Point and Interval Estimation Questions and Answers

Q: The following are some statements :(I) If T is a boundedly complete sufficient statistic for \(\theta\) and S is an ancillary statistic for \(\theta\), then T and S are independent.(II) An ancillary statistic may or may not be independent of a sufficient statistic.(III) It is possible to have variance smaller than that of the Cramer-Rao lower bound at some points of the parameter space.(IV) The parametric function need not be differentiable for the Cramer-Rao lower bound to hold.Which of the following options is true ?

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).

Q: Suppose {X1, X2, ......., Xn} are independent random variables, where Xi follows uniform U(0, \(i\theta\)) distribution for i = 1, 2, ......., n. Which of the following statements is true ?

The consistent estimator is max{X1, X2/2, ..., Xn/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.

Q: Suppose {X1, X2, ......., Xn} is a random sample from \(N(\mu, \sigma^{2})\) distribution. Then the approximate dispersion matrix of\[ \sqrt{n}\big((\bar{X}_{n},S_{n}^{2})'-(\mu,\sigma^{2})'\big) \]is :

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.

Q: If T is MLE of \(\theta\) and \(\Psi(\theta)\) is one-to-one function of \(\theta\), then \(\Psi(T)\) is MLE of \(\Psi(\theta)\). This follows from :

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.

Q: Let \((X_{1}, X_{2}, X_{3})\) be a random sample from B(1, p). Which one of the following statistics is not sufficient for p ?

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.

Q: Suppose that \(X_{1}, X_{2}, \ldots, X_{n}\) are i.i.d. random variables such that \(X_{1}\) has normal distribution with mean \(\theta\) and variance \(\theta \gt 0\). Then, which of the following is a minimal sufficient statistic for \(\theta\) ?

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.

Q: Suppose that \(x_{1} = -2\), \(x_{2} = 1\), \(x_{3} = 3\), \(x_{4} = -4\) be the observed values of a random sample from the c.d.f. with density,\[ \begin{aligned} f(x, \theta) &= \frac{e^{-x}}{e^{\theta} - e^{-\theta}}, \\ &-\theta \lt x \lt \theta, \; \theta \gt 0. \end{aligned} \]Then, the maximum likelihood estimator of \(\theta\) is :

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.

Q: If \(T_{1}\) is an unbiased estimator of \(\theta\) and T is a minimal sufficient statistic for \(\theta\), then which of the following statements is NOT true ? Suppose \(T^{*} = E[T_{1} \mid T]\).

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.

51.

The following are some statements :

(I) If T is a boundedly complete sufficient statistic for \(\theta\) and S is an ancillary statistic for \(\theta\), then T and S are independent.
(II) An ancillary statistic may or may not be independent of a sufficient statistic.
(III) It is possible to have variance smaller than that of the Cramer-Rao lower bound at some points of the parameter space.
(IV) The parametric function need not be differentiable for the Cramer-Rao lower bound to hold.

Which of the following options is true ?

A.(I), (II) and (III) only
B.(II), (III) and (IV) only
C.(I), (III) and (IV) only
D.(I), (II) and (IV) only

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

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).

Discuss in Forum

52.

Which of the following probability laws is not a member of an exponential family ?

A.\(f(x, \theta) = \theta(1 - x)^{\theta - 1}\), \(0 \lt x \lt 1\), \(\theta \gt 0\)
B.\(f(x, \theta) = \theta^{x}(1 - \theta)^{1 - x}\), \(x = 0, 1\), \(0 \lt \theta \lt 1\)
C.\(N(\theta, \theta^{2})\), \(\theta \in \mathbf{R}\)
D.\(N(\theta, \theta)\), \(\theta \gt 0\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

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 \(-x^{2}/(2\theta)\).

Discuss in Forum

53.

Suppose {X₁, X₂, ......., X_n} are independent random variables, where X_i follows uniform U(0, \(i\theta\)) distribution for i = 1, 2, ......., n. Which of the following statements is true ?

A.\(\dfrac{2}{n}\sum_{i=1}^{n} \dfrac{X_{i}}{i}\) is not consistent for \(\theta\)
B.Sample mean is consistent for \(\theta\)
C.\(\max\left\{X_{1},\frac{X_{2}}{2},\frac{X_{3}}{3},\ldots,\frac{X_{n}}{n}\right\}\) is consistent for \(\theta\)
D.\(X_{(n)}\) is consistent for \(\theta\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

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.

Discuss in Forum

54.

