Central Limit Theorem - Statistics Graduate Level

1. 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?
   Answer: $N(0,1/3)$ because the normalized squared weights converge to one-third 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.
2. 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 $\sqrt n(R_n-2)$?
   Answer: $N(0,4)$, obtained via the delta method applied to $g(x,y)=x/y$ at the point $(2,1)$ 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 $\Sigma = \begin{pmatrix}4 & 1\\1 & 1\end{pmatrix}$. The asymptotic variance is: $\nabla g^\top \Sigma \nabla g = (1,\,-2)\begin{pmatrix}4&1\\1&1\end{pmatrix}\begin{pmatrix}1\\-2\end{pmatrix} = (1,-2)\begin{pmatrix}2\\-1\end{pmatrix} = 2+2 = 4.$ 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.
3. 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 $n\bar X_n^2$?
   Answer: A distribution equal to $4\chi_1^2$ 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.
4. 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 $(S_n-n\mu)/\sqrt n$?
   Answer: $N(0,\sigma^2+\mu^2)$, because both sources of variation contribute 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 $X_i$.
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 $\sqrt n(\bar Y_n-\mu_N)$?
   Answer: $N(0,15)$, after applying the finite-population correction 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.
6. 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 $\bar X_n$?
   Answer: $n^{1/3}(\bar X_n-\mu)$ converges to a non-Gaussian stable distribution with index $3/2$ 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.
7. 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 $m\to\infty$?
   Answer: $N(0,1.75\sigma^2)$, because within-cluster covariances inflate the effective variance by a factor of $1+3\rho$ 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.
8. 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?
   Answer: $T_n\Rightarrow N(0,1)$ despite the infinite fourth moment 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.
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 $\sqrt n(\bar X_n-\mu)$?
   Answer: $N(0,3)$, obtained from the long-run variance summing all lags symmetrically 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.

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