Sampling Distributions and Limit Theorems Questions and Answers

Q: Suppose \(\{X_{n}\}\) and \(\{Y_{n}\}\) are sequences of random variables and X and Y are random variables such that as \(n \to \infty\), \(X_{n}\) converges in quadratic mean to X and \(Y_{n}\) converges in quadratic mean to Y.Consider the following statements :(I) The sequence \((X_{n} + Y_{n})\) converges in quadratic mean to (X + Y) as \(n \to \infty\)(II) \(\lim_{n \to \infty} P[|X_{n}Y_{n} - XY| \gt \varepsilon] = 0\), \(\forall \varepsilon \gt 0\)(III) \(\lim_{n \to \infty} E[e^{iX_{n}}] = E[e^{iX}]\)Which of the above are true ?

All three statements are true.(I) By Minkowski's inequality, \(\|(X_{n} + Y_{n}) - (X + Y)\|_{2}\) is at most \(\|X_{n} - X\|_{2} + \|Y_{n} - Y\|_{2} \to 0\), so the sum converges in quadratic mean.(II) Convergence in quadratic mean implies convergence in probability, and products of sequences converging in probability converge in probability to the product (continuous mapping), so \(X_{n}Y_{n} \to XY\) in probability.(III) Convergence in probability implies convergence in distribution, and \(e^{ix}\) is bounded and continuous, so \(E[e^{iX_{n}}] \to E[e^{iX}]\), the characteristic functions at \(t = 1\).Each partial option leaves out a true statement.

Q: Suppose \(X_{1}, X_{2}, \ldots, X_{n}\) are independent and identically distributed random variables each having exponential distribution with mean \(\theta\). Then the asymptotic distribution of \(\sqrt{n}\left(\dfrac{\overline{X}}{S} - 1\right)\) where \(S^{2} = \dfrac{1}{n}\sum_{1}^{n} X_{i}^{2} - (\overline{X})^{2}\) is :

The asymptotic distribution is N(0, 1).For the exponential with mean \(\theta\): \(\sigma^{2} = \theta^{2}\), \(\mu_{3} = 2\theta^{3}\), \(\mu_{4} = 9\theta^{4}\). By the delta method for \(g(m, v) = m/\sqrt{v}\) at \((\theta, \theta^{2})\), with gradient \((1/\theta, -1/(2\theta^{2}))\) and asymptotic covariance of \((\bar{X}, S^{2})\) equal to \(1/n\) times the matrix with rows \((\theta^{2}, 2\theta^{3})\) and \((2\theta^{3}, 8\theta^{4})\):\[ \begin{aligned} n\,\mathrm{Var} &\approx \frac{\theta^{2}}{\theta^{2}} - 2\cdot\frac{2\theta^{3}}{2\theta^{3}} + \frac{8\theta^{4}}{4\theta^{4}} \\ &= 1 - 2 + 2 = 1 \end{aligned} \]Since \(\bar{X}/S \to 1\), \(\sqrt{n}(\bar{X}/S - 1) \to N(0, 1)\). The ratio is scale-free, so \(\theta\) cannot appear in the limit, which rules out \(N(0, \theta)\) and \(N(0, \theta^{2})\); and the limit is centred at 0, not 1.

Q: Suppose \(\{X_{n}, n \ge 1\}\) is a sequence of independent and identically distributed random variables each following uniform U(2, 6) distribution. Which of the following is not true ?

The false statement is that \(X_{(n)}\) converges in law to an exponential distribution.For U(2, 6), \(P(X_{(n)} \le 6 - \epsilon) = \left(\frac{4 - \epsilon}{4}\right)^{n} \to 0\), so the maximum converges in probability to 6, and hence in law to the degenerate distribution at 6, not to an exponential. Likewise \(X_{(1)} \to 2\) in probability, and the sample median converges in probability to the population median 4.An exponential limit appears only after rescaling: \(n(6 - X_{(n)})\) converges in law to an exponential distribution with mean 4. Without the factor \(n\) the limit is a point mass. So the first, second and fourth statements are true and the third is the one that is not.

