Point and Interval Estimation Questions and Answers

Q: Let \(X_{1}, X_{2}, \ldots, X_{2n+1}\) be a random sample from a uniform distribution on the interval (\(\theta - 1\), \(\theta + 1\)). Let\(T_{1} = \overline{X}\) (sample mean on 2n + 1 observations)\(T_{2} = \tilde{X}\) (sample median on 2n + 1 observations)\(T_{3} = \dfrac{T_{1} + T_{2}}{2}\)be three estimators of \(\theta\). Then which of the following statements is correct ?

The correct statement is that \(T_{1}\) is consistent for \(\theta\).\(T_{1} = \overline{X}\) has mean \(\theta\) and variance \(\frac{1}{3(2n + 1)} \to 0\) (the uniform on an interval of length 2 has variance \(\frac{1}{3}\)), so it is consistent.The others fail. The sample median is also consistent for the centre \(\theta\) of a symmetric distribution, so option A is wrong. By symmetry about \(\theta\), the mean, the median and their average are all unbiased, so 'only \(T_{1}\)' is wrong. The sufficient statistic for this family is the pair (smallest, largest observation), and the median is not sufficient.

Q: Suppose \(\{X_{1}, X_{2}, \ldots, X_{n}\}\) is a random sample from an exponential distribution with mean \(\theta\). Then \(100(1 - \alpha)\%\) asymptotic confidence interval for \(\theta\) based on sample mean \(\overline{X}_{n}\) :

The interval is \(\Bigg(\frac{\sqrt{n}\,\overline{X}_{n}}{z_{1-\alpha/2} + \sqrt{n}},\) \(\frac{\sqrt{n}\,\overline{X}_{n}}{z_{\alpha/2} + \sqrt{n}}\Bigg)\), with \(z_{p}\) the lower p-quantile of N(0, 1).For an exponential with mean \(\theta\), the variance is \(\theta^{2}\), so by the central limit theorem \(\frac{\sqrt{n}(\overline{X}_{n} - \theta)}{\theta} \approx N(0, 1)\). Hence, approximately,\[ \begin{aligned} z_{\alpha/2} &\lt \sqrt{n}\left(\frac{\overline{X}_{n}}{\theta} - 1\right) \lt z_{1-\alpha/2} \\ \frac{\sqrt{n}\,\overline{X}_{n}}{z_{1-\alpha/2} + \sqrt{n}} &\lt \theta \lt \frac{\sqrt{n}\,\overline{X}_{n}}{z_{\alpha/2} + \sqrt{n}} \end{aligned} \]with probability about \(1 - \alpha\). Since \(z_{\alpha/2} \lt 0 \lt z_{1-\alpha/2}\), the larger quantile gives the lower limit.Option A has the limits in reverse order (its lower end exceeds its upper end), and option C drops the factor \(\sqrt{n}\) from the numerators. The interval can be determined, so D is wrong.

61.

Let \(X_{1}, X_{2}, \ldots, X_{2n+1}\) be a random sample from a uniform distribution on the interval (\(\theta - 1\), \(\theta + 1\)). Let

\(T_{1} = \overline{X}\) (sample mean on 2n + 1 observations)
\(T_{2} = \tilde{X}\) (sample median on 2n + 1 observations)
\(T_{3} = \dfrac{T_{1} + T_{2}}{2}\)

be three estimators of \(\theta\). Then which of the following statements is correct ?

A.\(T_{1}\) is consistent but \(T_{2}\) is not consistent for \(\theta\)
B.\(T_{1}\) is consistent for \(\theta\)
C.Only \(T_{1}\) is unbiased for \(\theta\)
D.\(T_{2}\) is sufficient statistic for \(\theta\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: B

The correct statement is that \(T_{1}\) is consistent for \(\theta\).

\(T_{1} = \overline{X}\) has mean \(\theta\) and variance \(\frac{1}{3(2n + 1)} \to 0\) (the uniform on an interval of length 2 has variance \(\frac{1}{3}\)), so it is consistent.

The others fail. The sample median is also consistent for the centre \(\theta\) of a symmetric distribution, so option A is wrong. By symmetry about \(\theta\), the mean, the median and their average are all unbiased, so 'only \(T_{1}\)' is wrong. The sufficient statistic for this family is the pair (smallest, largest observation), and the median is not sufficient.

