Central Limit Theorems
Y. Eddie Lu, Summer 2026
ECON 8002 course index · Lecture 9 of 9
Orientation
How to use this page
The Lindeberg proof is a chain, not a formula to memorize. At each step, name the assumption that permits the next arrow.
Answer every quiz before opening it, and rebuild the proof blueprint after the first pass.
How should I think about Lecture 9?
Many independent centered contributions
No contribution dominates total variance
\(\longrightarrow\)
After the right normalization, why does the sum acquire a Gaussian law?
\(\longrightarrow\)
Taylor-expand characteristic functions
Control large summands
Studentize for feasible inference
Course map
- Measure spaces
- Integration and Radon–Nikodym derivatives
- Distribution properties
- Conditioning
- Convergence modes
- Continuous mapping and Slutsky
- Laws of large numbers
- Weak limits and the delta method
- Central limit theorems and inference (current lecture)
Lecture 8’s Lévy-Cramér theorem identifies the Gaussian limit. Lecture 6’s CMT and Slutsky turn that limit into feasible test statistics.
Prerequisite retrieval
You should be able to state:
- \(\varphi_{U+V}(t)=\varphi_U(t)\varphi_V(t)\) for independent \(U,V\).
- Lévy-Cramér: pointwise characteristic-function convergence to a known law implies weak convergence.
- Cramér-Wold: every scalar projection determines vector weak convergence.
- WLLN for \(n^{-1}\sum Y_i\) and \(n^{-1}\sum Y_i^2\).
- Slutsky for a distributional numerator and a constant probability limit.
Triangular arrays and Lindeberg’s condition
Motivation: one normal theorem is not enough
The IID CLT treats identical contributions. Econometric sums often look like
\[S_n=\sum_{j=1}^{k_n}w_{nj}\varepsilon_{nj},\]
where weights, variances, and even the number of terms change with \(n\).
A triangular-array CLT asks for a common principle: after scaling by total standard deviation, no rare or individual contribution should control the sum.
Definition 32: triangular array
For each \(n\in\mathbb N\), let
\[X_{n1},\ldots,X_{n k_n}\]
be random variables on \((\Omega_n,\mathcal F_n,P_n)\), independent within row \(n\). The collection
\[\{X_{nj}:1\le j\le k_n,\ n\ge1\}\]
is a triangular array.
The distributions may vary across \(j\) and \(n\).
What independence does and does not mean
- Within a fixed row, \(X_{n1},\ldots,X_{n k_n}\) are independent.
- Independence between different rows is unnecessary: rows describe different statistics and may even live on different probability spaces.
- Identical distribution within a row is not assumed.
Do not silently replace “row-wise independent” with “IID.” The theorem is built to allow heteroskedastic summands.
Retrieval check.
If \(k_n=n\) and \(X_{nj}=w_{nj}\varepsilon_j\), which index selects the statistic and which selects a contribution to it?
\(n\) selects the row, hence the statistic or sample size. The index \(j\) selects one summand within that row.
The natural scale is total variance
Assume every summand has finite variance and define
\[ \sigma_{nj}^2=\operatorname{Var}(X_{nj}), \qquad \sigma_n^2=\sum_{j=1}^{k_n}\sigma_{nj}^2\in(0,\infty). \]
Row-wise independence gives
\[ \operatorname{Var}\!\left(\sum_{j=1}^{k_n}X_{nj}\right)=\sigma_n^2. \]
Thus \(\sigma_n\), not \(k_n\), is the correct general normalizing scale.
Theorem 9.1: Lindeberg CLT
Let \(\{X_{nj}\}\) be a row-wise independent triangular array with \(k_n\to\infty\) and \(0<\sigma_n^2<\infty\). If, for every \(\varepsilon>0\),
\[ \frac1{\sigma_n^2}\sum_{j=1}^{k_n} E\!\left[(X_{nj}-E[X_{nj}])^2 \mathbf1\{|X_{nj}-E[X_{nj}]|>\varepsilon\sigma_n\}\right]\to0, \]
then
\[ \frac1{\sigma_n}\sum_{j=1}^{k_n}(X_{nj}-E[X_{nj}]) \xrightarrow{d}N(0,1). \]
Lindeberg controls tail second moments
The expression
\[ E\!\left[(X_{nj}-E[X_{nj}])^2 \mathbf1\{|X_{nj}-E[X_{nj}]|>\varepsilon\sigma_n\}\right] \]
is the tail second-moment contribution from summand \(j\) when it is large relative to the row’s total standard deviation.
