The central limit theorem

Theorem

The central limit theorem

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Statement

Let \((X_n)_{n\ge 1}\) be a sequence of independent, identically distributed real random variables on a probability space \((\Omega,\mathcal F,\mathbb P)\) with \(\mathbb E[X_1]=\mu\) and \(0\lt \operatorname{Var}(X_1)=\sigma^2\lt \infty\). Let \(S_n=\sum_{k=1}^n X_k\) and define the standardised sum \[Z_n=\frac{S_n-n\mu}{\sigma\sqrt n}.\] Then \(Z_n\) converges in distribution to a standard normal random variable \(Z\sim N(0,1)\): for every \(z\in\mathbb R\), \[\lim_{n\to\infty}\mathbb P(Z_n\le z)=\Phi(z)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{z} e^{-t^2/2}\,dt.\] Equivalently, \(\mathbb E[f(Z_n)]\to\mathbb E[f(Z)]\) for every bounded continuous \(f:\mathbb R\to\mathbb R\).

Why it matters

The central limit theorem is the reason the normal distribution is ubiquitous: it says that regardless of the shape of the underlying distribution — so long as it has finite variance — the fluctuations of a sum of many independent contributions look Gaussian after the correct \(\sqrt n\) rescaling. This single fact underwrites the theory of confidence intervals, hypothesis tests, and the entire apparatus of classical statistical inference, none of which would have universal validity without it.

It is also the prototype for a much larger phenomenon — universality — in which a wide class of microscopically different systems (sums, but also random matrices, statistical-mechanical models, combinatorial statistics) converge to the same small family of limit laws. The Lindeberg–Lévy form proved here is the identically-distributed case; the Lindeberg–Feller theorem (MU-404, later theorem) removes the identical-distribution assumption at the cost of a uniform smallness condition.

Hypotheses

Independence of \(X_1,X_2,\dots\)Counterexample: let \(X_1\) be standard normal and \(X_k=X_1\) for all \(k\). Then \(S_n=nX_1\) and \(Z_n=\sqrt n\,X_1/\sigma\), whose distribution spreads out (variance \(n\)) rather than converging to \(N(0,1)\); perfect dependence destroys the averaging that produces Gaussianity. Identical distributionCounterexample: let \(X_k\) be independent with \(\operatorname{Var}(X_k)=k^2\) growing rapidly with \(k\). Then \(S_n\) is dominated by the last few terms, whose distribution need not be anything close to Gaussian even after rescaling; without identical (or at least comparably-sized) summands the "many small independent contributions" mechanism fails. (Lindeberg's condition is the correct general replacement.) Finite variance \(\sigma^2\lt\infty\)Counterexample: let \(X_1\) be Cauchy-distributed (or more generally in the domain of attraction of an \(\alpha\)-stable law with \(\alpha\lt 2\)). Then \(S_n/n\) already has the Cauchy distribution for every \(n\) (stability of Cauchy under summation), so no rescaling by \(\sqrt n\) produces a Gaussian limit; instead \(S_n/n\) converges to Cauchy itself. Heavy enough tails replace the Gaussian limit with a stable law of index \(\alpha\lt 2\). Existence of the mean \(\mu=\mathbb E[X_1]\)This is implied by finite second moment (Cauchy–Schwarz: \(\mathbb E|X_1|\le\sqrt{\mathbb E[X_1^2]}\)), so it is not an independent extra hypothesis here, but it is worth flagging that centring at \(\mu\) is only meaningful once \(\mu\) is known to exist.

Proof

\text{Reduce to mean }0,\text{ variance }1\text{: set } Y_k=\frac{X_k-\mu}{\sigma}, \qquad \frac{S_n-n\mu}{\sigma\sqrt n}=\frac{1}{\sqrt n}\sum_{k=1}^n Y_k = Z_n.