Discuss in Forum

62.

Which of the following distributions belong to the one parameter exponential family ?

A.Uniform \(U(0, \theta)\) distribution
B.Gamma distribution with scale parameter 3 and shape parameter \(\lambda \in \{2, 3, 4, 5, \ldots\}\)
C.\(N(\theta, \theta^{2})\) \(\theta \gt 0\)
D.Laplace distribution with location parameter 0 and scale parameter \(\theta \gt 0\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: D

The Laplace distribution with location 0 and scale \(\theta\) belongs to the one-parameter exponential family, so the answer is option D.

Its density is \(\frac{1}{2\theta}e^{-|x|/\theta}\), which can be written as \(h(x)\exp\{\eta(\theta)T(x) - A(\theta)\}\) with \(T(x) = |x|\), \(\eta(\theta) = -1/\theta\) and \(A(\theta) = \ln 2\theta\), and its support does not depend on \(\theta\).

The uniform \(U(0, \theta)\) fails because its support depends on \(\theta\). \(N(\theta, \theta^{2})\) has one parameter but two natural statistics (x and \(x^{2}\)), a curved exponential family and not a one-parameter one. The Gamma option is the one that comes closest: with the scale known it has the form \(\exp\{(\lambda - 1)\ln x - \ldots\}\), but the shape is restricted to the integers 2, 3, 4, ..., and the standard definition of a one-parameter family asks for a parameter space that contains an interval. On a looser definition that allows any parameter set the Gamma option would also qualify; the key chooses D.

Discuss in Forum

63.

Suppose \(\{X_{1}, X_{2}, \ldots, X_{n}\}\) is a random sample from an exponential distribution with mean \(\theta\). Then \(100(1 - \alpha)\%\) asymptotic confidence interval for \(\theta\) based on sample mean \(\overline{X}_{n}\) :

A.is \(\Bigg(\frac{\sqrt{n}\,\overline{X}_{n}}{z_{\alpha/2} + \sqrt{n}},\) \(\frac{\sqrt{n}\,\overline{X}_{n}}{z_{1-\alpha/2} + \sqrt{n}}\Bigg)\)
B.is \(\Bigg(\frac{\sqrt{n}\,\overline{X}_{n}}{z_{1-\alpha/2} + \sqrt{n}},\) \(\frac{\sqrt{n}\,\overline{X}_{n}}{z_{\alpha/2} + \sqrt{n}}\Bigg)\)
C.is \(\Bigg(\frac{\overline{X}_{n}}{z_{1-\alpha/2} + \sqrt{n}},\) \(\frac{\overline{X}_{n}}{z_{\alpha/2} + \sqrt{n}}\Bigg)\)
D.Cannot be determined

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: B

The interval is \(\Bigg(\frac{\sqrt{n}\,\overline{X}_{n}}{z_{1-\alpha/2} + \sqrt{n}},\) \(\frac{\sqrt{n}\,\overline{X}_{n}}{z_{\alpha/2} + \sqrt{n}}\Bigg)\), with \(z_{p}\) the lower p-quantile of N(0, 1).

For an exponential with mean \(\theta\), the variance is \(\theta^{2}\), so by the central limit theorem \(\frac{\sqrt{n}(\overline{X}_{n} - \theta)}{\theta} \approx N(0, 1)\). Hence, approximately,

\[ \begin{aligned} z_{\alpha/2} &\lt \sqrt{n}\left(\frac{\overline{X}_{n}}{\theta} - 1\right) \lt z_{1-\alpha/2} \\ \frac{\sqrt{n}\,\overline{X}_{n}}{z_{1-\alpha/2} + \sqrt{n}} &\lt \theta \lt \frac{\sqrt{n}\,\overline{X}_{n}}{z_{\alpha/2} + \sqrt{n}} \end{aligned} \]

with probability about \(1 - \alpha\). Since \(z_{\alpha/2} \lt 0 \lt z_{1-\alpha/2}\), the larger quantile gives the lower limit.

Option A has the limits in reverse order (its lower end exceeds its upper end), and option C drops the factor \(\sqrt{n}\) from the numerators. The interval can be determined, so D is wrong.

Point and Interval Estimation Questions and Answers

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