Point and Interval Estimation Questions and Answers

Q: Suppose that \(X_{1}\) and \(X_{2}\) are independent random variables having Bernoulli distribution with mean \(p\) and \(2p\) respectively, where \(0 \lt p \lt 1/2\). The following are two statements :(I) \((X_{1}, X_{2})\) is minimal sufficient for \(p\)(II) \(2X_{1} + 5X_{2}\) is minimal sufficient for \(p\)Which of the following is true ?

Both (I) and (II) are true.The joint pmf is \(p^{x_{1}}(1 - p)^{1 - x_{1}}(2p)^{x_{2}}(1 - 2p)^{1 - x_{2}}\). The four sample points give likelihoods \((1 - p)(1 - 2p)\), \(p(1 - 2p)\), \(2p(1 - p)\) and \(2p^{2}\). No two of these have a ratio free of \(p\); for instance \(p(1 - 2p)/[2p(1 - p)] = (1 - 2p)/(2(1 - p))\) depends on \(p\). So no two points can be merged, and the full sample \((X_{1}, X_{2})\) is minimal sufficient.The statistic \(2X_{1} + 5X_{2}\) takes the values 0, 2, 5 and 7 on the four points, one value per point, so it is a one-to-one function of \((X_{1}, X_{2})\) and is minimal sufficient too.The trap is to assume a minimal sufficient statistic must reduce the data. Here the sum \(X_{1} + X_{2}\) is not even sufficient, because it merges (1, 0) and (0, 1), whose likelihoods have different shapes. So the options denying (I) or (II) are wrong.

Q: Which of the following distributions does not belong to the Cramer family ?

The distribution outside the Cramer family is the exponential distribution with location parameter \(\theta\) and scale 1.The Cramer (regular) family requires, among other things, that the support of the density does not depend on the parameter, so that differentiation and integration can be interchanged and the Cramer-Rao bound applies. The density \(e^{-(x - \theta)}\) for \(x \ge \theta\) has support \([\theta, \infty)\), which moves with \(\theta\). This is the classic non-regular case, where the MLE \(X_{(1)}\) converges at rate \(n\), faster than the regular \(\sqrt{n}\).The Cauchy location family, \(N(\theta, \theta)\) and the Laplace scale family all have fixed support (the whole line) and densities smooth in \(\theta\), so they satisfy the regularity conditions.

Q: Consider the following statements :(I) The Fisher information \(I(\theta)\) associated with a logistic distribution with location parameter \(\theta\) and scale parameter 1 is 1/3.(II) The Fisher information \(I(\theta)\) associated with a Cauchy distribution with location parameter \(\theta\) and scale parameter 1 is 1/2.

Both (I) and (II) are true.For a location family \(f(x - \theta)\), the Fisher information is \(I = \int (f'/f)^{2} f\, dx\), the same for every \(\theta\).(I) For the standard logistic, \(f'/f = -(1 - 2F(x))\), and \(1 - 2F\) is uniform on (-1, 1) when X is logistic. So \(I = E[(1 - 2F)^{2}] = \int_{-1}^{1} \frac{u^{2}}{2}\, du = \frac{1}{3}\).(II) For the standard Cauchy, \(f'/f = -2x/(1 + x^{2})\), and\[ \begin{aligned} I &= \int_{-\infty}^{\infty} \frac{4x^{2}}{(1 + x^{2})^{2}} \cdot \frac{dx}{\pi(1 + x^{2})} \\ &= \frac{1}{2} \end{aligned} \]Both values are standard results, so the options denying either statement are wrong.

Q: Suppose that X1, X2, ........, Xn are i.i.d. random variables such that X1 has uniform distribution over (0, \(\theta\)), \(\theta \gt 0\). Then, which of the following statements is true ?(I) The conditional density \(f_{y}(\cdot)\) of \(X_{(1)} \mid X_{(n)} = y\) is given by :\[ f_{y}(x) = \begin{cases} (n-1) \cdot \frac{1}{y}\left(1 - \frac{x}{y}\right)^{n-2}, & \text{if } 0 \lt x \lt y \\ 0, & \text{otherwise} \end{cases} \](II) The conditional distribution of \(X_{(1)} \mid X_{(n)} = y\) does not depend on \(\theta\).

