Point and Interval Estimation Questions and Answers

Q: Let \(X_{1}\), \(X_{2}\), .......... \(X_{n}\) be a random sample of size \(n\) observed from Cauchy distribution with location parameter \(\theta\). Define \(T_{1}\) = sample mean and \(T_{2}\) = sample median. Then :

\(T_{2}\), the sample median, is asymptotically normal.For any continuous distribution with density \(f\) positive at the median, the sample median is asymptotically normal with mean the population median and variance \(1/(4nf(\theta)^{2})\). For the Cauchy, \(f(\theta) = 1/\pi\), so \(T_{2}\) is approximately \(N(\theta, \pi^{2}/(4n))\), and it is consistent.The Cauchy has no mean, and the sample mean \(T_{1}\) has exactly the same Cauchy distribution as a single observation, whatever \(n\). So \(T_{1}\) is neither consistent nor asymptotically normal, and \((T_{1} + T_{2})/2\) inherits its Cauchy tail. That rules out A, B and D.

Q: Let \(X_{1}\), \(X_{2}\), .......... \(X_{n}\) be the random sample observed from Poisson distribution with mean \(\lambda\). Then consistent estimator of \(1 - e^{-\lambda}\) is :

The consistent estimator is not unique.\(\bar{X}\) is consistent for \(\lambda\), so by continuity \(1 - e^{-\bar{X}}\) is consistent for \(1 - e^{-\lambda}\). But so is the proportion of non-zero observations, since \(P(X \gt 0) = 1 - e^{-\lambda}\) and the law of large numbers applies. And any consistent estimator plus a term tending to 0, such as \(1 - e^{-\bar{X}} + 1/n\), is again consistent.So consistency never singles out one estimator. The proportion of non-zero values shows that a consistent estimator need not be a function of \(\bar{X}\) only, and both examples are asymptotically normal, so D is false too.

Q: Let \(X_{1}\), \(X_{2}\), ........, \(X_{n}\) be a random sample of size \(n\) from Bernoulli distribution with parameter \(\theta\). Then variance of conditional expectation of \(\left(\dfrac{X_{1} + X_{2}}{2}\right)\) given \(\sum_{i=1}^{n} X_{i}\) is :

The variance is \(\theta(1 - \theta)/n\).Given \(T = \sum X_{i} = t\), the observations are exchangeable, so \(E(X_{1}|T) = E(X_{2}|T) = T/n\). Hence\[ \begin{aligned} E\left(\frac{X_{1} + X_{2}}{2}\,\Big|\,T\right) \\ &= \frac{T}{n} = \bar{X} \end{aligned} \]and \(Var(\bar{X}) = \theta(1 - \theta)/n\). This is Rao-Blackwellisation: conditioning the unbiased estimator \((X_{1} + X_{2})/2\), with variance \(\theta(1 - \theta)/2\), on the sufficient statistic reduces its variance to that of \(\bar{X}\).\(\theta(1 - \theta)\) is the variance of a single observation, 0 would need a constant, and \(\theta\) is the mean, not a variance.

Q: Let \(X_{1}, X_{2}, ......., X_{n}\) be a random sample from \(N(\theta, 1)\) where \(\theta \subset [a, b]\) and \(a \lt b\) are real numbers.Then which of the following statements about the Maximum Likelihood Estimator (MLE) of \(\theta\) is correct ?

The MLE of \(\theta\) exists but it is not \(\bar{X}\).The paper's \(\theta \subset [a, b]\) means the parameter is restricted to \([a, b]\). The normal log-likelihood is a downward parabola in \(\theta\) centred at \(\bar{X}\), so on \([a, b]\) it is maximised at the point of \([a, b]\) nearest \(\bar{X}\):\[ \hat{\theta} = \begin{cases} a, & \bar{X} \lt a \\ \bar{X}, & a \le \bar{X} \le b \\ b, & \bar{X} \gt b \end{cases} \]This always exists (a continuous function on a closed interval), but it equals \(\bar{X}\) only when \(\bar{X}\) falls in \([a, b]\), which is not guaranteed. It is also biased: when \(\theta = a\), \(\hat{\theta} \ge a\) always and \(\hat{\theta} \gt a\) with positive probability, so \(E\hat{\theta} \gt \theta\).

