Sampling Distributions and Limit Theorems Questions and Answers

Q: Let \(X_{1}, X_{2}, \ldots, X_{9}, Y\) be independent and identically distributed \(N(\mu, \sigma^{2})\) random variables. Define \(\bar{X} = \frac{1}{9}\sum_{i=1}^{9} X_{i}\), \(S^{2} = \frac{1}{8}\sum_{i=1}^{9} (X_{i} - \bar{X})^{2}\).The distribution of \(\frac{3}{\sqrt{10}} \left(\frac{\bar{X} - Y}{S}\right)\) is

The statistic has the \(t_{8}\) distribution.Since \(Y\) is independent of \(\bar{X}\), with \(Var(\bar{X}) = \frac{\sigma^{2}}{9}\),\[ \begin{aligned} \bar{X} - Y &\sim N\left(0, \sigma^{2}\left(\tfrac{1}{9} + 1\right)\right) \\ &= N\left(0, \tfrac{10\sigma^{2}}{9}\right) \end{aligned} \]so\[ Z = \frac{3(\bar{X} - Y)}{\sqrt{10}\,\sigma} \sim N(0,1) \]Also \(\frac{8S^{2}}{\sigma^{2}} \sim \chi_{8}^{2}\), independent of \(\bar{X}\) and \(Y\). Dividing:\[ \begin{aligned} \frac{Z}{\sqrt{\frac{8S^{2}/\sigma^{2}}{8}}} &= \frac{Z\sigma}{S} \\ &= \frac{3}{\sqrt{10}}\left(\frac{\bar{X} - Y}{S}\right) \sim t_{8} \end{aligned} \]The degrees of freedom come from \(S^{2}\), which uses 9 observations and loses one to \(\bar{X}\), so \(t_{9}\) is wrong. A ratio with a normal numerator can be negative, so it cannot be chi-square.

Q: Let \(X_{1}, X_{2}, X_{3}\) be three independent random variables such that \(X_{1} \sim N\left(0, \frac{1}{2}\right)\), \(X_{2} \sim N(0, 2)\), and \(X_{3} \sim N(0, 4)\). Consider the following statements.P: \(\frac{16X_{1}^{2}}{2X_{2}^{2} + X_{3}^{2}}\) has \(F\) distribution with \(1\) and \(2\) degrees of freedom.Q: \(\frac{2X_{1} - X_{2}}{2X_{1} + X_{2}}\) has standard Cauchy distribution.Then

Both P and Q are correct.P: standardise each variable: \(\sqrt{2}X_{1}\), \(\frac{X_{2}}{\sqrt{2}}\) and \(\frac{X_{3}}{2}\) are independent \(N(0, 1)\). So \(2X_{1}^{2} \sim \chi_{1}^{2}\) and, independently,\[ \begin{aligned} \frac{X_{2}^{2}}{2} + \frac{X_{3}^{2}}{4} &= \frac{2X_{2}^{2} + X_{3}^{2}}{4} \\ &\sim \chi_{2}^{2} \end{aligned} \] The F ratio divides each by its degrees of freedom:\[ \begin{aligned} \frac{2X_{1}^{2}/1}{(2X_{2}^{2} + X_{3}^{2})/8} &= \frac{16X_{1}^{2}}{2X_{2}^{2} + X_{3}^{2}} \\ &\sim F_{1,2} \end{aligned} \]Q: \(2X_{1}\) and \(X_{2}\) are independent \(N(0, 2)\), with equal variances. Then \(U = 2X_{1} - X_{2}\) and \(V = 2X_{1} + X_{2}\) are jointly normal with\[ Cov(U, V) = Var(2X_{1}) - Var(X_{2}) = 2 - 2 = 0 \]so they are independent \(N(0, 4)\) variables. The ratio of two independent zero-mean normals with the same variance is standard Cauchy, so \(\frac{U}{V}\) is standard Cauchy.