Both (I) and (II) are true.Given the maximum \(X_{(n)} = y\), the remaining n − 1 observations are independent and uniform on (0, y). The minimum of n − 1 such variables has\[ \begin{aligned} &P(X_{(1)} \gt x) \\ &= \left(1 - \frac{x}{y}\right)^{n-1},\quad 0 \lt x \lt y \end{aligned} \]Differentiating, its density is \((n - 1)\frac{1}{y}\left(1 - \frac{x}{y}\right)^{n-2}\), which is the density in (I).This density involves only y, not \(\theta\), which is (II). It reflects the fact that \(X_{(n)}\) is sufficient for \(\theta\), so the conditional distribution of any other statistic given it is free of \(\theta\).

Q: A random sample X1, X2, ....., Xn is drawn from a Poisson distribution with parameter \(\lambda\). Which one of the following is an unbiased estimator of \(\lambda^{2}\) ?

The unbiased estimator is (1/n) Σ Xi(Xi − 1).For a Poisson variable, the second factorial moment is\[ \begin{aligned} E[X(X - 1)] &= E(X^{2}) - E(X) \\ &= (\lambda + \lambda^{2}) - \lambda \\ &= \lambda^{2} \end{aligned} \]so the average of \(X_{i}(X_{i} - 1)\) has expectation \(\lambda^{2}\).The others are biased. \(E(\bar{X}^{2}) = \lambda^{2} + \lambda/n\), \(E(X_{i}^{2}) = \lambda^{2} + \lambda\), and \(E[X_{i}(X_{i} + 1)] = \lambda^{2} + 2\lambda\).

Q: Cramer-Rao lower bound of variance for estimating the parameter \(\theta\) of the distribution with pdf :\[ f(x,\theta)=\frac{1}{\pi}\,\frac{1}{1+(x-\theta)^{2}} \]where \(-\infty \lt x \lt \infty\) is :

The Cramer-Rao lower bound is 2/n.For the Cauchy location family, \(\log f = -\log\pi - \log(1 + (x - \theta)^{2})\), and the Fisher information in one observation is\[ \begin{aligned} I(\theta) &= E\left[\frac{4(X - \theta)^{2}}{(1 + (X - \theta)^{2})^{2}}\right] \\ &= \frac{1}{2} \end{aligned} \](the integral \(\frac{4}{\pi}\int \frac{u^{2}}{(1 + u^{2})^{3}}du\) equals \(\frac{1}{2}\)). For a sample of n, the information is \(n/2\), so the bound for an unbiased estimator of \(\theta\) is \(1/(n/2) = 2/n\).1/n would correspond to information 1 per observation, and 1/(2n) inverts the factor 2. \(2/n^{2}\) has the wrong power of n.

41.

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be i.i.d. Bernoulli \((p)\) random variables and \(S = \sum_{i=1}^{n} X_{i}\). Then unbiased estimator of \(p^{3}\) is :

A.\(\left(\frac{S}{n}\right)^{3}\)
B.\(\frac{S(S - 1)(S - 2)}{n(n - 1)(n - 2)}\)
C.\(\frac{(S - 1)(S - 2)(S - 3)}{(n - 1)(n - 2)(n - 3)}\)
D.None of the above

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: B

The unbiased estimator is \(\dfrac{S(S - 1)(S - 2)}{n(n - 1)(n - 2)}\).

\(S\) is binomial \((n, p)\), and its factorial moments are \(E[S(S - 1)(S - 2)] = n(n - 1)(n - 2)p^{3}\). Dividing by the constant gives an estimator with expectation \(p^{3}\). It is also the probability that three specified trials all succeed, \(X_{1}X_{2}X_{3}\), averaged over all triples, so it is the UMVUE by the Lehmann-Scheffe theorem.

\((S/n)^{3}\) is biased: \(E(S^{3})/n^{3}\) contains lower-order terms in \(p\) besides \(p^{3}\). The shifted form \((S - 1)(S - 2)(S - 3)/[(n - 1)(n - 2)(n - 3)]\) has the wrong factorial moment. Since an unbiased estimator exists, 'None of the above' is wrong.

