Point and Interval Estimation Questions and Answers

Q: Suppose \(\{X_{1}, X_{2}, ......, X_{n}\}\) are independent and identically distributed random variables with mean \(\mu\) and variance \(\sigma^{2}\). Suppose \(S_{n}^{2} = \dfrac{1}{n - 1}\sum_{i=1}^{n} (X_{i} - \bar{X}_{n})^{2}\). Then which of the following statements is not true ?

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.

Q: Let \(\tau(\theta)\) be a parametric function and T be an estimator of \(\tau(\theta)\). Let \(V_{\theta}(T)\) denotes variance of T. The Cramer-Rao bound gives a :

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.

Q: Let \(\{X_{1}, \ldots, X_{n}\}\) be a random sample from \(N(\mu_{1}, \sigma^{2})\) and \(\{Y_{1}, \ldots, Y_{m}\}\) be a random sample from \(N(\mu_{2}, \sigma^{2})\). Let\(T_{1} = \sum_{i=1}^{n} X_{i}^{2} + \sum_{j=1}^{m} Y_{i}^{2}\), \(T_{2} = \sum_{i=1}^{n} X_{i}\),\(T_{3} = \sum_{i=1}^{m} Y_{i}\), \(T_{4} = \sum_{i=1}^{n} X_{i}^{2}\), \(T_{5} = \sum_{i=1}^{m} Y_{i}^{2}\).Then which of the following is a jointly sufficient complete statistic for \((\mu_{1}, \mu_{2}, \sigma^{2})\) :

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.

Q: Suppose \(\{X_{1}, X_{2}, \ldots, X_{m}\}\) is a random sample from the distribution of X with mean \(\mu_{1}\) and variance \(\sigma_{1}^{2}\), \(\{y_{1}, \ldots, y_{n}\}\) is a random sample from the distribution of Y with mean \(\mu_{2}\) and variance \(\sigma_{2}^{2}\). The test statistic to test \(H_{0} : \sigma_{1}^{2} = \sigma_{2}^{2}\) against \(H_{1} : \sigma_{1}^{2} = \sigma_{2}^{2}\) is \(T_{n} = \dfrac{S_{X}^{2}}{S_{Y}^{2}}\), where \(S_{X}^{2}\) and \(S_{Y}^{2}\) are sample variances. Then the null distribution of \(T_{n}\) is :

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.)

Q: In the method of constructing confidence interval for an unknown parameter based on pivotal quantity Q, which of the following statements is correct ?

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.

Q: Suppose \(X_{1}, \ldots, X_{n}\) are independent random variables with probability density function\[ f(x_{i}, \theta) = \begin{cases} \dfrac{1}{2i\theta}, & \text{if } -i\theta \lt x_{i} \lt i\theta \\ 0, & \text{otherwise.} \end{cases} \]Then the minimal sufficient statistic for \(\theta\) is :

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.

Q: Suppose \(f(x, \theta)\) is a probability density function of X for which differentiation under integral sign is permissible.Then \(E\left(\dfrac{d}{d\theta}\log f(X, \theta)\right)\) is :

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.

Q: Suppose \(\{X_{1}, X_{2}, \ldots, X_{n}\}\) is a random sample of size \(n\) from the \(U(0, \theta)\) distribution. Then :

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 \(\theta\).)

Q: Suppose \(\{X_{1}, X_{2}, \ldots, X_{n}\}\) are independent and identically distributed random variables each having the probability density function :\(f(x) = \dfrac{\theta}{x^{2}},\ x \ge \theta\)Then :\[ \begin{aligned} &\sqrt{n}\left(X\left(\left[\tfrac{n}{4}\right] + 1\right) - \tfrac{4\theta}{3}\right) \\ &\xrightarrow{L} z, \end{aligned} \]

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).

Q: Let \(X_{1}, X_{2}, \ldots, X_{n}\) be iid \(f(x, \theta, \alpha)\), where :\[ \begin{aligned} &f(x; \theta, \alpha) \\ &= \frac{e^{-x/\theta}\, x^{\alpha-1}}{\Gamma(\alpha)\, \theta^{\alpha}}. \end{aligned} \]Then, \(\sqrt{n}(\log \bar{X} - \log \alpha\theta) \xrightarrow{d} z\), where \(z\) is distributed as :

\(z\) 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 \(\alpha = 1\).

21.

Suppose \(\{X_{1}, X_{2}, ......, X_{n}\}\) are independent and identically distributed random variables with mean \(\mu\) and variance \(\sigma^{2}\). Suppose \(S_{n}^{2} = \dfrac{1}{n - 1}\sum_{i=1}^{n} (X_{i} - \bar{X}_{n})^{2}\). Then which of the following statements is not true ?

A.\(S_{n}\) is unbiased for \(\sigma\)
B.\(S_{n}^{2}\) is unbiased for \(\sigma^{2}\)
C.\(S_{n}\) is consistent for \(\sigma\)
D.\(S_{n}^{2}\) is consistent for \(\sigma^{2}\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

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.

Discuss in Forum

22.

Let \(\tau(\theta)\) be a parametric function and T be an estimator of \(\tau(\theta)\). Let \(V_{\theta}(T)\) denotes variance of T. The Cramer-Rao bound gives a :

A.Lower bound for \(V_{\theta}(T)\) for any arbitrary estimator T of \(\tau(\theta)\)
B.Upper bound for \(V_{\theta}(T)\) for any arbitrary estimator T of \(\tau(\theta)\)
C.Lower bound for \(V_{\theta}(T)\) among all unbiased estimators T of \(\tau(\theta)\)
D.Upper bound for \(V_{\theta}(T)\) among all unbiased estimators T of \(\tau(\theta)\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

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.

