Statistics Graduate Level Quiz: Central Limit Theorem

9 questions · exam conditions

0:00

Central Limit TheoremQuestion 1 of 9

Let $X_1,X_2,\ldots$ be independent and identically distributed with $E[X_i]=0$ and $\operatorname{Var}(X_i)=4$ . No normality assumption is made. If $\bar X_n=n^{-1}\sum_{i=1}^n X_i$ , what is the limiting distribution of $n\bar X_n^2$ ?

A distribution equal to

4\chi_1^2

A distribution equal to

\chi_1^2

A normal distribution with mean

0

and variance

16

A degenerate distribution concentrated at

4

← Back to quizzes

Statistics Graduate Level Quiz

Statistics Graduate Level Quiz: Central Limit Theorem

Practice Central Limit Theorem 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 Central Limit Theorem, 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

A distribution equal to $4\chi_1^2$ (correct answer)
A distribution equal to $\chi_1^2$
A normal distribution with mean $0$ and variance $16$
A degenerate distribution concentrated at $4$

Explanation: Whenever you see a question involving

\bar{X}_n

raised to a power, your instinct should be to apply the Central Limit Theorem (CLT) combined with the Continuous Mapping Theorem (CMT). The CLT tells you that

\sqrt{n}\bar{X}_n \xrightarrow{d} N(0, \sigma^2)

, where

\sigma^2 = \text{Var}(X_i) = 4

. So

\sqrt{n}\bar{X}_n \xrightarrow{d} N(0,4)

. Now notice that

n\bar{X}_n^2 = \left(\sqrt{n}\bar{X}_n\right)^2

. Since squaring is a continuous function, the CMT lets you square the limiting distribution directly:

n\bar{X}_n^2 \xrightarrow{d} [N(0,4)]^2

. A

N(0,4)

random variable can be written as

2Z

where

Z \sim N(0,1)

, so

(2Z)^2 = 4Z^2 \sim 4\chi^2_1

. This confirms answer A is correct. Answer B,

\chi^2_1

, forgets to account for the variance. It would be correct only if

\sigma^2 = 1

, but here

\sigma^2 = 4

, so the scaling factor of 4 must appear. Answer C proposes a normal limiting distribution, but squaring a normal random variable produces a chi-squared-type distribution, not another normal. The CMT preserves the distribution under continuous transformations — it does not preserve normality. Answer D, a degenerate distribution at 4, would be the result of applying the Law of Large Numbers:

n\bar{X}_n^2

grows like

n

, not like a constant, so no degenerate limit exists. Key strategy: When you see

n\bar{X}_n^2

, immediately rewrite it as

(\sqrt{n}\bar{X}_n)^2

, apply CLT, then CMT. Always carry the variance

\sigma^2

through — it's the most common source of scaling errors on these problems.

Question 2

An experiment contains $m$ independent clusters, each with $4$ observations. Within every cluster, outcomes have common mean $\mu$ , common variance $\sigma^2$ , and pairwise correlation $\rho=0.25$ . Assume a central limit theorem applies to the independent cluster totals. If $\bar Y_m$ is the mean of all $4m$ outcomes, what is the limiting distribution of $\sqrt{4m}(\bar Y_m-\mu)$ as $m\to\infty$ ?

$N(0,\sigma^2)$ , treating all individual outcomes as independent and ignoring within-cluster correlation
$N(0,1.75\sigma^2)$ , because within-cluster covariances inflate the effective variance by a factor of $1+3\rho$ (correct answer)
$N(0,0.4375\sigma^2)$ , equal to the variance of a single cluster mean without rescaling by the sample size
$N(0,4\sigma^2)$ , treating each cluster total as a single perfectly correlated observation

Explanation: When you see clustered data with within-group correlation, your first instinct should be to work at the cluster level, not the individual observation level, because the CLT applies to independent units — and here, the independent units are clusters. Start by computing the variance of a single cluster total. Each cluster has 4 observations with variance

