Sampling Distributions and Limit Theorems Questions and Answers

Q: Let \(\{X_{n}, n \ge 1\}\) be a sequence of independent identically distributed random variables with finite (non-zero) fourth moment.Let \(Y_{n} = (X_{1}^{2} + ......... + X_{n}^{2})/n\)Which of the following is true ?As \(n \to \infty\), \(\sqrt{n}(Y_{n} - E[X_{1}^{2}])\).

The limit is a normal random variable.The \(X_{i}^{2}\) are i.i.d. with mean \(E[X_{1}^{2}]\) and variance \(\tau^{2} = E[X_{1}^{4}] - (E[X_{1}^{2}])^{2}\), which is finite because the fourth moment is. By the central limit theorem,\[ \begin{aligned} &\sqrt{n}\left(Y_{n} - E[X_{1}^{2}]\right) \\ &\xrightarrow{d} N(0, \tau^{2}) \end{aligned} \]It is standard normal only if \(\tau^{2} = 1\), which is not assumed, so B is wrong; C covers every \(\tau^{2}\) (including the degenerate normal when \(X_{1}^{2}\) is constant). A chi-square limit arises for \(n(\ldots)^{2}\)-type statistics, not this one, and degeneracy at 0 holds only in that special constant case.

Q: Suppose \(\{X_{n}\}\) is a sequence of random variables and X is a random variable such that \(P\left[\lim_{n\to\infty} X_{n} = X\right] = 1\).Then which of the following may not hold ?

The one that may not hold is \(\lim E[|X_{n} - X|^{2}] = 0\).Almost sure convergence implies convergence in probability (A) and in distribution (D), and a continuous function such as exp preserves almost sure convergence (B). It does not imply convergence in mean square.Example: on (0, 1) with Lebesgue measure, \(X_{n} = n\,1_{(0, 1/n)}\) tends to \(X = 0\) at every point, but \(E[X_{n}^{2}] = n^{2} \cdot \tfrac{1}{n} = n \to \infty\). Without a uniform integrability condition, the mean square need not converge.

Q: Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from \(U(\theta, \theta + 1)\). If \(X_{(1)} \lt X_{(2)} \lt \cdots \lt X_{(n)}\) are corresponding order statistics, then :

The answer is \((X_{(1)}, X_{(n)})\) is jointly sufficient for \(\theta\).The joint density is\[ \begin{aligned} &\prod_{i} 1_{\{\theta \lt x_{i} \lt \theta + 1\}} \\ &= 1_{\{x_{(n)} - 1 \lt \theta \lt x_{(1)}\}} \end{aligned} \]which depends on the data only through the smallest and largest observations. By the factorisation theorem, \((X_{(1)}, X_{(n)})\) is sufficient (indeed minimal sufficient).Neither order statistic alone is sufficient, because the likelihood is non-zero on an interval whose two ends come from both of them. Adding the constant 1 to a component does not lose information, so the last option is also false.

Q: Let \(\{X_{n}, n \ge 1\}\) be a sequence of independent identically distributed random variables with mean 0 and finite variance \(\sigma^{2} \gt 0\) :Let \(S_{n} = \dfrac{X_{1} + X_{2} + \cdots + X_{n^{2}}}{n^{2}}\)Then, which of the following is false ?

The false statement is \(S_{n}e^{S_{n}}\) converges to a normal random variable with mean 0 and variance \(\sigma^{2}\).\(S_{n}\) is the average of the first \(n^{2}\) terms of an iid sequence with mean 0, so by the strong law of large numbers \(S_{n} \to 0\) almost surely. By continuity, \(S_{n}e^{S_{n}} \to 0\) and \(e^{S_{n}} \to 1\) almost surely, and a.s. convergence implies convergence in probability. That makes A, C and D true.A degenerate limit 0 is not normal with variance \(\sigma^{2} \gt 0\). (A normal limit would appear only after rescaling, for \(\sqrt{n^{2}}\,S_{n} = nS_{n}\).)

