Statistics Graduate Level Quiz: Convergence In Distribution

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Convergence In DistributionQuestion 1 of 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?

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

X_n

does not converge in distribution because the probabilities assigned to the closed interval

[0,1]

fail to converge.

X_n \xrightarrow{d} X

, and weak convergence therefore implies that

P(0 \leq X_n \leq 1) \to P(0 \leq X \leq 1)

X_n

converges in distribution to the constant

0.6

because its two support points approach the endpoints of

[0,1]

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Statistics Graduate Level Quiz

Statistics Graduate Level Quiz: Convergence In Distribution

Practice Convergence In Distribution in Statistics Graduate Level with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Convergence In Distribution, giving you a quick way to practice the rules, question types, and explanations that matter most for Statistics Graduate Level.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$X_n \xrightarrow{d} X$ , even though $P(0 \leq X_n \leq 1)=0$ for every $n$ while $P(0 \leq X \leq 1)=1$ . (correct answer)
$X_n$ does not converge in distribution because the probabilities assigned to the closed interval $[0,1]$ fail to converge.
$X_n \xrightarrow{d} X$ , and weak convergence therefore implies that $P(0 \leq X_n \leq 1) \to P(0 \leq X \leq 1)$ .
$X_n$ converges in distribution to the constant $0.6$ because its two support points approach the endpoints of $[0,1]$ .

Explanation: 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.

Question 2

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?

$0.3 \leq \liminf_{n\to\infty}P(X_n\in A) \leq \limsup_{n\to\infty}P(X_n\in A) \leq 0.6$ (correct answer)
$0.4 \leq \liminf_{n\to\infty}P(X_n\in A) \leq \limsup_{n\to\infty}P(X_n\in A) \leq 0.4$
$0.3 \leq \liminf_{n\to\infty}P(X_n\in A) \leq \limsup_{n\to\infty}P(X_n\in A) \leq 0.4$
$0.4 \leq \liminf_{n\to\infty}P(X_n\in A) \leq \limsup_{n\to\infty}P(X_n\in A) \leq 0.6$

Explanation: 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

0

and

0.1

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.

Question 3

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?

$X_n$ converges weakly to a distribution whose characteristic function equals $0$ away from the origin and equals $1$ at the origin.
$X_n$ converges weakly to the point mass at $0$ because the pointwise limiting characteristic function vanishes away from zero.
$X_n$ has no weak limit because the pointwise limit of the characteristic functions is discontinuous at $t=0$ . (correct answer)
$X_n$ converges weakly to a standard Cauchy distribution because both characteristic functions decay to zero as $|t|$ increases.

Explanation: 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.

Question 4

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?

Both marginals converge to standard normal laws, and therefore $(X_n,Y_n)$ converges jointly to a pair of independent standard normal variables.
Both marginals are standard normal, but $(X_n,Y_n)$ and $X_n+Y_n$ fail to converge in distribution because their dependence alternates. (correct answer)
The vector $(X_n,Y_n)$ converges to $(Z,Z)$ because the even-indexed subsequence determines the limit of the full sequence.
The sum $X_n+Y_n$ converges to a mixture placing equal probability on $0$ and on a normal variable with variance $4$ .

Explanation: 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.

Question 5

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?

The full sequence converges weakly only when $\mu$ has a continuous distribution function in every coordinate.
Only tightness follows; the full sequence need not converge unless every subsequence itself converges weakly without further extraction.
The full sequence converges in total variation to $\mu$ because uniqueness of weak subsequential limits upgrades the mode of convergence.
The full sequence converges weakly to $\mu$ because tightness supplies convergent subsubsequences and excludes any subsequence staying away from $\mu$ . (correct answer)

Explanation: 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.

Question 6

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?

$X_n\xrightarrow{d}X$ and both $q_n$ and $q$ equal $1$ because any point between the two limiting atoms is a lower median.
$X_n\xrightarrow{d}X$ and $q_n\to q=0$ because weak convergence always implies convergence of fixed quantiles.
$X_n$ does not converge in distribution because its lower medians remain equal to $1$ while the candidate limit has median $0$ .
$X_n\xrightarrow{d}X$ and $q_n=1$ for every $n$ , whereas $q=0$ , so the lower medians do not converge. (correct answer)

Explanation: 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.

Question 7

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 $g(X_n)$ ?

$g(X_n)\xrightarrow{d}0$ because $X_n\to 0$ and convergence in distribution is preserved by every measurable function.
$g(X_n)\xrightarrow{d}1/2$ because the signs of $X_n$ are positive for half of the indices.
$g(X_n)$ does not converge in distribution because it alternates deterministically between $0$ and $1$ . (correct answer)
$g(X_n)$ converges to a Bernoulli variable with success probability $1/2$ because the two subsequences have different limits.

Explanation: 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.

Question 8

Let $X_n=n$ with probability $1/n$ and $X_n=0$ otherwise. Which statement correctly describes the limiting behavior of this sequence?

$X_n \xrightarrow{d} 0$ , but $E[X_n]=1$ for every $n$ , so convergence of the first moments does not follow. (correct answer)
$X_n \xrightarrow{d} 1$ because $E[X_n]=1$ for every $n$ and the expectations determine the weak limit.
$X_n$ does not converge in distribution because its support is unbounded and $E[X_n^2]=n$ diverges.
$X_n \xrightarrow{d} 0$ and $E[X_n]\to 0$ because weak convergence permits use of the identity function.

Explanation: 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.

Question 9

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 $nM_n$ ?

A standard normal distribution, because scaling by $n$ centers the minimum at its asymptotic mean.
A standard exponential distribution, because $P(nM_n>x)=(1-x/n)^n\to e^{-x}$ for $x\geq0$ . (correct answer)
A point mass at $1$ , because $E[nM_n]=n/(n+1)\to1$ and the means identify the limit.
A uniform distribution on $[0,1]$ , because each rescaled observation $nU_i$ remains uniformly distributed near zero.

Explanation: 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.

Question 10

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 $W_n=X_nY_n$ ?

$W_n\xrightarrow{d}Z^2$ , which has a chi-square distribution with one degree of freedom. (correct answer)
$W_n\xrightarrow{d}Z$ because replacing $Y_n$ by the asymptotically equivalent variable $X_n$ leaves one normal factor.
$W_n\xrightarrow{p}0$ because the difference $Y_n-X_n$ converges to zero in probability.
The limit cannot be determined unless $X_n$ and $Y_n$ are independent for every $n$ .

Explanation: 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.