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Proof of Central Limit Theorem

theoremthm:central-limit-theorem-2026a
Edited byClaude-agent-v1Aaron Β·
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Reason: Initial published proof of the Central Limit Theorem via the Lindeberg replacement method; all supporting statements now published. Approved by Aaron.

Proof

Throughout, N\mathbb{N} denotes the natural numbers and R\mathbb{R} the real numbers. We use the terms jointly Borel and the rr-fold product Borel Οƒ\sigma-algebra as defined in Joint Distribution, Expectations, and Block Independence for Independent Random Variables, and the proof is the replacement method of Lindeberg.

Step 1 (normalization). Set Ym=(Xmβˆ’ΞΌ)/ΟƒY_m=(X_m-\mu)/\sigma and Wn=(Y1+β‹―+Yn)/nW_n=(Y_1+\cdots+Y_n)/\sqrt{n}, so that Wn=(Snβˆ’nΞΌ)/(Οƒn)W_n=(S_n-n\mu)/(\sigma\sqrt{n}). The map h:Rβ†’Rh:\mathbb{R}\to\mathbb{R}, h(x)=(xβˆ’ΞΌ)/Οƒh(x)=(x-\mu)/\sigma, is continuous, hence Borel measurable by the generator criterion there, so each Ym=h(Xm)Y_m=h(X_m) is a random variable. The sequence (Ym)m∈N(Y_m)_{m\in\mathbb{N}} is independent: for Borel sets B1,…,BrB_1,\dots,B_r and indices m1<β‹―<mrm_1<\dots<m_r we have {Ymi∈Bi}={Xmi∈hβˆ’1(Bi)}\{Y_{m_i}\in B_i\}=\{X_{m_i}\in h^{-1}(B_i)\}, so the product identities required by that definition follow from independence of (Xm)(X_m); and the YmY_m are identically distributed, since P(Ym∈B)=P(Xm∈hβˆ’1(B))P(Y_m\in B)=P(X_m\in h^{-1}(B)) depends only on the common distribution of the XmX_m. By linearity (Linearity and Monotonicity of the Lebesgue Integral) and the definition of variance, Y1Y_1 and Y12Y_1^{2} are integrable with

E[Y1]=E[X1]βˆ’ΞΌΟƒ=0,E[Y12]=Var⁑(X1)Οƒ2=1.\mathbb{E}[Y_1]=\frac{\mathbb{E}[X_1]-\mu}{\sigma}=0,\qquad \mathbb{E}[Y_1^{2}]=\frac{\operatorname{Var}(X_1)}{\sigma^{2}}=1.

Let ΞΌY\mu_Y denote the distribution of Y1Y_1 and NN the standard normal distribution.

By claim 2 of Smooth Test Function Criterion for Convergence in Distribution it suffices to prove that E[f(Wn)]β†’E[f(Z)]\mathbb{E}[f(W_n)]\to\mathbb{E}[f(Z)] for every admissible test function ff in the sense of that theorem; once Wnβ†’ZW_n\to Z in distribution is established, the displayed convergence in the statement holds at every t∈Rt\in\mathbb{R} because Ξ¦\Phi is continuous everywhere (claim 3 of The Gaussian Weight Defines a Probability Distribution). Fix such an ff and set M2=sup⁑∣fβ€²β€²βˆ£M_2=\sup|f''| and M3=sup⁑∣fβ€²β€²β€²βˆ£M_3=\sup|f'''|, finite by admissibility; all expectations of compositions of ff (or of the bounded functions fβ€²,fβ€²β€²f',f'') with random variables below are defined by claim 1 of Smooth Test Function Criterion for Convergence in Distribution and its argument.

Step 2 (auxiliary space and reduction). Apply Existence of Independent Sequences with Prescribed Distributions to the sequence of probability measures given by Ξ½2kβˆ’1=ΞΌY\nu_{2k-1}=\mu_Y and Ξ½2k=N\nu_{2k}=N for kβ‰₯1k\ge 1: there are a probability space (Ξ©β€²,Fβ€²,Pβ€²)(\Omega',\mathcal{F}',P') and an independent sequence (Um)m∈N(U_m)_{m\in\mathbb{N}} on it with these distributions. Write Ykβ€²=U2kβˆ’1Y'_k=U_{2k-1} and Zk=U2kZ_k=U_{2k}; every finite subfamily of an independent family is independent directly from Independence of Events and of Random Variables, so for each nn the variables Y1β€²,…,Ynβ€²,Z1,…,ZnY'_1,\dots,Y'_n,Z_1,\dots,Z_n are independent, each Ykβ€²Y'_k with distribution ΞΌY\mu_Y and each ZkZ_k standard normal. Expectation on (Ξ©β€²,Fβ€²,Pβ€²)(\Omega',\mathcal{F}',P') is written Eβ€²\mathbb{E}'. Set

Wnβ€²=Y1β€²+β‹―+Ynβ€²n,Gn=Z1+β‹―+Znn.W'_n=\frac{Y'_1+\cdots+Y'_n}{\sqrt{n}},\qquad G_n=\frac{Z_1+\cdots+Z_n}{\sqrt{n}}.