Discuss in Forum

23.

Let \(\{X_{1}, \ldots, X_{n}\}\) be a random sample from \(N(\mu_{1}, \sigma^{2})\) and \(\{Y_{1}, \ldots, Y_{m}\}\) be a random sample from \(N(\mu_{2}, \sigma^{2})\). Let

\(T_{1} = \sum_{i=1}^{n} X_{i}^{2} + \sum_{j=1}^{m} Y_{i}^{2}\), \(T_{2} = \sum_{i=1}^{n} X_{i}\),
\(T_{3} = \sum_{i=1}^{m} Y_{i}\), \(T_{4} = \sum_{i=1}^{n} X_{i}^{2}\), \(T_{5} = \sum_{i=1}^{m} Y_{i}^{2}\).

Then which of the following is a jointly sufficient complete statistic for \((\mu_{1}, \mu_{2}, \sigma^{2})\) :

A.\((T_{2}, T_{3}, T_{4}, T_{5})\)
B.\((T_{3}, T_{4}, T_{5})\)
C.\((T_{2}, T_{3}, T_{1})\)
D.\((T_{2}, T_{3}, T_{5})\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

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.

Discuss in Forum

24.

Suppose \(\{X_{1}, X_{2}, \ldots, X_{m}\}\) is a random sample from the distribution of X with mean \(\mu_{1}\) and variance \(\sigma_{1}^{2}\), \(\{y_{1}, \ldots, y_{n}\}\) is a random sample from the distribution of Y with mean \(\mu_{2}\) and variance \(\sigma_{2}^{2}\). The test statistic to test \(H_{0} : \sigma_{1}^{2} = \sigma_{2}^{2}\) against \(H_{1} : \sigma_{1}^{2} = \sigma_{2}^{2}\) is \(T_{n} = \dfrac{S_{X}^{2}}{S_{Y}^{2}}\), where \(S_{X}^{2}\) and \(S_{Y}^{2}\) are sample variances. Then the null distribution of \(T_{n}\) is :

A.F if X and Y are independent and have normal distributions
B.F if X and Y have normal distributions
C.F if X and Y are independent
D.F if X and Y have normal distributions with equal mean

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

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.)

Discuss in Forum

25.

In the method of constructing confidence interval for an unknown parameter based on pivotal quantity Q, which of the following statements is correct ?

A.The distribution of Q is known and depends on the unknown parameters
B.The distribution of Q is known and free from the unknown parameters
C.The distribution of Q is not known but is free from the unknown parameters
D.The distribution of Q is known and normal

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

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.

Discuss in Forum

26.

Suppose \(X_{1}, \ldots, X_{n}\) are independent random variables with probability density function

\[ f(x_{i}, \theta) = \begin{cases} \dfrac{1}{2i\theta}, & \text{if } -i\theta \lt x_{i} \lt i\theta \\ 0, & \text{otherwise.} \end{cases} \]

Then the minimal sufficient statistic for \(\theta\) is :

A.\(\{X_{(1)}, \ldots X_{(n)}\}\)
B.\(\max\{|X_{1}|, |X_{2}|/2, \ldots |X_{n}|/n\}\)
C.\(\max\{|X_{1}, X_{2}/2, X_{3}/3, \ldots X_{n}/n\}\)
D.\(\min\{|X_{1}, X_{2}/2, X_{3}/3, \ldots X_{n}/n\}\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

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.

Discuss in Forum

27.

Suppose \(f(x, \theta)\) is a probability density function of X for which differentiation under integral sign is permissible.

Then \(E\left(\dfrac{d}{d\theta}\log f(X, \theta)\right)\) is :

A.equal to the Fisher information
B.less than zero
C.less than 1
D.equal to zero

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

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.

Discuss in Forum

28.

Suppose \(\{X_{1}, X_{2}, \ldots, X_{n}\}\) is a random sample of size \(n\) from the \(U(0, \theta)\) distribution. Then :

A.Sample mean \(\bar{X}_{n}\) is consistent for \(\theta\)
B.Sample median is cosistent for \(\theta\)
C.\(2\bar{X}_{n}\) is consistent for \(\theta\)
D.\(X_{(1)}\) is consistent for \(\theta\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

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 \(\theta\).)

Discuss in Forum

29.

Suppose \(\{X_{1}, X_{2}, \ldots, X_{n}\}\) are independent and identically distributed random variables each having the probability density function :

\(f(x) = \dfrac{\theta}{x^{2}},\ x \ge \theta\)

Then :

\[ \begin{aligned} &\sqrt{n}\left(X\left(\left[\tfrac{n}{4}\right] + 1\right) - \tfrac{4\theta}{3}\right) \\ &\xrightarrow{L} z, \end{aligned} \]

A.\(z \sim N(0, 16\ \theta/27)\)
B.\(z \sim N(0, \theta/3)\)
C.\(z \sim N(0, 16\ \theta^{2}/27)\)
D.\(z \sim N(0, 16\ \theta/27)\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

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).

Discuss in Forum

30.

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be iid \(f(x, \theta, \alpha)\), where :

\[ \begin{aligned} &f(x; \theta, \alpha) \\ &= \frac{e^{-x/\theta}\, x^{\alpha-1}}{\Gamma(\alpha)\, \theta^{\alpha}}. \end{aligned} \]

Then, \(\sqrt{n}(\log \bar{X} - \log \alpha\theta) \xrightarrow{d} z\), where \(z\) is distributed as :

A.\(N(0, 1)\)
B.\(N(0, 1/\alpha)\)
C.\(N(0, \alpha^{2}/\theta^{2})\)
D.\(N(0, \theta^{2}/\alpha^{2})\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

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 \(\alpha = 1\).

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