Convergence Mode Relationships - Statistics Graduate Level

1. Which convergence conclusion follows from this bound without any independence assumption?
   Answer: $X_n\xrightarrow{a.s.}X$, and hence convergence in probability and in distribution also hold. When a question gives you an explicit probability bound, your first instinct should be to check whether the Borel-Cantelli lemma applies — this is the bridge between probabilistic bounds and almost-sure convergence. Here, you're given $P(|X_n - X| > \varepsilon) \leq \frac{1}{n^2 \varepsilon^2}$. For any fixed $\varepsilon > 0$, sum these bounds over all $n$: $\sum_{n=1}^{\infty} P(|X_n - X| > \varepsilon) \leq \sum_{n=1}^{\infty} \frac{1}{n^2 \varepsilon^2} = \frac{\pi^2}{6\varepsilon^2} < \infty.$ Since the sum of probabilities is finite, the first Borel-Cantelli lemma guarantees — with no independence assumption — that $P(|X_n - X| > \varepsilon \text{ i.o.}) = 0$. Because this holds for every $\varepsilon > 0\$, you conclude$$X_n$ \xrightarrow{a.s.} X$, making **D** correct. Almost-sure convergence then implies convergence in probability and in distribution for free. **A** is wrong precisely because it invokes independence as a requirement. Borel-Cantelli's first lemma requires only summability of the probabilities — independence is needed only for the *second* lemma (the converse direction). **B** is wrong because the bound controls tail probabilities, not$$L^1$$norms; you cannot extract moment information from this alone. **C** is similarly wrong — the$\varepsilon^{-2}$factor comes from a Chebyshev-style setup but does not imply$$L^2$$convergence; that would require bounding$E[|$X_n$ - X|^2]$directly. A reliable strategy: whenever tail probabilities decay like$$n^{-2}$$$ or faster, immediately check summability and invoke Borel-Cantelli. This pattern appears frequently on graduate probability exams.
2. Suppose $E[(T_n-\theta)^2]\to 0$ and $Z_n\xrightarrow{d}N(0,1)$. Let $W_n=(T_n-\theta)Z_n$. Which conclusion is guaranteed?
   Answer: $W_n\xrightarrow{p}0$, although convergence in mean or almost surely need not hold. When you see a product of two random sequences—one converging to zero, one remaining stochastically bounded—your first instinct should be to reach for convergence in probability tools, specifically Slutsky-type reasoning. Here, $E[(T_n - \theta)^2] \to 0$ means $T_n \xrightarrow{p} \theta$, so $(T_n - \theta) \xrightarrow{p} 0$. Since $Z_n \xrightarrow{d} N(0,1)$, the sequence $\{Z_n\}$ is bounded in probability (stochastically bounded, i.e., $O_p(1)$): for any $\varepsilon > 0$, you can find $M$ such that $P(|Z_n| > M) < \varepsilon$ for all large $n$. A sequence converging in probability to zero, multiplied by an $O_p(1)$ sequence, converges in probability to zero. Thus $W_n = (T_n - \theta)Z_n \xrightarrow{p} 0$, confirming B. A is wrong because mean-square convergence of $T_n$ to $\theta$ does not imply almost-sure convergence of $W_n$. Almost-sure convergence requires path-by-path control that neither assumption provides, and the joint behavior without independence is uncontrolled. C is wrong because $Z_n \xrightarrow{d} N(0,1)$ does not imply uniform integrability of $\{Z_n\}$. Uniform integrability requires control over tail expectations, which weak convergence alone cannot guarantee. D is wrong because $L^2$ convergence of $W_n$ would require $E[W_n^2] = E[(T_n-\theta)^2 Z_n^2] \to 0$, which needs $E[Z_n^2]$ to be uniformly bounded—an assumption not given here. Study tip: Memorize that $o_p(1) \cdot O_p(1) = o_p(1)$. This "little-oh times big-oh" rule is your go-to for products of degenerating and stochastically bounded sequences, and it appears constantly in asymptotic statistics.
3. Suppose $A_n\xrightarrow{p}A$ and $B_n\xrightarrow{d}B$. In general, these marginal statements do not imply $A_n+B_n\xrightarrow{d}A+B$. Which additional condition is sufficient to make the displayed conclusion valid?
