Point and Interval Estimation Questions and Answers

Q: Let \(X_{1}, X_{2}, \ldots, X_{n}\) be iid \(U(0, \theta)\). Define \(T_{1} = \bar{X}\) and \(T_{2} = (X_{(1)} + X_{(n)})/2\). Where \(X_{(1)} = \min(X_{1}, \ldots X_{n})\), \(X_{(n)} = \max(X_{1} \ldots X_{n})\). Then :

The answer is both \(T_{1}\) and \(T_{2}\) are equally efficient when \(n = 2\).With two observations, \(X_{(1)} + X_{(2)} = X_{1} + X_{2}\), so \(T_{2} = (X_{(1)} + X_{(2)})/2 = \bar{X} = T_{1}\). They are the same statistic and have the same variance.For larger \(n\) the midrange is far better: \(\mathrm{Var}(T_{1}) = \theta^{2}/(12n)\), while \(\mathrm{Var}(T_{2}) = \theta^{2}/(2(n + 1)(n + 2))\), which falls like \(1/n^{2}\). So the efficiency of \(T_{1}\) relative to \(T_{2}\) is below 1 for \(n \gt 2\) and tends to 0, which rules out (B) and (D), and (C) fails at \(n = 2\), where the ratio is exactly 1.

Q: Let X be a single observation from \(\theta e^{-\theta x}I_{[0, \infty)}(x)\), \(\theta \gt 0\). Then (X, 2X) is a confidence interval for \(1/\theta\) with confidence coefficient :

The confidence coefficient is \(e^{-1/2} - e^{-1}\).The interval covers \(1/\theta\) when \(X \le 1/\theta \le 2X\), that is, when \(\frac{1}{2\theta} \le X \le \frac{1}{\theta}\). With \(P(X \gt x) = e^{-\theta x}\):\[ \begin{aligned} P\left(\tfrac{1}{2\theta} \le X \le \tfrac{1}{\theta}\right) &= e^{-\theta \cdot \frac{1}{2\theta}} - e^{-\theta \cdot \frac{1}{\theta}} \\ &= e^{-1/2} - e^{-1} \approx 0.239 \end{aligned} \]It does not depend on \(\theta\), as a confidence coefficient must not. \(e^{-1/2}\) and \(e^{-1}\) are the two tail probabilities separately, and 1/2 does not arise.

Q: Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from uniform \(\left(\theta - \frac{1}{2}, \theta + \frac{1}{2}\right)\) distribution. Consider the problem of testing \(H_{0} : \theta = -\frac{1}{2}\) Vs \(H_{1} : \theta = \frac{1}{2}\).Let \(X_{(1)} = \min\{X_{1}, X_{2}, \ldots, X_{n}\}\). Consider the test, reject \(H_{0}\) if \(X_{(1)} \gt 0\) and accept otherwise. Then the power (\(p\)) and size (\(s\)) of the test (\(p\), \(s\)) are given by :

The power and size are (1, 0).Under \(H_{0}\) (\(\theta = -\frac{1}{2}\)) the observations lie in \((-1, 0)\), so \(X_{(1)} \gt 0\) is impossible: the size, the probability of rejecting a true \(H_{0}\), is 0.Under \(H_{1}\) (\(\theta = \frac{1}{2}\)) the observations lie in \((0, 1)\), so \(X_{(1)} \gt 0\) with probability 1: the power is 1.Because the two supports do not overlap, this test makes no errors at all. (0, 0), (0, 1) and (1, 1) mix up which probability belongs to which hypothesis.

Q: Which of the following theorems is useful for obtaining a sufficient statistics ?

The theorem used to obtain sufficient statistics is the Neyman factorization theorem.It states that T(X) is sufficient for \(\theta\) if and only if the joint density factorises as \(f(x; \theta) = g(T(x); \theta)\, h(x)\), with \(h\) free of \(\theta\). Reading off such a factorisation is the standard way to find a sufficient statistic.The Neyman-Pearson lemma gives most powerful tests. Basu's theorem uses a complete sufficient statistic to prove independence from ancillary statistics, and the Rao-Blackwell theorem improves an estimator by conditioning on a sufficient statistic; both take sufficiency as given rather than finding it.