\(Y_1,\dots,Y_n\) are i.i.d. (functions of independent variables applied coordinatewise preserve independence) with \(\mathbb E[Y_1]=0\), \(\operatorname{Var}(Y_1)=1\); this is a harmless affine reparametrisation, so it suffices to prove the theorem for \(Y_k\) in place of \(X_k\). A

\varphi(t)=\mathbb E\left[e^{itY_1}\right],\qquad t\in\mathbb R,

the characteristic function of \(Y_1\) is well defined for every real \(t\) because \(|e^{itY_1}|=1\) is bounded, hence integrable against any probability measure. We will show \(\varphi_{Z_n}(t)\to e^{-t^2/2}\) for every \(t\), where \(e^{-t^2/2}\) is the characteristic function of \(N(0,1)\). A

\varphi \text{ is twice differentiable at } 0 \text{ with } \varphi(0)=1,\ \varphi'(0)=i\,\mathbb E[Y_1]=0,\ \varphi''(0)=i^2\mathbb E[Y_1^2]=-1.

Since \(\mathbb E[Y_1^2]=\operatorname{Var}(Y_1)=1\lt\infty\), the standard theorem on differentiating characteristic functions under the expectation sign (dominated convergence, using \(|Y_1|^2\) as an integrable dominating function for the second difference quotients) applies: for \(Y_1\in L^2\), \(\varphi\in C^2(\mathbb R)\) and \(\varphi^{(k)}(0)=i^k\mathbb E[Y_1^k]\) for \(k=1,2\). B

\varphi(t)=1-\frac{t^2}{2}+o(t^2)\qquad (t\to 0).

Taylor's theorem with Peano remainder applied to the real and imaginary parts of \(\varphi\) (each twice differentiable at \(0\) by Step 3) gives \(\varphi(t)=\varphi(0)+\varphi'(0)t+\tfrac12\varphi''(0)t^2+o(t^2)=1+0\cdot t-\tfrac12 t^2+o(t^2)\). This is the only place finiteness of the variance is used, and it is used essentially: without \(\mathbb E[Y_1^2]\lt\infty\) there is no quadratic Taylor expansion of \(\varphi\) at \(0\). C

\varphi_{Z_n}(t)=\mathbb E\!\left[\exp\!\left(it\,\frac{1}{\sqrt n}\sum_{k=1}^n Y_k\right)\right]=\prod_{k=1}^n \mathbb E\!\left[e^{i(t/\sqrt n)Y_k}\right]=\left(\varphi\!\left(\frac{t}{\sqrt n}\right)\right)^{n}.

The characteristic function of a sum of independent random variables is the product of their characteristic functions (independence \(\Rightarrow\) the joint expectation of a product of functions of separate coordinates factorises), and the \(Y_k\) are identically distributed so each factor equals \(\varphi(t/\sqrt n)\). B

n\log\varphi\!\left(\frac{t}{\sqrt n}\right) \longrightarrow -\frac{t^2}{2} \qquad (n\to\infty),\text{ for each fixed } t.

Fix \(t\). By Step 4, \(\varphi(t/\sqrt n)=1-\dfrac{t^2}{2n}+o\!\left(\dfrac1n\right)\) as \(n\to\infty\). Since \(\varphi(t/\sqrt n)\to\varphi(0)=1\) (continuity of \(\varphi\), itself immediate from dominated convergence \(e^{isY_1}\to e^{i0\cdot Y_1}\) as \(s\to0\)), for \(n\) large \(\varphi(t/\sqrt n)\) lies in a neighbourhood of \(1\) where the principal branch of \(\log\) is holomorphic and satisfies \(\log(1+w)=w+o(w)\) as \(w\to0\). Substituting \(w=-\tfrac{t^2}{2n}+o(1/n)\): \(\log\varphi(t/\sqrt n)=-\dfrac{t^2}{2n}+o\!\left(\dfrac1n\right)\), hence \(n\log\varphi(t/\sqrt n)=-\dfrac{t^2}{2}+o(1)\to-\dfrac{t^2}{2}\). C

\varphi_{Z_n}(t)=\exp\!\left(n\log\varphi\!\left(\frac{t}{\sqrt n}\right)\right)\longrightarrow \exp\!\left(-\frac{t^2}{2}\right)\qquad \text{for every fixed } t\in\mathbb R.