Q: \(\{X_{n}, n \ge 1\}\), \(\{Y_{n}, n \ge 1\}\) and Y are random variables defined on the same probability space. Suppose as \(n \to \infty\), \(X_{n}\) converges in probability to 3 and \(Y_{n}\) converges in distribution to Y. Then which of the following is true as \(n \to \infty\) ?

\(X_{n}Y_{n}^{2}\) need not converge in probability but converges in distribution to \(3Y^{2}\).Since \(X_{n} \to 3\), a constant, in probability, and \(Y_{n} \to Y\) in distribution, Slutsky's theorem gives \(X_{n}Y_{n}^{2} \xrightarrow{d} 3Y^{2}\) (using the continuous mapping theorem for \(y \mapsto y^{2}\)).Convergence in probability to \(3Y^{2}\) would need \(Y_{n}\) to be close to Y on the probability space, not just in distribution. Take \(X_{n} = 3\), Y standard normal and \(Y_{n} = Z\) for every \(n\), where Z is an independent standard normal. Then \(Y_{n} \to Y\) in distribution, since the laws are identical, but \(X_{n}Y_{n}^{2} - 3Y^{2} = 3(Z^{2} - Y^{2})\) is a fixed non-degenerate random variable, so it does not tend to 0 in probability.So option A is too strong, option C contradicts Slutsky, and option B denies the convergence in distribution that Slutsky guarantees.

Q: Let \(\{X_{n}, n \ge 1\}\) be independent and identically distributed random variables with mean 0 and variance 1. Which of the following is true as \(n \to \infty\) ?

\(\sum X_{i} / \sum X_{i}^{2}\) converges in probability to 0.Divide numerator and denominator by \(n\). By the weak law of large numbers, \(\frac{1}{n}\sum X_{i} \to E X = 0\) and \(\frac{1}{n}\sum X_{i}^{2} \to E X^{2} = 1\) in probability. The ratio of the two converges to \(0/1 = 0\).The others fail. By the central limit theorem \(\sum X_{i}/\sqrt{n} \to N(0, 1)\), a non-degenerate law. \(\sqrt{n}\sum X_{i} = n \cdot (\sum X_{i}/\sqrt{n})\) blows up rather than converging. And dividing the last ratio through by \(n\) gives \(\dfrac{\frac{1}{n}\sum X_{i}^{2}}{\sum X_{i}/\sqrt{n}}\), which behaves like \(1/Z\) with Z standard normal, which is random, not 1.

Q: Suppose \(\{X_{n}, n \ge 1\}\) is a sequence of random variables such that :\[ \begin{aligned} &P[X_{n} = 0] = 1 - \frac{1}{n^{3}} \text{ and} \\ &P[X_{n} = n] = \frac{1}{n^{3}},\ n \ge 1 \end{aligned} \]Which of the following is not true ?

The statement that is not true is Xn does not converge to 0 almost surely.Since \(\sum P(X_{n} \neq 0) = \sum \frac{1}{n^{3}} \lt \infty\), the Borel-Cantelli lemma says that with probability 1 only finitely many \(X_{n}\) are non-zero. So \(X_{n} \to 0\) almost surely, and the claim that it does not is false.The other modes hold. In quadratic mean:\[ \begin{aligned} E(X_{n}^{2}) &= n^{2} \cdot \frac{1}{n^{3}} \\ &= \frac{1}{n} \\ &\to 0 \end{aligned} \]and convergence in quadratic mean (or almost surely) implies convergence in probability and in law.

Q: Let \(\{X_{n}, n \ge 1\}\) and X be random variables defined on the same probability space. Which of the following statements is not always true as \(n \to \infty\) ?

The statement not always true is the one claiming almost sure convergence of eXn from convergence in probability and distribution.Convergence in probability (which already implies convergence in distribution) does not imply almost sure convergence. The 'typewriter' sequence of indicators of shrinking intervals sweeping across [0, 1] converges to 0 in probability but at no point almost surely; applying the continuous map \(e^{x}\) does not change that.The other three are instances of the continuous mapping theorem: a continuous function preserves almost sure convergence (and hence convergence in probability), convergence in probability (hence in distribution), and convergence in distribution.

Q: Let \(\{X_{n}, n \ge 1\}\) be a sequence of independent identically distributed random variables with mean '0' and variance '2'. Let :\[ S_{n} = \sum_{i=}^{n} X_{i}^{2}. \]Which of the following statements is true ?