\sigma^2

and pairwise correlation

\rho = 0.25

. The variance of the cluster total is

\text{Var}(T_i) = 4\sigma^2 + 2\binom{4}{2}\rho\sigma^2 = 4\sigma^2 + 12(0.25)\sigma^2 = 7\sigma^2

. The grand mean is

\bar{Y}_m = \frac{1}{4m}\sum_{i=1}^m T_i

, so

\text{Var}(\bar{Y}_m) = \frac{7\sigma^2}{16m}

. Rescaling by

\sqrt{4m}

gives

\text{Var}\!\left(\sqrt{4m}\,\bar{Y}_m\right) = 4m \cdot \frac{7\sigma^2}{16m} = \frac{7\sigma^2}{4} = 1.75\sigma^2

. This equals

\sigma^2(1 + 3\rho)

, confirming answer B. A is tempting but wrong — it treats all

4m

outcomes as independent, ignoring the

\binom{4}{2} = 6

covariance terms per cluster, which underestimates the variance. C computes

\text{Var}(\bar{T}_i/4) = 7\sigma^2/16

, the variance of one cluster mean, but forgets to multiply by

4m

from the rescaling — it confuses a single-cluster quantity with the limiting variance. D overcorrects by assuming perfect correlation (

\rho=1

) within clusters, yielding

4\sigma^2

, which inflates variance far beyond what

\rho=0.25

produces. The key study tip: always compute

\text{Var}(\text{cluster total})

explicitly, accounting for all pairwise covariances. The design effect

1 + (n-1)\rho

is worth memorizing — it directly captures how intra-cluster correlation inflates variance relative to the independent case.

Question 3

Let $X_1,X_2,\ldots$ be independent and identically distributed with Pareto tail $P(X_i>x)=x^{-3/2}$ for $x\ge1$ , and let $\mu=E[X_i]$ . Which statement correctly describes the asymptotic behavior of the sample mean $\bar X_n$ ?

$\sqrt n(\bar X_n-\mu)$ converges to a normal distribution, because the mean is finite
$n^{1/3}(\bar X_n-\mu)$ converges to a non-Gaussian stable distribution with index $3/2$ (correct answer)
$n^{1/3}(\bar X_n-\mu)$ converges to a standard normal distribution at the stable rate
$n^{2/3}(\bar X_n-\mu)$ converges to a non-Gaussian stable distribution with index $3/2$

Explanation: When you see a sample mean from a heavy-tailed distribution, your first instinct should be to check whether the variance is finite — because that determines whether the classical CLT applies or whether you need a generalized stable-law framework. Here,

P(X_i > x) = x^{-3/2}

, so the density is

f(x) = \frac{3}{2}x^{-5/2}

for

x \geq 1

. The mean

\mu = E[X_i]

is finite (the integral converges), but

E[X_i^2] = \int_1^\infty x^2 \cdot \frac{3}{2}x^{-5/2}\,dx = \int_1^\infty \frac{3}{2}x^{-1/2}\,dx = \infty

. No finite variance means the classical CLT fails. Instead, you apply the Generalized Central Limit Theorem: when the tail satisfies

P(X > x) \sim x^{-\alpha}

with

\alpha \in (1,2)

, the properly centered and scaled mean converges to a stable distribution with index $\alpha$ . Here

\alpha = 3/2

, so the stable index is

3/2

. The correct normalization is

n^{1/\alpha} = n^{2/3}

... wait — actually the sum

S_n

scales as

n^{1/\alpha}

, meaning

\bar{X}_n - \mu = S_n/n - \mu

scales as

n^{1/\alpha - 1} = n^{1/3-1} = n^{-2/3}

, so multiplying by

n^{2/3}

centers the convergence. That confirms B — wait, B uses

n^{1/3}

, while D uses

n^{2/3}

. The correct scaling factor is

n^{1/\alpha} / n = n^{1/\alpha - 1}

, so you need

n^{2/3}

to get a nondegenerate limit, making D the correct answer. A is wrong because finite mean alone does not guarantee the CLT — you need finite variance. B uses the wrong normalizing exponent

n^{1/3}

instead of

n^{2/3}

. C compounds two errors: wrong exponent and wrong limiting distribution (stable ≠ normal). Study tip: Memorize the two-step check — finite mean gives you a center; finite variance gives you the CLT rate

\sqrt{n}

. If variance is infinite but the tail index is

\alpha \in (1,2)

, the rate is

n^{1-1/\alpha}

and the limit is a non-Gaussian stable law.

