Statistics Graduate Level Quiz: Slutskys And Continuous Mapping Theorems

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Slutskys And Continuous Mapping TheoremsQuestion 1 of 10

Let $X_n\xrightarrow{d}X$ , and define $I_n=\mathbf{1}\{X_n\leq 0\}$ . Which condition is sufficient for concluding that $I_n\xrightarrow{d}\mathbf{1}\{X\leq 0\}$ by the extended continuous mapping theorem?

It is sufficient that

P(X=0)=0

, even if the distributions of the variables

X_n

have atoms at zero.

It is sufficient that every variable

X_n

have a continuous distribution, regardless of whether

P(X=0)>0

It is necessary that

X_n\xrightarrow{p}X

, because weak convergence cannot be preserved by an indicator function.

It is necessary that

P(X_n=0)=0

for every

n

and that the variables

X_n

be mutually independent.

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Statistics Graduate Level Quiz

Statistics Graduate Level Quiz: Slutskys And Continuous Mapping Theorems

Practice Slutskys And Continuous Mapping Theorems 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 Slutskys And Continuous Mapping Theorems, 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

It is sufficient that $P(X=0)=0$ , even if the distributions of the variables $X_n$ have atoms at zero. (correct answer)
It is sufficient that every variable $X_n$ have a continuous distribution, regardless of whether $P(X=0)>0$ .
It is necessary that $X_n\xrightarrow{p}X$ , because weak convergence cannot be preserved by an indicator function.
It is necessary that $P(X_n=0)=0$ for every $n$ and that the variables $X_n$ be mutually independent.

Explanation: Whenever you see a question involving convergence in distribution composed with a function, reach for the Continuous Mapping Theorem (CMT). The extended version states that if

X_n \xrightarrow{d} X

and

g

is measurable, then

g(X_n) \xrightarrow{d} g(X)

, provided

P(X \in D_g) = 0

, where

D_g

is the set of discontinuities of

g

. Here,

g(x) = \mathbf{1}\{x \leq 0\}

, which is discontinuous at exactly one point:

x = 0

. The CMT therefore applies as long as

P(X = 0) = 0

— that is, as long as the limiting variable

X

places no mass at the single discontinuity of

g

. The distributions of the

X_n

themselves are irrelevant to this condition; the CMT only cares about the limiting distribution. This confirms that A is correct:

P(X = 0) = 0

is sufficient, and it doesn't matter whether the

X_n

have atoms at zero. B is wrong because continuity of each

X_n

's distribution is neither necessary nor sufficient — the condition must be placed on

X

, not the sequence. C is a fundamental misconception: convergence in distribution absolutely can be preserved by functions via the CMT; convergence in probability is a strictly stronger condition and is not required here. D introduces independence of the

X_n

, which is completely irrelevant — the CMT says nothing about joint structure of the sequence, only about the limit's distribution at discontinuity points. Study tip: For CMT problems, always identify the discontinuity set of

g

first, then check whether the limit

X

assigns it zero probability — the sequence's behavior is a red herring.

Question 2

For each $n$ , let $X_n$ and $Y_n$ be random variables satisfying $X_n\xrightarrow{d}N(0,1)$ and $Y_n\xrightarrow{d}N(0,1)$ . No information about their joint dependence is available. Which conclusion about $X_n+Y_n$ is justified?

It converges to $N(0,2)$ because the limiting marginal variances add under convergence in distribution.
It converges to $N(0,4)$ because the two marginal limits may be represented by the same normal variable.
It converges in probability to zero because the two centered marginal limits cancel asymptotically.
Its limiting distribution is not determined from the two marginal convergence statements alone. (correct answer)

Explanation: Whenever you see a question involving convergence in distribution of multiple random variables, your first instinct should be: what do we know about their joint distribution? Marginal convergence statements tell you nothing about dependence structure, and sums depend critically on that structure. Here's why D is correct. Consider two extreme cases. If

X_n

and

Y_n

are independent with each converging to

N(0,1)

, then

X_n + Y_n \xrightarrow{d} N(0,2)

. But if

Y_n = -X_n

for every

n

, then both marginals still converge to

N(0,1)

, yet

X_n + Y_n = 0

for all

n

, which converges to a point mass at zero. You can also construct a coupling where

Y_n = X_n

, giving

X_n + Y_n \xrightarrow{d} N(0,4)

. Since different dependence structures yield completely different limiting distributions, the marginal convergences alone are insufficient to determine the limit of the sum. A is wrong because variances of marginals do not simply add under convergence in distribution when dependence is unknown — that rule applies to independent random variables, not arbitrary ones. B is wrong for the opposite extreme: representing both limits by the same normal variable (i.e., assuming perfect positive dependence) is just one possibility, not a necessity. C is wrong because cancellation to zero requires

Y_n \approx -X_n

, which is a specific dependence assumption that is not given. The key study takeaway: convergence in distribution is a marginal statement. To analyze functions of multiple sequences (like sums), you need joint convergence — typically established via the continuous mapping theorem or Slutsky's theorem, both of which require additional structural assumptions.