Lindeberg says that, for every fixed relative cutoff, all unusually large pieces together contribute a vanishing share of total variance.
Feller’s condition: no dominant variance
The Feller condition is
\[ \max_{1\le j\le k_n}\frac{\sigma_{nj}^2}{\sigma_n^2}\to0. \]
It says no single summand owns a positive limiting fraction of total variance.
Lindeberg implies Feller. Feller alone does not generally imply the CLT because many rare large values can still contribute material tail variance.
Retrieval check.
Which condition is stronger, Lindeberg or Feller?
Lindeberg is stronger. It implies Feller, but Feller controls only the largest individual variance share, not the combined contribution of large tails.
Lindeberg implies Feller
After centering, for any \(\varepsilon>0\),
\[ \frac{\sigma_{nj}^2}{\sigma_n^2} \le \varepsilon^2+ \frac1{\sigma_n^2} E[X_{nj}^2\mathbf1\{|X_{nj}|>\varepsilon\sigma_n\}]. \]
Taking the maximum over \(j\) and bounding the final term by the sum over the row gives
\[ \limsup_n\max_j\frac{\sigma_{nj}^2}{\sigma_n^2}\le\varepsilon^2. \]
Let \(\varepsilon\downarrow0\).
A dominating-summand nonexample
Let \(X_{n1}=Z\) with \(E[Z]=0\), \(\operatorname{Var}(Z)=1\), and let every other \(X_{nj}=0\). Then \(\sigma_n^2=1\) and
\[\max_j\frac{\sigma_{nj}^2}{\sigma_n^2}=1.\]
The standardized sum equals \(Z\) for every \(n\). It is not asymptotically normal unless \(Z\) was already normal.
Increasing the number of columns does nothing when one column carries all uncertainty.
Proof of the Lindeberg CLT
Lindeberg proof: the full map
- Center and scale the row so its total variance equals one.
- Approximate each characteristic function by \(1-t^2\sigma_{nj}^2/2\).
- Use Lindeberg to make the sum of Taylor remainders vanish.
- Use Feller and a product inequality to combine the approximations.
- Recognize \(e^{-t^2/2}\) and apply Lévy-Cramér.
Normalize without loss of generality
Define
\[Z_{nj}=\frac{X_{nj}-E[X_{nj}]}{\sigma_n}.\]
Then
\[E[Z_{nj}]=0,\qquad \sum_{j=1}^{k_n}\operatorname{Var}(Z_{nj})=1.\]
Write \(s_{nj}^2=\operatorname{Var}(Z_{nj})\). It is enough to prove
\[\sum_{j=1}^{k_n}Z_{nj}\xrightarrow{d}N(0,1).\]
Lindeberg now uses the fixed threshold \(\varepsilon\).
The characteristic-function Taylor bound
For real \(x,t\),
\[ \left|e^{itx}-\left(1+itx-\frac{t^2x^2}{2}\right)\right| \le\min\{|tx|^2,|tx|^3\}. \]
Because \(E[Z_{nj}]=0\),
\[ \left|\varphi_{Z_{nj}}(t)-\left(1-\frac{t^2s_{nj}^2}{2}\right)\right| \le E\min\{|tZ_{nj}|^2,|tZ_{nj}|^3\}. \]
Split small and large values
For fixed \(\varepsilon>0\),
\[ \begin{aligned} E\min\{|tZ_{nj}|^2,|tZ_{nj}|^3\} &\le E[|tZ_{nj}|^3\mathbf1\{|Z_{nj}|<\varepsilon\}]\\ &\quad+E[|tZ_{nj}|^2\mathbf1\{|Z_{nj}|\ge\varepsilon\}]\\ &\le \varepsilon|t|^3s_{nj}^2 +t^2E[Z_{nj}^2\mathbf1\{|Z_{nj}|\ge\varepsilon\}]. \end{aligned} \]
Small values use \(|Z|^3\le\varepsilon Z^2\); large values invoke Lindeberg.
Sum the Taylor errors
Summing over \(j\) and using \(\sum_js_{nj}^2=1\) gives
\[ \begin{aligned} &\sum_{j=1}^{k_n} \left|\varphi_{Z_{nj}}(t)-\left(1-\frac{t^2s_{nj}^2}{2}\right)\right|\\ &\qquad\le \varepsilon|t|^3 +t^2\sum_{j=1}^{k_n}E[Z_{nj}^2\mathbf1\{|Z_{nj}|\ge\varepsilon\}]. \end{aligned} \]
For fixed \(\varepsilon\), the second term tends to zero by Lindeberg.