Question 4

Let $X_1,X_2,\ldots$ be independent and identically distributed with $E[X_i]=0$ and $\operatorname{Var}(X_i)=1$ . Consider the unequally weighted statistic $T_n=n^{-3/2}\sum_{i=1}^n iX_i$ . Which limiting distribution follows from an appropriate triangular-array central limit theorem?

$N(0,1)$ because the observations have unit variance
$N(0,1/2)$ because the average normalized weight is one-half
$N(0,1/3)$ because the normalized squared weights converge to one-third (correct answer)
No normal limit, because the largest weight is not equal to the others

Explanation: When you see a statistic built from non-constant weights applied to i.i.d. variables, your instinct should be to reach for the Lindeberg-Feller triangular array CLT, not the classical CLT. The key question becomes: what variance does

T_n

actually accumulate, and do the Lindeberg conditions hold? Start by computing

\operatorname{Var}(T_n)

. Since the

X_i

are independent with unit variance:

\operatorname{Var}(T_n) = n^{-3} \sum_{i=1}^n i^2 \cdot \operatorname{Var}(X_i) = n^{-3} \sum_{i=1}^n i^2.

Using the identity

\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6} \sim \frac{n^3}{3}

, you get

\operatorname{Var}(T_n) \to \frac{1}{3}

. The Lindeberg condition is satisfied because the largest individual term

n^{-3/2} \cdot n \cdot X_n

contributes variance

n^{-1} \to 0

, so no single summand dominates. The triangular array CLT therefore gives

T_n \xrightarrow{d} N(0, 1/3)

, confirming C. Choice A is wrong because

\operatorname{Var}(X_i) = 1

tells you about individual terms, not the accumulated variance of the weighted sum — you still need to track the weights. Choice B confuses the average weight

\frac{1}{n}\sum i/n \to 1/2

with the correct quantity, which involves squared weights summed. Choice D is a misconception: unequal weights are perfectly compatible with a normal limit as long as Lindeberg's condition holds, which it does here because the maximum weight shrinks to zero relative to the total. Your study tip: always compute

\sum w_{n,i}^2

for weighted sums — this is what determines the limiting variance, and it's almost never the same as the average weight.

Question 5

A sequence of finite populations has sizes $N\to\infty$ , population means $\mu_N$ , and population variances defined by $N^{-1}\sum_{j=1}^N(Y_{Nj}-\mu_N)^2\to25$ . From each population, a simple random sample without replacement of size $n$ is drawn, with $n/N\to0.4$ . Assume the finite-population central limit theorem's no-dominant-unit condition holds. What is the limit of $\sqrt n(\bar Y_n-\mu_N)$ ?

$N(0,25)$ , because the sample mean uses $n$ observations
$N(0,15)$ , after applying the finite-population correction (correct answer)
$N(0,10)$ , because only forty percent of the population is sampled
A degenerate limit at $0$ , because sampling is without replacement

Explanation: When sampling without replacement from a finite population, the variance of the sample mean is not simply

\sigma^2/n

— it gets deflated by the finite population correction (FPC). Specifically, the variance of

\bar{Y}_n

under simple random sampling without replacement is:

\text{Var}(\bar{Y}_n) = \frac{\sigma_N^2}{n}\left(1 - \frac{n}{N}\right)

where

\sigma_N^2

is the population variance. By the finite-population CLT (under the no-dominant-unit condition),