Q: Let \(\{X_{n}, n \ge 1\}\) be a sequence of random variables such that \(P\left[\lim_{n\to\infty} X_{n} = X\right] = 1\) for some r.v. X. Then which of the following is not always correct ?

The statement not always correct is \(\lim E[\exp(X_{n})] = E[\exp(X)]\).Almost sure convergence carries over to continuous functions: \(\exp(X_{n}) \to \exp(X)\) a.s. (C). For bounded continuous functions such as \(\cos\) and \(e^{itx}\), the dominated convergence theorem then gives convergence of expectations (A, B).\(\exp\) is unbounded, so expectations need not converge. For example, on [0, 1] with Lebesgue measure let \(X_{n} = \log n\) on \([0, 1/n]\) and 0 elsewhere. Then \(X_{n} \to 0\) a.s., but \(E[e^{X_{n}}] = n \cdot \tfrac{1}{n} + (1 - \tfrac{1}{n}) \to 2 \ne e^{0} = 1\).

Q: Let \(\{X_{n}\}\) be a sequence of independent and uniformly bounded random variables, and \(E[X_{n}] = 0\), and \(\mathrm{Var}(X_{n}) = \dfrac{1}{(2n)}\) for each \(n\).Let \(S_{n} = X_{1} + \cdots + X_{n}\) and \(\sigma_{n}^{2} = \mathrm{Var}(S_{n})\)Then which of the following is not true ?

The statement that is not true is \(S_{n}/\sigma_{n}\) does not converge in distribution.Here \(\sigma_{n}^{2} = \sum_{k \le n} 1/(2k) \to \infty\), since the harmonic series diverges. For independent, uniformly bounded variables (\(|X_{k}| \le M\)) with \(\sigma_{n} \to \infty\), Lyapunov's condition holds:\[ \begin{aligned} &\frac{1}{\sigma_{n}^{3}}\sum_{k=1}^{n} E|X_{k}|^{3} \le \frac{M\sigma_{n}^{2}}{\sigma_{n}^{3}} \\ &= \frac{M}{\sigma_{n}} \to 0 \end{aligned} \]So (C) is true, and the central limit theorem gives \(S_{n}/\sigma_{n} \to N(0, 1)\) in distribution (B). Then \(S_{n}/\sigma_{n}^{2} = (S_{n}/\sigma_{n})/\sigma_{n} \to 0\) in probability (D). Only (A) contradicts these.

Q: Let \(\{X_{n}\}\) be a sequence of independent random variables such that \(E[X_{n}] = 0\) and \(\mathrm{Var}(X_{n}) = \dfrac{1}{n^{2}}\), \(n = 1, 2, \ldots\). Let \(E = \left\{\omega \mid \sum_{n} X_{n}(w)\ \text{converges}\right\}\).Then which of the following is always correct ?

Always \(P(E) = 1\).By Kolmogorov's one-series (two-series) theorem, a series of independent random variables with mean 0 converges almost surely when the sum of the variances is finite:\[ \begin{aligned} \sum_{n} \mathrm{Var}(X_{n}) &= \sum_{n} \frac{1}{n^{2}} \\ &= \frac{\pi^{2}}{6} \lt \infty \end{aligned} \]So \(\sum X_{n}\) converges with probability 1. (Kolmogorov's 0-1 law already says that P(E) is 0 or 1, since convergence of the series is a tail event; this rules out 1/2 and anything strictly between 0 and 1.)

Q: Suppose \(X_{1}, X_{2}, X_{3}\) are independent and identically distributed random variables each having exponential distribution with mean 1. Then the probability density function of sample range is given by,

The density of the range is \(h(r) = 2(1 - e^{-r})e^{-r}\), \(r \gt 0\).For \(n\) independent standard exponentials, the spacings of the order statistics are independent exponentials with rates \(n - 1, n - 2, \ldots\) after the minimum. So the range \(X_{(3)} - X_{(1)}\) is the sum of exponentials with rates 2 and 1, which is the maximum of two independent standard exponentials. Its distribution function is \((1 - e^{-r})^{2}\), and differentiating gives\[ h(r) = 2(1 - e^{-r})e^{-r}, \quad r \gt 0 \]\(e^{-r}\) is the density of one exponential, and the other options lack the factor 2, so they do not integrate to 1, or are defined on all of \(\Re\), where a range cannot be negative.