The order of limits matters
The preceding bound yields
\[ \limsup_{n\to\infty}\sum_j \left|\varphi_{Z_{nj}}(t)-\left(1-\frac{t^2s_{nj}^2}{2}\right)\right| \le\varepsilon|t|^3. \]
Now let \(\varepsilon\downarrow0\).
Do not write the right side as converging to zero with \(n\). First take the row limit for fixed truncation, then shrink the truncation threshold.
Retrieval check.
What controls the small-value term, and what controls the large-value term?
The inequality \(|Z|^3\le\varepsilon Z^2\) controls small values. Lindeberg’s condition controls the summed large-value variance.
Lemma 9.2: compare products
If \(|z_j|\le1\) and \(|w_j|\le1\) for \(j=1,\ldots,m\), then
\[ \left|\prod_{j=1}^mz_j-\prod_{j=1}^mw_j\right| \le\sum_{j=1}^m|z_j-w_j|. \]
Add and subtract the intermediate products \(z_1\cdots z_{r-1}w_r\cdots w_m\). Each remaining factor has modulus at most one.
Independence turns sums into products
Within row \(n\), the normalized variables are independent, so
\[ \varphi_{\sum_jZ_{nj}}(t)=\prod_{j=1}^{k_n}\varphi_{Z_{nj}}(t). \]
Lemma 9.2 and the vanished Taylor errors imply
\[ \prod_j\varphi_{Z_{nj}}(t) -\prod_j\left(1-\frac{t^2s_{nj}^2}{2}\right)\to0. \]
This is the only step where independence enters the proof.
Replace linear factors by exponentials
Feller gives \(\max_js_{nj}^2\to0\), so for large \(n\) all \(t^2s_{nj}^2/2\) are small. Using \(|e^{-x}-(1-x)|\le x^2/2\) near zero,
\[ \sum_j\left|e^{-t^2s_{nj}^2/2}- \left(1-\frac{t^2s_{nj}^2}{2}\right)\right| \le C_t\sum_js_{nj}^4. \]
But \(\sum_js_{nj}^4\le(\max_js_{nj}^2)\sum_js_{nj}^2\to0\).
The Gaussian characteristic function appears
Because \(\sum_js_{nj}^2=1\),
\[ \prod_{j=1}^{k_n}e^{-t^2s_{nj}^2/2} =e^{-t^2/2}. \]
Combining the two product approximations gives
\[ \varphi_{\sum_jZ_{nj}}(t)\to e^{-t^2/2}. \]
The limit is the characteristic function of \(N(0,1)\). Lévy-Cramér completes the proof of Theorem 9.1.
Quick quiz 4: where independence enters
Retrieval check.
Which displayed equality in the proof fails without row-wise independence?
The characteristic function of the row sum need not equal the product of the individual characteristic functions.
Scalar and multivariate IID corollaries
Corollary 9.3: scalar IID CLT
If \(X_1,X_2,\ldots\) are IID with
\[E[X_1]=\mu,\qquad0<\sigma^2=\operatorname{Var}(X_1)<\infty,\]
then
\[ \sqrt n\,\frac{\bar X_n-\mu}{\sigma} \xrightarrow{d}N(0,1). \]
This is Theorem 9.1 with \(k_n=n\) and \(X_{nj}=X_j\).
IID Lindeberg follows from finite variance
Here \(\sigma_n^2=n\sigma^2\), and Lindeberg becomes
\[ \sigma^{-2}E\!\left[(X_1-\mu)^2 \mathbf1\{|X_1-\mu|>\varepsilon\sigma\sqrt n\}\right]\to0. \]
The indicator tends to zero almost surely and the integrand is dominated by the integrable variable \((X_1-\mu)^2\). DCT proves the claim.
No finite third or fourth moment is required for the classical IID CLT.
Quick quiz 5: moment requirements
Retrieval check.
Does the IID CLT require \(E|X_1|^3<\infty\)?
No. A finite, positive variance is enough. Higher moments support rate refinements such as Berry-Esseen bounds, not the basic CLT.