   Answer: The limit $A$ is almost surely equal to a fixed constant, so Slutsky's theorem applies. When you see a question about combining convergence in probability with convergence in distribution, your first instinct should be to recall Slutsky's theorem: if $X_n \xrightarrow{d} X$ and $Y_n \xrightarrow{p} c$ for some constant $c$, then $X_n + Y_n \xrightarrow{d} X + c$. The key ingredient is that one limit must be a degenerate constant, not a random variable. This is precisely what makes C correct. If $A \equiv c$ almost surely, then $A_n \xrightarrow{p} c$, and Slutsky's theorem directly gives $A_n + B_n \xrightarrow{d} c + B = A + B$. The constant collapses the joint distribution problem — you no longer need to worry about the dependence structure between $A_n$ and $B_n$. A is wrong because finite variances say nothing about joint behavior. Convergence in distribution is a statement about marginal laws, not moments, and knowing $\text{Var}(A)$ and $\text{Var}(B)$ are finite tells you nothing about how $A_n + B_n$ behaves jointly. B is a tempting distractor: boundedness in probability (tightness) controls tail behavior but does not pin down the limit of $A_n + B_n$ without specifying what $A_n$ actually converges to. Continuity of $B$'s distribution is irrelevant here. D is wrong because convergence of expectations (first moments) is far weaker than convergence in distribution — you can construct counterexamples easily. As a study tip: whenever you see a mix of $\xrightarrow{p}$ and $\xrightarrow{d}$, immediately ask yourself "Is one limit a constant?" — that's the Slutsky trigger.
4. Suppose $X_n\xrightarrow{p}X$ and the family $\{(X_n-X)^2:n\geq 1\}$ is uniformly integrable. Which conclusion is strongest among those listed?
   Answer: $X_n\xrightarrow{L^2}X$, although almost-sure convergence of the full sequence need not hold. When you see uniform integrability (UI) paired with convergence in probability, your instinct should be to connect them through the Vitali Convergence Theorem: if $X_n \xrightarrow{p} X$ and $\{Y_n\}$ is UI, then $E[Y_n] \to E[Y]$. Here, the UI family is $\{(X_n - X)^2\}$, which converges in probability to $(X - X)^2 = 0$. By Vitali, $E[(X_n - X)^2] \to 0$, which is precisely the definition of $X_n \xrightarrow{L^2} X$. Since $L^2$ convergence implies $L^1$ convergence (by Jensen's or Cauchy-Schwarz), and $L^1$ implies convergence in distribution, answer C is the strongest valid conclusion. Answer A is too weak — you can conclude far more than mere convergence in distribution when UI of the squared differences is given. A undersells what UI buys you. Answer B correctly identifies $L^1$ convergence but claims $L^2$ is unattainable. This is the key trap: UI of $(X_n - X)^2$ specifically licenses $L^2$ convergence, not just $L^1$. B mistakes which family is UI. Answer D overclaims. Almost-sure convergence is a strictly stronger mode than convergence in probability, and uniform integrability provides no mechanism to upgrade $\xrightarrow{p}$ to $\xrightarrow{a.s.}$ — counterexamples like the "sliding bump" sequence confirm this. Study tip: Always track which family is declared UI. UI of $|X_n - X|^r$ paired with $\xrightarrow{p}$ gives $L^r$ convergence via Vitali — match the exponent to the moment.
5. Which statement correctly illustrates the distinction between convergence in distribution and convergence in probability?
   Answer: $X_n\xrightarrow{d}X$ and $E[X_n]\to E[X]$, but $X_n$ does not converge to $X$ in probability. When comparing convergence modes, the key question is always: what exactly must be small? Convergence in distribution only requires that CDFs agree in the limit — it says nothing about whether $X_n$ and $X$ are close as random variables on the same probability space. Convergence in probability requires $P(|X_n - X| > \varepsilon) \to 0$, which is a statement about their joint behavior. Here, every $X_n \sim N(0,1)$ and $X \sim N(0,1)$, so trivially $X_n \xrightarrow{d} X$ — the CDFs are identical at every step, not just in the limit. Expectations also agree: $E[X_n] = E[X] = 0$. But because $X_n$ and $X$ are mutually independent, $X_n - X \sim N(0,2)$ for every $n$, so $P(|X_n - X| > \varepsilon) = P(|N(0,2)| > \varepsilon)$, which is a fixed positive constant — never tending to zero. Thus $X_n \not\xrightarrow{p} X$, making B the correct statement. A is wrong because identical distributions and equal expectations are insufficient for convergence in probability; you need the joint distribution of $(X_n, X)$ to concentrate near zero difference, which independence destroys. C is wrong because $L^1$ convergence implies convergence in probability, and we just showed that fails here. Uniform integrability is a red herring — it helps with moment convergence, not pathwise closeness. D is wrong because independence has no bearing on distributional convergence; the marginal CDFs still match perfectly. Study tip: Whenever $X_n$ and $X$ are i.i.d., you always get $X_n \xrightarrow{d} X$ for free, but independence makes $X_n - X$ spread out, killing convergence in probability. This is the canonical counterexample to the converse of "$\xrightarrow{p}$ implies $\xrightarrow{d}$."
