Convergence in Distribution - Statistics Graduate Level

1. For $n\geq3$, let $X_n$ have probability $1/2-1/n$ at $0$, probability $2/n$ at $1$, and probability $1/2-1/n$ at $2$. Let $q_n=\inf\{x:F_n(x)\geq1/2\}$ be its lower median. If $X$ assigns probability $1/2$ to each of $0$ and $2$ and has lower median $q$, which statement is correct?
   Answer: $X_n\xrightarrow{d}X$ and $q_n=1$ for every $n$, whereas $q=0$, so the lower medians do not converge. This question tests two related but separable ideas: weak convergence (convergence in distribution) and convergence of quantiles — and the key lesson is that these do not always go hand in hand. To check weak convergence, recall that $X_n \xrightarrow{d} X$ if and only if $F_n(x) \to F(x)$ at every continuity point of $F$. Here $F_n$ places mass $1/2 - 1/n$ at 0, $2/n$ at 1, and $1/2 - 1/n$ at 2. As $n \to \infty$, the mass at 1 vanishes and the masses at 0 and 2 each approach $1/2$, matching $X$'s distribution exactly. The continuity points of $F$ are all $x \notin \{0, 2\}$, and convergence holds there, so $X_n \xrightarrow{d} X$. Now compute the lower medians. For each finite $n$, $F_n(0) = 1/2 - 1/n < 1/2$, so $x = 0$ doesn't qualify; $F_n(1) = 1/2 + 1/n \geq 1/2$, so $q_n = \inf\{x : F_n(x) \geq 1/2\} = 1$ for all $n$. For the limit $X$, $F(0) = 1/2 \geq 1/2$, so $q = 0$. Thus $q_n = 1 \not\to 0 = q$, confirming D. Choice A is wrong because $q \neq 1$; once the mass at 1 disappears in the limit, 1 is no longer a median. Choice B is wrong because weak convergence does not guarantee quantile convergence at every level — convergence of quantiles requires the quantile function to be continuous at the relevant level. Choice C is wrong because the distributional limit exists; the failure of median convergence doesn't prevent weak convergence. The key takeaway: weak convergence preserves quantiles only at levels where the limiting quantile function is continuous. Always check the limit's CDF directly rather than assuming quantile convergence automatically follows.
2. Let $\{\mu_n\}$ be a tight sequence of probability measures on $\mathbb{R}^k$. Suppose every weakly convergent subsequence of $\{\mu_n\}$ has the same weak limit $\mu$. Which conclusion follows?
   Answer: The full sequence converges weakly to $\mu$ because tightness supplies convergent subsubsequences and excludes any subsequence staying away from $\mu$. Whenever you see a question combining tightness and uniqueness of subsequential limits, you should immediately think of the classical "subsequence principle" for metric spaces: a sequence converges to a limit if and only if every subsequence has a further subsequence converging to that same limit. Here's the core argument for D. Prohorov's theorem tells you that every tight sequence of probability measures on $\mathbb{R}^k$ is relatively compact — meaning every subsequence has a weakly convergent subsubsequence. Now suppose the full sequence $\{\mu_n\}$ does not converge weakly to $\mu$. Then there exists a subsequence $\{\mu_{n_j}\}$ that stays bounded away from $\mu$ in some weak neighborhood. But by relative compactness, this subsequence itself has a further weakly convergent subsubsequence — and by hypothesis, its limit must be $\mu$. This is a contradiction. Therefore, $\mu_n \xrightarrow{w} \mu$. Tightness supplies the compactness, and the uniqueness assumption closes the argument. D is correct. A is wrong because weak convergence of the full sequence requires no smoothness conditions on $\mu$'s distribution function; that's a red herring about continuity. B is wrong because it misreads the role of tightness. Tightness does more than just guarantee tightness — via Prohorov, it gives relative compactness, which is exactly the engine of the proof above. C is wrong because uniqueness of weak subsequential limits gives you weak convergence, not total variation convergence. These are fundamentally different modes, and no such automatic upgrade exists. A useful study mantra: tightness = relative compactness (Prohorov) + unique subsequential limit = full sequence converges weakly. Memorize this two-ingredient recipe.