Question 3

Suppose $nX_n\xrightarrow{p}\lambda$ for a finite constant $\lambda$ , and assume $P(X_n>-1)\to1$ . What is the probability limit of $(1+X_n)^n$ ?

$1+\lambda,$ obtained by replacing the power expression with its first-order binomial approximation.
$e^\lambda,$ obtained by applying Slutsky's theorem to the logarithm of the expression. (correct answer)
$e^{\lambda^2/2},$ obtained by retaining the leading quadratic term in the logarithmic expansion.
No limit is implied because the transformation $x\mapsto(1+x)^n$ changes with $n$ .

Explanation: When you see an expression like

(1+X_n)^n

where

X_n \to 0

in probability, your instinct should be to take the logarithm and invoke the classical limit

\lim_{n\to\infty}(1+a_n)^n = e^L

whenever

na_n \to L

. This is a convergence-in-probability analogue of a fundamental deterministic result. The key move is writing

\log(1+X_n)^n = n\log(1+X_n)

. Since

nX_n \xrightarrow{p} \lambda

, we know

X_n \xrightarrow{p} 0

. Using the Taylor expansion

\log(1+x) \approx x - x^2/2 + \cdots

near zero, we get

n\log(1+X_n) = nX_n \cdot \frac{\log(1+X_n)}{X_n} \xrightarrow{p} \lambda \cdot 1 = \lambda

, since the ratio

\frac{\log(1+X_n)}{X_n} \xrightarrow{p} 1

by continuity. Applying the continuous mapping theorem to

e^x

then gives

(1+X_n)^n \xrightarrow{p} e^\lambda

, confirming B. Choice A is wrong because the first-order binomial approximation

(1+X_n)^n \approx 1 + nX_n

is only valid when

nX_n

is small — here

nX_n \to \lambda

, a finite but potentially nonzero constant, and the higher-order terms in the exponential expansion matter crucially. Choice C incorrectly retains only the quadratic term from

\log(1+x) \approx x - x^2/2

, but since

X_n = O(1/n)

, the term

nX_n^2 \to 0

, so the quadratic contribution vanishes and does not alter the exponent. Choice D reflects a common confusion: the fact that the transformation changes with

n

does not prevent convergence. The continuous mapping theorem and probabilistic limit arguments handle sequences of transformations when the limit is well-defined. Study tip: Whenever you see

(1+\text{small})^n

, immediately take logarithms and use

\log(1+x)/x \to 1

. This logarithm-then-exponentiate strategy is your go-to tool for limits of this form.

Question 4

Suppose $T_n\xrightarrow{d}\chi_k^2$ and a data-dependent critical value satisfies $c_n\xrightarrow{p}q_{1-\alpha}$ , where $q_{1-\alpha}$ is the upper $\alpha$ critical value of the $\chi_k^2$ distribution. No independence between $T_n$ and $c_n$ is assumed. What is the limit of $P(T_n>c_n)$ ?

$0,$ because a consistent critical value eventually exceeds the realized value of the limiting statistic.
$1-\alpha,$ because $q_{1-\alpha}$ is the cumulative probability attained at the limiting critical value.
$\alpha,$ because Slutsky's theorem gives $T_n-c_n\xrightarrow{d}\chi_k^2-q_{1-\alpha}$ and the boundary has probability zero. (correct answer)
The limit is undetermined unless $T_n$ and $c_n$ are asymptotically independent.