Boundary case: infinite variance
The preceding theorem does not apply when \(\operatorname{Var}(X_1)=\infty\). Heavy-tailed sums may require a different normalization and may converge to a stable nonnormal law.
“Many observations” does not by itself imply a Gaussian limit. The moment and dependence assumptions decide the correct theorem and scaling.
Corollary 9.4: multivariate IID CLT
Let \(X_i\in\mathbb R^k\) be IID with
\[E[X_i]=\mu\in\mathbb R^k,\qquad \Sigma=\operatorname{Var}(X_i)\]
finite. Then
\[ \frac1{\sqrt n}\sum_{i=1}^n(X_i-\mu) \xrightarrow{d}N_k(0,\Sigma). \]
\(\Sigma\) may be singular.
Multivariate proof by Cramér-Wold
For fixed \(c\in\mathbb R^k\),
\[ c^\top\frac1{\sqrt n}\sum_{i=1}^n(X_i-\mu) =\frac1{\sqrt n}\sum_{i=1}^n c^\top(X_i-\mu). \]
The scalar summands are IID with variance \(c^\top\Sigma c\). The scalar CLT gives
\[ \frac1{\sqrt n}\sum_i c^\top(X_i-\mu) \xrightarrow{d}N(0,c^\top\Sigma c). \]
Cramér-Wold yields the vector result.
Singular covariance is allowed
Let \(X_i=(Z_i,2Z_i)^\top\) with \(E[Z_i]=0\) and \(\operatorname{Var}(Z_i)=1\). Then
\[ \Sigma=\begin{pmatrix}1&2\\2&4\end{pmatrix} \]
is singular. The limiting normal vector lies on the line \(x_2=2x_1\).
If \(c^\top\Sigma c=0\), then \(c^\top(X_i-\mu)=0\) almost surely, and that projection converges to the degenerate normal directly.
Quick quiz 6: a zero-variance projection
Retrieval check.
Does the multivariate CLT fail if some nonzero \(c\) satisfies \(c^\top\Sigma c=0\)?
No. That projection is constant almost surely after centering and has a degenerate normal limit. Nonsingularity is not required.
Studentization and asymptotic normality
Example 21: why studentize?
Suppose \(Y_i\) are IID with \(E[Y_i]=0\) and \(0<\sigma^2=\operatorname{Var}(Y_i)<\infty\). Define
\[ \bar Y_n=\frac1n\sum_{i=1}^nY_i,\qquad S_n^2=\frac1{n-1}\sum_{i=1}^n(Y_i-\bar Y_n)^2. \]
The CLT uses the unknown \(\sigma\). Studentization replaces it by the observable \(S_n\) without changing the limiting law.
Correct sample-variance algebra
Expand around the sample mean:
\[ \sum_{i=1}^n(Y_i-\bar Y_n)^2 =\sum_{i=1}^nY_i^2-n\bar Y_n^2. \]
Therefore
\[ S_n^2=\frac{n}{n-1} \left(\frac1n\sum_{i=1}^nY_i^2-\bar Y_n^2\right). \]
This replaces the incorrect squared expression in the source notes.
Prove variance consistency
Finite variance gives \(E[Y_1^2]<\infty\). The WLLN and CMT imply
\[ \frac1n\sum_{i=1}^nY_i^2\xrightarrow{p}E[Y_1^2], \qquad \bar Y_n^2\xrightarrow{p}(E[Y_1])^2=0. \]
Since \(n/(n-1)\to1\),
\[S_n^2\xrightarrow{p}E[Y_1^2]-(E[Y_1])^2=\sigma^2.\]
No fourth moment is needed.
Studentization by CMT and Slutsky
CMT gives \(S_n\xrightarrow{p}\sigma>0\). The scalar CLT gives
\[\frac{\sqrt n\bar Y_n}{\sigma}\xrightarrow{d}N(0,1).\]
Write
\[ \frac{\sqrt n\bar Y_n}{S_n} =\frac{\sqrt n\bar Y_n}{\sigma}\frac{\sigma}{S_n}. \]
Since \(\sigma/S_n\xrightarrow{p}1\), Slutsky yields
\[\frac{\sqrt n\bar Y_n}{S_n}\xrightarrow{d}N(0,1).\]
Finite-sample zero denominator
If \(S_n=0\) in a finite sample, define the statistic arbitrarily on that event. Since \(S_n\xrightarrow{p}\sigma>0\), \(P(S_n=0)\to0\), so this convention does not change the limit.