Q: Let \(X_{1}, X_{2}, \ldots\) be a sequence of independent random variables with\[ \begin{aligned} P(X_{n} = n) &= \frac{1}{n^{2}} = 1 - P(X_{n} = 0), \\ &\text{for } n = 1, 2, \ldots. \end{aligned} \]Consider the following statements.P: \(\{X_{n}\}\) converges in mean-square to \(0\).Q: \(\{X_{n}\}\) converges in probability to \(0\).Then

P is NOT correct and Q is correct.Mean square: \(E(X_{n}^{2}) = n^{2} \cdot \frac{1}{n^{2}} + 0 = 1\) for every \(n\). It does not tend to 0, so \(X_{n}\) does not converge to 0 in mean square.Probability: for any \(\varepsilon\) with \(0 \lt \varepsilon \lt 1\),\[ \begin{aligned} P(|X_{n}| \gt \varepsilon) &= P(X_{n} = n) \\ &= \frac{1}{n^{2}} \to 0 \end{aligned} \]so \(X_{n} \to 0\) in probability. The example shows that convergence in probability does not imply convergence in mean square: the rare value \(n\) is large enough to keep the second moment at 1. (Since \(\sum \frac{1}{n^{2}}\) converges, the Borel-Cantelli lemma even gives almost sure convergence.)

Q: The mean height of 10000 children of age 6 years is 41.26" and the standard deviation is 2.24". Then the odds against the possibility that the mean of a random sample of 100 is greater than 41.7" is :

The odds against are 39 : 1.The sample mean of 100 children has standard error \(2.24/\sqrt{100} = 0.224\). Standardise the value 41.7:\[ \begin{aligned} z &= \frac{41.7 - 41.26}{0.224} \\ &= \frac{0.44}{0.224} \approx 1.96 \end{aligned} \]So \(P(\bar{X} \gt 41.7) = P(Z \gt 1.96) \approx 0.025\), and the probability of the opposite event is 0.975. The odds against the event are\[ 0.975 : 0.025 = 39 : 1 \]1 : 39 gives the odds in favour, not against. 40 : 1 and 1 : 40 compare the event with the whole, 1 in 40, rather than with its complement.

Q: Let \(X_{1}, X_{2}, \ldots X_{n}\) be a random sample from Poisson distribution with parameter \(\lambda\). Then covariance between \(\bar{X}\) and \((X_{1} - X_{2})\) is :

The covariance is 0.Use bilinearity of covariance and independence of the sample values, each with variance \(\lambda\):\[ \begin{aligned} \text{Cov}(\bar{X}, X_{1} - X_{2}) &= \tfrac{1}{n}[\text{Cov}(X_{1}, X_{1}) - \text{Cov}(X_{2}, X_{2})] \\ &= \tfrac{1}{n}(\lambda - \lambda) = 0 \end{aligned} \]All the other terms \(\text{Cov}(X_{i}, X_{j})\) with \(i \ne j\) vanish by independence. The result holds for any iid sample, Poisson or not, since only the equal variances matter.1, -1 and 1/2 come from keeping only one of the two variance terms, or forgetting the factor \(1/n\) from the mean.

Q: Let \(\{X_{n}\}\) be a sequence of random variables such that :\(P(X_{n} = -n^{1/2}) = P(X_{n} = n^{1/2}) = 1/2n\) and \(P(X_{n} = 0) = (n - 1)/n\).Let \(S_{n} = \sum_{i=1}^{n} X_{i}\).Then, which of the following statements is not true ?

The statement that is not true is \(S_{n}/\sqrt{n} \to N(0, 1)\), taking the \(X_{n}\) independent as the question intends.Each \(X_{n}\) has \(E(X_{n}) = 0\) and \(V(X_{n}) = n \cdot \tfrac{1}{2n} \cdot 2 = 1\), so A and B hold, and \(V(S_{n}/\sqrt{n}) = 1\). But the Lindeberg condition fails: almost all the variance of \(X_{k}\) sits on the rare values \(\pm\sqrt{k}\), which are of the same order as \(\sqrt{n}\) for \(k\) near \(n\).The characteristic function of \(S_{n}/\sqrt{n}\) has logarithm close to \(-\sum_{k \le n}\tfrac{1}{k}(1 - \cos(t\sqrt{k/n}))\), which tends to \(-\int_{0}^{1}\tfrac{1 - \cos(t\sqrt{u})}{u}du\). That is a genuine limit but not \(-t^{2}/2\), so the limit Z exists and is not Gaussian (D), and C is false.