Q: Let \(\{X_{n}\}\) be a sequence of random variables and X a random variable such that :\[ \sum_{n=1}^{\infty} P[|X_{n} - X| \gt \epsilon] \lt \infty \]for every \(\epsilon \gt 0\).Consider the statements :(I) \(P\left[\lim_{n \to \infty} X_{n} = X\right] = 1\)(II) \(P\left[\lim_{n \to \infty} X_{n}^{2} = X^{2}\right] = 1\)(III) \(P\left[\lim_{n \to \infty} X_{n} = X\right]\) only if \(\{X_{n}\}\) is a sequence of independent random variables.Which of the above are always correct ?

The statements always correct are (I) and (II) only.By the Borel-Cantelli lemma, \(\sum P[|X_{n} - X| \gt \epsilon] \lt \infty\) for every \(\epsilon\) means that, almost surely, \(|X_{n} - X| \gt \epsilon\) happens only finitely often. So \(X_{n} \to X\) almost surely (I). Since \(x \mapsto x^{2}\) is continuous, \(X_{n}^{2} \to X^{2}\) almost surely too (II).(III) is false: this half of Borel-Cantelli needs no independence. Independence matters only for the converse, the second Borel-Cantelli lemma. So options including (III), or 'none of the three', are wrong.

Q: Suppose the sequence of random variables \(\{X_{n}\}\) converges in distribution to X. Then which of the following is not always true ?

The statement not always true is \(\lim_{n} E[X_{n}] = E[X]\).Convergence in distribution does not control expectations without uniform integrability. Let \(X_{n} = n\) with probability \(1/n\) and 0 otherwise: \(X_{n} \to 0\) in distribution, but \(E[X_{n}] = 1\) for every \(n\).The others always hold. (B) is Fatou's lemma in the form valid for convergence in distribution (via Skorokhod's representation). (C) holds because \(|x|/(1 + |x|)\) is bounded and continuous, which is exactly what convergence in distribution handles. (D) holds because a sequence converging in distribution is tight, so \(X_{n}/\sqrt{n} \to 0\) in probability.

11.

Let \(\{X_{n}, n \ge 1\}\) be a sequence of independent identically distributed random variables with finite (non-zero) fourth moment.

Let \(Y_{n} = (X_{1}^{2} + ......... + X_{n}^{2})/n\)

Which of the following is true ?

As \(n \to \infty\), \(\sqrt{n}(Y_{n} - E[X_{1}^{2}])\).

A.Converges in distribution to a chi-square r.v. with 1 degree of freedom.
B.Converges in distribution to a standard normal r.v.
C.Converges in distribution to a normal r.v.
D.Converges to a r.v. degenerate at zero.

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: C

The limit is a normal random variable.

The \(X_{i}^{2}\) are i.i.d. with mean \(E[X_{1}^{2}]\) and variance \(\tau^{2} = E[X_{1}^{4}] - (E[X_{1}^{2}])^{2}\), which is finite because the fourth moment is. By the central limit theorem,

\[ \begin{aligned} &\sqrt{n}\left(Y_{n} - E[X_{1}^{2}]\right) \\ &\xrightarrow{d} N(0, \tau^{2}) \end{aligned} \]

It is standard normal only if \(\tau^{2} = 1\), which is not assumed, so B is wrong; C covers every \(\tau^{2}\) (including the degenerate normal when \(X_{1}^{2}\) is constant). A chi-square limit arises for \(n(\ldots)^{2}\)-type statistics, not this one, and degeneracy at 0 holds only in that special constant case.

Discuss in Forum

12.

Suppose \(\{X_{n}\}\) is a sequence of random variables and X is a random variable such that \(P\left[\lim_{n\to\infty} X_{n} = X\right] = 1\).

Then which of the following may not hold ?