Sn/n converges to 2 almost surely.The variables \(X_{i}^{2}\) are i.i.d. with mean\[ E(X_{i}^{2}) = Var(X_{i}) + [E(X_{i})]^{2} = 2 + 0 = 2 \]By the strong law of large numbers, \(\frac{1}{n}\sum_{i=1}^{n} X_{i}^{2} \to 2\) almost surely. (The paper prints the lower limit as 'i =', meaning i = 1.)So the limit is 2, not 0. \(S_{n}/\sqrt{n}\) behaves like \(2\sqrt{n}\) and tends to infinity, not 0, and \(\sqrt{n}S_{n}/2\) also diverges, so it cannot converge to a standard normal.

21.

Suppose \(\{X_{n}\}\) and \(\{Y_{n}\}\) are sequences of random variables and X and Y are random variables such that as \(n \to \infty\), \(X_{n}\) converges in quadratic mean to X and \(Y_{n}\) converges in quadratic mean to Y.

Consider the following statements :

(I) The sequence \((X_{n} + Y_{n})\) converges in quadratic mean to (X + Y) as \(n \to \infty\)
(II) \(\lim_{n \to \infty} P[|X_{n}Y_{n} - XY| \gt \varepsilon] = 0\), \(\forall \varepsilon \gt 0\)
(III) \(\lim_{n \to \infty} E[e^{iX_{n}}] = E[e^{iX}]\)

Which of the above are true ?

A.All the three
B.(I) and (II) only
C.(I) and (III) only
D.(II) and (III) only

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: A

All three statements are true.

(I) By Minkowski's inequality, \(\|(X_{n} + Y_{n}) - (X + Y)\|_{2}\) is at most \(\|X_{n} - X\|_{2} + \|Y_{n} - Y\|_{2} \to 0\), so the sum converges in quadratic mean.

(II) Convergence in quadratic mean implies convergence in probability, and products of sequences converging in probability converge in probability to the product (continuous mapping), so \(X_{n}Y_{n} \to XY\) in probability.

(III) Convergence in probability implies convergence in distribution, and \(e^{ix}\) is bounded and continuous, so \(E[e^{iX_{n}}] \to E[e^{iX}]\), the characteristic functions at \(t = 1\).

Each partial option leaves out a true statement.

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22.

Suppose \(X_{1}, X_{2}, \ldots, X_{n}\) are independent and identically distributed random variables each having exponential distribution with mean \(\theta\). Then the asymptotic distribution of \(\sqrt{n}\left(\dfrac{\overline{X}}{S} - 1\right)\) where \(S^{2} = \dfrac{1}{n}\sum_{1}^{n} X_{i}^{2} - (\overline{X})^{2}\) is :

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

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: A

The asymptotic distribution is N(0, 1).

For the exponential with mean \(\theta\): \(\sigma^{2} = \theta^{2}\), \(\mu_{3} = 2\theta^{3}\), \(\mu_{4} = 9\theta^{4}\). By the delta method for \(g(m, v) = m/\sqrt{v}\) at \((\theta, \theta^{2})\), with gradient \((1/\theta, -1/(2\theta^{2}))\) and asymptotic covariance of \((\bar{X}, S^{2})\) equal to \(1/n\) times the matrix with rows \((\theta^{2}, 2\theta^{3})\) and \((2\theta^{3}, 8\theta^{4})\):

\[ \begin{aligned} n\,\mathrm{Var} &\approx \frac{\theta^{2}}{\theta^{2}} - 2\cdot\frac{2\theta^{3}}{2\theta^{3}} + \frac{8\theta^{4}}{4\theta^{4}} \\ &= 1 - 2 + 2 = 1 \end{aligned} \]

Since \(\bar{X}/S \to 1\), \(\sqrt{n}(\bar{X}/S - 1) \to N(0, 1)\). The ratio is scale-free, so \(\theta\) cannot appear in the limit, which rules out \(N(0, \theta)\) and \(N(0, \theta^{2})\); and the limit is centred at 0, not 1.

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23.