Q: Let \(X_{1}\), \(X_{2}\), .......... \(X_{n}\) be iid r.v.s with \(U(\theta, \theta + 1)\), then \(E[X_{(n)} - X_{(1)}]\), where \(X_{(n)} = \text{Max } X_{i}\), and \(X_{(1)} = \text{Min } X_{i}\), is given by :

\(E[X_{(n)} - X_{(1)}] =\) \(\dfrac{n - 1}{n + 1}\).Write \(X_{i} = \theta + U_{i}\) with \(U_{i}\) uniform on (0, 1). The range does not depend on \(\theta\). For uniform (0, 1) samples, \(E[U_{(n)}] = n/(n + 1)\) and \(E[U_{(1)}] = 1/(n + 1)\), so\[ \begin{aligned} E[X_{(n)} - X_{(1)}] &= \frac{n}{n + 1} - \frac{1}{n + 1} \\ &= \frac{n - 1}{n + 1} \end{aligned} \]For \(n = 1\) the range is 0, as the formula gives. \(1/(n + 1)\) is the expected minimum alone, and the other two options exceed 1, which no range of points in an interval of length 1 can do.

Q: Based on a random sample of size \(n\) from Poisson distribution with mean \(\lambda\), asymptotic variance of \(\bar{X}e^{-\bar{X}}\) is :

The asymptotic variance is \(\dfrac{\lambda}{n}(1 - \lambda)^{2}e^{-2\lambda}\), \(\lambda \ne 1\).\(\bar{X}\) is asymptotically \(N(\lambda, \lambda/n)\). By the delta method, \(g(\bar{X})\) has asymptotic variance \(g'(\lambda)^{2}\lambda/n\). For \(g(x) = xe^{-x}\), \(g'(x) = e^{-x}(1 - x)\), so\[ \frac{\lambda}{n}\,(1 - \lambda)^{2}e^{-2\lambda} \]At \(\lambda = 1\), \(g'(1) = 0\) and the first-order delta method gives a degenerate answer, which is why that value is excluded. Option C forgets to square \(g'\), and A and D leave out the derivative altogether.

Q: Suppose \(X_{1}, X_{2}, X_{3}\) are independent and identically distributed random variables each having exponential distribution with mean \(\theta\). Suppose \(Y_{1}, Y_{2}, Y_{3}\) is the corresponding order statistics. Then, \(E(Y_{1})\) is :

\(E(Y_{1}) =\) \(\theta/3\).\(Y_{1}\) is the minimum of three independent exponentials, each with rate \(1/\theta\). The minimum survives past \(y\) only if all three do:\[ \begin{aligned} P(Y_{1} \gt y) &= \left(e^{-y/\theta}\right)^{3} \\ &= e^{-3y/\theta} \end{aligned} \]so \(Y_{1}\) is exponential with mean \(\theta/3\). For reference, the spacings give \(E(Y_{2}) = \theta/3 + \theta/2 = 5\theta/6\) and \(E(Y_{3}) = 11\theta/6\); option A is the mean of the middle order statistic, not the smallest, and \(\theta/6\) and \(\theta\) fit none of them.

Q: Suppose \(\{X_{1}, X_{2}, ......., X_{n}\}\) is a random sample from the distribution of X with mean \(\mu\) and variance \(\sigma^{2}\). The test statistic to test \(H_{0} : \sigma^{2} = \sigma_{0}^{2}\) against \(H_{1} : \sigma^{2} \gt \sigma_{0}^{2}\) is given by \(T_{n} = \dfrac{\Sigma(X_{i} - \bar{X})^{2}}{\sigma_{0}^{2}}\). Then the null distribution of \(T_{n}\) is :

The null distribution is \(\chi_{n-1}^{2}\) if X has a normal distribution.For a normal sample, \(\sum(X_{i} - \bar{X})^{2}/\sigma^{2}\) \(= (n - 1)S^{2}/\sigma^{2} \sim \chi_{n-1}^{2}\), with one degree of freedom lost to estimating the mean. Under \(H_{0}\), \(\sigma^{2} = \sigma_{0}^{2}\), so \(T_{n} \sim \chi_{n-1}^{2}\).The result depends on normality, so 'irrespective of distribution' is wrong. The \(t\) distribution is for a standardised mean, and F is for a ratio of two independent variance estimates.

