Sampling Distributions and Limit Theorems Questions and Answers

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.

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.

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.

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.

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.

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.

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 \(\pi\).

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.

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.

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.