Q: Let \(T_{1}\) be a \(100 \cdot \alpha\%\) lower confidence limit for \(\theta\) and \(T_{2}\) be a \(100 \cdot \alpha\%\) upper confidence limit for \(\theta\). Let \(P[T_{1} \lt T_{2}] = 1\) and \(\dfrac{1}{2} \lt \alpha \lt 1\). Then a \(100(2\alpha - 1)\%\) confidence limit for \(\theta\) is :

The \(100(2\alpha - 1)\%\) confidence interval is \([T_{1}, T_{2})\).By definition \(P(T_{1} \le \theta) = \alpha\) and \(P(\theta \lt T_{2}) = \alpha\), so the two ways the interval can miss \(\theta\) each have probability \(1 - \alpha\). They cannot happen together, since \(T_{1} \lt T_{2}\) almost surely. Hence\[ \begin{aligned} P(T_{1} \le \theta \lt T_{2}) &= 1 - 2(1 - \alpha) \\ &= 2\alpha - 1 \end{aligned} \]The condition \(\alpha \gt 1/2\) makes this positive. Dividing the limits by \(\alpha\) or scaling them changes the coverage in an uncontrolled way, so options B, C and D have no stated coverage.

Q: Which of the following theorems is useful for obtaining uniformly minimum variance unbiased estimator of a parametric function ?

The theorem is Rao-Blackwell-Lehmann-Scheffe.Rao-Blackwell says that conditioning an unbiased estimator on a sufficient statistic never increases its variance. Lehmann-Scheffe adds that if the sufficient statistic is also complete, the resulting unbiased function of it is the unique UMVUE. Together they give the standard recipe for a UMVUE.Rao-Blackwell alone only improves an estimator; it does not guarantee the minimum. Neyman-Pearson concerns most powerful tests, and Basu's theorem concerns the independence of complete sufficient and ancillary statistics.

Q: X is a normal \((\mu, \sigma^{2})\) random variable and a 95% confidence interval for \(\mu\) is constructed based on a sample of size \(n\). Suppose the length of the interval is \(L_{1}\). If instead the variance of X is \(\sigma_{1}^{2} \gt \sigma^{2}\) the length of the 95% confidence interval constructed using the same technique will be :

The interval will be larger than \(L_{1}\).The usual 95% interval for \(\mu\) with known variance is \(\bar{x} \pm 1.96\sigma/\sqrt{n}\), whose length is\[ L = \frac{2 \times 1.96\,\sigma}{\sqrt{n}} \]With the same \(n\) and confidence level, the length is proportional to \(\sigma\). Replacing \(\sigma\) by \(\sigma_{1} \gt \sigma\) multiplies it by \(\sigma_{1}/\sigma \gt 1\). More spread in the data means less precision about the mean, so the interval cannot be equal or shorter.

Q: Suppose \(X_{1}\) has exponential distribution with mean \(\theta\), \(X_{2}\) has exponential distribution with mean \(\theta/2\) and \(X_{1}\) and \(X_{2}\) are independent. Which of the following statements is not correct ?

The statement that is not correct is \(X_{1} + X_{2}\) is sufficient for \(\theta\).The joint density is\[ \begin{aligned} &\frac{1}{\theta}e^{-x_{1}/\theta}\cdot\frac{2}{\theta}e^{-2x_{2}/\theta} \\ &= \frac{2}{\theta^{2}}e^{-(x_{1} + 2x_{2})/\theta} \end{aligned} \]By the factorisation theorem, \(T = X_{1} + 2X_{2}\) is sufficient, and since this is a one-parameter exponential family with a natural parameter ranging over an interval, T is also complete. \(E(T) = \theta + 2 \cdot \theta/2 = 2\theta\), so \(T/2\) is unbiased.\(X_{1} + X_{2}\) weights the two observations equally, but they carry information about \(\theta\) in the ratio 1 : 2, so it is not sufficient.

Q: Suppose \(\{X_{1}, X_{2}, ......, X_{n}\}\) is a random sample from Poisson \(P(\theta)\) distribution, \(\theta \gt 0\). Which of the following statements is not correct ?

The statement that is not correct is \(\bar{X}_{n}^{2}/S_{n}^{2}\) is unbiased for \(\theta\).For the Poisson distribution the mean and variance both equal \(\theta\). So \(E\bar{X}_{n} = \theta\) and \(E S_{n}^{2} = \theta\) (the sample variance with divisor \(n - 1\) is unbiased for the variance), and any weighted average such as \(0.3\bar{X}_{n} + 0.7S_{n}^{2}\) is unbiased as well.The ratio \(\bar{X}_{n}^{2}/S_{n}^{2}\) is a non-linear function of the two statistics and is undefined whenever \(S_{n}^{2} = 0\), that is, whenever all observations are equal, which has positive probability for Poisson data. So it has no finite expectation equal to \(\theta\), and it is the one statement that fails.