Q: Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from the density \(f(x, \theta) = c(\theta)\, h(x)\, e^{\pi(\theta)d(x)}\), \(\theta \in \Theta\).Let \(T(\underline{x}) = \sum_{i=1}^{n} d(x_{i})\).Then which of the following statements is not true ?

The statement that is not true is that T(X) is unbiased for \(\theta\).The density is a one-parameter exponential family, so \(T = \sum d(X_{i})\) is sufficient (factorisation) and, when the natural parameter space contains an interval, complete (A). By the Lehmann-Scheffe theorem a complete sufficient statistic is the UMVUE of its own expectation (B). Its distribution depends on \(\theta\) (D).But \(E[T] = nE[d(X)]\), a function of \(\theta\) that is generally not \(\theta\) itself. For example, for the Poisson family \(d(x) = x\) and \(E[T] = n\theta\). So T is not, in general, unbiased for \(\theta\).

Q: X is a normal (\(\mu\), \(\sigma^{2}\)) random variable and a 95% confidence interval for \(\mu\) based on a random sample of size \(n\) on X is given by (15, 25]. Hence it can be interpreted that :

The statement that fits a confidence interval is option D: the interval covers the true value of \(\mu\) with probability 0.95.Here \(\mu\) is a fixed, unknown number and the endpoints are computed from the random sample. The 95% belongs to the method: over many repeated samples, about 95% of the intervals built this way cover \(\mu\). Option D says this, with the interval as the thing that covers. Strictly, once the numbers (15, 25] have been computed, that one interval either contains \(\mu\) or does not; D is the usual textbook reading of the coverage property, and the key follows it.Option B turns \(\mu\) into a random variable that falls in a fixed interval with probability 0.95, which is the Bayesian credible-interval reading and not what a confidence interval says. Option C states certainty, and option A picks the midpoint, which the interval does not identify.

Q: Suppose \(X_{1}, \ldots, X_{n}\) are independent and identically distributed random variables each having \(U(-\theta, \theta)\) distribution. Then the minimal sufficient statistic for \(\theta\) is given by :

The minimal sufficient statistic is \(\max\{-X_{(1)}, X_{(n)}\}\).The joint density is \((2\theta)^{-n}\) when every \(|x_{i}| \le \theta\), i.e. when \(\max_{i}|x_{i}| \le \theta\), and 0 otherwise. By the factorisation theorem, \(\max|X_{i}| = \max\{-X_{(1)}, X_{(n)}\}\) is sufficient, and the likelihood ratio of two samples is free of \(\theta\) exactly when this maximum is the same, so it is minimal.\(-X_{(1)}\) or \(X_{(n)}\) alone captures only one end of the sample, and the likelihood depends on whichever of them is larger. The minimum of the two is not sufficient.

Q: Suppose \(\{X_{1}, X_{2}, \ldots, X_{n}\}\) are independent and identically distributed random variables each having \(U(-\theta, \theta)\) r.vs. Then the maximum likelihood estimator of \(\theta\) is :

The MLE of \(\theta\) is \(\max\{|X_{1}|, \ldots, |X_{n}|\}\).The likelihood \(L(\theta) = (2\theta)^{-n}\) for \(\theta \ge \max|x_{i}|\), and 0 otherwise. It decreases in \(\theta\), so it is maximised at the smallest admissible value, \(\hat{\theta} = \max|X_{i}|\).\(X_{(n)}\) or \(-X_{(1)}\) alone can be smaller than \(\max|X_{i}|\), and then the likelihood at that value is 0 because some observation lies outside \((-\theta, \theta)\). \((X_{(1)} + X_{(n)})/4\) is not even guaranteed to be positive.

Q: Suppose \(\{X_{1}, X_{2}, \ldots, X_{n}\}\) is a random sample of size \(n\) from uniform \(U(\theta - 1, \theta + 1)\) distribution. Then which of the following is not true ?

The statement that is not true is that \(X_{(n)}\) is consistent for \(\theta\).The upper end of the support is \(\theta + 1\), and the sample maximum converges to it: \(X_{(n)} \to \theta + 1\) in probability. So \(X_{(n)}\) is consistent for \(\theta + 1\), not for \(\theta\), and \(X_{(n)} - 1\) is consistent for \(\theta\) (D).The distribution is symmetric about \(\theta\), so the sample mean (by the law of large numbers) and the sample median (which converges to the population median \(\theta\)) are both consistent, making B and C true.

