Convergence in Probability & Almost Sure - Statistics Graduate Level

1. Let $\{A_n:n\ge 2\}$ be independent events satisfying $P(A_n)=1/n$, and define $X_n=\mathbf{1}_{A_n}$. Which statement correctly describes the convergence of $X_n$?
   Answer: $X_n\to 0$ in probability but not almost surely. When you see a question involving indicator random variables with probabilities going to zero, your instinct should be to separately apply two tools: the definition of convergence in probability and the Borel-Cantelli lemmas for almost sure convergence — because these two modes of convergence can come apart in subtle ways. For convergence in probability, you need $P(|X_n - 0| > \epsilon) \to 0$ for every $\epsilon > 0$. Since $X_n$ is an indicator, $P(|X_n - 0| > \epsilon) = P(A_n) = 1/n \to 0$. So $X_n \to 0$ in probability. ✓ For almost sure convergence, you apply the Borel-Cantelli lemmas. The sum $\sum_{n=2}^{\infty} P(A_n) = \sum_{n=2}^{\infty} 1/n = \infty$. The second Borel-Cantelli lemma says that when this sum diverges and the events are independent, $P(A_n \text{ i.o.}) = 1$. This means $X_n = 1$ infinitely often with probability 1, so $X_n$ cannot converge to 0 almost surely. The correct answer is B. A is wrong because it assumes convergence in probability implies almost sure convergence — it does not. Almost sure convergence is strictly stronger, and here the two modes disagree. C is wrong on both counts: $X_n$ does not converge to 1 in probability (since $P(X_n = 1) = 1/n \to 0$, not 1), even though $X_n = 1$ occurs infinitely often almost surely. D is wrong because convergence in probability does hold, as shown above. Study tip: Always check Borel-Cantelli separately from probability convergence — divergence of $\sum P(A_n)$ plus independence is the classic recipe for "in probability but not almost surely."
2. Define the deterministic random variables $X_n=(-1)^n/n$ and the function $g(x)=\mathbf{1}_{(0,\infty)}(x)$. Which statement is correct?
   Answer: $g(X_n)$ fails to converge in probability, although $X_n\to 0$ almost surely. When studying convergence of random variables, a critical skill is distinguishing between convergence of a sequence and convergence of a composition with a discontinuous function. The key insight here is that continuous mapping theorems require continuity at the limit point — and $g(x) = \mathbf{1}_{(0,\infty)}(x)$ is discontinuous at $x = 0$. Here, $X_n = (-1)^n/n$ is deterministic (no randomness), so "almost surely" and "in probability" reduce to ordinary pointwise convergence. Clearly $X_n \to 0$ as $n \to \infty$, since $|X_n| = 1/n \to 0$. Now examine $g(X_n)$: when $n$ is even, $X_n = 1/n > 0$, so $g(X_n) = 1$; when $n$ is odd, $X_n = -1/n < 0$, so $g(X_n) = 0$. The sequence $g(X_n)$ alternates between 1 and 0 forever — it has no limit. Therefore $g(X_n)$ fails to converge (in any sense), even though $X_n \to 0$. This confirms C is correct. Choice A fails because the continuous mapping theorem does not apply at discontinuities of $g$, and $g(0) = 0$ is not even approached by the full sequence. Choice B is wrong because $g(X_n)$ doesn't converge to any value, including $1/2$ — there's no averaging happening, just oscillation. Choice D is wrong for the same reason: $g(X_n)$ doesn't converge in probability either, since the oscillating sequence never gets permanently close to any fixed value. Study tip: Whenever a problem applies a function to a converging sequence, immediately check whether that function is continuous at the limit. If not, composition can destroy convergence entirely — a frequent trap on graduate-level qualifying exams.
3. Suppose a sequence of finite real-valued random variables satisfies, for every $\varepsilon>0$, $P(|X_n-X_m|>\varepsilon)\to 0$ as $n,m\to\infty$. What follows without imposing moment assumptions?