Q: Suppose \(\{X_{1}, X_{2}, ......, X_{n}\}\) is a random sample from \(U(\theta, \theta + 1)\). Which of the following statistic is not a maximum likelihood estimator of \(\theta\) ?

The statistic that is not an MLE is \(X_{(n)}\).The likelihood for \(U(\theta, \theta + 1)\) is 1 when \(\theta \le X_{(1)}\) and \(\theta + 1 \ge X_{(n)}\), and 0 otherwise. So every \(\theta\) in \([X_{(n)} - 1, X_{(1)}]\) maximises it: the MLE is not unique.\(X_{(1)}\) and \(X_{(n)} - 1\) are the end points of that interval, and the midrange minus 0.5 is its midpoint, so all three are MLEs. \(X_{(n)}\) exceeds \(X_{(1)}\) whenever \(n \ge 2\), so it lies outside the interval, where the likelihood is 0.

11.

Let \(X_{1}\), \(X_{2}\), .......... \(X_{n}\) be a random sample of size \(n\) observed from Cauchy distribution with location parameter \(\theta\). Define \(T_{1}\) = sample mean and \(T_{2}\) = sample median. Then :

A.\(T_{1}\) and \(T_{2}\) are consistent estimator of \(\theta\)
B.\(T_{1}\) is asymptotically normal
C.\(T_{2}\) is asymptotically normal
D.\((T_{1} + T_{2})/2\) is asymptotically normal

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: C

\(T_{2}\), the sample median, is asymptotically normal.

For any continuous distribution with density \(f\) positive at the median, the sample median is asymptotically normal with mean the population median and variance \(1/(4nf(\theta)^{2})\). For the Cauchy, \(f(\theta) = 1/\pi\), so \(T_{2}\) is approximately \(N(\theta, \pi^{2}/(4n))\), and it is consistent.

The Cauchy has no mean, and the sample mean \(T_{1}\) has exactly the same Cauchy distribution as a single observation, whatever \(n\). So \(T_{1}\) is neither consistent nor asymptotically normal, and \((T_{1} + T_{2})/2\) inherits its Cauchy tail. That rules out A, B and D.

Discuss in Forum

12.

Let \(X_{1}\), \(X_{2}\), .......... \(X_{n}\) be the random sample observed from Poisson distribution with mean \(\lambda\). Then consistent estimator of \(1 - e^{-\lambda}\) is :

A.unique
B.not unique
C.function of \(\bar{X}\) only
D.not asymptotically normally distributed

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

The consistent estimator is not unique.

\(\bar{X}\) is consistent for \(\lambda\), so by continuity \(1 - e^{-\bar{X}}\) is consistent for \(1 - e^{-\lambda}\). But so is the proportion of non-zero observations, since \(P(X \gt 0) = 1 - e^{-\lambda}\) and the law of large numbers applies. And any consistent estimator plus a term tending to 0, such as \(1 - e^{-\bar{X}} + 1/n\), is again consistent.

So consistency never singles out one estimator. The proportion of non-zero values shows that a consistent estimator need not be a function of \(\bar{X}\) only, and both examples are asymptotically normal, so D is false too.

Discuss in Forum

13.

Let \(X_{1}\), \(X_{2}\), ........, \(X_{n}\) be a random sample of size \(n\) from Bernoulli distribution with parameter \(\theta\). Then variance of conditional expectation of \(\left(\dfrac{X_{1} + X_{2}}{2}\right)\) given \(\sum_{i=1}^{n} X_{i}\) is :

A.0
B.\(\theta\)
C.\(\theta(1 - \theta)\)
D.\(\theta(1 - \theta)/n\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: D

The variance is \(\theta(1 - \theta)/n\).

Given \(T = \sum X_{i} = t\), the observations are exchangeable, so \(E(X_{1}|T) = E(X_{2}|T) = T/n\). Hence

\[ \begin{aligned} E\left(\frac{X_{1} + X_{2}}{2}\,\Big|\,T\right) \\ &= \frac{T}{n} = \bar{X} \end{aligned} \]

and \(Var(\bar{X}) = \theta(1 - \theta)/n\). This is Rao-Blackwellisation: conditioning the unbiased estimator \((X_{1} + X_{2})/2\), with variance \(\theta(1 - \theta)/2\), on the sufficient statistic reduces its variance to that of \(\bar{X}\).

\(\theta(1 - \theta)\) is the variance of a single observation, 0 would need a constant, and \(\theta\) is the mean, not a variance.

Discuss in Forum

14.

