Sampling Distributions and Limit Theorems Questions and Answers

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.

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.

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.

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 \(\sqrt{n^{2}}\,S_{n} = nS_{n}\).)

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 \(E[e^{X_{n}}] = n \cdot \tfrac{1}{n} + (1 - \tfrac{1}{n}) \to 2 \ne e^{0} = 1\).

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.

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.)

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.

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.

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.