A.\(\lim_{n\to\infty} P[|X_{n} - X| \gt 1/2] = 0\)
B.\(P\left[\lim_{n\to\infty} \exp(X_{n}) = \exp(X)\right] = 1\)
C.\(\lim_{n\to\infty} E[|X_{n} - X|^{2}] = 0\)
D.\(\lim_{n\to\infty} P[X_{n} \le x] = P[X \le x]\) at all continuity points \(x\) of \(P[X \le x]\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: C

The one that may not hold is \(\lim E[|X_{n} - X|^{2}] = 0\).

Almost sure convergence implies convergence in probability (A) and in distribution (D), and a continuous function such as exp preserves almost sure convergence (B). It does not imply convergence in mean square.

Example: on (0, 1) with Lebesgue measure, \(X_{n} = n\,1_{(0, 1/n)}\) tends to \(X = 0\) at every point, but \(E[X_{n}^{2}] = n^{2} \cdot \tfrac{1}{n} = n \to \infty\). Without a uniform integrability condition, the mean square need not converge.

Discuss in Forum

13.

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from \(U(\theta, \theta + 1)\). If \(X_{(1)} \lt X_{(2)} \lt \cdots \lt X_{(n)}\) are corresponding order statistics, then :

A.\((X_{(1)}, X_{(n)})\) is jointly sufficient for \(\theta\)
B.\(X_{(1)}\) is sufficient for \(\theta\)
C.\(X_{(n)}\) is sufficient for \(\theta\)
D.\((X_{(1)}, X_{(n)}, + 1)\) is not jointly sufficient for \(\theta\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: A

The answer is \((X_{(1)}, X_{(n)})\) is jointly sufficient for \(\theta\).

The joint density is

\[ \begin{aligned} &\prod_{i} 1_{\{\theta \lt x_{i} \lt \theta + 1\}} \\ &= 1_{\{x_{(n)} - 1 \lt \theta \lt x_{(1)}\}} \end{aligned} \]

which depends on the data only through the smallest and largest observations. By the factorisation theorem, \((X_{(1)}, X_{(n)})\) is sufficient (indeed minimal sufficient).

Neither order statistic alone is sufficient, because the likelihood is non-zero on an interval whose two ends come from both of them. Adding the constant 1 to a component does not lose information, so the last option is also false.

Discuss in Forum

14.

Let \(\{X_{n}, n \ge 1\}\) be a sequence of independent identically distributed random variables with mean 0 and finite variance \(\sigma^{2} \gt 0\) :

Let \(S_{n} = \dfrac{X_{1} + X_{2} + \cdots + X_{n^{2}}}{n^{2}}\)

Then, which of the following is false ?

A.\(P\left[\lim_{n\to\infty} S_{n}e^{S_{n}} = 0\right] = 1\)
B.As \(n \to \infty\), \(S_{n}e^{S_{n}}\) converges to a normal random variable with mean '0' and variance \(\sigma^{2}\)
C.As \(n \to \infty\), \(S_{n}\) converges to '0' in probability
D.As \(n \to \infty\), \(e^{S_{n}}\) converges to 1 with probability 1

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: B

The false statement is \(S_{n}e^{S_{n}}\) converges to a normal random variable with mean 0 and variance \(\sigma^{2}\).

\(S_{n}\) is the average of the first \(n^{2}\) terms of an iid sequence with mean 0, so by the strong law of large numbers \(S_{n} \to 0\) almost surely. By continuity, \(S_{n}e^{S_{n}} \to 0\) and \(e^{S_{n}} \to 1\) almost surely, and a.s. convergence implies convergence in probability. That makes A, C and D true.

A degenerate limit 0 is not normal with variance \(\sigma^{2} \gt 0\). (A normal limit would appear only after rescaling, for \(\sqrt{n^{2}}\,S_{n} = nS_{n}\).)

Discuss in Forum

15.

Let \(\{X_{n}, n \ge 1\}\) be a sequence of random variables such that \(P\left[\lim_{n\to\infty} X_{n} = X\right] = 1\) for some r.v. X. Then which of the following is not always correct ?