\sqrt{n}(\bar{Y}_n - \mu_N)

converges in distribution to

N(0,\, \sigma_N^2(1-f))

, where

f = n/N

is the sampling fraction. Here,

\sigma_N^2 \to 25

and

f \to 0.4

, so the limiting variance is

25(1-0.4) = 25(0.6) = 15

. The correct answer is B: the limiting distribution is

N(0,15)

. Choice A ignores the FPC entirely, as if the draws were independent (i.e., sampling with replacement). When

n/N

converges to a positive constant, this omission is not asymptotically negligible — you must account for the correction. Choice C computes

25 \times 0.4 = 10

, mistakenly multiplying by the sampling fraction itself rather than its complement

(1-f)

. This confuses what's removed from variance with what's retained. Choice D is a tempting trap: without replacement does reduce variance, but it does not collapse the distribution to a point mass unless

f \to 1

(a census). Here

f \to 0.4 < 1

, so a genuine normal limit persists. Study tip: Always ask whether

n/N

converges to a positive constant. If yes, the FPC factor

(1-f)

is asymptotically material and must multiply the variance in your CLT statement.

Question 6

Suppose $X_1,X_2,\ldots$ are independent and identically distributed with mean $\mu$ and variance $0<\sigma^2<\infty$ , but $E[(X_1-\mu)^4]=\infty$ . Let $S_n^2=(n-1)^{-1}\sum_{i=1}^n(X_i-\bar X_n)^2$ . Which conclusion about $T_n=\sqrt n(\bar X_n-\mu)/S_n$ is justified?

$T_n\Rightarrow N(0,1)$ despite the infinite fourth moment (correct answer)
$T_n$ has exactly a $t_{n-1}$ distribution for every $n$
$T_n$ has no limiting distribution because its denominator is inconsistent
$T_n\Rightarrow N(0,1)$ only if the fourth moment is finite

Explanation: When you see a question about the Student's t-statistic under non-normal assumptions, your instinct should be to think carefully about what each piece of the statistic requires to converge — the numerator and denominator separately. The key insight is that

T_n = \sqrt{n}(\bar{X}_n - \mu)/S_n

is a ratio. The numerator

\sqrt{n}(\bar{X}_n - \mu)

converges in distribution to

N(0, \sigma^2)

by the Central Limit Theorem, which requires only finite variance — not finite fourth moment. The denominator

S_n^2

is a consistent estimator of

\sigma^2

under the same condition: by the Law of Large Numbers,

S_n^2 \xrightarrow{p} \sigma^2

whenever

\sigma^2 < \infty

. Applying Slutsky's theorem, the ratio converges to

N(0,1)

. The fourth moment plays no role in either argument. So A is correct. B is wrong because the exact

t_{n-1}

distribution requires normality of the

X_i

, not just finite variance. Without normality, the t-distribution is only a large-sample approximation. C is wrong because the premise is false —

S_n^2

is consistent under finite variance; inconsistency would require

\sigma^2 = \infty

. D is the most tempting trap: it conflates what's needed for the CLT with what's needed for other results (like the delta method or variance of

S_n^2

). Finite fourth moment is sufficient for

\text{Var}(S_n^2) < \infty

, but it is not necessary for

S_n^2 \xrightarrow{p} \sigma^2

. Your study tip: always decompose a statistic into its parts and ask what each part needs to converge. The CLT and LLN require only two moments — don't impose stronger conditions than necessary.

Question 7

Let $N_n\sim\operatorname{Poisson}(n)$ , and independently let $X_1,X_2,\ldots$ be independent and identically distributed with $E[X_i]=\mu$ and $\operatorname{Var}(X_i)=\sigma^2<\infty$ . Define the random sum $S_n=\sum_{i=1}^{N_n}X_i$ , with an empty sum equal to $0$ . What is the limiting distribution of $(S_n-n\mu)/\sqrt n$ ?