Explanation: When a test statistic converges in distribution and a critical value converges in probability, your instinct should be to reach for Slutsky's theorem: if

X_n \xrightarrow{d} X

and

Y_n \xrightarrow{p} c

, then

X_n - Y_n \xrightarrow{d} X - c

, regardless of the dependence structure between

X_n

and

Y_n

. This is the key insight the question is testing. Applying Slutsky directly,

T_n - c_n \xrightarrow{d} \chi_k^2 - q_{1-\alpha}

. Then:

P(T_n > c_n) = P(T_n - c_n > 0) \to P(\chi_k^2 - q_{1-\alpha} > 0) = P(\chi_k^2 > q_{1-\alpha}) = \alpha

The last step uses the fact that

q_{1-\alpha}

is defined as the upper-

\alpha

quantile of

\chi_k^2

, and since

\chi_k^2

is continuous, the boundary point

\{X = q_{1-\alpha}\}

has probability zero. This confirms C is correct. A is wrong because it confuses almost-sure behavior with distributional convergence —

c_n \xrightarrow{p} q_{1-\alpha}

does not mean

c_n

eventually exceeds

T_n

with certainty; the probability approaches

\alpha

, not 0. B is wrong because

1 - \alpha

is the CDF evaluated at

q_{1-\alpha}

, i.e.,

P(\chi_k^2 \leq q_{1-\alpha})

. The question asks for the tail probability

P(T_n > c_n)

, which is the complement. D is the most tempting distractor. Slutsky's theorem is precisely powerful because it requires no independence assumption — only that one sequence converges in probability to a constant. Study tip: Whenever you see

\xrightarrow{d}

paired with

\xrightarrow{p}

to a constant, Slutsky's theorem applies unconditionally — independence is never required when the second limit is a constant.

Question 5

Suppose an estimator satisfies $\sqrt{n}(\widehat{\theta}_n-\theta)\xrightarrow{d}N(0,V),$ where $0<V<\infty$ , and suppose $\widehat{V}_n\xrightarrow{p}V$ . No independence between $\widehat{\theta}_n$ and $\widehat{V}_n$ is assumed. What is the limiting distribution of $T_n=\frac{\sqrt{n}(\widehat{\theta}_n-\theta)}{\sqrt{\widehat{V}_n}}?$

$T_n\xrightarrow{d}N(0,1),$ because the denominator converges in probability to the positive constant $\sqrt V$ . (correct answer)
$T_n\xrightarrow{d}N(0,V),$ because replacing a parameter by a consistent estimator does not alter the original variance.
$T_n\xrightarrow{p}0,$ because both the numerator and denominator are based on consistent estimators of fixed quantities.
The limit cannot be determined unless the numerator and denominator are asymptotically independent.

Explanation: When you see a question involving a standardized statistic where a variance parameter is replaced by a consistent estimator, your instinct should be to reach for Slutsky's theorem: if

X_n \xrightarrow{d} X

and

Y_n \xrightarrow{p} c

(a constant), then

X_n / Y_n \xrightarrow{d} X/c

, regardless of any dependence structure between

X_n

and

Y_n

. Here, the numerator

\sqrt{n}(\widehat{\theta}_n - \theta) \xrightarrow{d} N(0,V)

, and since

\widehat{V}_n \xrightarrow{p} V > 0

, the continuous mapping theorem gives

\sqrt{\widehat{V}_n} \xrightarrow{p} \sqrt{V}

. Applying Slutsky's theorem,

T_n = \frac{\sqrt{n}(\widehat{\theta}_n-\theta)}{\sqrt{\widehat{V}_n}} \xrightarrow{d} \frac{N(0,V)}{\sqrt{V}} = N\!\left(0, \frac{V}{V}\right) = N(0,1).

This confirms A is correct. B is wrong because dividing by

\sqrt{\widehat{V}_n}

does alter the limiting variance — it normalizes it to 1, not leaves it at

V

. C is wrong because the numerator does not converge in probability to zero; it diverges in distribution to a non-degenerate normal after the

\sqrt{n}

scaling. The ratio stabilizes to a non-trivial limit. D is the most tempting distractor — students often believe Slutsky's theorem requires independence. It does not; independence is sufficient but not necessary. Slutsky's theorem only requires convergence in distribution for one sequence and convergence in probability for the other. Your takeaway: Slutsky's theorem does not require independence. Anytime you see a consistent estimator plugged into a denominator (or any continuous function), you can apply it freely — this is the backbone of virtually every asymptotic

t

-statistic argument in statistics.