Let \(X_{1}, X_{2}, ......., X_{n}\) be a random sample from \(N(\theta, 1)\) where \(\theta \subset [a, b]\) and \(a \lt b\) are real numbers.

Then which of the following statements about the Maximum Likelihood Estimator (MLE) of \(\theta\) is correct ?

A.MLE of \(\theta\) does not exist.
B.MLE of \(\theta\) is \(\bar{X}\).
C.MLE of \(\theta\) exists but it is not \(\bar{X}\).
D.MLE of \(\theta\) is an unbiased estimator of \(\theta\).

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: C

The MLE of \(\theta\) exists but it is not \(\bar{X}\).

The paper's \(\theta \subset [a, b]\) means the parameter is restricted to \([a, b]\). The normal log-likelihood is a downward parabola in \(\theta\) centred at \(\bar{X}\), so on \([a, b]\) it is maximised at the point of \([a, b]\) nearest \(\bar{X}\):

\[ \hat{\theta} = \begin{cases} a, & \bar{X} \lt a \\ \bar{X}, & a \le \bar{X} \le b \\ b, & \bar{X} \gt b \end{cases} \]

This always exists (a continuous function on a closed interval), but it equals \(\bar{X}\) only when \(\bar{X}\) falls in \([a, b]\), which is not guaranteed. It is also biased: when \(\theta = a\), \(\hat{\theta} \ge a\) always and \(\hat{\theta} \gt a\) with positive probability, so \(E\hat{\theta} \gt \theta\).

Discuss in Forum

15.

Let \(T_{1}\) be a \(100 \cdot \alpha\%\) lower confidence limit for \(\theta\) and \(T_{2}\) be a \(100 \cdot \alpha\%\) upper confidence limit for \(\theta\). Let \(P[T_{1} \lt T_{2}] = 1\) and \(\dfrac{1}{2} \lt \alpha \lt 1\). Then a \(100(2\alpha - 1)\%\) confidence limit for \(\theta\) is :

A.\([T_{1}, T_{2})\)
B.\(\left[\dfrac{T_{1}}{\alpha}, \dfrac{T_{2}}{\alpha}\right]\)
C.\(\left[\dfrac{T_{2}}{\alpha}, \dfrac{T_{1}}{\alpha}\right]\)
D.\([(2\alpha - 1)T_{1}, \alpha T_{2}]\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: A

The \(100(2\alpha - 1)\%\) confidence interval is \([T_{1}, T_{2})\).

By definition \(P(T_{1} \le \theta) = \alpha\) and \(P(\theta \lt T_{2}) = \alpha\), so the two ways the interval can miss \(\theta\) each have probability \(1 - \alpha\). They cannot happen together, since \(T_{1} \lt T_{2}\) almost surely. Hence

\[ \begin{aligned} P(T_{1} \le \theta \lt T_{2}) &= 1 - 2(1 - \alpha) \\ &= 2\alpha - 1 \end{aligned} \]

The condition \(\alpha \gt 1/2\) makes this positive. Dividing the limits by \(\alpha\) or scaling them changes the coverage in an uncontrolled way, so options B, C and D have no stated coverage.

Discuss in Forum

16.

Which of the following theorems is useful for obtaining uniformly minimum variance unbiased estimator of a parametric function ?

A.Neyman-Pearson theorem
B.Basu's theorem
C.Rao-Blackwell theorem
D.Rao-Blackwell-Lehmann-Scheffe theorem

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: D

The theorem is Rao-Blackwell-Lehmann-Scheffe.

Rao-Blackwell says that conditioning an unbiased estimator on a sufficient statistic never increases its variance. Lehmann-Scheffe adds that if the sufficient statistic is also complete, the resulting unbiased function of it is the unique UMVUE. Together they give the standard recipe for a UMVUE.

Rao-Blackwell alone only improves an estimator; it does not guarantee the minimum. Neyman-Pearson concerns most powerful tests, and Basu's theorem concerns the independence of complete sufficient and ancillary statistics.

Discuss in Forum

17.

X is a normal \((\mu, \sigma^{2})\) random variable and a 95% confidence interval for \(\mu\) is constructed based on a sample of size \(n\). Suppose the length of the interval is \(L_{1}\). If instead the variance of X is \(\sigma_{1}^{2} \gt \sigma^{2}\) the length of the 95% confidence interval constructed using the same technique will be :

A.equal to \(L_{1}\)
B.smaller than \(L_{1}\)
C.larger than \(L_{1}\)
D.larger or smaller than \(L_{1}\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: C

The interval will be larger than \(L_{1}\).