6. Which conclusion is valid on the underlying probability space?
   Answer: There exists a finite random variable $T$ such that $T_n\xrightarrow{p}T$, and the limit is unique almost surely. Whenever you see a question about Cauchy sequences in probability, think of the analogy to metric space completeness — the key question is whether the probability space is "complete enough" to guarantee a limit exists. The space of random variables under convergence in probability is indeed complete in this sense. Formally, if $\{T_n\}$ is Cauchy in probability, you can extract a subsequence $T_{n_k}$ that converges almost surely to some measurable random variable $T$. From that almost-sure convergence along the subsequence, you can show the full sequence satisfies $T_n \xrightarrow{p} T$. Moreover, probability limits are unique almost surely: if $T_n \xrightarrow{p} T$ and $T_n \xrightarrow{p} T'$, then $P(T \neq T') = 0$. This confirms D is correct. A is wrong because the Cauchy-in-probability condition does not guarantee almost-sure convergence of the full sequence — only convergence in probability. Almost-sure convergence is strictly stronger and requires additional structure or arguments beyond the Cauchy criterion alone. B is wrong because it imposes unnecessary conditions. Uniform integrability and almost-sure boundedness are relevant for $L^1$ or $L^\infty$ convergence, not for the existence of a probability limit. No such extra assumption is needed here. C is wrong in a subtle but important way: it confuses weak convergence (distributional limits) with convergence in probability. A Cauchy-in-probability sequence guarantees a limit on the same probability space, not merely a distributional limit. As a study tip: remember that "Cauchy in probability $\Rightarrow$ convergence in probability" is the probabilistic analogue of completeness in $\mathbb{R}$, and always distinguish between almost-sure, in-probability, and distributional convergence — they are a favorite source of traps on graduate exams.
7. Which statement correctly characterizes the convergence of $X_n$ to zero?
   Answer: $X_n\to 0$ almost surely and in probability, but not in $L^1$. Whenever you see a question about modes of convergence, your instinct should be to check each mode independently — they don't imply one another in general, and this sequence is a classic illustration of why. Start with almost sure convergence. Fix a realization $u \in (0,1)$. For large enough $n$, specifically once $n > 1/u$, we have $u > 1/n$, so $X_n(u) = 0$. Since every $u \in (0,1)$ eventually satisfies this, $X_n \to 0$ almost surely. So A's claim of a.s. convergence is correct. Now check **$L^1$ convergence**. Compute $E[X_n] = n \cdot P(U \leq 1/n) = n \cdot (1/n) = 1$ for all $n$. Since $E[|X_n - 0|] = 1 \not\to 0$, there is no $L^1$ convergence. This confirms A's claim that $L^1$ convergence fails. For convergence in probability: $P(|X_n| > \varepsilon) = P(U \leq 1/n) = 1/n \to 0$ for any $\varepsilon < n$. So $X_n \to 0$ in probability — consistent with A. Now address the wrong answers. B is wrong because it claims $L^1$ convergence, which fails as shown above. C is doubly wrong: it denies convergence in probability (which does hold) and falsely grants $L^1$ convergence. D is wrong on two counts: $X_n$ does converge in probability, and the reasoning about expectation is a non-sequitur — having constant expectation doesn't prevent convergence in distribution. The takeaway: **a.s. convergence does not imply $L^1$ convergence**, and vice versa. This sequence — where probability mass collapses onto a shrinking set but with compensating magnitude — is the canonical counterexample to memorize.
8. Suppose that every subsequence $\{X_{n_k}\}$ contains a further subsequence $\{X_{n_{k_j}}\}$ such that $X_{n_{k_j}}\xrightarrow{a.s.}X$. What is the strongest conclusion that is necessarily valid?