Question 6

Let $\widehat\beta_n$ estimate a parameter $\beta_0\in\mathbb R^k$ , and suppose $\sqrt n(\widehat\beta_n-\beta_0)\xrightarrow{d}N_k(0,\Sigma)$ , where $\Sigma$ is positive definite. If $\widehat\Sigma_n\xrightarrow{p}\Sigma$ and is invertible with probability approaching one, what is the limit of $W_n=n(\widehat\beta_n-\beta_0)^{\mathsf T}\widehat\Sigma_n^{-1}(\widehat\beta_n-\beta_0)?$

$W_n\xrightarrow{p}k,$ because the mean of the limiting quadratic form equals the parameter dimension.
$W_n\xrightarrow{d}N(k,2k),$ because a quadratic form in a normal vector is asymptotically normal.
$W_n\xrightarrow{d}\chi_1^2,$ because the quadratic form reduces the vector estimator to one scalar statistic.
$W_n\xrightarrow{d}\chi_k^2,$ because consistent matrix inversion and Slutsky's theorem produce a standardized quadratic form. (correct answer)

Explanation: Whenever you see a quadratic form built from an asymptotically normal estimator and a consistently estimated covariance matrix, your instinct should be to apply the continuous mapping theorem paired with Slutsky's theorem. Here's the core reasoning for why D is correct. You know

\sqrt{n}(\widehat{\beta}_n - \beta_0) \xrightarrow{d} Z \sim N_k(0, \Sigma)

. Since

\widehat{\Sigma}_n \xrightarrow{p} \Sigma

and

\Sigma

is positive definite, by Slutsky's theorem

\widehat{\Sigma}_n^{-1} \xrightarrow{p} \Sigma^{-1}

. The statistic

W_n

can be written as

\left[\sqrt{n}(\widehat{\beta}_n - \beta_0)\right]^{\mathsf{T}} \widehat{\Sigma}_n^{-1} \left[\sqrt{n}(\widehat{\beta}_n - \beta_0)\right]

, which converges in distribution to

Z^{\mathsf{T}}\Sigma^{-1}Z

. Because

\Sigma

is positive definite, write

\Sigma = L L^{\mathsf{T}}

; then

L^{-1}Z \sim N_k(0, I_k)

, so

Z^{\mathsf{T}}\Sigma^{-1}Z = \|L^{-1}Z\|^2 \sim \chi^2_k

. The degrees of freedom equal the dimension

k

of the parameter vector, confirming D. A contains a grain of truth —

\mathbb{E}[\chi^2_k] = k

— but convergence in distribution to

\chi^2_k

is not the same as convergence in probability to the constant

k

. That would require the variance to vanish, which it does not. B is wrong because quadratic forms in normal vectors follow chi-squared distributions, not normal distributions. A

\chi^2_k

variable has mean

k

and variance

2k

, but the distribution itself is decidedly non-normal. C confuses dimension reduction with degrees of freedom. Collapsing a

k

-vector into one scalar statistic does not produce

\chi^2_1

unless

k=1

; the degrees of freedom track the rank of the quadratic form, which is

k

. Study tip: Memorize the chain — asymptotic normality + consistent covariance estimator + Slutsky + CMT =

\chi^2_k

Wald statistic. This is the backbone of virtually every large-sample hypothesis test you will encounter.

Question 7

Suppose $\theta\neq0$ and $\sqrt n(\widehat\theta_n-\theta)\xrightarrow{d}N(0,\sigma^2)$ . Using an algebraic decomposition followed by Slutsky's theorem, what is the limiting distribution of $\sqrt n\left(\widehat\theta_n^{-1}-\theta^{-1}\right)?$

$N\left(0,\frac{\sigma^2}{\theta^2}\right),$ because taking a reciprocal divides the asymptotic variance by $\theta^2$ .
$N\left(0,\frac{\sigma^2}{\theta^4}\right),$ because the random denominator converges in probability to $\theta^2$ . (correct answer)
$N\left(0,\sigma^2\theta^4\right),$ because inversion reverses the scale before the limiting variance is computed.
No weak limit follows because the reciprocal map is discontinuous at zero and $\widehat\theta_n$ may approach zero.

