Central Limit Theorems

Write $g(x)=e^{-x^2\mathrm{/}2}\mathrm{/}\sqrt{2\pi }$ and let $Z$ be the coordinate under its probability law. By [F1]–[F2], this is a probability law with density $g$. For any measurable complex $h$ with $\int \mathrm{\mid }h\mathrm{\mid }g<\mathrm{\infty }$, apply [F11] to the positive and negative parts of $\mathrm{Re}h$ and $\mathrm{Im}h$. The definitions in [F19] then give $\mathbb{E}h(Z)=\int h(x)g(x)dx$; in particular, every density-integral identity used below is covered. The derivative rules give $g\mathrm{\prime }(x)=-xg(x)$. All compact-interval functions below are continuously differentiable, so [F6]–[F9] apply also to each real and imaginary component. AC is used through the normal-law construction and the countable-choice Riemann/Lebesgue bridge; no sequence of arbitrary witnesses is selected.

For $R\ge 1$, the series in [F20] has nonnegative terms at $R^2\mathrm{/}2$, so $e^{R^2\mathrm{/}2}\ge (R^2\mathrm{/}2{)}^2\mathrm{/}2=R^4\mathrm{/}8$. By [F18], $e^{-R^2\mathrm{/}2}=1\mathrm{/}{e}^{R^2\mathrm{/}2}$; hence $0<g(R)\le 8\mathrm{/}(\sqrt{2\pi }{R}^4)$ and both $g(R)$ and $Rg(R)$ tend to zero. For each positive integer $R$, FTC on each half interval gives ${\int }_{-R}^R\mathrm{\mid }x\mathrm{\mid }g(x)dx=2[g(0)-g(R)]$. Thus MCT proves $\mathbb{E}\mathrm{\mid }Z\mathrm{\mid }=2g(0)<\mathrm{\infty }$. Also ${\int }_{-R}^{R}xg(x)dx=g(-R)-g(R)=0$; DCT with majorant $\mathrm{\mid }x\mathrm{\mid }g(x)$ proves $\mathbb{E}Z=0$. Integration by parts with $u=x,v=g$ gives ${\int }_{-R}^R{x}^{2}g(x)dx={\int }_{-R}^{R}g(x)dx-2Rg(R)$. MCT and [F2] now give $\mathbb{E}{Z}^2=1$. All truncations here use the explicit integers R.

By the finite first moment and [F12], ${\phi }_Z\mathrm{\prime }(t)=i\int x{e}^{itx}g(x)dx$. On $[-R,R]$, componentwise integration by parts, using [F14]–[F15], gives $\int x{e}^{itx}g(x)dx=-[e^{itx}g(x){]}_{-R}^R+it\int e^{itx}g(x)dx$. The boundary has modulus at most $2g(R)\to 0$. The two integrands are dominated respectively by $\mathrm{\mid }x\mathrm{\mid }g(x)$ and $g(x)$, already integrable. DCT therefore gives ${\phi }_Z\mathrm{\prime }(t)=-t{\phi }_Z(t)$ for every t, with no improper differentiation left unjustified.

By the real-component product and chain rules, $H(t)=e^{t^2\mathrm{/}2}{\phi }_Z(t)$ has derivative zero. Apply [F16] to its real and imaginary parts on every real interval. Since $H(0)=\int g=1$, $H(t)=1$ for all t, hence ${\phi }_Z(t)=e^{-t^2\mathrm{/}2}$. For $X=m+\sigma Z$ in law, [F17] yields $e^{imt}{\phi }_Z(\sigma t)$, and [F18] combines the exponents. Its moments follow by expanding the finite integrable expressions: $\mathbb{E}X=m+\sigma \mathbb{E}Z=m$ and $\mathbb{E}(X-m{)}^2={\sigma }^2\mathbb{E}{Z}^2={\sigma }^2$. For $\sigma =0$ the variable equals m almost surely and both formulas give the Dirac law directly.

Taylor at zero through degree two for cosine and sine, with third derivatives bounded by one, gives $\mathrm{\mid }\cos u-1+u^2\mathrm{/}2\mathrm{\mid }\le \mathrm{\mid }u{\mathrm{\mid }}^3\mathrm{/}6$ and $\mathrm{\mid }\sin u-u\mathrm{\mid }\le \mathrm{\mid }u{\mathrm{\mid }}^3\mathrm{/}6$. Euler form and the triangle inequality give $\mathrm{\mid }r(u)\mathrm{\mid }\le \mathrm{\mid }u{\mathrm{\mid }}^3\mathrm{/}3$. Taylor through degree one gives $\mathrm{\mid }\cos u-1\mathrm{\mid }\le u^2\mathrm{/}2$ and $\mathrm{\mid }\sin u-u\mathrm{\mid }\le u^2\mathrm{/}2$, hence $\mathrm{\mid }{e}^{iu}-1-iu\mathrm{\mid }\le u^2$. Adding $u^2\mathrm{/}2$ bounds $\mathrm{\mid }r(u)\mathrm{\mid }\le 3u^2\mathrm{/}2\le 4u^2$ for all real u. At u=0 all remainders vanish.