$N(0,\sigma^2)$ , because only the summand variability contributes
$N(0,\mu^2)$ , because only the random count contributes
$N(0,\sigma^2+\mu^2)$ , because both sources of variation contribute (correct answer)
$N(0,\sigma^2\mu^2)$ , because the two variances multiply

Explanation: When you encounter a random sum

S_n = \sum_{i=1}^{N_n} X_i

where the number of terms is itself random, you need to account for two independent sources of variability: randomness in the individual

X_i

values, and randomness in the count

N_n

. The key tool here is the law of total variance:

\operatorname{Var}(S_n) = E[\operatorname{Var}(S_n \mid N_n)] + \operatorname{Var}(E[S_n \mid N_n])

. Conditioning on

N_n

E[S_n \mid N_n] = N_n \mu

and

\operatorname{Var}(S_n \mid N_n) = N_n \sigma^2

. Since

N_n \sim \operatorname{Poisson}(n)

, we have

E[N_n] = \operatorname{Var}(N_n) = n

. Plugging in:

\operatorname{Var}(S_n) = E[N_n \sigma^2] + \operatorname{Var}(N_n \mu) = n\sigma^2 + n\mu^2

. Also,

E[S_n] = n\mu

, so

(S_n - n\mu)/\sqrt{n}

has variance

\sigma^2 + \mu^2

. A CLT argument (via characteristic functions or Wald's identity extensions) confirms the limit is

N(0, \sigma^2 + \mu^2)

, making C correct. Choice A captures only the within-group variance

\sigma^2

, ignoring that the random count

N_n

fluctuates around

n

with variance

n

, contributing

\mu^2

per unit. Choice B keeps only

\mu^2

, the contribution from count randomness, and discards the summand variability

\sigma^2

. Choice D multiplies the two variance components, which has no probabilistic justification — variances from independent additive sources add, they do not multiply. A useful rule of thumb: for Poisson random sums, the asymptotic variance is always

\sigma^2 + \mu^2

, reflecting that Poisson randomness contributes exactly

\mu^2

regardless of the distribution of

X_i

Question 8

Independent and identically distributed pairs $(X_i,Y_i)$ satisfy $E[X_i]=2$ , $E[Y_i]=1$ , $\operatorname{Var}(X_i)=4$ , $\operatorname{Var}(Y_i)=1$ , and $\operatorname{Cov}(X_i,Y_i)=1$ . For the ratio estimator $R_n=\bar X_n/\bar Y_n$ , what is the limiting distribution of $\sqrt n(R_n-2)$ ?

$N(0,4)$ , obtained via the delta method applied to $g(x,y)=x/y$ at the point $(2,1)$ (correct answer)
$N(0,8)$ , obtained by ignoring the covariance and summing only the scaled numerator and denominator variances
$N(0,2)$ , obtained by accounting for the covariance term only once instead of twice in the quadratic form
$N(0,5)$ , obtained by linearizing the ratio as the unweighted difference $\bar X_n - \bar Y_n$

Explanation: Whenever you see a ratio estimator problem, your first instinct should be the delta method: if

\sqrt{n}(\bar{\mathbf{Z}}_n - \boldsymbol{\mu}) \xrightarrow{d} N(0, \Sigma)

, then

\sqrt{n}(g(\bar{\mathbf{Z}}_n) - g(\boldsymbol{\mu})) \xrightarrow{d} N(0, \nabla g^\top \Sigma \nabla g)

. For

g(x,y) = x/y

evaluated at

\boldsymbol{\mu} = (2,1)

, the gradient is

\nabla g = (1/y,\, -x/y^2)\big|_{(2,1)} = (1, -2)

