Point and Interval Estimation Questions and Answers

Answer: A

The statement that is not true is \(S_{n}\) is unbiased for \(\sigma\).

\(S_{n}^{2}\) with divisor \(n - 1\) is unbiased for \(\sigma^{2}\), but the square root is strictly concave, so by Jensen's inequality

\[ E(S_{n}) = E\sqrt{S_{n}^{2}} \lt \sqrt{E S_{n}^{2}} = \sigma \]

unless \(S_{n}\) is constant. So \(S_{n}\) underestimates \(\sigma\) on average.

The other three are true: B is the reason for the \(n - 1\) divisor, \(S_{n}^{2} \to \sigma^{2}\) in probability by the law of large numbers, and \(S_{n} \to \sigma\) then follows because the square root is continuous. Unbiasedness is not preserved by non-linear maps, but consistency is.

Answer: C

The Cramer-Rao bound is a lower bound for \(V_{\theta}(T)\) among all unbiased estimators T of \(\tau(\theta)\).

Under the regularity conditions, every unbiased estimator T of \(\tau(\theta)\) satisfies

\[ V_{\theta}(T) \ge \frac{[\tau'(\theta)]^{2}}{nI(\theta)} \]

where \(I(\theta)\) is the Fisher information per observation. The derivation uses \(E_{\theta}(T) = \tau(\theta)\), so the bound applies to unbiased estimators, not to arbitrary ones: a constant estimator has variance 0, below any positive bound. It bounds the variance from below, not above.

Answer: C

The complete sufficient statistic is \((T_{2}, T_{3}, T_{1})\).

The joint density of the two samples is an exponential family whose exponent is

\[ \begin{aligned} &\frac{\mu_{1}}{\sigma^{2}}\sum X_{i} + \frac{\mu_{2}}{\sigma^{2}}\sum Y_{j} \\ &- \frac{1}{2\sigma^{2}}\left(\sum X_{i}^{2} + \sum Y_{j}^{2}\right) \end{aligned} \]

with three natural parameters, and the parameter space contains an open set in \(\mathbf{R}^{3}\). So the statistic \((\sum X_{i}, \sum Y_{j}, \sum X_{i}^{2} + \sum Y_{j}^{2})\), which is \((T_{2}, T_{3}, T_{1})\), is complete and sufficient.

Because the variance is common, the squares enter only through their sum \(T_{1}\). \((T_{2}, T_{3}, T_{4}, T_{5})\) is sufficient but not complete (four statistics for three parameters), and the other options leave out a needed sum.

Answer: A

The null distribution is F if X and Y are independent and have normal distributions.

For normal samples, \((m - 1)S_{X}^{2}/\sigma_{1}^{2}\) and \((n - 1)S_{Y}^{2}/\sigma_{2}^{2}\) are chi-square variables. If the samples are independent, their ratio, each divided by its degrees of freedom, has an F distribution; under \(H_{0} : \sigma_{1}^{2} = \sigma_{2}^{2}\) the variances cancel and \(S_{X}^{2}/S_{Y}^{2} \sim F_{m-1, n-1}\).

Both conditions are needed: normality gives the chi-square laws, and independence makes the ratio F. Equal means are not needed. (The printed alternative repeats '=' where '\(\ne\)' is meant; it does not affect the null distribution.)

Answer: B

The correct statement is the distribution of Q is known and free from the unknown parameters.

A pivotal quantity is a function of the data and the parameter whose distribution does not depend on any unknown parameter. That is what lets us find constants with \(P(a \le Q \le b) = 1 - \alpha\) and then invert the inequality for the parameter. For example, \((\bar{X} - \mu)/(S/\sqrt{n})\) has a t distribution whatever \(\mu\) and \(\sigma\) are.

If its distribution depended on the parameters, or were unknown, no such constants could be found. It need not be normal: t, chi-square and F pivots are common.

Answer: B

The minimal sufficient statistic is \(\max\{|X_{1}|, |X_{2}|/2, \ldots, |X_{n}|/n\}\).

The joint density is

\[ \begin{aligned} &\prod_{i} \frac{1}{2i\theta}1_{\{|x_{i}| \lt i\theta\}} \\ &= \frac{1}{(2\theta)^{n} n!}1_{\{\max_{i} |x_{i}|/i \lt \theta\}} \end{aligned} \]

which depends on the data only through \(T = \max_{i}|X_{i}|/i\). By factorisation T is sufficient, and since the likelihood as a function of \(\theta\) is determined exactly by T, it is minimal.

The order statistics in (A) are sufficient but not minimal (and ignore that the \(X_{i}\) have different scales). Options C and D drop the absolute values (as printed) or take the minimum, which loses the information that matters.

Answer: D

The expectation is equal to zero.

Differentiate the identity \(\int f(x, \theta)\,dx = 1\) under the integral sign:

\[ \begin{aligned} 0 &= \int \frac{\partial f}{\partial\theta}\,dx = \int \frac{\partial \log f}{\partial\theta}f\,dx \\ &= E\left(\frac{d}{d\theta}\log f(X, \theta)\right) \end{aligned} \]

So the score has mean 0. The Fisher information is the variance of the score, \(E[(d\log f/d\theta)^{2}]\), not its mean. 'Less than zero' and 'less than 1' are not the exact value.

Answer: C

The consistent estimator is \(2\bar{X}_{n}\).

For \(U(0, \theta)\), \(E(X) = \theta/2\). By the law of large numbers \(\bar{X}_{n} \to \theta/2\) in probability, so \(2\bar{X}_{n} \to \theta\): it is consistent (and unbiased, with variance \(\theta^{2}/(3n) \to 0\)).

The others converge to the wrong value: \(\bar{X}_{n}\) and the sample median both tend to \(\theta/2\), and the minimum \(X_{(1)}\) tends to 0. (The maximum \(X_{(n)}\), not offered, would also be consistent for \(\theta\).)

Answer: C

The limit is \(z \sim N(0, 16\theta^{2}/27)\).

\(X([n/4] + 1)\) is the sample lower quartile. Its population counterpart \(\xi\) solves \(F(\xi) = 1 - \theta/\xi = 1/4\), so \(\xi = 4\theta/3\). A sample \(p\)-quantile is asymptotically normal with variance \(p(1 - p)/f(\xi)^{2}\):

\[ \begin{aligned} f(\xi) &= \frac{\theta}{\xi^{2}} = \frac{9}{16\theta} \\ \frac{(1/4)(3/4)}{f(\xi)^{2}} &= \frac{3}{16}\cdot\frac{256\theta^{2}}{81} = \frac{16\theta^{2}}{27} \end{aligned} \]

The variance must scale with \(\theta^{2}\), since \(\theta\) is a scale parameter, which rules out the forms in \(\theta\) (two of which are printed identically).

Answer: B

\(z\) is distributed as \(N(0, 1/\alpha)\).

X is gamma with shape \(\alpha\) and scale \(\theta\), so \(E(X) = \alpha\theta\) and \(\mathrm{Var}(X) = \alpha\theta^{2}\). By the central limit theorem \(\sqrt{n}(\bar{X} - \alpha\theta) \to N(0, \alpha\theta^{2})\). Apply the delta method with \(g(u) = \log u\), \(g'(\alpha\theta) = 1/(\alpha\theta)\):

\[ \begin{aligned} \mathrm{Var}(z) &= \frac{\alpha\theta^{2}}{(\alpha\theta)^{2}} \\ &= \frac{1}{\alpha} \end{aligned} \]

The log removes the scale \(\theta\) entirely, which is why the answer cannot involve \(\theta\) as in (C) and (D), and it is 1 only when \(\alpha = 1\).