3. Let $X_n=n$ with probability $1/n$ and $X_n=0$ otherwise. Which statement correctly describes the limiting behavior of this sequence?
   Answer: $X_n \xrightarrow{d} 0$, but $E[X_n]=1$ for every $n$, so convergence of the first moments does not follow. When you see a question about convergence in distribution alongside moment behavior, your first instinct should be to separate these two concepts — they operate independently, and conflating them is exactly the trap this question sets. To find the weak limit, compute the CDF of $X_n$. Since $X_n = n$ with probability $1/n$ and $X_n = 0$ with probability $1 - 1/n$, for any fixed $x \geq 0$ we have $P(X_n \leq x) = 1 - 1/n$ for large enough $n$ (once $n > x$), which converges to $1$. This is precisely the CDF of a point mass at zero, so $X_n \xrightarrow{d} 0$. Meanwhile, $E[X_n] = n \cdot (1/n) + 0 \cdot (1 - 1/n) = 1$ for every $n$, yet $E[0] = 0$. The first moment does not converge to the moment of the limiting distribution — a perfectly valid situation, since weak convergence does not guarantee convergence of moments. Answer A captures both facts correctly and is the right choice. Answer B is wrong because expectations do not determine weak limits. The fact that $E[X_n] = 1$ says nothing about the distributional limit being $\delta_1$; you must examine the full CDF. Answer C is wrong because an unbounded support or a diverging second moment does not prevent convergence in distribution. The CDF criterion is all that matters for weak convergence. Answer D is wrong because it claims $E[X_n] \to 0$, which is false — the expectation stays fixed at 1. Weak convergence does not allow you to pass limits through the identity function here. Study tip: Always check moment convergence separately from distributional convergence — they are logically independent, and exam questions routinely exploit the gap between them.
4. Let $U_1,\ldots,U_n$ be independent uniform random variables on $[0,1]$, and let $M_n=\min(U_1,\ldots,U_n)$. What is the weak limit of $nM_n$?
   Answer: A standard exponential distribution, because $P(nM_n>x)=(1-x/n)^n\to e^{-x}$ for $x\geq0$. When studying extreme order statistics, the key move is to find the limiting distribution by examining the survival function of the rescaled variable and recognizing a familiar limit. For $M_n = \min(U_1, \ldots, U_n)$, note that $P(M_n > t) = (1-t)^n$ for $t \in [0,1]$. Now rescale: set $x = nt$, so $t = x/n$, and compute $P(nM_n > x) = \left(1 - \frac{x}{n}\right)^n$. As $n \to \infty$, this converges to the iconic limit $e^{-x}$ for $x \geq 0$. That is precisely the survival function of an Exponential(1) distribution, confirming B is correct. This is a classic instance of the Poisson approximation to rare events: the minimum of $n$ uniforms, after scaling by $n$, converges weakly to a standard exponential. A is wrong because the minimum concentrates near zero, not near its mean, and normality requires a completely different regime (CLT for sums, not extremes). C is a subtle trap: yes, $E[nM_n] = n/(n+1) \to 1$, but convergence of means does not imply weak convergence to a point mass — weak limits are determined by entire distributions, not single moments. D is wrong because $nU_i$ is not uniform on $[0,1]$; multiplying by $n\$ stretches the distribution, and the minimum of stretched variables follows exponential, not uniform, behavior. As a strategy, whenever you see the minimum (or maximum) of$n$i.i.d. variables rescaled by$n$, immediately write out the survival function and apply the limit$\left(1 - x/n\right)^n \to $e^{-x}$$$ — it almost always points to an exponential limit.
5. Suppose $X_n\xrightarrow{d}Z$, where $Z\sim N(0,1)$, and suppose $Y_n-X_n\xrightarrow{p}0$. No independence assumptions are imposed. What is the limiting distribution of $W_n=X_nY_n$?
