Statistics Graduate Level Quiz: Convergence Mode Relationships

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Convergence Mode RelationshipsQuestion 1 of 10

Let $X,X_1,X_2,\ldots$ be mutually independent random variables, each having the standard normal distribution.

Which statement correctly illustrates the distinction between convergence in distribution and convergence in probability?

X_n\xrightarrow{p}X

because

X_n

and

X

have identical distributions and equal expectations.

X_n\xrightarrow{d}X

and

E[X_n]\to E[X]

, but

X_n

does not converge to

X

in probability.

X_n\xrightarrow{L^1}X

because the common normal distribution makes the sequence uniformly integrable.

X_n

fails to converge in distribution to

X

because independence prevents pathwise agreement.

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Statistics Graduate Level Quiz

Statistics Graduate Level Quiz: Convergence Mode Relationships

Practice Convergence Mode Relationships 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 Mode Relationships, 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

Let $X,X_1,X_2,\ldots$ be mutually independent random variables, each having the standard normal distribution.

Which statement correctly illustrates the distinction between convergence in distribution and convergence in probability?

$X_n\xrightarrow{p}X$ because $X_n$ and $X$ have identical distributions and equal expectations.
$X_n\xrightarrow{d}X$ and $E[X_n]\to E[X]$ , but $X_n$ does not converge to $X$ in probability. (correct answer)
$X_n\xrightarrow{L^1}X$ because the common normal distribution makes the sequence uniformly integrable.
$X_n$ fails to converge in distribution to $X$ because independence prevents pathwise agreement.

Explanation: 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}

Question 2

For random variables $X_n$ and $X$ on a common probability space, suppose that for every $\varepsilon>0$ and every positive integer $n$ , $P(|X_n-X|>\varepsilon)\leq 1/(n^2\varepsilon^2)$ .

Which convergence conclusion follows from this bound without any independence assumption?

$X_n\xrightarrow{p}X$ , but almost-sure convergence requires independence of the events $\{|X_n-X|>\varepsilon\}$ and cannot be deduced from this bound alone.
$X_n\xrightarrow{L^1}X$ , but the bound is insufficient to establish almost-sure convergence.
$X_n\xrightarrow{L^2}X$ , because the $\varepsilon^{-2}$ factor in the probability bound implies control over second moments.
$X_n\xrightarrow{a.s.}X$ , and hence convergence in probability and in distribution also hold. (correct answer)

Explanation: 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.

Question 3

A sequence of estimators is known to be Cauchy in probability: for every $\varepsilon>0$ , $P(|T_n-T_m|>\varepsilon)\to 0$ as $m,n\to\infty$ .

Which conclusion is valid on the underlying probability space?

There exists a finite random variable $T$ such that $T_n\xrightarrow{a.s.}T$ , with no additional assumptions.
A probability limit exists only if $\{T_n\}$ is additionally uniformly integrable or almost surely bounded.
There exists a distributional limit for $T_n$ , but no random variable on the same space need be a probability limit.
There exists a finite random variable $T$ such that $T_n\xrightarrow{p}T$ , and the limit is unique almost surely. (correct answer)

Explanation: 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

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.

Question 4

An estimator and an auxiliary statistic are computed from the same data, so no independence assumption is available.

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?

$W_n\xrightarrow{a.s.}0$ , because mean-square convergence can be combined with weak convergence.
$W_n\xrightarrow{p}0$ , although convergence in mean or almost surely need not hold. (correct answer)
$W_n\xrightarrow{L^1}0$ , because convergence in distribution makes the sequence $\{Z_n\}$ uniformly integrable.
$W_n\xrightarrow{L^2}0$ , even without independence or higher-moment assumptions on $Z_n$ .

Explanation: 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

\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.

Question 5

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?

The random variables $A$ and $B$ both have finite variances, with no restriction on their joint dependence.
The sequence $\{A_n\}$ is bounded in probability, while $B$ has a continuous distribution.
The limit $A$ is almost surely equal to a fixed constant, so Slutsky's theorem applies. (correct answer)
The expectations of $A_n$ and $B_n$ converge separately to those of $A$ and $B$ .

Explanation: 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.

Question 6

Let $A_1,A_2,\ldots$ be independent events satisfying $P(A_n)=1/n$ , and define $X_n=\mathbf{1}_{A_n}$ .

Which description of the convergence of $X_n$ to zero is correct?

$X_n\to 0$ in probability and in every $L^r$ with $r>0$ , but not almost surely. (correct answer)
$X_n\to 0$ almost surely and in probability, but not in any $L^r$ with $r\geq 1$ .
$X_n\to 0$ only in distribution, because the divergent series of event probabilities prevents probability convergence.
$X_n\to 0$ in probability and almost surely, but only in $L^r$ when $0<r<1$ .

Explanation: 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

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.

Question 7

A sequence of simulation-based estimators is defined on a common probability space.

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?

$X_n\xrightarrow{d}X$ , but neither first- nor second-mean convergence can be concluded.
$X_n\xrightarrow{L^1}X$ , but convergence in $L^2$ is not ensured by uniform integrability.
$X_n\xrightarrow{L^2}X$ , although almost-sure convergence of the full sequence need not hold. (correct answer)
$X_n\xrightarrow{a.s.}X$ and $X_n\xrightarrow{L^2}X$ , because uniform integrability strengthens probability convergence.

Explanation: 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}

\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.

Question 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?

The original sequence converges to $X$ almost surely, since every subsequence has an almost-surely convergent refinement.
The original sequence converges to $X$ in probability, but almost-sure or $L^1$ convergence need not follow. (correct answer)
The original sequence converges to $X$ in distribution only, because the exceptional null sets may depend on the subsequence.
The original sequence is Cauchy in $L^1$ , although it need not converge to $X$ in probability.

Explanation: 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).

Question 9

Let $U$ be uniformly distributed on $(0,1)$ , and define $X_n=n\mathbf{1}_{\{U\leq 1/n\}}$ .

Which statement correctly characterizes the convergence of $X_n$ to zero?

$X_n\to 0$ almost surely and in probability, but not in $L^1$ . (correct answer)
$X_n\to 0$ in $L^1$ and in probability, but not almost surely.
$X_n\to 0$ almost surely and in $L^1$ , but not in probability.
$X_n\to 0$ in probability but not in distribution, because its expectation remains equal to one for all $n$ .

Explanation: 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.

Question 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?

$X_n\xrightarrow{a.s.}c$ , but convergence in $L^1$ may fail when the sequence is not uniformly integrable.
$X_n\xrightarrow{p}c$ , but almost-sure and $L^1$ convergence are not guaranteed. (correct answer)
$X_n\xrightarrow{L^1}c$ , but convergence in probability can fail because the variables may be dependent.
Only $X_n\xrightarrow{d}c$ follows; convergence in probability requires all variables to share one distribution.

Explanation: 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.