Discuss in Forum

42.

Suppose \(\{X_{1}, X_{2}, \ldots, X_{n}\}\), \(n \ge 2\) is a random sample from Poisson distribution with mean \(\theta\). If \(T = \sum_{i=1}^{n} X_{i}\), then the MVUE of \(e^{-\theta}\) is :

A.\(\left(\frac{n}{n - 1}\right)^{T}\)
B.\(\left(\frac{n - 1}{n}\right)^{T}\)
C.\(\frac{T}{n - T}\)
D.\(\frac{n - T}{T}\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: B

The MVUE of \(e^{-\theta}\) is \(\left(\frac{n - 1}{n}\right)^{T}\).

\(e^{-\theta} = P(X_{1} = 0)\), so the indicator \(I(X_{1} = 0)\) is unbiased. T is complete and sufficient, so by Rao-Blackwell and Lehmann-Scheffe the MVUE is \(E[I(X_{1} = 0) \mid T]\). Given \(T = t\), \(X_{1}\) is binomial \((t, 1/n)\), so

\[ \begin{aligned} P(X_{1} = 0 \mid T = t) &= \left(1 - \frac{1}{n}\right)^{t} \\ &= \left(\frac{n - 1}{n}\right)^{t} \end{aligned} \]

Check of unbiasedness: \(E[s^{T}] = e^{n\theta(s - 1)}\) with \(s = (n - 1)/n\) gives \(e^{-\theta}\).

\((n/(n - 1))^{T}\) has expectation \(e^{n\theta/(n - 1)}\), and the ratio forms involving \(T/(n - T)\) are not unbiased and are undefined for some values of T.

Discuss in Forum

43.

Suppose that \(X_{1}\) and \(X_{2}\) are independent random variables having Bernoulli distribution with mean \(p\) and \(2p\) respectively, where \(0 \lt p \lt 1/2\). The following are two statements :

(I) \((X_{1}, X_{2})\) is minimal sufficient for \(p\)
(II) \(2X_{1} + 5X_{2}\) is minimal sufficient for \(p\)

Which of the following is true ?

A.Only (I) is true
B.Only (II) is true
C.Both (I) and (II) are true
D.Neither (I) nor (II) is true

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: C

Both (I) and (II) are true.

The joint pmf is \(p^{x_{1}}(1 - p)^{1 - x_{1}}(2p)^{x_{2}}(1 - 2p)^{1 - x_{2}}\). The four sample points give likelihoods \((1 - p)(1 - 2p)\), \(p(1 - 2p)\), \(2p(1 - p)\) and \(2p^{2}\). No two of these have a ratio free of \(p\); for instance \(p(1 - 2p)/[2p(1 - p)] = (1 - 2p)/(2(1 - p))\) depends on \(p\). So no two points can be merged, and the full sample \((X_{1}, X_{2})\) is minimal sufficient.

The statistic \(2X_{1} + 5X_{2}\) takes the values 0, 2, 5 and 7 on the four points, one value per point, so it is a one-to-one function of \((X_{1}, X_{2})\) and is minimal sufficient too.

The trap is to assume a minimal sufficient statistic must reduce the data. Here the sum \(X_{1} + X_{2}\) is not even sufficient, because it merges (1, 0) and (0, 1), whose likelihoods have different shapes. So the options denying (I) or (II) are wrong.

Discuss in Forum

44.

Suppose \(\{X_{1}, X_{2}, \ldots, X_{n}\}\) is a random sample from \(N(\theta, 1)\) distribution. Which of the following is an MVUE of \(\theta^{2}\) ?

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

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: D

The MVUE of \(\theta^{2}\) is \(\overline{X}_{n}^{2} - \frac{1}{n}\).

\(\overline{X}_{n} \sim N(\theta, 1/n)\), so

\[ \begin{aligned} E(\overline{X}_{n}^{2}) &= \text{Var}(\overline{X}_{n}) + (E\overline{X}_{n})^{2} \\ &= \frac{1}{n} + \theta^{2} \end{aligned} \]

Subtracting \(1/n\) removes the bias. \(\overline{X}_{n}\) is complete and sufficient, and \(\overline{X}_{n}^{2} - 1/n\) is a function of it, so by the Lehmann-Scheffe theorem it is the minimum variance unbiased estimator.