Let \(X_{n}\) be asymptotically normally distributed with mean \(\mu\) and variance \(\sigma_{n}^{2}\), such that \(\sigma_{n}^{2} \to 0\) as \(n \to \infty\). Which of the following statements is not true ?

A.For \(\mu \neq 0\), \(1/X_{n} \sim \mathrm{AN}(1/\mu, \sigma_{n}^{2}/\mu^{2})\)
B.For any \(\mu\), \(e^{X_{n}} \sim \mathrm{AN}(e^{\mu}, e^{2\mu}\sigma_{n}^{2})\)
C.For \(\mu = 0\), \(\log|X_{n}|\sigma_{n}| \xrightarrow{d} \log|N(0, 1)|\)
D.For \(\mu \neq 0\), \(\log|X_{n}| \sim \mathrm{AN}(\log|\mu|, \sigma_{n}^{2}/\mu^{2})\)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: A

The statement that is not true is option A, \(1/X_{n} \sim AN(1/\mu, \sigma_{n}^{2}/\mu^{2})\).

By the delta method, \(g(X_{n}) \sim AN(g(\mu), [g'(\mu)]^{2}\sigma_{n}^{2})\) when \(g'(\mu) \neq 0\). For \(g(x) = 1/x\), \(g'(\mu) = -1/\mu^{2}\), so the correct variance is \(\sigma_{n}^{2}/\mu^{4}\), not \(\sigma_{n}^{2}/\mu^{2}\).

The others are right: \(g = e^{x}\) gives variance \(e^{2\mu}\sigma_{n}^{2}\) (B), and \(g = \log|x|\) gives \(\sigma_{n}^{2}/\mu^{2}\) (D). Option C, as printed 'log|X_n|σ_n|', evidently means \(\log|X_{n}/\sigma_{n}|\): when \(\mu = 0\), \(X_{n}/\sigma_{n} \to N(0, 1)\) and the continuous map \(\log|x|\) carries this over.

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24.

Suppose \(\{X_{n}, n \ge 1\}\) is a sequence of independent and identically distributed random variables each following uniform U(2, 6) distribution. Which of the following is not true ?

A.\(X_{(n)}\) converges to 6 in probability
B.\(X_{(1)}\) converges to 2 in probability
C.\(X_{(n)}\) converges in law to exponential distribution
D.\(M_{n}\) converges in probability to 4, where \(M_{n}\) is the median of \(\{X_{1}, X_{2}, \ldots, X_{n}\}\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: C

The false statement is that \(X_{(n)}\) converges in law to an exponential distribution.

For U(2, 6), \(P(X_{(n)} \le 6 - \epsilon) = \left(\frac{4 - \epsilon}{4}\right)^{n} \to 0\), so the maximum converges in probability to 6, and hence in law to the degenerate distribution at 6, not to an exponential. Likewise \(X_{(1)} \to 2\) in probability, and the sample median converges in probability to the population median 4.

An exponential limit appears only after rescaling: \(n(6 - X_{(n)})\) converges in law to an exponential distribution with mean 4. Without the factor \(n\) the limit is a point mass. So the first, second and fourth statements are true and the third is the one that is not.

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25.

\(\{X_{n}, n \ge 1\}\), \(\{Y_{n}, n \ge 1\}\) and Y are random variables defined on the same probability space. Suppose as \(n \to \infty\), \(X_{n}\) converges in probability to 3 and \(Y_{n}\) converges in distribution to Y. Then which of the following is true as \(n \to \infty\) ?

A.\(X_{n} Y_{n}^{2}\) converges in probability and in distribution to \(3Y^{2}\)
B.\(X_{n} Y_{n}^{2}\) need not converge in probability nor in distribution
C.\(X_{n} Y_{n}^{2}\) converges in probability but not in distribution
D.\(X_{n} Y_{n}^{2}\) need not converge in probability but converges in distribution to \(3Y^{2}\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: D

\(X_{n}Y_{n}^{2}\) need not converge in probability but converges in distribution to \(3Y^{2}\).

Since \(X_{n} \to 3\), a constant, in probability, and \(Y_{n} \to Y\) in distribution, Slutsky's theorem gives \(X_{n}Y_{n}^{2} \xrightarrow{d} 3Y^{2}\) (using the continuous mapping theorem for \(y \mapsto y^{2}\)).