For $0<\mathrm{\mid }t\mathrm{\mid }\le 1\mathrm{/}n$, the first step gives $\mathrm{\mid }r(tX)\mathrm{\mid }\mathrm{/}{t}^2\le X^2\min (\mathrm{\mid }X\mathrm{\mid }\mathrm{/}(3n),4)$. This is a measurable nonnegative sequence tending pointwise to zero, bounded by the integrable $4X^2$. DCT therefore makes its expectations tend to zero. The bound is uniform over all such t, so $\mathbb{E}r(tX)=o(t^2)$ as a genuine two-sided real limit, without choosing a sequence of frequencies or requiring $\mathbb{E}\mathrm{\mid }X{\mathrm{\mid }}^3<\mathrm{\infty }$. Also $\mathrm{\mid }X\mathrm{\mid }\le 1+X^2$ gives integrability of the linear term.

By linearity, ${\phi }_X(t)=1+it\mathbb{E}X-t^2\mathbb{E}{X}^2\mathrm{/}2+\mathbb{E}r(tX)$. Insert the stated moments and the preceding remainder limit. If $\sigma =0$, the same majorants have zero integral, so the remainder vanishes and the formula still holds. At t=0 the defining expectation is one. Every limiting integrand was explicitly specified; no choice principle enters.

Put $w_{n,k}=z_{n,k}-1$. For $\mathrm{\mid }w\mathrm{\mid }\le 1\mathrm{/}2$, absolute convergence gives $\mathrm{\mid }{e}^w-1-w\mathrm{\mid }\le {\sum }_{j\ge 2}\mathrm{\mid }w{\mathrm{\mid }}^j\mathrm{/}j!\le \mathrm{\mid }w{\mathrm{\mid }}^2{\sum }_{j\ge 2}(1\mathrm{/}2{)}^{j-2}\mathrm{/}2=\mathrm{\mid }w{\mathrm{\mid }}^2$. Also $\mathrm{\mid }1+w\mathrm{\mid }\le 1+\mathrm{\mid }w\mathrm{\mid }\le e^{\mathrm{\mid }w\mathrm{\mid }}$ and $\mathrm{\mid }{e}^w\mathrm{\mid }\le {\sum }_{j\ge 0}\mathrm{\mid }w{\mathrm{\mid }}^j\mathrm{/}j!=e^{\mathrm{\mid }w\mathrm{\mid }}$. The real series is nonnegative term by term, which proves the middle inequality without a logarithm.

For arbitrary finite lists a,b of length r, subtracting successive mixed products gives ${\prod }_{k=1}^r{a}_k-{\prod }_{k=1}^r{b}_k={\sum }_{j=1}^r(a_j-b_j)({\prod }_{k<j}{a}_k)({\prod }_{k>j}{b}_k)$. The cancellation follows by expanding each difference; for r=0 both products are one and the sum zero. Apply it with $a_k=1+w_{n,k}$ and $b_k=e^{w_{n,k}}$. For all sufficiently large n every modulus of w is at most 1/2. Each pair of partial products is bounded by $\exp ({\sum }_{k\mathrm{\ne }j}\mathrm{\mid }{w}_{n,k}\mathrm{\mid })\le e^M$, so the difference of the row products has modulus at most $e^M{\sum }_j\mathrm{\mid }{w}_{n,j}{\mathrm{\mid }}^2$.

The bound tends to zero by hypothesis. Repeated exponential addition identifies the comparison product with $\exp ({\sum }_k{w}_{n,k})$, including empty and one-factor rows. If M=0 then all w vanish and equality is exact in every row. All products and telescoping sums are finite; no branch of logarithm or choice principle is used.

Set $Y_k=(X_k-m)\mathrm{/}\sigma$. The affine functions are Borel; independence is preserved because an event concerning Y_k is the corresponding inverse-image event concerning X_k. Their laws agree, and linearity gives $\mathbb{E}{Y}_k=0$, $\mathbb{E}{Y}_k^2=1$. The normalized sum in the statement is $n^{-1\mathrm{/}2}{\sum }_{k=1}^n{Y}_k$. The positive finite variance makes every division well defined.