Quick quiz 7: theorem chain
Retrieval check.
Name the tools establishing \(S_n^2\xrightarrow{p}\sigma^2\), \(S_n\xrightarrow{p}\sigma\), and the final studentized limit.
WLLN plus algebra and CMT; then CMT for the square root; then the scalar CLT and Slutsky for the studentized statistic.
Exact \(t\) law versus asymptotic normality
If the \(Y_i\) are normal, the studentized mean has an exact finite-sample \(t_{n-1}\) distribution. Under general IID sampling with finite positive variance, the result proved here is only
\[\frac{\sqrt n\bar Y_n}{S_n}\xrightarrow{d}N(0,1).\]
Studentization does not create an exact \(t\) distribution under arbitrary sampling. It creates an asymptotically pivotal statistic.
Example 22: standardized weak convergence
Let \(Y_n\) have c.d.f. \(F_{Y_n}\), and let \(\mu_n\in\mathbb R\) and \(\sigma_n>0\) be deterministic. Suppose
\[Z_n=\frac{Y_n-\mu_n}{\sigma_n}\xrightarrow{d}N(0,1).\]
Then, for every fixed \(x\),
\[F_{Z_n}(x)\to\Phi(x).\]
The question is whether the normal approximation remains valid uniformly over \(x\).
Pólya’s theorem supplies uniformity
If c.d.f.s \(F_n\) converge pointwise to a continuous c.d.f. \(F\), then
\[\sup_{x\in\mathbb R}|F_n(x)-F(x)|\to0.\]
Since the normal c.d.f. \(\Phi\) is continuous,
\[ \sup_x|F_{Z_n}(x)-\Phi(x)|\to0. \]
Continuity of the limiting c.d.f. is essential for this upgrade.
Return to the unstandardized statistic
Because
\[ F_{Z_n}(x)=F_{Y_n}(\mu_n+\sigma_nx), \]
Pólya’s theorem is equivalent to
\[ \sup_{y\in\mathbb R} \left|F_{Y_n}(y)- \Phi\!\left(\frac{y-\mu_n}{\sigma_n}\right)\right|\to0. \]
The normal c.d.f. is therefore a uniform approximation to the entire finite-sample c.d.f.
Moving thresholds need not converge
For any deterministic real sequence \(y_n\),
\[ \left|F_{Y_n}(y_n)- \Phi\!\left(\frac{y_n-\mu_n}{\sigma_n}\right)\right| \le \sup_y\left|F_{Y_n}(y)- \Phi\!\left(\frac{y-\mu_n}{\sigma_n}\right)\right| \to0. \]
No limit for \(y_n\) is required because the approximation is uniform.
Quick quiz 8: why moving thresholds work
Retrieval check.
Why can \(y_n\) vary arbitrarily with \(n\) in the preceding conclusion?
Pólya controls the supremum over every threshold. The error at a particular moving threshold is bounded by that uniform error.
Definition 33: asymptotic normality, repaired
\(Y_n\) is asymptotically normal with centering \(\mu_n\) and scaling \(\sigma_n>0\) if
\[\frac{Y_n-\mu_n}{\sigma_n}\xrightarrow{d}N(0,1).\]
Equivalently, by Pólya,
\[ \sup_y\left|F_{Y_n}(y)- \Phi\!\left(\frac{y-\mu_n}{\sigma_n}\right)\right|\to0. \]
Why the source’s sequential wording is too weak
Approximation at one selected threshold sequence \(y_n\) does not characterize an entire distribution. The displayed sequential criterion is correct only if it holds for every deterministic sequence \(\{y_n\}\).
The standardized weak-convergence definition is cleaner and directly names the centering and scale.
Never define asymptotic normality through accuracy at a single cutoff.
Quick quiz 9: repair the definition
Retrieval check.
What quantifier is missing if asymptotic normality is stated using a sequence of cutoffs \(y_n\)?
The approximation must hold for every deterministic real sequence \(\{y_n\}\), equivalently uniformly over all real cutoffs.
Applications and theorem selection
Finance application: a diversified weighted sum
Let
\[R_n=\sum_{j=1}^{k_n}w_{nj}\varepsilon_j,\]
where the shocks are independent, centered, and have variances \(v_j\). Then
\[\sigma_n^2=\sum_jw_{nj}^2v_j.\]
A necessary no-dominance check is
\[ \max_j\frac{w_{nj}^2v_j}{\sum_\ell w_{n\ell}^2v_\ell}\to0. \]
Lindeberg additionally controls extreme weighted shocks.