   Answer: $W_n\xrightarrow{d}Z^2$, which has a chi-square distribution with one degree of freedom. When you see a product of two random sequences, your instinct should be to ask: can I replace one factor with something simpler using a convergence result? That's exactly the tool needed here — the Slutsky-type continuous mapping argument. The key insight is that $Y_n = X_n + (Y_n - X_n)$. Since $Y_n - X_n \xrightarrow{p} 0$ and $X_n \xrightarrow{d} Z$, Slutsky's theorem tells you that $(X_n, Y_n - X_n) \xrightarrow{d} (Z, 0)$, which means $Y_n \xrightarrow{d} Z$ as well. More importantly, the joint vector $(X_n, Y_n) \xrightarrow{d} (Z, Z)$ — both components converge to the same limiting random variable. Applying the continuous mapping theorem to the product function $g(x,y) = xy$ gives $W_n = X_n Y_n \xrightarrow{d} Z \cdot Z = Z^2$, which is exactly a chi-squared distribution with one degree of freedom. So A is correct. B is subtly wrong: substituting $X_n$ for $Y_n$ gives $X_n^2$, not $X_n$. The product of two asymptotically standard normal variables is $Z^2$, not $Z$. C confuses the difference $Y_n - X_n \to 0$ with the product $X_n Y_n \to 0$. The difference vanishing says nothing about the magnitude of the product; each factor can still grow or fluctuate. D is a common trap. Independence is not required here — Slutsky's theorem and the continuous mapping theorem apply without any independence assumption. Study tip: Whenever you see $Y_n - X_n \xrightarrow{p} 0$, mentally replace $Y_n$ with $X_n$ in any continuous function of the joint vector, then apply the continuous mapping theorem.
6. Define the deterministic random variables $X_n=(-1)^n/n$ and the function $g(x)=\mathbf{1}\{x>0\}$. Which statement best describes the limiting behavior of $g(X_n)$?
   Answer: $g(X_n)$ does not converge in distribution because it alternates deterministically between $0$ and $1$. When you see a question about applying a function to a convergent sequence of random variables, your first instinct might be to invoke the continuous mapping theorem — but that theorem has a critical requirement: the function must be continuous (or at least continuous at the limit point). This question is testing exactly that boundary condition. Here, $X_n = (-1)^n/n \to 0$, and $g(x) = \mathbf{1}\{x > 0\}$ is discontinuous at $x = 0$. When $n$ is even, $X_n > 0$, so $g(X_n) = 1$. When $n$ is odd, $X_n < 0$, so $g(X_n) = 0$. The sequence $g(X_n)$ therefore alternates deterministically: $0, 1, 0, 1, \ldots$ For $g(X_n)$ to converge in distribution to some constant $c$, we'd need $P(g(X_n) \leq t) \to \mathbf{1}\{c \leq t\}$ for all continuity points $t$. But $P(g(X_n) \leq 0.5)$ alternates between 1 and 0, so no such limit exists. Answer C is correct. Answer A fails because the continuous mapping theorem does not apply to discontinuous functions at the limit point — continuity at $x=0$ is precisely what $g$ violates. Answer B is tempting but wrong: "half the indices are positive" is informal heuristic reasoning, not a valid distributional limit. No single distribution is approached. Answer D similarly misuses the language of convergence — alternating subsequences with different limits is the reason convergence fails, not a pathway to a Bernoulli limit. Your strategy: whenever you see the continuous mapping theorem invoked, immediately check whether the function is continuous at the limit point. If not, analyze subsequences directly to determine whether a distributional limit can exist.
7. Suppose $X_n \xrightarrow{d} X$, where $X$ has probabilities $0.2$, $0.3$, $0.1$, and $0.4$ at the points $0$, $1/2$, $1$, and $2$, respectively. For $A=(0,1]$, which is the strongest bound guaranteed solely by weak convergence?