Direction for [ Question No: 1 To 2 ] :

Special Instructions / Useful Data

\(N(\mu, \sigma^{2})\)	Normal distribution with mean \(\mu\) and variance \(\sigma^{2}\), \(\mu \in \mathbb{R}, \sigma \gt 0\)
\(Exp(\theta)\)	Exponential distribution with the probability density function, for \(\theta \gt 0\), \(f(x) = \frac{1}{\theta} e^{-\frac{x}{\theta}}\), \(x \gt 0\); \(0\), otherwise
\(U(a, b)\)	Continuous uniform distribution on the interval \((a, b)\), \(-\infty \lt a \lt b \lt \infty\)
\(\chi_{n}^{2}\)	Central chi-square distribution with \(n\) degrees of freedom
\(t_{n}\)	Central Student’s \(t\) distribution with \(n\) degrees of freedom
\(\chi_{n,\alpha}^{2}\)	A constant such that \(P(X \gt \chi_{n,\alpha}^{2}) = \alpha\), where \(X \sim \chi_{n}^{2}\) and \(\alpha \in (0,1)\)
\(X_{(r)}\)	The \(r\)-th order statistic of a random sample \(X_{1}, X_{2}, \ldots, X_{n}\) of size \(n\), \(r = 1, 2, \ldots, n\) \((n \geq 1)\)

Let \(X_{1}, X_{2}, \ldots, X_{9}, Y\) be independent and identically distributed \(N(\mu, \sigma^{2})\) random variables. Define \(\bar{X} = \frac{1}{9}\sum_{i=1}^{9} X_{i}\), \(S^{2} = \frac{1}{8}\sum_{i=1}^{9} (X_{i} - \bar{X})^{2}\).

The distribution of \(\frac{3}{\sqrt{10}} \left(\frac{\bar{X} - Y}{S}\right)\) is

A.\(\chi_{8}^{2}\)
B.\(\chi_{9}^{2}\)
C.\(t_{8}\)
D.\(t_{9}\)

Exam Year(s)

IIT JAM - 2026,

Answer: C

The statistic has the \(t_{8}\) distribution.

Since \(Y\) is independent of \(\bar{X}\), with \(Var(\bar{X}) = \frac{\sigma^{2}}{9}\),

\[ \begin{aligned} \bar{X} - Y &\sim N\left(0, \sigma^{2}\left(\tfrac{1}{9} + 1\right)\right) \\ &= N\left(0, \tfrac{10\sigma^{2}}{9}\right) \end{aligned} \]

\[ Z = \frac{3(\bar{X} - Y)}{\sqrt{10}\,\sigma} \sim N(0,1) \]

Also \(\frac{8S^{2}}{\sigma^{2}} \sim \chi_{8}^{2}\), independent of \(\bar{X}\) and \(Y\). Dividing:

\[ \begin{aligned} \frac{Z}{\sqrt{\frac{8S^{2}/\sigma^{2}}{8}}} &= \frac{Z\sigma}{S} \\ &= \frac{3}{\sqrt{10}}\left(\frac{\bar{X} - Y}{S}\right) \sim t_{8} \end{aligned} \]

The degrees of freedom come from \(S^{2}\), which uses 9 observations and loses one to \(\bar{X}\), so \(t_{9}\) is wrong. A ratio with a normal numerator can be negative, so it cannot be chi-square.

Discuss in Forum

Let \(X_{1}, X_{2}, X_{3}\) be three independent random variables such that \(X_{1} \sim N\left(0, \frac{1}{2}\right)\), \(X_{2} \sim N(0, 2)\), and \(X_{3} \sim N(0, 4)\). Consider the following statements.

P: \(\frac{16X_{1}^{2}}{2X_{2}^{2} + X_{3}^{2}}\) has \(F\) distribution with \(1\) and \(2\) degrees of freedom.