\(\overline{X}_{n}^{2}\) is biased by \(1/n\). \(\overline{X}_{n}^{2} - 1\) over-corrects (it subtracts the variance of one observation, not of the mean), and \((\sum X_{i})^{2}/n = n\overline{X}_{n}^{2}\) has expectation \(1 + n\theta^{2}\).

Discuss in Forum

45.

Which of the following distributions does not belong to the Cramer family ?

A.Exponential distribution with location parameter \(\theta \neq 0\) and scale parameter 1
B.Cauchy distribution with location parameter \(\theta\) and scale parameter 1
C.\(N(\theta, \theta)\ \theta \gt 0\)
D.Laplace distribution with location parameter 0 and scale parameter \(\theta\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: A

The distribution outside the Cramer family is the exponential distribution with location parameter \(\theta\) and scale 1.

The Cramer (regular) family requires, among other things, that the support of the density does not depend on the parameter, so that differentiation and integration can be interchanged and the Cramer-Rao bound applies. The density \(e^{-(x - \theta)}\) for \(x \ge \theta\) has support \([\theta, \infty)\), which moves with \(\theta\). This is the classic non-regular case, where the MLE \(X_{(1)}\) converges at rate \(n\), faster than the regular \(\sqrt{n}\).

The Cauchy location family, \(N(\theta, \theta)\) and the Laplace scale family all have fixed support (the whole line) and densities smooth in \(\theta\), so they satisfy the regularity conditions.

Discuss in Forum

46.

Suppose \(\{X_{1}, X_{2}, \ldots, X_{n}\}\) is a random sample from a Laplace distribution with p.d.f. \(f(x, \theta)\) given by :

\[ \begin{aligned} f(x, \theta) &= \frac{1}{2\theta} \\ &\quad \exp\left(-\frac{|x|}{\theta}\right), \end{aligned} \]

\(x \in R\), \(\theta \gt 0\).

Which of the following estimators is consistent for \(\theta\) ?

A.\(\sum_{i=1}^{n} \frac{|X_{i}|}{n}\)
B.\(\left(\frac{1}{n} \sum_{i=1}^{n} X_{i}^{2}\right)^{1/2}\)
C.Sample mean
D.Sample median

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: A

The consistent estimator is \(\sum_{i=1}^{n} \frac{|X_{i}|}{n}\).

For this Laplace density, \(|X|\) is exponential with mean \(\theta\): \(E|X| = \int_{-\infty}^{\infty} \frac{|x|}{2\theta} e^{-|x|/\theta}\, dx = \theta\). By the weak law of large numbers the sample mean of the \(|X_{i}|\) converges in probability to \(\theta\). It is in fact the MLE of \(\theta\).

The others converge to the wrong value. \(E X^{2} = 2\theta^{2}\), so \(\left(\frac{1}{n}\sum X_{i}^{2}\right)^{1/2} \to \sqrt{2}\,\theta\). The distribution is symmetric about 0, so the sample mean and the sample median both converge to 0, not \(\theta\).

Discuss in Forum

47.

Consider the following statements :

(I) The Fisher information \(I(\theta)\) associated with a logistic distribution with location parameter \(\theta\) and scale parameter 1 is 1/3.
(II) The Fisher information \(I(\theta)\) associated with a Cauchy distribution with location parameter \(\theta\) and scale parameter 1 is 1/2.

A.Only (I) is true
B.Only (II) is true
C.Both (I) and (II) are true
D.Both (I) and (II) are not true

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: C

Both (I) and (II) are true.

For a location family \(f(x - \theta)\), the Fisher information is \(I = \int (f'/f)^{2} f\, dx\), the same for every \(\theta\).

(I) For the standard logistic, \(f'/f = -(1 - 2F(x))\), and \(1 - 2F\) is uniform on (-1, 1) when X is logistic. So \(I = E[(1 - 2F)^{2}] = \int_{-1}^{1} \frac{u^{2}}{2}\, du = \frac{1}{3}\).