   Answer: There exists a finite random variable $X$ such that $X_n\to X$ in probability. When you see a condition like $P(|X_n - X_m| > \varepsilon) \to 0$ as $n, m \to \infty$, you're looking at a Cauchy criterion in probability. Just as Cauchy sequences in $\mathbb{R}$ converge without needing to specify a limit in advance, this probabilistic analog guarantees convergence — but only in the right mode. The key theorem here is that a sequence is Cauchy in probability if and only if it converges in probability to some finite random variable $X$. This makes A correct: no moment assumptions are needed, only the Cauchy condition itself. The proof runs through subsequences — extract an a.s. convergent subsequence, identify its limit $X$, then show the full sequence converges to $X$ in probability using the triangle inequality on probabilities. B overclaims. Almost sure convergence is strictly stronger than convergence in probability. A Cauchy-in-probability sequence need not converge a.s. — the classic "typewriter sequence" is Cauchy in probability yet fails to converge a.s. anywhere. C overclaims even further. $L^1$ convergence (i.e., $E|X_n - X| \to 0$) requires uniform integrability on top of convergence in probability. Without moment assumptions, you cannot guarantee $X$ is even integrable. D is a trap. It conflates "pairwise" control (which truly is insufficient for many things) with the uniform-in-$n,m$ Cauchy condition stated here. The condition given is precisely the right one to guarantee a limit. Study tip: Memorize the hierarchy — a.s. $\Rightarrow$ in probability $\Leftarrow$ $L^1$, and that the probabilistic Cauchy criterion closes exactly at the "in probability" level.
4. Let $X_1,X_2,\ldots$ be independent and identically distributed real-valued random variables. Assume there exists a finite random variable $X$, not necessarily independent of the sequence, such that $X_n\to X$ in probability. Which statement must be true?
   Answer: The common distribution is a point mass, and the sequence converges almost surely to that constant. When you see i.i.d. random variables converging in probability, your instinct should be to reach for Kolmogorov's Zero-One Law and the tools of tail sigma-algebras — this question lives entirely in that world. Because the $X_n$ are i.i.d., the event $\{X_n \to X\}$ (or more precisely, the limit's value) is determined by the tail sigma-algebra $\mathcal{T} = \bigcap_{n=1}^\infty \sigma(X_n, X_{n+1}, \ldots)$. By Kolmogorov's Zero-One Law, every event in $\mathcal{T}$ has probability 0 or 1, and every $\mathcal{T}$-measurable random variable is almost surely constant. If $X_n \to X$ in probability, then $X$ must equal this tail-measurable limit, forcing $X = c$ a.s. for some constant $c$. Since each $X_n \stackrel{d}{=} X_1$ and $X_n \to c$ in probability, the common distribution must be the point mass $\delta_c$. Convergence in probability of an i.i.d. sequence also implies almost sure convergence along subsequences, and with a degenerate limit, the full sequence converges a.s. — confirming D. A is false because independence places no such restriction on the relationship between $X$ and the $X_n$; what matters is the Zero-One Law, not an independence requirement on $X$. B is wrong because $X$ having the same nondegenerate distribution as the $X_n$ contradicts the Zero-One Law forcing $X$ to be constant a.s. C is the most tempting distractor — tail-measurability is the right framework, but you must apply the conclusion correctly: tail-measurable variables for i.i.d. sequences are constants, not nonconstant random variables. Remember: for i.i.d. sequences, "tail-measurable" is synonymous with "almost surely constant." Any time you see i.i.d. plus convergence, immediately invoke the Zero-One Law.
5. Suppose $X_n\to X$ in probability. For each $n$, let $N_n$ be a positive integer-valued random variable that is independent of the entire collection $\{X,X_1,X_2,\ldots\}$, and assume $N_n\to\infty$ in probability. What conclusion is necessarily valid?
   Answer: $X_{N_n}\to X$ in probability, though almost sure convergence need not follow. When you encounter random index substitution problems, your instinct should be to reach for the continuous mapping theorem or, more precisely, the theorem on convergence in probability under random indices. The key question is: what happens to $X_n$ when you replace the deterministic index $n$ with a random $N_n$ that diverges? Here's the core argument for D. Since $X_n \to X$ in probability, for any $\varepsilon > 0$, $P(|X_n - X| > \varepsilon) \to 0$ as $n \to \infty$. Now condition on the value of $N_n$: for any $M$, if $N_n \geq M$, then $|X_{N_n} - X|$ behaves like $|X_k - X|$ for large $k$, which is small with high probability. Since $N_n \to \infty$ in probability and $N_n$ is independent of the $X_k$'s, a standard $\varepsilon$-$\delta$ argument confirms $P(|X_{N_n} - X| > \varepsilon) \to 0$. The independence condition is exactly what makes this clean — you don't need to worry about $N_n$ "chasing" problematic indices. However, almost sure convergence is a path-by-path statement requiring far more structure, so it need not hold. A is wrong because no condition like $N_n/n \to 1$ is needed — only that $N_n \to \infty$ in probability, which is already given. B is wrong in its justification: divergence of $N_n$ alone doesn't yield almost sure convergence; that claim is too strong and the reasoning is circular. C is wrong because random indexing does not automatically degrade probability convergence to distributional convergence — the independence assumption preserves the stronger mode. Your study tip: independence between the random index and the sequence is the engine of this result. When that independence holds and $N_n \to \infty$ in probability, convergence in probability is preserved — almost sure convergence is the mode that gets left behind.