Convergence in probability to \(3Y^{2}\) would need \(Y_{n}\) to be close to Y on the probability space, not just in distribution. Take \(X_{n} = 3\), Y standard normal and \(Y_{n} = Z\) for every \(n\), where Z is an independent standard normal. Then \(Y_{n} \to Y\) in distribution, since the laws are identical, but \(X_{n}Y_{n}^{2} - 3Y^{2} = 3(Z^{2} - Y^{2})\) is a fixed non-degenerate random variable, so it does not tend to 0 in probability.

So option A is too strong, option C contradicts Slutsky, and option B denies the convergence in distribution that Slutsky guarantees.

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26.

Let \(\{X_{n}, n \ge 1\}\) be independent and identically distributed random variables with mean 0 and variance 1. Which of the following is true as \(n \to \infty\) ?

A.\(\sum_{i=1}^{n} X_{i} / \sqrt{n}\) converges in distribution to a degenerate random variable
B.\(\sum_{i=1}^{n} X_{i} / \sum_{i=1}^{n} X_{i}^{2}\) converges in probability to 0
C.\(\sqrt{n} \sum_{i=1}^{n} X_{i}\) converges in distribution to a non-zero random variable
D.\(\dfrac{\sum_{i=1}^{n} X_{i}^{2}}{\sqrt{n} \sum_{i=1}^{n} X_{i}}\) converges in probability to 1

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: B

\(\sum X_{i} / \sum X_{i}^{2}\) converges in probability to 0.

Divide numerator and denominator by \(n\). By the weak law of large numbers, \(\frac{1}{n}\sum X_{i} \to E X = 0\) and \(\frac{1}{n}\sum X_{i}^{2} \to E X^{2} = 1\) in probability. The ratio of the two converges to \(0/1 = 0\).

The others fail. By the central limit theorem \(\sum X_{i}/\sqrt{n} \to N(0, 1)\), a non-degenerate law. \(\sqrt{n}\sum X_{i} = n \cdot (\sum X_{i}/\sqrt{n})\) blows up rather than converging. And dividing the last ratio through by \(n\) gives \(\dfrac{\frac{1}{n}\sum X_{i}^{2}}{\sum X_{i}/\sqrt{n}}\), which behaves like \(1/Z\) with Z standard normal, which is random, not 1.

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27.

Suppose \(\{X_{1}, X_{2}, \ldots, X_{n}\}\) is a random sample from normal \(N(\mu, \sigma^{2})\) distribution. Suppose \(M_{n}\) is the sample median. Which of the following is true ?

A.\(\sqrt{n}(M_{n} - \mu) \xrightarrow{d} Z\) where \(Z \sim N(0, 2\sigma^{2})\)
B.\(\sqrt{n}(M_{n} - \mu) \xrightarrow{d} Z\) where \(Z \sim N\left(0, \frac{\pi\sigma^{2}}{2}\right)\)
C.\(\sqrt{n}(M_{n} - \mu) \xrightarrow{d} Z\) where \(Z \sim N\left(0, \frac{\pi}{2\sigma^{2}}\right)\)
D.\(\sqrt{n}(M_{n} - \mu) \xrightarrow{d} Z\) where \(Z \sim N\left(0, \frac{\sigma^{2}}{2\pi}\right)\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: B

\(\sqrt{n}(M_{n} - \mu) \xrightarrow{d} N\left(0, \frac{\pi\sigma^{2}}{2}\right)\).

For a sample median from a density \(f\) with median \(m\), \(\sqrt{n}(M_{n} - m) \to N\left(0, \frac{1}{4f(m)^{2}}\right)\). For \(N(\mu, \sigma^{2})\), \(m = \mu\) and \(f(\mu) = \frac{1}{\sigma\sqrt{2\pi}}\), so

\[ \begin{aligned} \frac{1}{4f(\mu)^{2}} &= \frac{2\pi\sigma^{2}}{4} \\ &= \frac{\pi\sigma^{2}}{2} \end{aligned} \]

The asymptotic variance \(\pi\sigma^{2}/2 \approx 1.57\sigma^{2}\) is larger than the \(\sigma^{2}\) of the sample mean, which is why the median has asymptotic relative efficiency \(2/\pi \approx 0.64\) at the normal. \(2\sigma^{2}\) is not the right constant, \(\pi/(2\sigma^{2})\) inverts \(\sigma^{2}\), and \(\sigma^{2}/(2\pi)\) inverts the \(\pi\).