Fix a nonzero real t and write $w_n={\phi }_{Y_1}(t\mathrm{/}\sqrt{n})-1$. By [F1], $n{w}_n\to -t^2\mathrm{/}2$. Therefore $w_n\to 0$, $n\mathrm{\mid }{w}_n\mathrm{\mid }$ is bounded, and $n\mathrm{\mid }{w}_n{\mathrm{\mid }}^2=(n\mathrm{\mid }{w}_n\mathrm{\mid })\mathrm{\mid }{w}_n\mathrm{\mid }\to 0$. The bounds extend to the finitely many early n since their values are finite. Apply [F2] to the row with n copies of $1+w_n$. Its product differs from $e^{n{w}_n}$ by a quantity tending to zero. Continuity of the exponential yields the limit $e^{-t^2\mathrm{/}2}$. At t=0 each factor is exactly one.

By [F3], row independence identifies that product with the characteristic function of the normalized sum. By [F4] its limit is the standard-normal transform, continuous at zero and equal to one there. Thus [F5] proves the claimed convergence of laws. AC is inherited only through the normal-law and Levy-criterion suppliers; the iid sequence is given and no new copies are constructed. This theorem excludes sigma=0 because its displayed normalization divides by sigma.

For an omega-indexed family $(A_n)$ of nonempty sets, its image $\mathcal{S}=\{A_n:n\in \mathbb{N}\}$ is a set of nonempty sets. By [F1] choose $c$ with $c(A)\in A$ for every $A\in \mathcal{S}$. Then $b(n)=c(A_n)$ is a function and $b(n)\in A_n$, which is [F2]. Repeated sets use the same selected value and cause no ambiguity. For an empty index family the empty function already suffices; singleton fibers force their unique value.

For the stated serial R, every fiber $R[x]=\{y\in X:xRy\}$ is nonempty. AC applied to the set of all these fibers gives c with $c(R[x])\in R[x]$. Define the self-map $s(x)=c(R[x])$ on X. No recursively changing choice is being assumed: s is now one fixed function chosen from a fixed set of nonempty fibers. It satisfies $xRs(x)$ for every x.

Apply [F4] to X, the prescribed a and s to obtain $x_0=a$ and $x_{n+1}=s(x_n)$ for every natural n. The defining property of s gives $x_{n}R{x}_{n+1}$, precisely [F3]. If X has one element and R is serial, this is its constant sequence. Empty X is excluded by the prescribed a. The uses of AC are exactly the two fixed-family selections in steps 1.1 and 1.2; recursion itself requires no further choice. This proves the two implications from AC, not either converse.

By [F2], AC supplies the countable and dependent choice required in [F1]. Apply that result to the two-point probability with masses 1-p and p to obtain iid Bernoulli variables $I_k$. The law of $C_n={\sum }_{k=1}^n{I}_k$ is Bin(n,p) by [F5]. In particular it agrees with the specified law of B_n for each n; equality persists under the displayed affine standardization.

By [F3], $\mathbb{E}{I}_k=p$ and $\mathrm{Var}(I_k)=p(1-p)>0$. The variables are bounded, hence their second moments are finite. [F4] gives $(C_n-np)\mathrm{/}\sqrt{np(1-p)}\Rightarrow N(0,1)$. Equality of laws in step 1.1 transfers the conclusion to B_n. The excluded p=0,1 and n=0 would make the denominator zero; no claim using that denominator is made there. Neither a finite-n error estimate nor continuity correction follows from this limit theorem.

Fix $\epsilon >0$. For each entry split its square on $\mathrm{\mid }{X}_{n,k}\mathrm{\mid }\le \epsilon$ and its complement. The first integral is at most ${\epsilon }^2\mathbb{P}(\mathrm{\mid }{X}_{n,k}\mathrm{\mid }\le \epsilon )\le {\epsilon }^2$. The second is one nonnegative term of $L_n(\epsilon )$. Centering identifies variance with second moment, hence $0\le {\max }_k\mathrm{Var}(X_{n,k})\le {\epsilon }^2+L_n(\epsilon )$. The maximum exists because the row is finite and nonempty.

For any $\eta >0$, take $0<\epsilon <\sqrt{\eta \mathrm{/}2}$. Lindeberg gives an index after which $L_n(\epsilon )<\eta \mathrm{/}2$. The preceding bound is then less than eta, proving convergence to zero. Values at the cutoff stay in the small part, zero entries satisfy the bound, and a single variance-one summand in every row would contradict this conclusion and hence could not satisfy Lindeberg. The proof uses only given finite rows and explicit inequalities, not independence or choice.