   Answer: $0.3 \leq \liminf_{n\to\infty}P(X_n\in A) \leq \limsup_{n\to\infty}P(X_n\in A) \leq 0.6$ When you see weak convergence (convergence in distribution) paired with a question about probabilities of sets, your key tool is the Portmanteau theorem: for $X_n \xrightarrow{d} X$, $P(X \in A^\circ) \leq \liminf_{n\to\infty} P(X_n \in A) \leq \limsup_{n\to\infty} P(X_n \in A) \leq P(X \in \bar{A})$ where $A^\circ$ is the interior and $\bar{A}$ is the closure of $A$. Equality to $P(X \in A)$ is only guaranteed when $A$ is a continuity set — meaning $P(X \in \partial A) = 0$. For $A = (0,1]$, the interior is $A^\circ = (0,1)$ and the closure is $\bar{A} = [0,1]$. The boundary is $\partial A = \{0, 1\}$. Since $X$ places mass $0.2$ at $0$ and $0.1$ at $1$, we have $P(X \in \partial A) = 0.3 > 0$, so $A$ is not a continuity set, and the bound is genuinely an interval. Now compute: $P(X \in A^\circ) = P(X \in (0,1)) = P(X = 1/2) = 0.3$, and $P(X \in \bar{A}) = P(X \in [0,1]) = 0.2 + 0.3 + 0.1 = 0.6$. This gives the bound $0.3 \leq \liminf \leq \limsup \leq 0.6$, confirming A. Choice B claims exact convergence to $0.4$, which would require $A$ to be a continuity set — it isn't. Choice C uses $0.4$ as the upper bound, incorrectly excluding the boundary mass at $0$. Choice D uses $0.4$ as the lower bound, incorrectly including the mass at $0$ in the interior calculation. Your strategy: always check $\partial A$ for point masses. If $P(X \in \partial A) > 0$, the Portmanteau bounds will be strict, and you must compute interior and closure probabilities separately.
8. Let $X_n$ be uniformly distributed on $[-n,n]$. Its characteristic function is $\varphi_n(t)=\sin(nt)/(nt)$ for $t\neq 0$ and $\varphi_n(0)=1$. Since $\varphi_n(t)\to 0$ for each fixed $t\neq 0$, which conclusion is valid?
   Answer: $X_n$ has no weak limit because the pointwise limit of the characteristic functions is discontinuous at $t=0$. When studying weak convergence, the key theorem to internalize is Lévy's Continuity Theorem: a sequence of distributions converges weakly if and only if its characteristic functions converge pointwise to a function that is continuous at $t = 0$. That continuity condition is doing enormous work here. For $X_n \sim \text{Uniform}[-n, n]$, you correctly compute $\varphi_n(t) = \sin(nt)/(nt) \to 0$ for every fixed $t \neq 0$, while $\varphi_n(0) = 1$ always. The pointwise limit is therefore $\varphi(t) = \mathbf{1}\{t = 0\}$ — a function that jumps discontinuously from 1 to 0 at the origin. Because this limit is not continuous at $t = 0$, Lévy's theorem tells you no weak limit exists. The sequence is "escaping to infinity," spreading probability mass over larger and larger intervals until no finite distribution can capture it. Answer C is correct. Answer A fails because the function $\mathbf{1}\{t=0\}$ is not a valid characteristic function — all characteristic functions must be continuous everywhere, so no distribution corresponds to it. Answer B makes the opposite error: the point mass at 0 has characteristic function $\varphi(t) = 1$ for all $t$, not just at the origin. A vanishing pointwise limit is the signature of mass escaping to infinity, not concentrating at zero. Answer D is a superficial distractor — the Cauchy characteristic function $e^{-|t|}$ also decays, but sharing a qualitative feature with $\sin(nt)/(nt)$ is not a valid convergence argument. Study tip: Whenever characteristic functions converge pointwise, always check continuity of the limit at $t = 0$ before claiming weak convergence — this single condition is the difference between convergence and escape to infinity.