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28.

Suppose \(\{X_{n}, n \ge 1\}\) is a sequence of random variables such that :

\[ \begin{aligned} &P[X_{n} = 0] = 1 - \frac{1}{n^{3}} \text{ and} \\ &P[X_{n} = n] = \frac{1}{n^{3}},\ n \ge 1 \end{aligned} \]

Which of the following is not true ?

A.\(X_{n}\) converges to 0 in probability
B.\(X_{n}\) converges to 0 in quadratic mean
C.\(X_{n}\) does not converge to 0 almost surely
D.\(X_{n}\) converges to 0 in law

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: C

The statement that is not true is X_n does not converge to 0 almost surely.

Since \(\sum P(X_{n} \neq 0) = \sum \frac{1}{n^{3}} \lt \infty\), the Borel-Cantelli lemma says that with probability 1 only finitely many \(X_{n}\) are non-zero. So \(X_{n} \to 0\) almost surely, and the claim that it does not is false.

The other modes hold. In quadratic mean:

\[ \begin{aligned} E(X_{n}^{2}) &= n^{2} \cdot \frac{1}{n^{3}} \\ &= \frac{1}{n} \\ &\to 0 \end{aligned} \]

and convergence in quadratic mean (or almost surely) implies convergence in probability and in law.

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29.

Let \(\{X_{n}, n \ge 1\}\) and X be random variables defined on the same probability space. Which of the following statements is not always true as \(n \to \infty\) ?

A.If \(X_{n}\) converges to X in probability and in distribution then \(e^{X_{n}}\) converges to \(e^{X}\) almost surely
B.If \(X_{n}\) converges to X almost surely then \(e^{X_{n}}\) converges to \(e^{X}\) in probability
C.If \(X_{n}\) converges to X in distribution then \(e^{X_{n}}\) converges to \(e^{X}\) in distribution
D.If \(X_{n}\) converges to X in probability then \(e^{X_{n}}\) converges to \(e^{X}\) in distribution

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: A

The statement not always true is the one claiming almost sure convergence of e^X_n from convergence in probability and distribution.

Convergence in probability (which already implies convergence in distribution) does not imply almost sure convergence. The 'typewriter' sequence of indicators of shrinking intervals sweeping across [0, 1] converges to 0 in probability but at no point almost surely; applying the continuous map \(e^{x}\) does not change that.

The other three are instances of the continuous mapping theorem: a continuous function preserves almost sure convergence (and hence convergence in probability), convergence in probability (hence in distribution), and convergence in distribution.

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30.

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

\[ S_{n} = \sum_{i=}^{n} X_{i}^{2}. \]

Which of the following statements is true ?

A.\(\dfrac{S_{n}}{n}\) converges to 0 almost surely as \(n \to \infty\)
B.\(\dfrac{S_{n}}{n}\) converges to 2 almost surely as \(n \to \infty\)
C.\(\dfrac{S_{n}}{\sqrt{n}}\) converges to 0 in probability as \(n \to \infty\)
D.\(\sqrt{n}\dfrac{S_{n}}{2}\) converges in distribution to a standard normal random variable as \(n \to \infty\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: B

S_n/n converges to 2 almost surely.

The variables \(X_{i}^{2}\) are i.i.d. with mean

\[ E(X_{i}^{2}) = Var(X_{i}) + [E(X_{i})]^{2} = 2 + 0 = 2 \]

By the strong law of large numbers, \(\frac{1}{n}\sum_{i=1}^{n} X_{i}^{2} \to 2\) almost surely. (The paper prints the lower limit as 'i =', meaning i = 1.)

So the limit is 2, not 0. \(S_{n}/\sqrt{n}\) behaves like \(2\sqrt{n}\) and tends to infinity, not 0, and \(\sqrt{n}S_{n}/2\) also diverges, so it cannot converge to a standard normal.

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