Q: If one observation X from Bernoulli distribution with parameter \(p \in \left[\frac{1}{3}, \frac{2}{3}\right]\) is obtained, then the MLE of \(p\) is :

The MLE is \(\frac{X + 1}{3}\).The likelihood of one Bernoulli observation is \(p^{X}(1 - p)^{1 - X}\). If \(X = 1\), it is \(p\), increasing in \(p\), so on \([1/3, 2/3]\) it is maximised at the right end \(p = 2/3\). If \(X = 0\), it is \(1 - p\), decreasing, so it is maximised at \(p = 1/3\).The formula \((X + 1)/3\) gives exactly 1/3 for \(X = 0\) and 2/3 for \(X = 1\).The unrestricted MLE X takes the values 0 and 1, which lie outside the parameter space. \((2X + 1)/3\) gives 1 when \(X = 1\), also outside, and \((X - 1)/3\) is negative or zero, never in \([1/3, 2/3]\).

31.

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be iid \(U(0, \theta)\). Define \(T_{1} = \bar{X}\) and \(T_{2} = (X_{(1)} + X_{(n)})/2\). Where \(X_{(1)} = \min(X_{1}, \ldots X_{n})\), \(X_{(n)} = \max(X_{1} \ldots X_{n})\). Then :

A.both \(T_{1}\) and \(T_{2}\) are equally efficient when \(n = 2\)
B.\(ARE(T_{1}, T_{2}) \gt 1\) when \(n \gt 2\)
C.\(ARE(T_{1}, T_{2}) \lt 1\) for all \(n\)
D.\(ARE(T_{1}, T_{2}) \not\to 0\) as \(n \to \infty\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: A

The answer is both \(T_{1}\) and \(T_{2}\) are equally efficient when \(n = 2\).

With two observations, \(X_{(1)} + X_{(2)} = X_{1} + X_{2}\), so \(T_{2} = (X_{(1)} + X_{(2)})/2 = \bar{X} = T_{1}\). They are the same statistic and have the same variance.

For larger \(n\) the midrange is far better: \(\mathrm{Var}(T_{1}) = \theta^{2}/(12n)\), while \(\mathrm{Var}(T_{2}) = \theta^{2}/(2(n + 1)(n + 2))\), which falls like \(1/n^{2}\). So the efficiency of \(T_{1}\) relative to \(T_{2}\) is below 1 for \(n \gt 2\) and tends to 0, which rules out (B) and (D), and (C) fails at \(n = 2\), where the ratio is exactly 1.

Discuss in Forum

32.

Let X be a single observation from \(\theta e^{-\theta x}I_{[0, \infty)}(x)\), \(\theta \gt 0\). Then (X, 2X) is a confidence interval for \(1/\theta\) with confidence coefficient :

A.\(\dfrac{1}{2}\)
B.\(e^{-1}\)
C.\(e^{-\frac{1}{2}}\)
D.\(e^{-\frac{1}{2}} - e^{-1}\)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: D

The confidence coefficient is \(e^{-1/2} - e^{-1}\).

The interval covers \(1/\theta\) when \(X \le 1/\theta \le 2X\), that is, when \(\frac{1}{2\theta} \le X \le \frac{1}{\theta}\). With \(P(X \gt x) = e^{-\theta x}\):

\[ \begin{aligned} P\left(\tfrac{1}{2\theta} \le X \le \tfrac{1}{\theta}\right) &= e^{-\theta \cdot \frac{1}{2\theta}} - e^{-\theta \cdot \frac{1}{\theta}} \\ &= e^{-1/2} - e^{-1} \approx 0.239 \end{aligned} \]

It does not depend on \(\theta\), as a confidence coefficient must not. \(e^{-1/2}\) and \(e^{-1}\) are the two tail probabilities separately, and 1/2 does not arise.

Discuss in Forum

33.

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from uniform \(\left(\theta - \frac{1}{2}, \theta + \frac{1}{2}\right)\) distribution. Consider the problem of testing \(H_{0} : \theta = -\frac{1}{2}\) Vs \(H_{1} : \theta = \frac{1}{2}\).