Exponentiation is continuous on \(\mathbb C\), so Step 6's convergence of the exponent passes through \(\exp\) unchanged; the right-hand side is exactly the characteristic function of \(N(0,1)\), namely \(\varphi_Z(t)=e^{-t^2/2}\). A

\varphi_{Z_n}(t)\to\varphi_Z(t)\ \forall t\in\mathbb R \implies Z_n \xrightarrow{d} Z.

Lévy's continuity theorem: if characteristic functions converge pointwise everywhere to a function that is itself the characteristic function of a probability distribution (and \(\varphi_Z\) is continuous at \(0\), automatic here since \(\varphi_Z\) is entire), then the corresponding laws converge weakly. This is the one deep input from measure-theoretic probability that is cited rather than reproved here — it rests on tightness of \((Z_n)\) (obtained from continuity of the limit \(\varphi_Z\) at \(0\) via Lévy's tightness criterion) plus a subsequential weak-limit argument using Helly selection. C

Z_n\xrightarrow{d} Z\sim N(0,1) \iff \mathbb P(Z_n\le z)\to\Phi(z) \text{ at every continuity point } z \text{ of } \Phi.

Since \(\Phi\) is continuous everywhere on \(\mathbb R\) (it has a bounded density), weak convergence of \(Z_n\) to \(Z\) is equivalent, by definition, to convergence of CDFs at every \(z\in\mathbb R\); this recovers exactly the statement claimed. A

Result

\frac{S_n-n\mu}{\sigma\sqrt n}\ \xrightarrow{\ d\ }\ N(0,1)\qquad\text{as } n\to\infty

Reading. If you add up \(n\) independent copies of the same random quantity, subtract off the expected total \(n\mu\), and divide by the standard deviation of the sum \(\sigma\sqrt n\), the resulting standardised random variable looks, for large \(n\), like a standard bell curve — no matter what the original distribution of a single \(X_k\) looked like, as long as it had finite variance.

Scope. Applies to any i.i.d. sequence with finite second moment: discrete or continuous, symmetric or skewed, bounded or unbounded. It is a statement about convergence of distributions (weak convergence), not about convergence of the random variables themselves, and it says nothing about the rate of convergence (that is the province of the Berry–Esseen theorem) nor about the tails far from the centre (large-deviation theory).

Corollaries & converses

Approximate normality of sample means. Writing \(\bar X_n=S_n/n\), the theorem gives \(\bar X_n\approx N(\mu,\sigma^2/n)\) for large \(n\), the basis of the standard error \(\sigma/\sqrt n\) and of Wald-type confidence intervals.
De Moivre–Laplace theorem is the special case \(X_k\sim\text{Bernoulli}(p)\): binomial proportions are asymptotically normal.
Slutsky-type extensions. Combined with a consistent variance estimator \(\hat\sigma_n\to\sigma\) in probability, \(\dfrac{S_n-n\mu}{\hat\sigma_n\sqrt n}\) is also asymptotically \(N(0,1)\) (Slutsky's theorem), which is what makes the \(t\)-statistic asymptotically valid even though \(\sigma\) is unknown.
Converse fails. Weak convergence of \(Z_n\) to \(N(0,1)\) does not imply the summands were i.i.d. with finite variance — many non-i.i.d. or heavy-tailed-but-truncated schemes also produce a Gaussian limit under other rescalings, so "the limit is Gaussian" cannot be reverse-engineered into "the hypotheses of Lindeberg–Lévy held." The theorem gives a sufficient, not necessary, route to asymptotic normality.
No converse recovering the summand law. Two different distributions for \(X_1\) with the same mean and variance produce the same Gaussian limit; the CLT is deliberately lossy and cannot be inverted to recover the distribution of \(X_1\) from the limiting law of \(Z_n\).