9. Let $Z$ be standard normal. For every $n$, define $X_n=Z$ and define $Y_n=Z$ when $n$ is even but $Y_n=-Z$ when $n$ is odd. Which conclusion is correct?
   Answer: Both marginals are standard normal, but $(X_n,Y_n)$ and $X_n+Y_n$ fail to converge in distribution because their dependence alternates. Convergence in distribution of a sequence requires that the CDFs converge for every fixed index going to infinity — not just along a subsequence. Keep that principle in mind here. For every $n$, both $X_n = Z$ and $Y_n = \pm Z$ are individually standard normal, since $-Z \sim N(0,1)$ whenever $Z \sim N(0,1)$. So both marginal sequences are constant (always standard normal) — they trivially "converge" marginally. But the joint distribution of $(X_n, Y_n)$ alternates: when $n$ is even, $(X_n, Y_n) = (Z, Z)$, which is a perfectly correlated pair; when $n$ is odd, $(X_n, Y_n) = (Z, -Z)$, which is perfectly anti-correlated. Because these two joint distributions are distinct and the sequence keeps switching between them, no single limiting joint distribution exists. Similarly, $X_n + Y_n$ alternates between $2Z$ (even) and $0$ (odd) — two different distributions — so it also fails to converge in distribution. This confirms B. Choice A commits the classic error of concluding that convergence of marginals implies convergence of the joint distribution — it does not, and independence is an additional assumption you cannot derive from marginals alone. Choice C is wrong because a limit of a full sequence cannot be determined by one subsequence alone; the odd subsequence converges to $(Z, -Z)$, contradicting the even-subsequence limit. Choice D describes something like the "Cesàro average" of the alternating distributions, which is a mixture, but distributional convergence requires the sequence itself to settle — alternation prevents that. The key study tip: marginal convergence never guarantees joint convergence, and a sequence that oscillates between two distinct distributions diverges, full stop.
10. For each positive integer $n$, let $X_n$ place probability $0.4$ at $-1/n$ and probability $0.6$ at $1+1/n$. Let $X$ place probability $0.4$ at $0$ and probability $0.6$ at $1$. Which statement is correct?
    Answer: $X_n \xrightarrow{d} X$, even though $P(0 \leq X_n \leq 1)=0$ for every $n$ while $P(0 \leq X \leq 1)=1$. Convergence in distribution is defined through CDFs, not through probabilities of arbitrary sets — and this question tests whether you know exactly where that definition has teeth and where it doesn't. Recall that $X_n \xrightarrow{d} X$ if and only if $F_{X_n}(t) \to F_X(t)$ at every continuity point of $F_X$. The CDF of $X$ has jumps at $0$ and $1$, so its continuity points are all $t \notin \{0, 1\}$. For any such $t$, you can verify directly: if $t < 0$, both CDFs equal $0$; if $0 < t < 1$, $F_{X_n}(t) = 0.4 \to 0.4 = F_X(t)$; if $t > 1$, both equal $1$. Convergence holds at every continuity point, so $X_n \xrightarrow{d} X$. Meanwhile, $P(0 \leq X_n \leq 1) = 0$ for every $n$ (since the mass sits at $-1/n$ and $1+1/n$, outside $[0,1]$), yet $P(0 \leq X \leq 1) = 1$. This is perfectly legal — weak convergence only guarantees convergence of probabilities for sets whose boundary has zero probability under the limit. The set $[0,1]$ has boundary $\{0,1\}$, which carries all the probability mass of $X$, so the guarantee fails here. This makes A correct. B is wrong because it conflates probability convergence on a specific set with the definition of distributional convergence — these are different things. C is wrong because it asserts a conclusion that weak convergence does not guarantee for sets with probability mass on their boundary. D is wrong and essentially nonsensical; the limit is a two-point distribution, not a constant. Your takeaway: weak convergence says nothing about $P(X_n \in B) \to P(X \in B)$ unless $P(X \in \partial B) = 0$. Always check the boundary condition.

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