A.\(\lim_{n\to\infty} E[\cos(X_{n})] = E[\cos(X)]\)
B.\(\lim_{n\to\infty} E[\exp(itX_{n})] = E[\exp(itX)]\)
C.\(P\left[\lim_{n\to\infty} \exp(X_{n}) = \exp(X)\right] = 1\)
D.\(\lim_{n\to\infty} E[\exp(X_{n})] = E[\exp(X)]\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: D

The statement not always correct is \(\lim E[\exp(X_{n})] = E[\exp(X)]\).

Almost sure convergence carries over to continuous functions: \(\exp(X_{n}) \to \exp(X)\) a.s. (C). For bounded continuous functions such as \(\cos\) and \(e^{itx}\), the dominated convergence theorem then gives convergence of expectations (A, B).

\(\exp\) is unbounded, so expectations need not converge. For example, on [0, 1] with Lebesgue measure let \(X_{n} = \log n\) on \([0, 1/n]\) and 0 elsewhere. Then \(X_{n} \to 0\) a.s., but \(E[e^{X_{n}}] = n \cdot \tfrac{1}{n} + (1 - \tfrac{1}{n}) \to 2 \ne e^{0} = 1\).

Discuss in Forum

16.

Let \(\{X_{n}\}\) be a sequence of independent and uniformly bounded random variables, and \(E[X_{n}] = 0\), and \(\mathrm{Var}(X_{n}) = \dfrac{1}{(2n)}\) for each \(n\).

Let \(S_{n} = X_{1} + \cdots + X_{n}\) and \(\sigma_{n}^{2} = \mathrm{Var}(S_{n})\)

Then which of the following is not true ?

A.\(\dfrac{S_{n}}{\sigma_{n}}\) does not converge in distribution
B.\(\dfrac{S_{n}}{\sigma_{n}}\) converges in distribution to a normal random variable
C.\(\lim_{n\to\infty} \sum_{k=1}^{n} \frac{1}{\sigma_{n}^{3}} E[|X_{k}|^{3}] = 0\)
D.\(\lim_{n\to\infty} \dfrac{S_{n}}{\sigma_{n}^{2}} = 0\) in probability

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: A

The statement that is not true is \(S_{n}/\sigma_{n}\) does not converge in distribution.

Here \(\sigma_{n}^{2} = \sum_{k \le n} 1/(2k) \to \infty\), since the harmonic series diverges. For independent, uniformly bounded variables (\(|X_{k}| \le M\)) with \(\sigma_{n} \to \infty\), Lyapunov's condition holds:

\[ \begin{aligned} &\frac{1}{\sigma_{n}^{3}}\sum_{k=1}^{n} E|X_{k}|^{3} \le \frac{M\sigma_{n}^{2}}{\sigma_{n}^{3}} \\ &= \frac{M}{\sigma_{n}} \to 0 \end{aligned} \]

So (C) is true, and the central limit theorem gives \(S_{n}/\sigma_{n} \to N(0, 1)\) in distribution (B). Then \(S_{n}/\sigma_{n}^{2} = (S_{n}/\sigma_{n})/\sigma_{n} \to 0\) in probability (D). Only (A) contradicts these.

Discuss in Forum

17.

Let \(\{X_{n}\}\) be a sequence of independent random variables such that \(E[X_{n}] = 0\) and \(\mathrm{Var}(X_{n}) = \dfrac{1}{n^{2}}\), \(n = 1, 2, \ldots\). Let \(E = \left\{\omega \mid \sum_{n} X_{n}(w)\ \text{converges}\right\}\).

Then which of the following is always correct ?

A.\(0 \lt P(E) \lt 1\)
B.\(P(E) = \dfrac{1}{2}\)
C.\(P(E) = 0\)
D.\(P(E) = 1\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: D

Always \(P(E) = 1\).

By Kolmogorov's one-series (two-series) theorem, a series of independent random variables with mean 0 converges almost surely when the sum of the variances is finite:

\[ \begin{aligned} \sum_{n} \mathrm{Var}(X_{n}) &= \sum_{n} \frac{1}{n^{2}} \\ &= \frac{\pi^{2}}{6} \lt \infty \end{aligned} \]

So \(\sum X_{n}\) converges with probability 1. (Kolmogorov's 0-1 law already says that P(E) is 0 or 1, since convergence of the series is a tail event; this rules out 1/2 and anything strictly between 0 and 1.)