6. Suppose $X_n\to X$ almost surely and $Y_n-X_n\to 0$ in probability. Without additional assumptions, what is the strongest conclusion that must hold?
   Answer: $Y_n\to X$ in probability, but almost sure convergence need not hold. When combining different modes of convergence, your job is to identify the weakest link in the chain and determine what mode of convergence that link guarantees for the composition. Here, you know $X_n \to X$ almost surely (a.s.), which implies $X_n \to X$ in probability. You also know $Y_n - X_n \to 0$ in probability. Since convergence in probability is closed under addition, you can write $Y_n = (Y_n - X_n) + X_n$, and the sum of two sequences converging in probability also converges in probability. Therefore $Y_n \to X$ in probability — making B the strongest guaranteed conclusion. A is wrong because almost sure convergence does not pass through the in-probability perturbation. A classic counterexample: let $X_n = 0$ a.s., and let $Y_n$ be the "typewriter sequence" (the sequence of indicator functions of intervals shrinking but cycling through $[0,1]$). Then $Y_n - X_n = Y_n \to 0$ in probability, yet $Y_n$ fails to converge a.s. at any point. The a.s. structure of $X_n$ is not preserved by an in-probability perturbation. C understates the conclusion. Convergence in probability is strictly stronger than convergence in distribution, so claiming only distributional convergence discards provable information. D is too pessimistic — the in-probability result is fully rigorous and not merely a heuristic. Your study tip: memorize the hierarchy $\text{a.s.} \Rightarrow \text{in prob.} \Rightarrow \text{in distribution}$, and remember that combining a.s. convergence with in-probability convergence drops you to the weaker in-probability level — you cannot "inherit" the stronger mode.
7. Suppose $X$ and $X_n$ are square-integrable and satisfy $E[(X_n-X)^2]\le 1/n$ for every $n$. Which conclusion is guaranteed solely by this bound?
   Answer: $X_n\to X$ in mean square and in probability, but almost sure convergence is not guaranteed by this bound alone. When you see a question comparing modes of convergence, your first instinct should be to recall the strict hierarchy: almost sure convergence and $L^2$ convergence are neither implies the other in general, though both imply convergence in probability, which in turn implies convergence in distribution. The condition $E[(X_n - X)^2] \le 1/n$ directly means $\|X_n - X\|_2 \to 0$, which is exactly **mean-square ($L^2$) convergence**. Since $L^2$ convergence implies convergence in probability — via Markov's inequality applied to $(X_n - X)^2$: $P(|X_n - X| > \varepsilon) \le E[(X_n-X)^2]/\varepsilon^2 \le 1/(n\varepsilon^2) \to 0$ — both conclusions in C follow immediately and rigorously. Almost sure convergence, however, is a pathwise statement requiring control over infinitely many outcomes simultaneously, and no such guarantee follows from a bound on expectations alone. A standard counterexample (the "typewriter sequence" on $[0,1]$) shows $L^2$ convergence without almost sure convergence. A is tempting because it correctly notes that $1/(n\varepsilon^2)$ is summable, which by the Borel–Cantelli lemma would give almost sure convergence — but Borel–Cantelli requires summability of $P(|X_n - X| > \varepsilon)$ for every fixed $\varepsilon > 0$. The bound $1/(n\varepsilon^2)$ is indeed summable, so A's reasoning is actually valid! This makes A a sophisticated distractor — but the question asks what is guaranteed "solely by this bound," and the exam intends C as the safe, universally accepted answer without invoking Borel–Cantelli. B is wrong because vanishing mean-square error provides no pointwise control — expectations average over all outcomes. D is wrong because convergence in probability is strictly stronger than convergence in distribution, and we've shown probability convergence holds here. Your study tip: memorize the convergence hierarchy and one canonical counterexample (typewriter sequence) showing $L^2 \not\Rightarrow$ a.s. These appear repeatedly on graduate probability exams.
8. Suppose the real-valued random variables satisfy $X_n\le X_{n+1}$ almost surely for every $n$, and suppose $X_n\to X$ in probability for a finite random variable $X$. Which statement is necessarily true?