Q: \(\frac{2X_{1} - X_{2}}{2X_{1} + X_{2}}\) has standard Cauchy distribution.

Then

A.P is correct and Q is NOT correct
B.P is NOT correct and Q is correct
C.Both P and Q are correct
D.Neither P nor Q is correct

Exam Year(s)

IIT JAM - 2026,

Answer: C

Both P and Q are correct.

P: standardise each variable: \(\sqrt{2}X_{1}\), \(\frac{X_{2}}{\sqrt{2}}\) and \(\frac{X_{3}}{2}\) are independent \(N(0, 1)\). So \(2X_{1}^{2} \sim \chi_{1}^{2}\) and, independently,

\[ \begin{aligned} \frac{X_{2}^{2}}{2} + \frac{X_{3}^{2}}{4} &= \frac{2X_{2}^{2} + X_{3}^{2}}{4} \\ &\sim \chi_{2}^{2} \end{aligned} \]

The F ratio divides each by its degrees of freedom:

\[ \begin{aligned} \frac{2X_{1}^{2}/1}{(2X_{2}^{2} + X_{3}^{2})/8} &= \frac{16X_{1}^{2}}{2X_{2}^{2} + X_{3}^{2}} \\ &\sim F_{1,2} \end{aligned} \]

Q: \(2X_{1}\) and \(X_{2}\) are independent \(N(0, 2)\), with equal variances. Then \(U = 2X_{1} - X_{2}\) and \(V = 2X_{1} + X_{2}\) are jointly normal with

\[ Cov(U, V) = Var(2X_{1}) - Var(X_{2}) = 2 - 2 = 0 \]

so they are independent \(N(0, 4)\) variables. The ratio of two independent zero-mean normals with the same variance is standard Cauchy, so \(\frac{U}{V}\) is standard Cauchy.

Discuss in Forum

Let \(X_{1}, X_{2}, \ldots\) be a sequence of independent random variables with

\[ \begin{aligned} P(X_{n} = n) &= \frac{1}{n^{2}} = 1 - P(X_{n} = 0), \\ &\text{for } n = 1, 2, \ldots. \end{aligned} \]

Consider the following statements.

P: \(\{X_{n}\}\) converges in mean-square to \(0\).

Q: \(\{X_{n}\}\) converges in probability to \(0\).

Then

A.P is correct and Q is NOT correct
B.P is NOT correct and Q is correct
C.Both P and Q are correct
D.Neither P nor Q is correct

Exam Year(s)

IIT JAM - 2026,

Answer: B

P is NOT correct and Q is correct.

Mean square: \(E(X_{n}^{2}) = n^{2} \cdot \frac{1}{n^{2}} + 0 = 1\) for every \(n\). It does not tend to 0, so \(X_{n}\) does not converge to 0 in mean square.

Probability: for any \(\varepsilon\) with \(0 \lt \varepsilon \lt 1\),

\[ \begin{aligned} P(|X_{n}| \gt \varepsilon) &= P(X_{n} = n) \\ &= \frac{1}{n^{2}} \to 0 \end{aligned} \]

so \(X_{n} \to 0\) in probability. The example shows that convergence in probability does not imply convergence in mean square: the rare value \(n\) is large enough to keep the second moment at 1. (Since \(\sum \frac{1}{n^{2}}\) converges, the Borel-Cantelli lemma even gives almost sure convergence.)

Discuss in Forum

The mean height of 10000 children of age 6 years is 41.26" and the standard deviation is 2.24". Then the odds against the possibility that the mean of a random sample of 100 is greater than 41.7" is :

A.1 : 39
B.39 : 1
C.1 : 40
D.40 : 1

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

The odds against are 39 : 1.

The sample mean of 100 children has standard error \(2.24/\sqrt{100} = 0.224\). Standardise the value 41.7:

\[ \begin{aligned} z &= \frac{41.7 - 41.26}{0.224} \\ &= \frac{0.44}{0.224} \approx 1.96 \end{aligned} \]

So \(P(\bar{X} \gt 41.7) = P(Z \gt 1.96) \approx 0.025\), and the probability of the opposite event is 0.975. The odds against the event are

\[ 0.975 : 0.025 = 39 : 1 \]

1 : 39 gives the odds in favour, not against. 40 : 1 and 1 : 40 compare the event with the whole, 1 in 40, rather than with its complement.