Let \(X_{(1)} = \min\{X_{1}, X_{2}, \ldots, X_{n}\}\). Consider the test, reject \(H_{0}\) if \(X_{(1)} \gt 0\) and accept otherwise. Then the power (\(p\)) and size (\(s\)) of the test (\(p\), \(s\)) are given by :

A.(0, 0)
B.(0, 1)
C.(1, 0)
D.(1, 1)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: C

The power and size are (1, 0).

Under \(H_{0}\) (\(\theta = -\frac{1}{2}\)) the observations lie in \((-1, 0)\), so \(X_{(1)} \gt 0\) is impossible: the size, the probability of rejecting a true \(H_{0}\), is 0.

Under \(H_{1}\) (\(\theta = \frac{1}{2}\)) the observations lie in \((0, 1)\), so \(X_{(1)} \gt 0\) with probability 1: the power is 1.

Because the two supports do not overlap, this test makes no errors at all. (0, 0), (0, 1) and (1, 1) mix up which probability belongs to which hypothesis.

Discuss in Forum

34.

Which of the following theorems is useful for obtaining a sufficient statistics ?

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

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: B

The theorem used to obtain sufficient statistics is the Neyman factorization theorem.

It states that T(X) is sufficient for \(\theta\) if and only if the joint density factorises as \(f(x; \theta) = g(T(x); \theta)\, h(x)\), with \(h\) free of \(\theta\). Reading off such a factorisation is the standard way to find a sufficient statistic.

The Neyman-Pearson lemma gives most powerful tests. Basu's theorem uses a complete sufficient statistic to prove independence from ancillary statistics, and the Rao-Blackwell theorem improves an estimator by conditioning on a sufficient statistic; both take sufficiency as given rather than finding it.

Discuss in Forum

35.

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from the density \(f(x, \theta) = c(\theta)\, h(x)\, e^{\pi(\theta)d(x)}\), \(\theta \in \Theta\).

Let \(T(\underline{x}) = \sum_{i=1}^{n} d(x_{i})\).

Then which of the following statements is not true ?

A.\(T(\underline{X})\) is a complete sufficient statistics for \(\theta\)
B.\(T(\underline{X})\) is uniformly minimum variance unbiased estimator of its expected value
C.\(T(\underline{X})\) is unbiased for \(\theta\)
D.The probability distribution depends on \(\theta\)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: C

The statement that is not true is that T(X) is unbiased for \(\theta\).

The density is a one-parameter exponential family, so \(T = \sum d(X_{i})\) is sufficient (factorisation) and, when the natural parameter space contains an interval, complete (A). By the Lehmann-Scheffe theorem a complete sufficient statistic is the UMVUE of its own expectation (B). Its distribution depends on \(\theta\) (D).

But \(E[T] = nE[d(X)]\), a function of \(\theta\) that is generally not \(\theta\) itself. For example, for the Poisson family \(d(x) = x\) and \(E[T] = n\theta\). So T is not, in general, unbiased for \(\theta\).

Discuss in Forum

36.

X is a normal (\(\mu\), \(\sigma^{2}\)) random variable and a 95% confidence interval for \(\mu\) based on a random sample of size \(n\) on X is given by (15, 25]. Hence it can be interpreted that :

A.The true value of \(\mu\) is 20
B.The true value of \(\mu\) lies in (15, 25] with probability 0.95
C.The true value of \(\mu\) is between 15 and 25
D.The interval (15, 25] covers the true value of \(\mu\) with probability 0.95

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: D

The statement that fits a confidence interval is option D: the interval covers the true value of \(\mu\) with probability 0.95.

Here \(\mu\) is a fixed, unknown number and the endpoints are computed from the random sample. The 95% belongs to the method: over many repeated samples, about 95% of the intervals built this way cover \(\mu\). Option D says this, with the interval as the thing that covers. Strictly, once the numbers (15, 25] have been computed, that one interval either contains \(\mu\) or does not; D is the usual textbook reading of the coverage property, and the key follows it.

Option B turns \(\mu\) into a random variable that falls in a fixed interval with probability 0.95, which is the Bayesian credible-interval reading and not what a confidence interval says. Option C states certainty, and option A picks the midpoint, which the interval does not identify.

Discuss in Forum

37.