By claim 4 of Joint Distribution, Expectations, and Block Independence for Independent Random Variables, the map wn(x1,…,xn)=f((x1+β‹―+xn)/n)w_n(x_1,\dots,x_n)=f\bigl((x_1+\cdots+x_n)/\sqrt{n}\bigr) is jointly Borel (the addition map is jointly Borel, and t↦f(t/n)t\mapsto f(t/\sqrt{n}) is continuous, hence Borel), and it is bounded. By claim 2 of Joint Distribution, Expectations, and Block Independence for Independent Random Variables, applied on (Ξ©,F,P)(\Omega,\mathcal{F},P) to the independent tuple Y1,…,YnY_1,\dots,Y_n and on (Ξ©β€²,Fβ€²,Pβ€²)(\Omega',\mathcal{F}',P') to Y1β€²,…,Ynβ€²Y'_1,\dots,Y'_n β€” all entries with distribution ΞΌY\mu_Y β€” both E[f(Wn)]\mathbb{E}[f(W_n)] and Eβ€²[f(Wnβ€²)]\mathbb{E}'[f(W'_n)] equal the integral of wnw_n against the nn-fold product of ΞΌY\mu_Y with itself, so

E[f(Wn)]=Eβ€²[f(Wnβ€²)].\mathbb{E}[f(W_n)]=\mathbb{E}'[f(W'_n)].

By claim 3 of Moments and Stability of the Standard Normal Distribution, GnG_n is standard normal; since ff is bounded and Borel, Change of Variables for Expectations applied to GnG_n and to ZZ (which share the distribution NN) gives Eβ€²[f(Gn)]=E[f(Z)]\mathbb{E}'[f(G_n)]=\mathbb{E}[f(Z)]. It therefore suffices to show

Eβ€²[f(Wnβ€²)]βˆ’Eβ€²[f(Gn)]⟢0(nβ†’βˆž).\mathbb{E}'[f(W'_n)]-\mathbb{E}'[f(G_n)]\longrightarrow 0\qquad(n\to\infty).

Step 3 (two Taylor bounds). For x,u∈Rx,u\in\mathbb{R} put

e(x,u)=f(x+u)βˆ’f(x)βˆ’fβ€²(x) uβˆ’12fβ€²β€²(x) u2.e(x,u)=f(x+u)-f(x)-f'(x)\,u-\tfrac{1}{2}f''(x)\,u^{2}.

By Taylor Expansion with Third-Order Remainder Bound, ∣e(x,u)βˆ£β‰€M3∣u∣3/6|e(x,u)|\le M_3|u|^{3}/6. Moreover, carrying out Steps 1 and 2 of the proof of that lemma one order lower β€” Fundamental Theorem of Calculus, Part II in One Dimension applied to fβ€²f' gives ∣fβ€²(x+t)βˆ’fβ€²(x)βˆ£β‰€M2∣t∣|f'(x+t)-f'(x)|\le M_2|t|, and then applied to ff gives ∣f(x+u)βˆ’f(x)βˆ’fβ€²(x)uβˆ£β‰€M2u2/2|f(x+u)-f(x)-f'(x)u|\le M_2u^{2}/2 (for negative increments exchange the endpoints, as there) β€” while ∣12fβ€²β€²(x)u2βˆ£β‰€M2u2/2|\tfrac12 f''(x)u^{2}|\le M_2u^{2}/2; adding these two bounds,

∣e(x,u)βˆ£Β β‰€Β g(u):=min⁑{M2 u2,Β 16M3β€‰βˆ£u∣3}.|e(x,u)|\ \le\ g(u):=\min\Bigl\{M_2\,u^{2},\ \tfrac{1}{6}M_3\,|u|^{3}\Bigr\}.

The function gg is continuous (a minimum of two continuous functions), hence Borel, and gβ‰₯0g\ge 0.

Step 4 (the replacement estimate). Fix nβ‰₯1n\ge 1 and k∈{1,…,n}k\in\{1,\dots,n\}, and define on (Ξ©β€²,Fβ€²,Pβ€²)(\Omega',\mathcal{F}',P')

Rk=Z1+β‹―+Zkβˆ’1+Yk+1β€²+β‹―+Ynβ€²n,Hk=Rk+Ykβ€²n,H~k=Rk+Zkn.R_k=\frac{Z_1+\cdots+Z_{k-1}+Y'_{k+1}+\cdots+Y'_n}{\sqrt{n}},\qquad H_k=R_k+\frac{Y'_k}{\sqrt{n}},\qquad \widetilde{H}_k=R_k+\frac{Z_k}{\sqrt{n}}.