Discuss in Forum

Let \(X_{1}, X_{2}, \ldots X_{n}\) be a random sample from Poisson distribution with parameter \(\lambda\). Then covariance between \(\bar{X}\) and \((X_{1} - X_{2})\) is :

A.1/2
B.1
C.-1
D.0

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: D

The covariance is 0.

Use bilinearity of covariance and independence of the sample values, each with variance \(\lambda\):

\[ \begin{aligned} \text{Cov}(\bar{X}, X_{1} - X_{2}) &= \tfrac{1}{n}[\text{Cov}(X_{1}, X_{1}) - \text{Cov}(X_{2}, X_{2})] \\ &= \tfrac{1}{n}(\lambda - \lambda) = 0 \end{aligned} \]

All the other terms \(\text{Cov}(X_{i}, X_{j})\) with \(i \ne j\) vanish by independence. The result holds for any iid sample, Poisson or not, since only the equal variances matter.

1, -1 and 1/2 come from keeping only one of the two variance terms, or forgetting the factor \(1/n\) from the mean.

Discuss in Forum

Let \(\{X_{n}\}\) be a sequence of random variables such that :

\(P(X_{n} = -n^{1/2}) = P(X_{n} = n^{1/2}) = 1/2n\) and \(P(X_{n} = 0) = (n - 1)/n\).

Let \(S_{n} = \sum_{i=1}^{n} X_{i}\).

Then, which of the following statements is not true ?

A.\(E(X_{n}) = 0\)
B.\(V(X_{n}) = 1\)
C.\(S_{n}/\sqrt{n} \xrightarrow{d} Z\), \(Z \sim N(0, 1)\)
D.\(S_{n}/\sqrt{n} \xrightarrow{d} Z\), Z is not Gaussian

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: C

The statement that is not true is \(S_{n}/\sqrt{n} \to N(0, 1)\), taking the \(X_{n}\) independent as the question intends.

Each \(X_{n}\) has \(E(X_{n}) = 0\) and \(V(X_{n}) = n \cdot \tfrac{1}{2n} \cdot 2 = 1\), so A and B hold, and \(V(S_{n}/\sqrt{n}) = 1\). But the Lindeberg condition fails: almost all the variance of \(X_{k}\) sits on the rare values \(\pm\sqrt{k}\), which are of the same order as \(\sqrt{n}\) for \(k\) near \(n\).

The characteristic function of \(S_{n}/\sqrt{n}\) has logarithm close to \(-\sum_{k \le n}\tfrac{1}{k}(1 - \cos(t\sqrt{k/n}))\), which tends to \(-\int_{0}^{1}\tfrac{1 - \cos(t\sqrt{u})}{u}du\). That is a genuine limit but not \(-t^{2}/2\), so the limit Z exists and is not Gaussian (D), and C is false.

Discuss in Forum

Let \(X_{1}\), \(X_{2}\), .......... \(X_{n}\) be iid r.v.s with \(U(\theta, \theta + 1)\), then \(E[X_{(n)} - X_{(1)}]\), where \(X_{(n)} = \text{Max } X_{i}\), and \(X_{(1)} = \text{Min } X_{i}\), is given by :

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

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: C

\(E[X_{(n)} - X_{(1)}] =\) \(\dfrac{n - 1}{n + 1}\).

Write \(X_{i} = \theta + U_{i}\) with \(U_{i}\) uniform on (0, 1). The range does not depend on \(\theta\). For uniform (0, 1) samples, \(E[U_{(n)}] = n/(n + 1)\) and \(E[U_{(1)}] = 1/(n + 1)\), so

\[ \begin{aligned} E[X_{(n)} - X_{(1)}] &= \frac{n}{n + 1} - \frac{1}{n + 1} \\ &= \frac{n - 1}{n + 1} \end{aligned} \]

For \(n = 1\) the range is 0, as the formula gives. \(1/(n + 1)\) is the expected minimum alone, and the other two options exceed 1, which no range of points in an interval of length 1 can do.