(II) For the standard Cauchy, \(f'/f = -2x/(1 + x^{2})\), and

\[ \begin{aligned} I &= \int_{-\infty}^{\infty} \frac{4x^{2}}{(1 + x^{2})^{2}} \cdot \frac{dx}{\pi(1 + x^{2})} \\ &= \frac{1}{2} \end{aligned} \]

Both values are standard results, so the options denying either statement are wrong.

Discuss in Forum

48.

Suppose that X₁, X₂, ........, X_n are i.i.d. random variables such that X₁ has uniform distribution over (0, \(\theta\)), \(\theta \gt 0\). Then, which of the following statements is true ?

(I) The conditional density \(f_{y}(\cdot)\) of \(X_{(1)} \mid X_{(n)} = y\) is given by :

\[ f_{y}(x) = \begin{cases} (n-1) \cdot \frac{1}{y}\left(1 - \frac{x}{y}\right)^{n-2}, & \text{if } 0 \lt x \lt y \\ 0, & \text{otherwise} \end{cases} \]

(II) The conditional distribution of \(X_{(1)} \mid X_{(n)} = y\) does not depend on \(\theta\).

A.Neither (I) nor (II) is true
B.Both (I) and (II) are true
C.Only (I) is true
D.Only (II) is true

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: B

Both (I) and (II) are true.

Given the maximum \(X_{(n)} = y\), the remaining n − 1 observations are independent and uniform on (0, y). The minimum of n − 1 such variables has

\[ \begin{aligned} &P(X_{(1)} \gt x) \\ &= \left(1 - \frac{x}{y}\right)^{n-1},\quad 0 \lt x \lt y \end{aligned} \]

Differentiating, its density is \((n - 1)\frac{1}{y}\left(1 - \frac{x}{y}\right)^{n-2}\), which is the density in (I).

This density involves only y, not \(\theta\), which is (II). It reflects the fact that \(X_{(n)}\) is sufficient for \(\theta\), so the conditional distribution of any other statistic given it is free of \(\theta\).

Discuss in Forum

49.

A random sample X₁, X₂, ....., X_n is drawn from a Poisson distribution with parameter \(\lambda\). Which one of the following is an unbiased estimator of \(\lambda^{2}\) ?

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

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: C

The unbiased estimator is (1/n) Σ X_i(X_i − 1).

For a Poisson variable, the second factorial moment is

\[ \begin{aligned} E[X(X - 1)] &= E(X^{2}) - E(X) \\ &= (\lambda + \lambda^{2}) - \lambda \\ &= \lambda^{2} \end{aligned} \]

so the average of \(X_{i}(X_{i} - 1)\) has expectation \(\lambda^{2}\).

The others are biased. \(E(\bar{X}^{2}) = \lambda^{2} + \lambda/n\), \(E(X_{i}^{2}) = \lambda^{2} + \lambda\), and \(E[X_{i}(X_{i} + 1)] = \lambda^{2} + 2\lambda\).

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50.

Cramer-Rao lower bound of variance for estimating the parameter \(\theta\) of the distribution with pdf :

\[ f(x,\theta)=\frac{1}{\pi}\,\frac{1}{1+(x-\theta)^{2}} \]

where \(-\infty \lt x \lt \infty\) is :

A.\(\dfrac{2}{n}\)
B.\(\dfrac{1}{n}\)
C.\(\dfrac{2}{n^{2}}\)
D.\(\dfrac{1}{2n}\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: A

The Cramer-Rao lower bound is 2/n.

For the Cauchy location family, \(\log f = -\log\pi - \log(1 + (x - \theta)^{2})\), and the Fisher information in one observation is

\[ \begin{aligned} I(\theta) &= E\left[\frac{4(X - \theta)^{2}}{(1 + (X - \theta)^{2})^{2}}\right] \\ &= \frac{1}{2} \end{aligned} \]

(the integral \(\frac{4}{\pi}\int \frac{u^{2}}{(1 + u^{2})^{3}}du\) equals \(\frac{1}{2}\)). For a sample of n, the information is \(n/2\), so the bound for an unbiased estimator of \(\theta\) is \(1/(n/2) = 2/n\).

1/n would correspond to information 1 per observation, and 1/(2n) inverts the factor 2. \(2/n^{2}\) has the wrong power of n.

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