Fails without

Infinite variance: if \(X_1\) is Student-\(t\) with 2 degrees of freedom (finite mean, infinite variance), the naive standardisation \(S_n/\sqrt n\) does not converge to any non-degenerate law with the classical scaling; one instead needs a different normalising sequence and the limit (if it exists) is a non-Gaussian stable law, per the Lévy generalised CLT.
Dependence: for a stationary but strongly dependent sequence such as \(X_k=\rho X_{k-1}+\varepsilon_k\) with \(\rho\) close to 1 (or the degenerate case \(X_k\equiv X_1\) from the Hypotheses section), the correlations do not decay fast enough (or at all) for the variance of \(S_n\) to scale like \(n\); \(\operatorname{Var}(S_n)\) can grow like \(n^2\), and \(S_n/\sqrt n\) diverges rather than stabilising to \(N(0,1)\).
Non-identical distribution without Lindeberg's condition: if \(X_k\) takes value \(\pm 2^k\) with tiny probability \(2^{-2k}\) and \(0\) otherwise (arranged to have finite variance), a single extreme summand can dominate \(S_n\) for every \(n\), and the limiting law of \(Z_n\) can fail to be Gaussian because no individual summand becomes asymptotically negligible relative to the sum — the content of the Lindeberg condition that this construction violates.

Common errors

Concluding \(S_n\) itself is "approximately \(N(n\mu,n\sigma^2)\)" and then treating this as an exact statement for small \(n\) — the CLT is an \(n\to\infty\) limit theorem with no built-in rate; how large \(n\) must be depends on the shape (especially skewness/kurtosis) of \(X_1\)'s distribution.
Applying the theorem to a single fixed \(n\) drawn from a wildly skewed or heavy-tailed distribution (e.g. lognormal with large variance, or income data) and expecting the normal approximation to already be accurate — the theorem gives no explicit control on the approximation error without further tools like Berry–Esseen.
Forgetting to check finite variance: applying the CLT to distributions such as Pareto with tail index \(\alpha\le 2\) without verifying \(\mathbb E[X_1^2]\lt\infty\).
Confusing the CLT (a statement about the distribution of a rescaled sum) with the law of large numbers (a statement about \(\bar X_n\) itself converging, in probability/a.s., to the constant \(\mu\)) — these are different limit regimes with different scalings (\(\sqrt n\) versus \(n\)).
Assuming independence is dispensable "because \(n\) is large" — dependence, however weak-looking, must actually decay for any CLT (even a dependent-data CLT) to hold; large \(n\) alone does not manufacture asymptotic independence.

Discussion

The characteristic-function proof given above is essentially the one found in Lindeberg (1922) and Lévy's subsequent sharpening, and it exemplifies a technique — pass to Fourier/characteristic functions, where sums of independent variables become products (and hence sums of logarithms), then use continuity theorems to transfer convergence back to distributions — that recurs throughout probability, from the Poisson limit theorem to the theory of infinitely divisible laws. The essential arithmetic fact underlying the proof is that a second-order Taylor expansion of \(\log\varphi\) near \(0\) has no linear term (because \(\mathbb E[Y_1]=0\)) and a quadratic coefficient fixed exactly by the variance; rescaling by \(\sqrt n\) is precisely the scaling under which \(n\) independent quadratic contributions of size \(1/n\) each sum to a finite, nonzero limit.

Historically the theorem's roots go back to de Moivre's 1733 approximation of binomial probabilities and Laplace's early-19th-century extensions, but the modern, general statement and rigorous proof via characteristic functions are due to Lyapunov (1901, using a stronger third-moment condition) and then Lindeberg (1922), who found the essentially minimal condition (named after him) for triangular arrays of independent, not-necessarily-identical summands. The identically-distributed case proved here is sometimes attributed separately to Lévy.