. The covariance matrix of

(X_i, Y_i)

is $$\Sigma = \begin{pmatrix}4 & 1\1 & 1\end{pmatrix}

$$\nabla g^\top \Sigma \nabla g = (1,\,-2)\begin{pmatrix}4&1\\1&1\end{pmatrix}\begin{pmatrix}1\\-2\end{pmatrix} = (1,-2)\begin{pmatrix}2\\-1\end{pmatrix} = 2+2 = 4.$$ So $$\sqrt{n}(R_n - 2) \xrightarrow{d} N(0,4)$$, confirming **A is correct**. **B** inflates the variance to 8 by summing $$\text{Var}(X)/\mu_Y^2 + \text{Var}(Y)\mu_X^2/\mu_Y^4 = 4 + 4 = 8$$ while completely ignoring the cross-covariance term $$-2\text{Cov}(X,Y)/\mu_Y^2 = -4$$, which subtracts 4 from the total. **C** gets 2 by subtracting the covariance contribution only once ($$8 - 2 = 6$$, or miscounting to get 2), misapplying the quadratic form where the off-diagonal must be counted twice due to symmetry. **D** linearizes $$R_n \approx \bar{X}_n - \bar{Y}_n$$, which ignores the proper first-order Taylor expansion and the scaling by $$\mu_Y$$; the correct linearization is $$R_n \approx \mu_X/\mu_Y + (1/\mu_Y)(\bar{X}_n - \mu_X) - (\mu_X/\mu_Y^2)(\bar{Y}_n - \mu_Y)$$. **Study tip:** Always write out $$\nabla g$$ explicitly and carefully compute the full quadratic form $$\nabla g^\top \Sigma \nabla g$$—off-diagonal covariance terms are the most common source of error in delta method problems.

Question 9

A stationary Gaussian time series has mean $\mu$ , marginal variance $1$ , and autocovariance $\operatorname{Cov}(X_t,X_{t+h})=(1/2)^{|h|}$ . Let $\bar X_n=n^{-1}\sum_{t=1}^nX_t$ . What is the limiting distribution of $\sqrt n(\bar X_n-\mu)$ ?

$N(0,1)$ , using only the marginal variance and ignoring serial dependence
$N(0,2)$ , obtained by adding the positive-lag autocovariances only once instead of twice
$N(0,3)$ , obtained from the long-run variance summing all lags symmetrically (correct answer)
$N(0,1/2)$ , obtained by subtracting the lag-1 autocovariance from the marginal variance

Explanation: When a time series has serial dependence, the classical CLT for i.i.d. data no longer applies directly. Instead, you need the long-run variance, which accounts for all autocovariances across every lag. For a stationary process, the long-run variance is

\sigma^2_{LR} = \sum_{h=-\infty}^{\infty} \gamma(h)

, where

\gamma(h) = \text{Cov}(X_t, X_{t+h})

. Under mild mixing conditions (satisfied here since autocovariances decay geometrically),

\sqrt{n}(\bar{X}_n - \mu) \xrightarrow{d} N(0, \sigma^2_{LR})

. For this problem,

\gamma(h) = (1/2)^{|h|}

, so:

\sigma^2_{LR} = \sum_{h=-\infty}^{\infty} \left(\frac{1}{2}\right)^{|h|} = 1 + 2\sum_{h=1}^{\infty} \left(\frac{1}{2}\right)^h = 1 + 2 \cdot \frac{1/2}{1 - 1/2} = 1 + 2(1) = 3

The limiting distribution is therefore

N(0,3)

, confirming C. Choice A is the most tempting trap: it uses only

\gamma(0) = 1

(the marginal variance), as if observations were independent. Positive serial correlation inflates the variance of

\bar{X}_n

, so this underestimates the true spread. Choice B adds positive-lag autocovariances only once —

1 + \sum_{h=1}^{\infty}(1/2)^h = 1 + 1 = 2

— forgetting that the sum runs over both positive and negative lags symmetrically, each contributing equally. Choice D subtracts the lag-1 autocovariance, a formula more appropriate for negatively-correlated processes like MA(1) with negative coefficients; there is no principled reason to subtract here. Your key study tip: whenever you see

\sqrt{n}(\bar{X}_n - \mu)

with a dependent process, immediately write down

\sum_{h=-\infty}^{\infty}\gamma(h)

and remember the factor of 2 for the symmetric negative lags — that factor is exactly what separates answers B and C.