Suppose {X₁, X₂, ......., X_n} is a random sample from \(N(\mu, \sigma^{2})\) distribution. Then the approximate dispersion matrix of

\[ \sqrt{n}\big((\bar{X}_{n},S_{n}^{2})'-(\mu,\sigma^{2})'\big) \]

is :

A.\(\begin{pmatrix}\sigma^{2}&0\\0&\sigma^{4}\end{pmatrix}\)
B.\(\begin{pmatrix}\sigma^{2}&0\\0&2\sigma^{4}\end{pmatrix}\)
C.\(\begin{pmatrix}\frac{1}{\sigma^{2}}&0\\0&\frac{1}{\sigma^{4}}\end{pmatrix}\)
D.\(\begin{pmatrix}\frac{1}{\sigma^{2}}&0\\0&\frac{1}{2\sigma^{4}}\end{pmatrix}\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

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.

Discuss in Forum

55.

Suppose X ~ B (1, p) distribution, \(0 \lt p \lt 1\). On the basis of a random sample of size n from the distribution of X, we want to test \(H_{0} : p = p_{0}\) against the alternative \(H_{1} : p \neq p_{0}\), where \(p_{0}\) is a specified constant. Then asymptotic null distribution of which of the following test statistics is N(0, 1) ?

A.\(\dfrac{n(\bar{X}_{n} - p_{0})}{\sqrt{\bar{X}_{n}(1 - \bar{X}_{n})}}\)
B.\(\dfrac{\bar{X}_{n} - p_{0}}{\sqrt{\bar{X}_{n}(1 - \bar{X}_{n})}}\)
C.\(\dfrac{\sqrt{n}(\bar{X}_{n} - p_{0})}{\sqrt{p_{0}(1 - p_{0})}}\)
D.\(\dfrac{\bar{X}_{n} - p_{0}}{\sqrt{p_{0}(1 - p_{0})}}\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

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 \(\sqrt{n}\).)

Discuss in Forum

56.

If T is MLE of \(\theta\) and \(\Psi(\theta)\) is one-to-one function of \(\theta\), then \(\Psi(T)\) is MLE of \(\Psi(\theta)\). This follows from :

A.Invariance property of MLE
B.Asymptotic property of MLE
C.Consistency property of MLE
D.Regularity condition of MLE

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

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.

Discuss in Forum

57.

Let \((X_{1}, X_{2}, X_{3})\) be a random sample from B(1, p). Which one of the following statistics is not sufficient for p ?

A.\(X_{1} + X_{2} + X_{3}\)
B.\((X_{1}, X_{2} + X_{3})\)
C.\((X_{1} + X_{2}, X_{3})\)
D.\(X_{1} - X_{2} + X_{3}\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

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.

Discuss in Forum

58.

Suppose that \(X_{1}, X_{2}, \ldots, X_{n}\) are i.i.d. random variables such that \(X_{1}\) has normal distribution with mean \(\theta\) and variance \(\theta \gt 0\). Then, which of the following is a minimal sufficient statistic for \(\theta\) ?

A.\(\overline{X}\)
B.\(\sum_{i=1}^{n} (X_{i} - \overline{X})^{2}\)
C.\(\left(\overline{X}, \sum_{i=1}^{n} (X_{i} - \overline{X})^{2}\right)\)
D.\(\sum_{i=1}^{n} X_{i}^{2}\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

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.

Discuss in Forum

59.

Suppose that \(x_{1} = -2\), \(x_{2} = 1\), \(x_{3} = 3\), \(x_{4} = -4\) be the observed values of a random sample from the c.d.f. with density,

\[ \begin{aligned} f(x, \theta) &= \frac{e^{-x}}{e^{\theta} - e^{-\theta}}, \\ &-\theta \lt x \lt \theta, \; \theta \gt 0. \end{aligned} \]

Then, the maximum likelihood estimator of \(\theta\) is :

A.3
B.0.5
C.4
D.1.5

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

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.

Discuss in Forum

60.

If \(T_{1}\) is an unbiased estimator of \(\theta\) and T is a minimal sufficient statistic for \(\theta\), then which of the following statements is NOT true ? Suppose \(T^{*} = E[T_{1} \mid T]\).

A.\(\mathrm{Var}(T^{*}) \le \mathrm{Var}(U)\) for every unbiased estimator U of \(\theta\) and T a complete sufficient statistic
B.\(T^{*}\) represents the procedure of Rao-Blackwellisation
C.\(\mathrm{Var}(T^{*}) \le \mathrm{Var}(U)\), for every unbiased estimator U of \(\theta\)
D.\(\mathrm{Var}(T^{*}) \le \mathrm{Var}(T_{1})\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

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.

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