Discuss in Forum

18.

Suppose \(X_{1}, X_{2}, X_{3}\) are independent and identically distributed random variables each having exponential distribution with mean 1. Then the probability density function of sample range is given by,

A.\(h(r) = e^{-r},\ r \gt 0\)
B.\(h(r) = re^{-r},\ r \in \Re\)
C.\(h(r) = (1 - e^{-r})e^{-r},\ r \in \Re\)
D.\(h(r) = 2(1 - e^{-r})e^{-r},\ r \gt 0\)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: D

The density of the range is \(h(r) = 2(1 - e^{-r})e^{-r}\), \(r \gt 0\).

For \(n\) independent standard exponentials, the spacings of the order statistics are independent exponentials with rates \(n - 1, n - 2, \ldots\) after the minimum. So the range \(X_{(3)} - X_{(1)}\) is the sum of exponentials with rates 2 and 1, which is the maximum of two independent standard exponentials. Its distribution function is \((1 - e^{-r})^{2}\), and differentiating gives

\[ h(r) = 2(1 - e^{-r})e^{-r}, \quad r \gt 0 \]

\(e^{-r}\) is the density of one exponential, and the other options lack the factor 2, so they do not integrate to 1, or are defined on all of \(\Re\), where a range cannot be negative.

Discuss in Forum

19.

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

\[ \sum_{n=1}^{\infty} P[|X_{n} - X| \gt \epsilon] \lt \infty \]

for every \(\epsilon \gt 0\).

Consider the statements :

(I) \(P\left[\lim_{n \to \infty} X_{n} = X\right] = 1\)
(II) \(P\left[\lim_{n \to \infty} X_{n}^{2} = X^{2}\right] = 1\)
(III) \(P\left[\lim_{n \to \infty} X_{n} = X\right]\) only if \(\{X_{n}\}\) is a sequence of independent random variables.

Which of the above are always correct ?

A.(I) only
B.(I) and (II) only
C.(III) only
D.None of the three

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: B

The statements always correct are (I) and (II) only.

By the Borel-Cantelli lemma, \(\sum P[|X_{n} - X| \gt \epsilon] \lt \infty\) for every \(\epsilon\) means that, almost surely, \(|X_{n} - X| \gt \epsilon\) happens only finitely often. So \(X_{n} \to X\) almost surely (I). Since \(x \mapsto x^{2}\) is continuous, \(X_{n}^{2} \to X^{2}\) almost surely too (II).

(III) is false: this half of Borel-Cantelli needs no independence. Independence matters only for the converse, the second Borel-Cantelli lemma. So options including (III), or 'none of the three', are wrong.

Discuss in Forum

20.

Suppose the sequence of random variables \(\{X_{n}\}\) converges in distribution to X. Then which of the following is not always true ?

A.\(\lim_{n} E[X_{n}] = E[X]\)
B.\(E[|X|] \le \lim_{n} \inf E[|X_{n}|]\)
C.\(\lim_{n} E\left[\frac{|X_{n}|}{1 + |X_{n}|}\right]\) = \(E\left[\frac{|X|}{1 + |X|}\right]\)
D.\(\lim_{n} \frac{X_{n}}{\sqrt{n}} = 0\) in probability

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: A

The statement not always true is \(\lim_{n} E[X_{n}] = E[X]\).

Convergence in distribution does not control expectations without uniform integrability. Let \(X_{n} = n\) with probability \(1/n\) and 0 otherwise: \(X_{n} \to 0\) in distribution, but \(E[X_{n}] = 1\) for every \(n\).

The others always hold. (B) is Fatou's lemma in the form valid for convergence in distribution (via Skorokhod's representation). (C) holds because \(|x|/(1 + |x|)\) is bounded and continuous, which is exactly what convergence in distribution handles. (D) holds because a sequence converging in distribution is tight, so \(X_{n}/\sqrt{n} \to 0\) in probability.

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