   Answer: The original sequence converges to $X$ in probability, but almost-sure or $L^1$ convergence need not follow. When you see a condition like "every subsequence contains a further almost-surely convergent subsequence," you should immediately recognize this as the subsequence characterization of convergence in probability. This is a fundamental theorem in probability theory worth knowing cold. The key result is: $X_n \xrightarrow{p} X$ if and only if every subsequence $\{X_{n_k}\}$ contains a further subsequence $\{X_{n_{k_j}}\}$ such that $X_{n_{k_j}} \xrightarrow{a.s.} X$. The condition in the problem is precisely this characterization, so the strongest valid conclusion is convergence in probability — making B correct. Why can't you conclude almost-sure convergence (choice A)? Because the null sets on which convergence fails may differ across subsequences. You cannot take a union over uncountably many subsequences and preserve measure zero. A classic counterexample is the "typewriter sequence" on $[0,1]$: it converges in probability to 0 but fails to converge almost surely at any point. Choice C is too weak. Convergence in probability is strictly stronger than convergence in distribution, so stopping at distributional convergence undersells what the hypothesis actually gives you. Choice D is off-target entirely. The hypothesis says nothing about $L^1$ structure; you need uniform integrability to connect almost-sure convergence to $L^1$ convergence. Being Cauchy in $L^1$ is neither implied nor relevant here. Study tip: Memorize the subsequence criterion as a two-way bridge — it's both a way to prove convergence in probability (find the sub-subsequence) and a way to disprove it (exhibit a subsequence with no such refinement).
9. Which description of the convergence of $X_n$ to zero is correct?
   Answer: $X_n\to 0$ in probability and in every $L^r$ with $r>0$, but not almost surely. When analyzing convergence of indicator random variables, you need to check four modes simultaneously: almost sure (a.s.), in probability, in $L^r$, and in distribution. Each has a distinct criterion, and they don't always align. Almost sure convergence requires asking whether $X_n(\omega) \to 0$ for almost every sample point. By the second Borel-Cantelli lemma, since the $A_n$ are independent and $\sum P(A_n) = \sum 1/n = \infty$, infinitely many $A_n$ occur with probability 1. This means $X_n = 1$ infinitely often almost surely, so $X_n \not\to 0$ a.s. In probability, you need $P(|X_n| > \epsilon) \to 0$. Since $P(X_n = 1) = 1/n \to 0$, convergence in probability holds trivially. **In $L^r$**, note that $\|X_n\|_r^r = E[|X_n|^r] = P(A_n) = 1/n \to 0$ for every $r > 0$. So $X_n \to 0$ in $L^r$ for all positive $r$. This confirms A is correct. Choice B is wrong because $X_n$ does not converge a.s. (Borel-Cantelli rules this out), and it does converge in every $L^r$. Choice C confuses Borel-Cantelli's conclusion: divergence of $\sum P(A_n)$ kills a.s. convergence, not convergence in probability. Choice D incorrectly restricts $L^r$ convergence to $r < 1$, ignoring that $E[X_n^r] = 1/n \to 0$ for all $r > 0$. Study tip: Memorize both Borel-Cantelli lemmas as a pair — the first (convergent series) gives a.s. convergence to zero; the second (divergent series + independence) gives a.s. non-convergence. They're a frequent source of traps on graduate probability exams.
10. A sequence of random variables satisfies $X_n\xrightarrow{d}c$, where $c$ is a finite constant. Which statement most accurately describes what follows without additional assumptions?
    Answer: $X_n\xrightarrow{p}c$, but almost-sure and $L^1$ convergence are not guaranteed. When you see convergence in distribution to a constant, that's a special case worth memorizing because it's strictly stronger than typical distributional convergence. The key theorem: if $X_n \xrightarrow{d} c$ where $c$ is a constant (a degenerate distribution), then $X_n \xrightarrow{p} c$. Here's the intuition — for any $\epsilon > 0$, the CDF of the limit is $F(x) = \mathbf{1}_{x \geq c}$, so $P(|X_n - c| > \epsilon) = P(X_n > c+\epsilon) + P(X_n \leq c-\epsilon) \to 0 + 0 = 0$ by pointwise convergence of CDFs at continuity points. This makes B correct: convergence in probability is guaranteed, but almost-sure and $L^1$ convergence require additional structure. A is wrong because convergence in distribution — even to a constant — does not imply almost-sure convergence. You can construct sequences that converge in probability (and hence in distribution) but fail to converge almost surely (e.g., the "typewriter sequence"). C is wrong on two counts: $L^1$ convergence is not guaranteed without uniform integrability, and convergence in probability is not blocked by dependence — the argument above makes no independence assumption. D is wrong because it invents a false requirement. Convergence in probability from $X_n \xrightarrow{d} c$ holds regardless of whether the variables share a common distribution. Study tip: Memorize this hierarchy exception — convergence in distribution to a constant upgrades to convergence in probability, but stops there. $L^1$ and a.s. convergence still need uniform integrability or monotonicity arguments, respectively.

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