Quick quiz 10: diversification is not a count
Retrieval check.
Can \(k_n\to\infty\) while Feller’s condition fails?
Yes. One asset can retain a fixed fraction of portfolio variance even while arbitrarily many negligible assets are added.
Econometric application: score sums
Many estimators admit an expansion
\[ \sqrt n(\hat\theta_n-\theta_0) =A_n^{-1}\frac1{\sqrt n}\sum_{i=1}^n\psi_{ni}+o_p(1). \]
- A CLT handles the score sum.
- An LLN gives \(A_n\xrightarrow{p}A\).
- CMT gives \(A_n^{-1}\xrightarrow{p}A^{-1}\) when \(A\) is nonsingular.
- Slutsky produces the estimator’s limiting distribution.
This is the standard inference architecture behind regression and GMM.
Dependence is a theorem-selection warning
The Lindeberg theorem here assumes independence within each row. Common factors, serial dependence, and clustered observations violate that assumption.
Do not “approximately” invoke an independent-array CLT. Use a dependence-robust CLT with its own mixing, martingale, or cluster conditions.
The normal limit may still be true, but this theorem no longer proves it.
Quick quiz 11: diagnose the failed assumption
Retrieval check.
Suppose \(X_{nj}=\lambda_jF_n+\varepsilon_{nj}\), where the common factor \(F_n\) appears in every summand. Which assumption of Theorem 9.1 fails?
Row-wise independence fails because every pair of summands shares the same random factor.
Synthesis and retrieval
Assumption audit
| Assumption | Job | Failure without it |
|---|---|---|
| Row-wise independence | factors the c.f. of the sum | product proof collapses |
| Finite row variances | defines \(\sigma_n\) | classical normalization may fail |
| Lindeberg for every \(\varepsilon\) | removes large-tail variance | rare jumps can dominate |
| \(\sigma^2>0\) in IID CLT | nondegenerate scaling | denominator is zero |
| Finite \(E[Y^2]\) | WLLN for \(Y_i^2\) | sample variance need not be consistent |
| Continuous limit c.d.f. | Pólya uniformity | moving-threshold control may fail |
Common errors and false converses
- A triangular array is not necessarily IID.
- \(k_n\to\infty\) does not imply no summand dominates.
- Feller does not generally imply Lindeberg.
- The truncation limit must follow the row limit.
- Finite third moments are not required for the IID CLT.
- A singular covariance matrix is allowed in the vector CLT.
- Sample-variance consistency uses \(n^{-1}\sum Y_i^2-\bar Y_n^2\), not a squared difference of those terms.
- Studentization is not an exact finite-sample \(t\) result outside normal sampling.
- One threshold sequence cannot define asymptotic normality.
Proof blueprint
Lindeberg
center and scale \(\to\) truncate \(\to\) control Taylor errors
Combine
independence \(\to\) product lemma \(\to e^{-t^2/2}\)
Inference
CLT numerator \(+\) consistent scale \(\to\) Slutsky
Exit ticket 1: state Lindeberg precisely
For every \(\varepsilon>0\), the sum of truncated centered second moments above \(\varepsilon\sigma_n\), divided by total variance \(\sigma_n^2\), must converge to zero.
Exit ticket 2: rebuild the characteristic-function proof
Normalize; Taylor-expand each characteristic function; use the small/large split and Lindeberg; use independence, Lemma 9.2, and Feller to combine factors; recognize \(e^{-t^2/2}\) and apply Lévy-Cramér.
Exit ticket 3: prove studentization
Use \(S_n^2=\frac n{n-1}(n^{-1}\sum_iY_i^2-\bar Y_n^2)\xrightarrow{p}\sigma^2\), then CMT gives \(S_n\xrightarrow{p}\sigma\). Multiply the CLT statistic by \(\sigma/S_n\xrightarrow{p}1\) and apply Slutsky.
Mastery checklist
You should now be able to:
- define a triangular array and distinguish row-wise independence from IID;
- state and interpret Lindeberg and Feller conditions;
- rebuild the characteristic-function proof of Theorem 9.1;
- derive scalar and multivariate IID CLTs;
- handle zero-variance projections correctly;
- prove sample-variance consistency with only a finite second moment;
- distinguish exact \(t\) inference from asymptotic studentization;
- define asymptotic normality through standardized weak convergence.