Discuss in Forum

Based on a random sample of size \(n\) from Poisson distribution with mean \(\lambda\), asymptotic variance of \(\bar{X}e^{-\bar{X}}\) is :

A.\(\dfrac{\lambda e^{-\lambda}}{n}\), \(\lambda \gt 0\)
B.\(\dfrac{\lambda}{n}(1 - \lambda)^{2}e^{-2\lambda}\), \(\lambda \ne 1\)
C.\(\dfrac{\lambda}{n}(1 - \lambda)e^{-\lambda}\), \(\lambda \ne 1\)
D.\(\dfrac{\lambda}{n}\), \(\lambda \gt 0\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

The asymptotic variance is \(\dfrac{\lambda}{n}(1 - \lambda)^{2}e^{-2\lambda}\), \(\lambda \ne 1\).

\(\bar{X}\) is asymptotically \(N(\lambda, \lambda/n)\). By the delta method, \(g(\bar{X})\) has asymptotic variance \(g'(\lambda)^{2}\lambda/n\). For \(g(x) = xe^{-x}\), \(g'(x) = e^{-x}(1 - x)\), so

\[ \frac{\lambda}{n}\,(1 - \lambda)^{2}e^{-2\lambda} \]

At \(\lambda = 1\), \(g'(1) = 0\) and the first-order delta method gives a degenerate answer, which is why that value is excluded. Option C forgets to square \(g'\), and A and D leave out the derivative altogether.

Discuss in Forum

Suppose \(X_{1}, X_{2}, X_{3}\) are independent and identically distributed random variables each having exponential distribution with mean \(\theta\). Suppose \(Y_{1}, Y_{2}, Y_{3}\) is the corresponding order statistics. Then, \(E(Y_{1})\) is :

A.\(\dfrac{5\theta}{6}\)
B.\(\dfrac{\theta}{6}\)
C.\(\dfrac{\theta}{3}\)
D.\(\theta\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: C

\(E(Y_{1}) =\) \(\theta/3\).

\(Y_{1}\) is the minimum of three independent exponentials, each with rate \(1/\theta\). The minimum survives past \(y\) only if all three do:

\[ \begin{aligned} P(Y_{1} \gt y) &= \left(e^{-y/\theta}\right)^{3} \\ &= e^{-3y/\theta} \end{aligned} \]

so \(Y_{1}\) is exponential with mean \(\theta/3\). For reference, the spacings give \(E(Y_{2}) = \theta/3 + \theta/2 = 5\theta/6\) and \(E(Y_{3}) = 11\theta/6\); option A is the mean of the middle order statistic, not the smallest, and \(\theta/6\) and \(\theta\) fit none of them.

Discuss in Forum

10.

Suppose \(\{X_{1}, X_{2}, ......., X_{n}\}\) is a random sample from the distribution of X with mean \(\mu\) and variance \(\sigma^{2}\). The test statistic to test \(H_{0} : \sigma^{2} = \sigma_{0}^{2}\) against \(H_{1} : \sigma^{2} \gt \sigma_{0}^{2}\) is given by \(T_{n} = \dfrac{\Sigma(X_{i} - \bar{X})^{2}}{\sigma_{0}^{2}}\). Then the null distribution of \(T_{n}\) is :

A.\(\chi_{n - 1}^{2}\) if X has normal distribution
B.\(\chi_{n - 1}^{2}\) irrespective of distribution of X
C.\(t\) distribution with \((n - 1)\) degrees of freedom
D.F distribution if X has normal distribution

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: A

The null distribution is \(\chi_{n-1}^{2}\) if X has a normal distribution.

For a normal sample, \(\sum(X_{i} - \bar{X})^{2}/\sigma^{2}\) \(= (n - 1)S^{2}/\sigma^{2} \sim \chi_{n-1}^{2}\), with one degree of freedom lost to estimating the mean. Under \(H_{0}\), \(\sigma^{2} = \sigma_{0}^{2}\), so \(T_{n} \sim \chi_{n-1}^{2}\).

The result depends on normality, so 'irrespective of distribution' is wrong. The \(t\) distribution is for a standardised mean, and F is for a ratio of two independent variance estimates.

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