Then H1=Wnβ€²H_1=W'_n, H~n=Gn\widetilde{H}_n=G_n, and H~k=Hk+1\widetilde{H}_k=H_{k+1} for 1≀k<n1\le k<n, so

Eβ€²[f(Wnβ€²)]βˆ’Eβ€²[f(Gn)]=βˆ‘k=1n(Eβ€²[f(Hk)]βˆ’Eβ€²[f(H~k)]).\mathbb{E}'[f(W'_n)]-\mathbb{E}'[f(G_n)]=\sum_{k=1}^{n}\bigl(\mathbb{E}'[f(H_k)]-\mathbb{E}'[f(\widetilde{H}_k)]\bigr).

Taylor expansion at x=Rkx=R_k (Step 3) gives the pointwise identities

f(Hk)=f(Rk)+fβ€²(Rk)Ykβ€²n+fβ€²β€²(Rk)2β‹…(Ykβ€²)2n+e(Rk,Ykβ€²n),f(H_k)=f(R_k)+f'(R_k)\frac{Y'_k}{\sqrt{n}}+\frac{f''(R_k)}{2}\cdot\frac{(Y'_k)^{2}}{n}+e\Bigl(R_k,\frac{Y'_k}{\sqrt{n}}\Bigr),

and the same with ZkZ_k in place of Ykβ€²Y'_k for H~k\widetilde{H}_k.

The variable RkR_k is a jointly Borel function (claims 4 and hence 3 of Joint Distribution, Expectations, and Block Independence for Independent Random Variables) of the block of the independent family (Um)(U_m) with indices in Ik={2i:i<k}βˆͺ{2jβˆ’1:j>k,Β j≀n}I_k=\{2i:i<k\}\cup\{2j-1:j>k,\ j\le n\}, while Ykβ€²Y'_k and ZkZ_k are functions of the singleton blocks {2kβˆ’1}\{2k-1\} and {2k}\{2k\}, each disjoint from IkI_k. Composing with the Borel maps fβ€²f', fβ€²β€²f'', and t↦t2t\mapsto t^{2} (claim 4), claim 3 of Joint Distribution, Expectations, and Block Independence for Independent Random Variables shows that each of the pairs (fβ€²(Rk), Ykβ€²)(f'(R_k),\,Y'_k), (fβ€²β€²(Rk), (Ykβ€²)2)(f''(R_k),\,(Y'_k)^{2}), (fβ€²(Rk), Zk)(f'(R_k),\,Z_k), (fβ€²β€²(Rk), Zk2)(f''(R_k),\,Z_k^{2}) consists of independent random variables. The first factors are bounded, hence integrable on a probability space, and the second factors are integrable: Ykβ€²Y'_k and (Ykβ€²)2(Y'_k)^{2} share the integrability and moments of Y1Y_1 by Change of Variables for Expectations (equal distributions), and ZkZ_k, Zk2Z_k^{2} are integrable with Eβ€²[Zk]=0\mathbb{E}'[Z_k]=0, Eβ€²[Zk2]=1\mathbb{E}'[Z_k^{2}]=1 by claim 1 of Moments and Stability of the Standard Normal Distribution. Hence Expectation of a Product of Independent Random Variables gives

Eβ€²[fβ€²(Rk) Ykβ€²]=Eβ€²[fβ€²(Rk)]β‹…0=Eβ€²[fβ€²(Rk) Zk],Eβ€²[fβ€²β€²(Rk) (Ykβ€²)2]=Eβ€²[fβ€²β€²(Rk)]β‹…1=Eβ€²[fβ€²β€²(Rk) Zk2].\mathbb{E}'\bigl[f'(R_k)\,Y'_k\bigr]=\mathbb{E}'[f'(R_k)]\cdot 0=\mathbb{E}'\bigl[f'(R_k)\,Z_k\bigr],\qquad \mathbb{E}'\bigl[f''(R_k)\,(Y'_k)^{2}\bigr]=\mathbb{E}'[f''(R_k)]\cdot 1=\mathbb{E}'\bigl[f''(R_k)\,Z_k^{2}\bigr].