The CLT sits inside a larger classification theorem: the only possible non-degenerate limit laws for appropriately rescaled sums of i.i.d. random variables are the stable distributions, of which the Gaussian (index \(\alpha=2\), the unique one with finite variance) is a single, especially tractable member. Summands with infinite variance but regularly varying tails of index \(\alpha\in(0,2)\) are drawn instead into the domain of attraction of an \(\alpha\)-stable law, with normalisation \(n^{1/\alpha}\) rather than \(\sqrt n\); this is the correct sense in which "the CLT fails" for heavy tails — it is replaced, not simply broken.

A subtlety worth flagging precisely: convergence in distribution (\(Z_n\xrightarrow{d}Z\)) is strictly weaker than convergence in probability or almost surely, and indeed \(Z_n\) does not converge pointwise or in probability to any random variable at all — for a fixed sample path, \(Z_n(\omega)\) generally keeps fluctuating forever with no limit. The CLT is a statement purely about the shape of the law of \(Z_n\) stabilising, not about the trajectory of any single realisation; conflating the two is the deepest common misconception surrounding the theorem, and is precisely what separates it conceptually from the (strong) law of large numbers, which is a genuine pathwise statement.

Worked examples

\text{Fair-die sum: } X_k\sim\text{Uniform}\{1,\dots,6\}\text{ i.i.d., } n=180.\quad \mu=\tfrac{7}{2},\ \ \sigma^2=\mathbb E[X_1^2]-\mu^2=\tfrac{91}{6}-\tfrac{49}{4}=\tfrac{35}{12}.

Direct computation: \(\mathbb E[X_1^2]=\tfrac16(1+4+9+16+25+36)=\tfrac{91}{6}\), so \(\sigma^2=\tfrac{91}{6}-\left(\tfrac72\right)^2=\tfrac{182-147}{12}=\tfrac{35}{12}\approx 2.9167\). A

\mathbb P(S_{180}\gt 660)=\mathbb P\!\left(Z_{180}\gt\frac{660-180(3.5)}{\sqrt{180\cdot 35/12}}\right)=\mathbb P\!\left(Z_{180}\gt\frac{30}{\sqrt{525}}\right)=\mathbb P(Z_{180}\gt 1.309).

Standardise using \(n\mu=630\) and \(\sigma\sqrt n=\sqrt{180\cdot 35/12}=\sqrt{525}\approx 22.913\); this is exactly the transformation \(Z_n=(S_n-n\mu)/(\sigma\sqrt n)\) from the theorem. A

\mathbb P(S_{180}\gt 660)\ \approx\ 1-\Phi(1.309)\ \approx\ 0.0953.

Apply the theorem's conclusion with \(n=180\) treated as large enough for the Gaussian approximation, reading \(1-\Phi(1.309)\) from the standard normal table; the true value (from the exact but unwieldy sum-of-dice distribution) is close to this. B

\mathbb P(S_{180}\gt 660)\approx 9.5\%

Reading. Rolling a fair die 180 times, the probability the total exceeds 660 is estimated at about 9.5%, obtained purely from the mean and variance of a single roll via the CLT, without ever computing the exact (and computationally heavy) distribution of a sum of 180 discrete uniforms.

\text{Polling: } X_k\sim\text{Bernoulli}(p),\ p=0.5\ \text{unknown true proportion}, \ n=1000\ \text{i.i.d. respondents.}

Here \(\mu=p\), \(\sigma^2=p(1-p)\); this is the De Moivre–Laplace special case of the CLT corollary noted above. A

\hat p_n=\frac{S_n}{n},\qquad \frac{\hat p_n-p}{\sqrt{p(1-p)/n}}=\frac{S_n-np}{\sigma\sqrt n}=Z_n\ \xrightarrow{d}\ N(0,1).

