end{aligned} \]

the second because \((n - 1)S_{n}^{2}/\sigma^{2}\) is \(\chi^{2}_{n-1}\) with variance \(2(n - 1)\), giving \(Var(S_{n}^{2}) = 2\sigma^{4}/(n - 1)\).

\(\sigma^{4}\) misses the factor 2, and the matrices of reciprocals are information-type quantities, not the dispersion of the estimators.

Answer: C

The statistic with an asymptotic N(0, 1) null distribution is \(\sqrt{n}(\bar{X}_{n} - p_{0})/\sqrt{p_{0}(1 - p_{0})}\).

Under \(H_{0}\), \(\bar{X}_{n}\) is the mean of n Bernoulli(\(p_{0}\)) variables, with mean \(p_{0}\) and variance \(p_{0}(1 - p_{0})/n\). By the central limit theorem

\[ \begin{aligned} &\frac{\sqrt{n}(\bar{X}_{n} - p_{0})}{\sqrt{p_{0}(1 - p_{0})}} \\ &\to N(0, 1) \end{aligned} \]

The options without \(\sqrt{n}\) shrink to 0, and the one with n in place of \(\sqrt{n}\) blows up, so neither has a N(0, 1) limit. (Replacing \(p_{0}\) by \(\bar{X}_{n}\) under the root would also work, but only together with the factor \(\sqrt{n}\).)

Answer: A

It follows from the invariance property of MLE.

If T maximises the likelihood \(L(\theta)\) and \(\Psi\) is one-to-one, reparametrise by \(\eta = \Psi(\theta)\). The likelihood in \(\eta\) is \(L(\Psi^{-1}(\eta))\), which is largest when \(\Psi^{-1}(\eta) = T\), that is \(\eta = \Psi(T)\). So the MLE of \(\Psi(\theta)\) is \(\Psi(T)\).

This is an exact property for every sample size. Asymptotic properties, consistency and regularity conditions concern the behaviour of the MLE as n grows and are not what this result rests on.

Answer: D

The statistic that is not sufficient is \(X_{1} - X_{2} + X_{3}\).

For a Bernoulli sample, the joint probability is \(p^{\sum x_{i}}(1 - p)^{3 - \sum x_{i}}\), so \(T = X_{1} + X_{2} + X_{3}\) is sufficient, and so is any statistic from which T can be computed: \((X_{1}, X_{2} + X_{3})\) and \((X_{1} + X_{2}, X_{3})\) both determine T by adding their components.

\(X_{1} - X_{2} + X_{3}\) does not determine T. The samples (1, 1, 0) and (0, 0, 0) both give the value 0, but their totals are 2 and 0, and their probabilities \(p^{2}(1 - p)\) and \((1 - p)^{3}\) have a ratio that depends on p. So the conditional distribution given this statistic depends on p, and it is not sufficient.

Answer: D

The minimal sufficient statistic is \(\sum_{i=1}^{n} X_{i}^{2}\).

With mean and variance both \(\theta\), the joint density is

\[ \begin{aligned} &(2\pi\theta)^{-n/2}\exp\left(-\frac{\sum (x_{i} - \theta)^{2}}{2\theta}\right) \\ &= (2\pi\theta)^{-n/2}\exp\left(-\frac{\sum x_{i}^{2}}{2\theta} + \sum x_{i} - \frac{n\theta}{2}\right) \end{aligned} \]

The term \(e^{\sum x_{i}}\) does not involve \(\theta\), so it factors out, and the data enter the \(\theta\)-dependent part only through \(\sum x_{i}^{2}\). It is a one-parameter exponential family in \(\sum x_{i}^{2}\), which is minimal sufficient.

\(\overline{X}\) and \(\sum (X_{i} - \overline{X})^{2}\) alone are not sufficient. The pair in option C is sufficient (it determines \(\sum X_{i}^{2}\)) but not minimal, since it carries more than is needed.

Answer: C

The maximum likelihood estimate of \(\theta\) is 4.

The likelihood is

\[ L(\theta) = e^{-\sum x_{i}}\,(e^{\theta} - e^{-\theta})^{-4} \]

provided every observation lies in \((-\theta, \theta)\), that is \(\theta \gt \max |x_{i}| = 4\); otherwise it is 0. The denominator \(e^{\theta} - e^{-\theta} = 2\sinh\theta\) increases with \(\theta\), so L decreases in \(\theta\) and is largest at the smallest admissible value. The MLE is \(\max |x_{i}| = |-4| = 4\).

3 is the largest observation, which ignores the negative value \(-4\) that also has to lie inside \((-\theta, \theta)\). 0.5 and 1.5 are not admissible at all, since they exclude several observations.

Answer: C

The statement that is NOT true is \(\mathrm{Var}(T^{*}) \le \mathrm{Var}(U)\) for every unbiased estimator U of \(\theta\) (without completeness).

Rao-Blackwellisation, \(T^{*} = E[T_{1} \mid T]\), always improves on the estimator it starts from: \(T^{*}\) is unbiased and \(\mathrm{Var}(T^{*}) \le \mathrm{Var}(T_{1})\). That makes options B and D true.

Beating every unbiased estimator needs more. By the Lehmann-Scheffe theorem, if T is complete as well as sufficient, \(T^{*}\) is the unique UMVUE, which is option A. Minimal sufficiency alone does not give completeness, and then different unbiased starting points can lead to different conditional expectations, so option C can fail.