The usual 95% interval for \(\mu\) with known variance is \(\bar{x} \pm 1.96\sigma/\sqrt{n}\), whose length is

\[ L = \frac{2 \times 1.96\,\sigma}{\sqrt{n}} \]

With the same \(n\) and confidence level, the length is proportional to \(\sigma\). Replacing \(\sigma\) by \(\sigma_{1} \gt \sigma\) multiplies it by \(\sigma_{1}/\sigma \gt 1\). More spread in the data means less precision about the mean, so the interval cannot be equal or shorter.

Discuss in Forum

18.

Suppose \(X_{1}\) has exponential distribution with mean \(\theta\), \(X_{2}\) has exponential distribution with mean \(\theta/2\) and \(X_{1}\) and \(X_{2}\) are independent. Which of the following statements is not correct ?

A.\(X_{1} + X_{2}\) is sufficient for \(\theta\)
B.\(X_{1} + 2X_{2}\) is sufficient for \(\theta\)
C.\(X_{1} + 2X_{2}\) is complete for \(\theta\)
D.\((X_{1} + 2X_{2})/2\) is unbiased for \(\theta\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: A

The statement that is not correct is \(X_{1} + X_{2}\) is sufficient for \(\theta\).

The joint density is

\[ \begin{aligned} &\frac{1}{\theta}e^{-x_{1}/\theta}\cdot\frac{2}{\theta}e^{-2x_{2}/\theta} \\ &= \frac{2}{\theta^{2}}e^{-(x_{1} + 2x_{2})/\theta} \end{aligned} \]

By the factorisation theorem, \(T = X_{1} + 2X_{2}\) is sufficient, and since this is a one-parameter exponential family with a natural parameter ranging over an interval, T is also complete. \(E(T) = \theta + 2 \cdot \theta/2 = 2\theta\), so \(T/2\) is unbiased.

\(X_{1} + X_{2}\) weights the two observations equally, but they carry information about \(\theta\) in the ratio 1 : 2, so it is not sufficient.

Discuss in Forum

19.

Suppose \(\{X_{1}, X_{2}, ......, X_{n}\}\) is a random sample from Poisson \(P(\theta)\) distribution, \(\theta \gt 0\). Which of the following statements is not correct ?

A.Sample mean \(\bar{X}_{n}\) is unbiased for \(\theta\)
B.Sample variance \(S_{n}^{2}\) is unbiased for \(\theta\)
C.\(0.3\bar{X}_{n} + 0.7S_{n}^{2}\) is unbiased for \(\theta\)
D.\(\bar{X}_{n}^{2}/S_{n}^{2}\) is unbiased for \(\theta\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: D

The statement that is not correct is \(\bar{X}_{n}^{2}/S_{n}^{2}\) is unbiased for \(\theta\).

For the Poisson distribution the mean and variance both equal \(\theta\). So \(E\bar{X}_{n} = \theta\) and \(E S_{n}^{2} = \theta\) (the sample variance with divisor \(n - 1\) is unbiased for the variance), and any weighted average such as \(0.3\bar{X}_{n} + 0.7S_{n}^{2}\) is unbiased as well.

The ratio \(\bar{X}_{n}^{2}/S_{n}^{2}\) is a non-linear function of the two statistics and is undefined whenever \(S_{n}^{2} = 0\), that is, whenever all observations are equal, which has positive probability for Poisson data. So it has no finite expectation equal to \(\theta\), and it is the one statement that fails.

Discuss in Forum

20.

Suppose \(\{X_{1}, X_{2}, ......, X_{n}\}\) is a random sample from \(U(\theta, \theta + 1)\). Which of the following statistic is not a maximum likelihood estimator of \(\theta\) ?

A.\(X_{(1)}\)
B.\(X_{(n)} - 1\)
C.\(X_{(n)}\)
D.\(\dfrac{X_{(1)} + X_{(n)}}{2} - 0.5\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: C

The statistic that is not an MLE is \(X_{(n)}\).

The likelihood for \(U(\theta, \theta + 1)\) is 1 when \(\theta \le X_{(1)}\) and \(\theta + 1 \ge X_{(n)}\), and 0 otherwise. So every \(\theta\) in \([X_{(n)} - 1, X_{(1)}]\) maximises it: the MLE is not unique.

\(X_{(1)}\) and \(X_{(n)} - 1\) are the end points of that interval, and the midrange minus 0.5 is its midpoint, so all three are MLEs. \(X_{(n)}\) exceeds \(X_{(1)}\) whenever \(n \ge 2\), so it lies outside the interval, where the likelihood is 0.

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