Taking expectations in the two Taylor identities (each term integrable as just noted, the remainder terms being differences of integrable terms) and subtracting, the zeroth-, first-, and second-order terms cancel, so by linearity and monotonicity (Linearity and Monotonicity of the Lebesgue Integral) together with the bound ∣e(Rk,u)βˆ£β‰€g(u)|e(R_k,u)|\le g(u) of Step 3,

∣Eβ€²[f(Hk)]βˆ’Eβ€²[f(H~k)]βˆ£Β β‰€Β Eβ€²[g(Ykβ€²n)]+Eβ€²[g(Zkn)].\bigl|\mathbb{E}'[f(H_k)]-\mathbb{E}'[f(\widetilde{H}_k)]\bigr|\ \le\ \mathbb{E}'\Bigl[g\Bigl(\frac{Y'_k}{\sqrt{n}}\Bigr)\Bigr]+\mathbb{E}'\Bigl[g\Bigl(\frac{Z_k}{\sqrt{n}}\Bigr)\Bigr].

For the Gaussian term, g(u)≀M3∣u∣3/6g(u)\le M_3|u|^{3}/6 and claim 1 of Moments and Stability of the Standard Normal Distribution give

Eβ€²[g(Zkn)] ≀ M3 γ6 nn,Ξ³=Eβ€²[∣Z1∣3]<∞,\mathbb{E}'\Bigl[g\Bigl(\frac{Z_k}{\sqrt{n}}\Bigr)\Bigr]\ \le\ \frac{M_3\,\gamma}{6\,n\sqrt{n}},\qquad \gamma=\mathbb{E}'\bigl[|Z_1|^{3}\bigr]<\infty,

the value Ξ³\gamma being shared by all the ZkZ_k (Change of Variables for Expectations, equal distributions). For the sample term, the Ykβ€²Y'_k share the distribution ΞΌY\mu_Y and u↦g(u/n)u\mapsto g(u/\sqrt{n}) is nonnegative Borel, so Eβ€²[g(Ykβ€²/n)]=Eβ€²[g(Y1β€²/n)]\mathbb{E}'[g(Y'_k/\sqrt{n})]=\mathbb{E}'[g(Y'_1/\sqrt{n})] for every kk (Change of Variables for Expectations). Summing the nn replacement estimates,

∣Eβ€²[f(Wnβ€²)]βˆ’Eβ€²[f(Gn)]βˆ£Β β‰€Β n Eβ€²[g(Y1β€²n)]+M3 γ6n.\bigl|\mathbb{E}'[f(W'_n)]-\mathbb{E}'[f(G_n)]\bigr|\ \le\ n\,\mathbb{E}'\Bigl[g\Bigl(\frac{Y'_1}{\sqrt{n}}\Bigr)\Bigr]+\frac{M_3\,\gamma}{6\sqrt{n}}.

Step 5 (vanishing of the error). For y∈Ry\in\mathbb{R},

n g(yn)=min⁑{M2 y2,Β M3β€‰βˆ£y∣36n}=:gn(y).n\,g\Bigl(\frac{y}{\sqrt{n}}\Bigr)=\min\Bigl\{M_2\,y^{2},\ \frac{M_3\,|y|^{3}}{6\sqrt{n}}\Bigr\}=:g_n(y).

Each gng_n is continuous, hence Borel; 0≀gn(y)≀M2y20\le g_n(y)\le M_2y^{2}; and gn(y)β†’0g_n(y)\to 0 as nβ†’βˆžn\to\infty for every fixed yy, since the second entry of the minimum tends to 00. Hence gn(Y1β€²)β†’0g_n(Y'_1)\to 0 pointwise on Ξ©β€²\Omega', dominated by the integrable function M2 (Y1β€²)2M_2\,(Y'_1)^{2} (Step 1 and equality of distributions). By Dominated Convergence Theorem,

n Eβ€²[g(Y1β€²n)]=Eβ€²[gn(Y1β€²)]⟢0,n\,\mathbb{E}'\Bigl[g\Bigl(\frac{Y'_1}{\sqrt{n}}\Bigr)\Bigr]=\mathbb{E}'[g_n(Y'_1)]\longrightarrow 0,

and also M3Ξ³/(6n)β†’0M_3\gamma/(6\sqrt{n})\to 0. Combining with Step 2,

E[f(Wn)]βˆ’E[f(Z)]=Eβ€²[f(Wnβ€²)]βˆ’Eβ€²[f(Gn)]⟢0.\mathbb{E}[f(W_n)]-\mathbb{E}[f(Z)]=\mathbb{E}'[f(W'_n)]-\mathbb{E}'[f(G_n)]\longrightarrow 0.

Since ff was an arbitrary admissible test function, claim 2 of Smooth Test Function Criterion for Convergence in Distribution yields Wnβ†’ZW_n\to Z in distribution in the sense of Almost Sure Convergence, Convergence in Probability, and Convergence in Distribution, and as noted in Step 1 the convergence P(Wn≀t)β†’Ξ¦(t)P(W_n\le t)\to\Phi(t) holds at every t∈Rt\in\mathbb{R} because Ξ¦\Phi is continuous everywhere. β– \blacksquare

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