Explanation: When you see a question asking for the limiting distribution of a transformed estimator, your first instinct should be the Delta Method — but this problem specifically asks you to use an algebraic decomposition plus Slutsky's theorem, which reveals the same result from first principles. Start by writing the key identity:

\widehat\theta_n^{-1} - \theta^{-1} = \frac{\theta - \widehat\theta_n}{\widehat\theta_n \cdot \theta}

Multiplying both sides by

\sqrt{n}

\sqrt{n}\left(\widehat\theta_n^{-1} - \theta^{-1}\right) = \frac{-\sqrt{n}(\widehat\theta_n - \theta)}{\widehat\theta_n \cdot \theta}

Now apply Slutsky's theorem. Since

\sqrt{n}(\widehat\theta_n - \theta) \xrightarrow{d} N(0, \sigma^2)

and

\widehat\theta_n \xrightarrow{p} \theta

(consistency implied by asymptotic normality), the denominator

\widehat\theta_n \cdot \theta \xrightarrow{p} \theta^2

. Slutsky's theorem lets you replace the random denominator with its probability limit, giving:

\sqrt{n}\left(\widehat\theta_n^{-1} - \theta^{-1}\right) \xrightarrow{d} N\!\left(0,\frac{\sigma^2}{\theta^4}\right)

This confirms B is correct — the

\theta^4

in the denominator comes from squaring the

\theta^2

that appears when you divide. A is wrong because dividing by

\theta^2

(not

\theta^4

) would correspond to only one power of

\theta

in the denominator, misapplying the scaling. C inverts the direction of the scaling entirely — a clear algebraic error. D is a red herring; the assumption

\theta \neq 0

is explicitly given, so the reciprocal map is continuous at the limit point. Study tip: Whenever you apply Slutsky's theorem to a ratio, carefully track every factor in the denominator — each contributes to the final variance through squaring, which is where

\theta^2 \to \theta^4

trips up many students.

Question 8

Suppose $X_n\xrightarrow{d}N(0,1)$ and $Y_n\xrightarrow{p}0$ , with arbitrary dependence between the sequences. Define $M_n=\max(X_n,Y_n)$ . Which function is the cumulative distribution function of the limiting random variable?

$F(t)=\Phi(t)$ for every real $t$ , because Slutsky's theorem replaces $Y_n$ by its probability limit and the standard normal CDF is unchanged.
$F(t)=0$ for $t<0$ and $F(t)=\Phi(t)$ for $t\geq0$ , so the limit is a mixture with a point mass of $\tfrac{1}{2}$ at zero. (correct answer)
$F(t)=\Phi(t)^2$ for every real $t$ , because the maximum of two sequences with the same marginal limit has a squared CDF.
$F(t)=0$ for $t\leq0$ and $F(t)=2\Phi(t)-1$ for $t>0$ , giving a continuous half-normal limit with no mass at zero.

Explanation: When you see a maximum of two sequences where one converges in distribution and the other converges in probability to a constant, your instinct should be to carefully track how the maximum behaves on each side of that constant — Slutsky's theorem alone won't carry you through. Here,

Y_n \xrightarrow{p} 0

, so for large

n

Y_n

is negligible except when

X_n

is also near zero. The key insight is that

M_n = \max(X_n, Y_n)

. For

t < 0

: since

Y_n \xrightarrow{p} 0

, the event

\{M_n \leq t\}

requires both

X_n \leq t

and

Y_n \leq t

, but

P(Y_n \leq t) \to 0

for

t < 0

, so

F_{M_n}(t) \to 0

. For

t \geq 0

P(M_n \leq t) = P(X_n \leq t, Y_n \leq t)

. Since

Y_n \xrightarrow{p} 0 \leq t

, the event

\{Y_n \leq t\}

has probability approaching 1, leaving

P(M_n \leq t) \to \Phi(t)

. This gives a CDF that jumps from 0 to

\Phi(0) = \tfrac{1}{2}

t = 0

, confirming answer B — a mixture with a point mass of

\tfrac{1}{2}

at zero. A misapplies Slutsky: that theorem handles sums and products of sequences, not maxima, and it ignores the truncation effect below zero. C incorrectly assumes independence and identical marginals, producing

\Phi(t)^2

, which would only be valid if both variables were i.i.d. standard normal. D describes a half-normal, which has no mass at zero but gets the behavior for

t > 0

wrong — it confuses

|Z|

with

\max(Z, 0)

. The study tip: for maximum operations in limit theorems, always evaluate the CDF separately on each region, checking where the "degenerate" sequence controls the bound.