Divide the theorem's standardised sum through by \(n\) inside numerator and the corresponding \(\sqrt n\) in the denominator to re-express it in terms of the sample proportion \(\hat p_n\) rather than the raw sum \(S_n\); this is purely algebraic rearrangement of the definition of \(Z_n\), so the same convergence in distribution holds. B

\mathbb P\!\left(|\hat p_n-p|\le 1.96\sqrt{\tfrac{p(1-p)}{n}}\right)\approx 0.95 \ \Longrightarrow\ \text{95\% CI: } \hat p_n\pm 1.96\sqrt{\tfrac{\hat p_n(1-\hat p_n)}{n}}.

Use \(\mathbb P(|Z|\le1.96)\approx 0.95\) for \(Z\sim N(0,1)\) (from Step 2's limit) and substitute the consistent plug-in estimator \(\hat p_n\) for the unknown \(p\) inside the standard error, valid by Slutsky's theorem (Corollaries section) since \(\hat p_n\to p\) in probability by the law of large numbers. C

\hat p_n \pm 1.96\sqrt{\hat p_n(1-\hat p_n)/n}

Reading. With \(n=1000\) survey respondents, if \(\hat p_n=0.52\) then the standard 95% confidence interval is \(0.52\pm1.96\sqrt{0.52\cdot0.48/1000}\approx 0.52\pm 0.031\), i.e. roughly \([48.9\%,55.1\%]\) — the entire machinery of margin-of-error reporting in polling rests on this application of the CLT.