Suppose \(X_{1}, \ldots, X_{n}\) are independent and identically distributed random variables each having \(U(-\theta, \theta)\) distribution. Then the minimal sufficient statistic for \(\theta\) is given by :

A.\(-X_{(1)}\)
B.\(X_{(n)}\)
C.\(\max\{-X_{(1)}, X_{(n)}\}\)
D.\(\min\{-X_{(1)}, X_{(n)}\}\)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: C

The minimal sufficient statistic is \(\max\{-X_{(1)}, X_{(n)}\}\).

The joint density is \((2\theta)^{-n}\) when every \(|x_{i}| \le \theta\), i.e. when \(\max_{i}|x_{i}| \le \theta\), and 0 otherwise. By the factorisation theorem, \(\max|X_{i}| = \max\{-X_{(1)}, X_{(n)}\}\) is sufficient, and the likelihood ratio of two samples is free of \(\theta\) exactly when this maximum is the same, so it is minimal.

\(-X_{(1)}\) or \(X_{(n)}\) alone captures only one end of the sample, and the likelihood depends on whichever of them is larger. The minimum of the two is not sufficient.

Discuss in Forum

38.

Suppose \(\{X_{1}, X_{2}, \ldots, X_{n}\}\) are independent and identically distributed random variables each having \(U(-\theta, \theta)\) r.vs. Then the maximum likelihood estimator of \(\theta\) is :

A.\(X_{(n)}\)
B.\((X_{(1)} + X_{(n)})/4\)
C.\(-X_{(1)}\)
D.\(\max\{|X_{1}|, \ldots, |X_{n}|\}\)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: D

The MLE of \(\theta\) is \(\max\{|X_{1}|, \ldots, |X_{n}|\}\).

The likelihood \(L(\theta) = (2\theta)^{-n}\) for \(\theta \ge \max|x_{i}|\), and 0 otherwise. It decreases in \(\theta\), so it is maximised at the smallest admissible value, \(\hat{\theta} = \max|X_{i}|\).

\(X_{(n)}\) or \(-X_{(1)}\) alone can be smaller than \(\max|X_{i}|\), and then the likelihood at that value is 0 because some observation lies outside \((-\theta, \theta)\). \((X_{(1)} + X_{(n)})/4\) is not even guaranteed to be positive.

Discuss in Forum

39.

Suppose \(\{X_{1}, X_{2}, \ldots, X_{n}\}\) is a random sample of size \(n\) from uniform \(U(\theta - 1, \theta + 1)\) distribution. Then which of the following is not true ?

A.\(X_{(n)}\) is consistent for \(\theta\)
B.Sample median is consistent for \(\theta\)
C.Sample mean is consistent for \(\theta\)
D.\(X_{(n)} - 1\) is consistent for \(\theta\)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: A

The statement that is not true is that \(X_{(n)}\) is consistent for \(\theta\).

The upper end of the support is \(\theta + 1\), and the sample maximum converges to it: \(X_{(n)} \to \theta + 1\) in probability. So \(X_{(n)}\) is consistent for \(\theta + 1\), not for \(\theta\), and \(X_{(n)} - 1\) is consistent for \(\theta\) (D).

The distribution is symmetric about \(\theta\), so the sample mean (by the law of large numbers) and the sample median (which converges to the population median \(\theta\)) are both consistent, making B and C true.

Discuss in Forum

40.

If one observation X from Bernoulli distribution with parameter \(p \in \left[\frac{1}{3}, \frac{2}{3}\right]\) is obtained, then the MLE of \(p\) is :

A.X
B.\(\frac{2X + 1}{3}\)
C.\(\frac{X + 1}{3}\)
D.\(\frac{X - 1}{3}\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: C

The MLE is \(\frac{X + 1}{3}\).

The likelihood of one Bernoulli observation is \(p^{X}(1 - p)^{1 - X}\). If \(X = 1\), it is \(p\), increasing in \(p\), so on \([1/3, 2/3]\) it is maximised at the right end \(p = 2/3\). If \(X = 0\), it is \(1 - p\), decreasing, so it is maximised at \(p = 1/3\).

The formula \((X + 1)/3\) gives exactly 1/3 for \(X = 0\) and 2/3 for \(X = 1\).

The unrestricted MLE X takes the values 0 and 1, which lie outside the parameter space. \((2X + 1)/3\) gives 1 when \(X = 1\), also outside, and \((X - 1)/3\) is negative or zero, never in \([1/3, 2/3]\).

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