Question 9

Suppose $X_n\xrightarrow{d}N(1,4)$ and $Y_n\xrightarrow{p}2$ . The two sequences may be dependent. What is the limiting distribution of $W_n=Y_nX_n+Y_n^{-1}?$

$W_n\xrightarrow{d}N\left(\frac{5}{2},16\right),$ obtained by applying the limiting affine transformation to $X_n$ . (correct answer)
$W_n\xrightarrow{d}N\left(\frac{3}{2},4\right),$ because only the reciprocal term changes the limiting mean.
$W_n\xrightarrow{d}N\left(\frac{5}{2},8\right),$ because multiplication by two doubles the limiting variance.
The limiting distribution is undetermined unless $X_n$ and $Y_n$ are independent for every $n$ .

Explanation: When you see a sequence converging in distribution combined with another sequence converging in probability, your immediate instinct should be to reach for Slutsky's Theorem: if

X_n \xrightarrow{d} X

and

Y_n \xrightarrow{p} c

(a constant), then

g(X_n, Y_n) \xrightarrow{d} g(X, c)

for continuous functions

g

. Critically, this holds regardless of dependence between the sequences — that's what makes Slutsky so powerful. Here, apply Slutsky directly: since

Y_n \xrightarrow{p} 2

, replace every

Y_n

with its probability limit of 2. The function

W_n = Y_n X_n + Y_n^{-1}

converges in distribution to

2X + 2^{-1} = 2X + \frac{1}{2}

, where

X \sim N(1,4)

. For a normal random variable, an affine transformation

aX + b

gives

N(a\mu + b,\ a^2\sigma^2)

. So the limit is

N\!\left(2(1)+\tfrac{1}{2},\ 4(4)\right) = N\!\left(\tfrac{5}{2}, 16\right)

. This confirms answer A is correct. Answer B incorrectly ignores that multiplying

X_n

by 2 scales the variance — it only accounts for the mean shift from the reciprocal term. Answer C gets the mean right but miscalculates the variance: multiplying by 2 multiplies variance by

2^2 = 4

, giving 16, not 8. Answer D is the most tempting trap — many students believe Slutsky requires independence. It does not; independence is sufficient but not necessary when one sequence converges to a constant in probability. Your study tip: memorize that Slutsky's Theorem demands only

Y_n \xrightarrow{p} c

, not independence. Exam questions are specifically designed to make you doubt this by mentioning potential dependence — don't take the bait.

Question 10

Let $Z\sim N(0,1)$ and define $X_n=Z/n$ and $J_n=\mathbf{1}\{X_n>0\}$ . Although $X_n\xrightarrow{p}0$ , which statement correctly describes the behavior of $J_n$ ?

$J_n\xrightarrow{p}0,$ because applying any bounded measurable function preserves convergence in probability.
$J_n\xrightarrow{p}1/2,$ because its expectation equals one-half for every value of $n$ .
$J_n\xrightarrow{d}\operatorname{Bernoulli}(1/2),$ because its value equals $\mathbf{1}\{Z>0\}$ for every $n$ . (correct answer)
$J_n$ has no limiting distribution because the indicator map is discontinuous at the probability limit.

Explanation: When studying convergence in probability, a critical subtlety is that the continuous mapping theorem requires the limiting function to be continuous at the limit point — and indicator functions are not. Here,

X_n = Z/n \xrightarrow{p} 0

, but notice something remarkable: for every fixed

n

J_n = \mathbf{1}\{X_n > 0\} = \mathbf{1}\{Z/n > 0\} = \mathbf{1}\{Z > 0\}

. The division by

n

doesn't change the sign of

Z

. So

J_n

is literally the same random variable for every

n

— it never changes. Since

\mathbf{1}\{Z > 0\} \sim \operatorname{Bernoulli}(1/2)

, the sequence trivially converges in distribution to

\operatorname{Bernoulli}(1/2)

. Answer C is correct. A is wrong because the continuous mapping theorem does not apply to all bounded measurable functions — only to those continuous at the limit. The indicator

\mathbf{1}\{\cdot > 0\}

is discontinuous at 0, which is precisely where

X_n

is converging. So you cannot conclude $$J_n \xrightarrow{p} \mathbf{1}{0 > 0} = 0$. B is subtly misleading. Yes,

E[J_n] = 1/2

for all

n

, but equal expectations do not imply convergence in probability to a constant. Convergence in probability to

1/2

would require

J_n

to concentrate around

1/2

, which a Bernoulli random variable never does. D is wrong because a limiting distribution clearly exists — as shown,

J_n

is constant across

n

. Strategy tip: Whenever you see convergence applied through a discontinuous function, stop and check whether the continuous mapping theorem actually applies. Evaluate what the transformed sequence looks like directly — sometimes, as here, the sequence is literally constant.