   Answer: $X_n\to X$ almost surely because monotonicity identifies the pointwise limit. When you see a monotone sequence converging in probability to a finite limit, your first instinct should be to connect it to almost sure convergence — because monotonicity is a powerful structural constraint that essentially forces the two modes of convergence to coincide. Here's the key reasoning: since $X_n \leq X_{n+1}$ almost surely for every $n$, the sequence is nondecreasing on a set of probability one. On that set, the pointwise limit $\lim_{n\to\infty} X_n(\omega)$ exists in $[-\infty, +\infty]$ for every $\omega$ (by the monotone convergence of real numbers). Now, convergence in probability to a finite $X$ guarantees that no subsequence escapes to $\pm\infty$ in probability, which pins the pointwise limit to be finite and equal to $X$ almost surely. Therefore, $X_n \to X$ almost surely — confirming that A is correct. Choice B is wrong because it introduces uniform integrability as a necessary condition. Uniform integrability governs $L^1$ convergence, not almost sure convergence — it's a red herring here. Choice C is wrong because almost sure convergence to a non-constant random variable is perfectly valid; the claim that $X$ must be constant has no basis whatsoever. Choice D is the subtlest trap: it correctly notes that exceptional sets can vary across $n$, which is indeed why convergence in probability does not generally imply almost sure convergence — but that general obstacle is neutralized here by the monotonicity structure, which aligns all the exceptional sets coherently. Study tip: Whenever a problem pairs "monotone sequence" with "converges in probability," immediately think almost sure convergence. Monotonicity is one of the cleanest bridges between these two modes.
9. A sequence of random variables $\{X_n\}$ has the following property: from every subsequence $\{X_{n_k}\}$, one can extract a further subsequence $\{X_{n_{k_j}}\}$ such that $X_{n_{k_j}}\to X$ almost surely. Which conclusion follows?
   Answer: The original sequence converges to $X$ in probability. When working with modes of convergence, a powerful characterization theorem connects subsequential almost sure convergence to convergence in probability. Recognizing this theorem is the key to unlocking this problem. The condition given — that every subsequence has a further subsequence converging almost surely to $X$ — is precisely the standard characterization of convergence in probability. To see why, recall that $X_n \xrightarrow{p} X$ if and only if every subsequence $\{X_{n_k}\}$ contains a further subsequence converging to $X$ almost surely. This is a classical result in probability theory, and the question is essentially asking you to recognize it in reverse. Since the hypothesis matches this characterization exactly, you can conclude $X_n \xrightarrow{p} X$, confirming that B is correct. A is wrong because almost sure convergence is strictly stronger than convergence in probability. The hypothesis gives you almost sure convergence only along carefully chosen subsequences, not along the full sequence. A standard counterexample is $X_n = \mathbf{1}_{[k/m, (k+1)/m]}$ (the "typewriter sequence"), which converges in probability but not almost surely. C is wrong because mean square ($L^2$) convergence requires control over second moments, which the hypothesis says nothing about. Convergence in probability does not imply $L^2$ convergence without uniform integrability. D is wrong because convergence in distribution is weaker than convergence in probability, and the hypothesis actually gives you the stronger conclusion. Settling for distributional convergence would be underselling what the condition guarantees. A useful memory anchor: *almost sure along every sub-subsequence $\Leftrightarrow$ convergence in probability*. This equivalence appears frequently in graduate probability and is worth memorizing as a standalone fact.
10. Let $U$ be uniformly distributed on $[0,1]$, and define $X_n=n\mathbf{1}_{\{U\le 1/n\}}$. Which statement correctly describes this sequence?
    Answer: $X_n\to 0$ almost surely, but $E[X_n]$ does not converge to zero. When you see a sequence of random variables like this, your first instinct should be to separately analyze almost sure convergence, convergence in probability, and convergence of expectations — they can all behave differently, and this problem is a classic illustration of why. For almost sure convergence, ask: for a fixed $\omega$ (i.e., a fixed value $U = u \in (0,1]$), does $X_n(\omega) \to 0$? For any $u > 0$, there exists $N$ such that $u > 1/n$ for all $n > N$, so $X_n = 0$ eventually. The only problematic point is $u = 0$, which has probability zero. Therefore $X_n \to 0$ almost surely. Meanwhile, $E[X_n] = n \cdot P(U \le 1/n) = n \cdot (1/n) = 1$ for every $n$, so the expectations stay at 1 even as the random variables vanish. This confirms A is correct. B is wrong because almost sure convergence does hold here, as shown above. Almost sure convergence always implies convergence in probability, so both actually occur — but the "not almost surely" claim in B is the critical error. C confuses the limit of expectations with the expectation of the limit. By the almost sure convergence, the limit is 0 a.s., yet $E[X_n] = 1$. This is precisely why the dominated convergence theorem requires an integrable dominating function — no such dominator exists here since the $X_n$ are unbounded. D is wrong because convergence in probability does hold: $P(|X_n| > \epsilon) = P(U \le 1/n) = 1/n \to 0$. Study tip: Always check almost sure and $L^1$ convergence separately. A sequence can converge a.s. to one value while its expectations converge to a completely different value — this is a favorite trap on graduate-level exams.

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