Problems

State precisely why the hypothesis "\(\operatorname{Var}(X_1)\lt\infty\)" cannot simply be replaced by "\(\mathbb E[X_1]\) exists" in the Lindeberg–Lévy CLT, giving a specific counterexample.
Solution
The proof's only use of finite variance is Step 4, the second-order Taylor expansion of \(\varphi\) at \(0\), which requires \(\varphi''(0)=-\mathbb E[Y_1^2]\) to exist and be finite, i.e. \(\operatorname{Var}(Y_1)\lt\infty\). If only the mean exists (finite variance fails), \(\varphi\) may fail to be twice differentiable at \(0\), and the quadratic term in the exponent of \(\varphi_{Z_n}\) is absent or infinite. Concretely, take \(X_1\) symmetric with \(\mathbb P(|X_1|\gt x)\sim c/x^2\) for large \(x\) (finite mean by symmetry, e.g. \(\mathbb E[X_1]=0\), but \(\mathbb E[X_1^2]=\infty\), a member of the domain of attraction of a symmetric 2-stable-type boundary case with a slowly varying correction, or more simply a symmetrised Pareto with tail index \(1\lt\alpha\lt2\)). Then \(S_n/\sigma\sqrt n\) does not converge to \(N(0,1)\); the correct normalisation is \(n^{1/\alpha}\), and the limit (if any) is an \(\alpha\)-stable law, not Gaussian.
Let \(X_1,\dots,X_n\) be i.i.d. \(\text{Exponential}(\lambda)\) (rate \(\lambda\), so \(\mu=1/\lambda\), \(\sigma^2=1/\lambda^2\)). Using the CLT, estimate \(\mathbb P(S_{100}\le 90)\) when \(\lambda=1\).
Solution
\(\mu=1,\ \sigma^2=1,\ n=100\). \(Z_{100}=\dfrac{S_{100}-100}{\sqrt{100}}=\dfrac{S_{100}-100}{10}\). \(\mathbb P(S_{100}\le90)=\mathbb P(Z_{100}\le -1)\approx\Phi(-1)\approx 0.1587\). (The exact answer, from the Gamma\((100,1)\) distribution of \(S_{100}\), is close to this; the approximation is imperfect here because Exponential\((1)\) is right-skewed, illustrating the "common error" of trusting the normal approximation too far into the tail for skewed summands at only moderate \(n\).)
Explain, using the proof's Step 5, why the characteristic-function method would break down if the \(X_k\) were independent but not identically distributed with wildly different variances \(\sigma_k^2\), and state (without proof) the extra condition that repairs it.
Solution
Step 5 relies on every factor in the product \(\prod_k \mathbb E[e^{i(t/\sqrt n)Y_k}]\) being the same function \(\varphi(t/\sqrt n)\), which needs identical distribution. If the \(X_k\) are merely independent, one instead gets \(\prod_{k=1}^n \varphi_k(t/s_n)\) where \(s_n^2=\sum_k\sigma_k^2\); the second-order expansion argument still works termwise but Step 6's conclusion \(n\log\varphi(t/\sqrt n)\to -t^2/2\) needs replacing by \(\sum_k\log\varphi_k(t/s_n)\to -t^2/2\), which requires that no single \(\sigma_k^2/s_n^2\) dominate the sum as \(n\to\infty\) (asymptotic negligibility of each summand) — precisely the content of the Lindeberg condition (or the simpler Lyapunov condition using higher moments), the extra hypothesis needed for the non-identically-distributed (Lindeberg–Feller) CLT.
Prove, without redoing the characteristic-function argument, that if \(Z_n\xrightarrow{d}N(0,1)\) as in the theorem, then \(Z_n^2\xrightarrow{d}\chi^2_1\) (chi-squared with 1 degree of freedom).
Solution
The map \(g(x)=x^2\) is continuous on \(\mathbb R\). By the continuous mapping theorem (a standard consequence of weak convergence: if \(Z_n\xrightarrow{d}Z\) and \(g\) is continuous, then \(g(Z_n)\xrightarrow{d}g(Z)\), provable via the portmanteau theorem applied to \(\mathbb E[f(g(Z_n))]\) for bounded continuous \(f\), since \(f\circ g\) is again bounded continuous), \(Z_n^2\xrightarrow{d}Z^2\). By definition, the square of a standard normal random variable has the \(\chi^2_1\) distribution. Hence \(Z_n^2\xrightarrow{d}\chi^2_1\).
A sequence \(X_1,X_2,\dots\) is i.i.d. with \(\mathbb E[X_1]=0\), \(\operatorname{Var}(X_1)=1\), and, additionally, \(\mathbb E[X_1^4]=\kappa\lt\infty\). Show that this extra hypothesis lets you upgrade Step 4 of the proof to a sharper expansion \(\varphi(t)=1-\tfrac{t^2}{2}+O(t^3)\) near \(0\) (assuming also \(\mathbb E|X_1|^3\lt\infty\)), and explain briefly why this is the kind of strengthening Lyapunov's/Berry–Esseen's refinements exploit.
Solution
If \(\mathbb E|X_1|^3\lt\infty\), \(\varphi\) is three-times continuously differentiable (again by dominated convergence, dominating the third difference quotient by \(|X_1|^3\)), with \(\varphi'''(0)=i^3\mathbb E[X_1^3]\), and Taylor's theorem with Lagrange (not just Peano) remainder gives \(\varphi(t)=1-\tfrac{t^2}{2}+\dfrac{\varphi'''(\xi)}{6}t^3\) for some \(\xi\) between \(0\) and \(t\), so \(\varphi(t)=1-\tfrac{t^2}2+O(t^3)\) as \(t\to0\), with an explicit (not merely "\(o\)") error controlled by \(\mathbb E|X_1|^3\). Feeding this sharper expansion through Steps 6–7 in place of the Peano-remainder version yields an explicit, quantitative bound on \(\sup_z|\mathbb P(Z_n\le z)-\Phi(z)|\) of order \(C\,\mathbb E|X_1|^3/(\sigma^3\sqrt n)\) — exactly the content of the Berry–Esseen theorem, which is what turns the CLT's purely qualitative convergence into a convergence rate.