Put $Y_{n,k}=X_{n,k}\mathrm{/}{s}_n$ and $v_{n,k}=\mathbb{E}{Y}_{n,k}^2$. By [F1], the normalized row has total variance one, remains centered and independent, and its tail sum $L_n(\epsilon )$ tends to zero. Fix real t and write $w_{n,k}={\phi }_{Y_{n,k}}(t)-1$. The centering and scalar bound in [F2] give $\mathrm{\mid }{w}_{n,k}\mathrm{\mid }=\mathrm{\mid }\mathbb{E}(e^{it{Y}_{n,k}}-1-it{Y}_{n,k})\mathrm{\mid }\le t^2{v}_{n,k}$, using [F8]–[F9]. Consequently ${\sum }_k\mathrm{\mid }{w}_{n,k}\mathrm{\mid }\le t^2$, ${\max }_k\mathrm{\mid }{w}_{n,k}\mathrm{\mid }\le t^2{\max }_k{v}_{n,k}\to 0$, and ${\sum }_k\mathrm{\mid }{w}_{n,k}{\mathrm{\mid }}^2\le t^4{\max }_k{v}_{n,k}\to 0$ by [F3].

With r as in [F2], linearity gives ${\sum }_k{w}_{n,k}=-t^2\mathrm{/}2+{\sum }_k\mathbb{E}r(t{Y}_{n,k})$. On $\mathrm{\mid }{Y}_{n,k}\mathrm{\mid }\le \epsilon$ the cubic bound gives $\mathrm{\mid }r(t{Y}_{n,k})\mathrm{\mid }\le \mathrm{\mid }t{\mathrm{\mid }}^3\epsilon Y_{n,k}^2\mathrm{/}3$; on the complement the quadratic bound gives $\mathrm{\mid }r(t{Y}_{n,k})\mathrm{\mid }\le 4t^2{Y}_{n,k}^2$. Therefore $\mathrm{\mid }{\sum }_k\mathbb{E}r(t{Y}_{n,k})\mathrm{\mid }\le \mathrm{\mid }t{\mathrm{\mid }}^3\epsilon \mathrm{/}3+4t^2{L}_n(\epsilon )$. First take limsup in n, then let the arbitrary positive epsilon decrease to zero. This proves ${\sum }_k{w}_{n,k}\to -t^2\mathrm{/}2$, without interchanging an unbounded number of unquantified little-o terms.

All hypotheses of [F4] were verified in step 1.1, so ${\prod }_k(1+w_{n,k})-\exp ({\sum }_k{w}_{n,k})\to 0$. Step 2.1 makes its limit $e^{-t^2\mathrm{/}2}$. By [F5] this product is the row-sum transform, and by [F6] its limit is the transform of N(0,1), continuous at zero. [F7] proves the result. At t=0 all factors and the limit are exactly one. AC is inherited in [F6]–[F7]; rows on different probability spaces cause no difficulty because only their laws are compared.

Fix real t and put $v_{n,k}=\mathbb{E}{X}_{n,k}^2$ and $w_{n,k}=\mathbb{E}(e^{it{X}_{n,k}}-1)$. Centering, [F1], [F9] and [F10] give $\mathrm{\mid }{w}_{n,k}\mathrm{\mid }\le t^2{v}_{n,k}$. Thus ${\sum }_k\mathrm{\mid }{w}_{n,k}\mathrm{\mid }\le t^2$, ${\max }_k\mathrm{\mid }{w}_{n,k}\mathrm{\mid }\to 0$, and ${\sum }_k\mathrm{\mid }{w}_{n,k}{\mathrm{\mid }}^2\le t^4{\max }_k{v}_{n,k}\to 0$. By [F2]–[F5] and the assumed normal limit, $\exp ({\sum }_k{w}_{n,k})\to e^{-t^2\mathrm{/}2}$. This inference uses the forward continuity theorem only, not Lindeberg sufficiency.

Define the nonnegative function $d_t(x)=t^2{x}^2\mathrm{/}2-(1-\cos (tx))$. Nonnegativity follows from the cosine Taylor bound in [F1]; also $d_t(x)\le t^2{x}^2\mathrm{/}2$ because $\cos (tx)\le 1$. Hence $D_n(t):={\sum }_k\mathbb{E}{d}_t(X_{n,k})$ is finite and nonnegative. Total variance one and [F9] give $\mathrm{Re}{\sum }_k{w}_{n,k}=-t^2\mathrm{/}2+D_n(t)$. Exponential addition and Euler modulus show $\mathrm{\mid }\exp ({\sum }_k{w}_{n,k})\mathrm{\mid }=e^{-t^2\mathrm{/}2+D_n(t)}$. Step 1.1 therefore implies $e^{D_n(t)}\to 1$. Since the nonnegative series gives $0\le D_n(t)\le e^{D_n(t)}-1$, we obtain $D_n(t)\to 0$. No complex logarithm or subsequence of measures is needed.

Now fix any $\epsilon >0$ and take the single frequency $t=4\mathrm{/}\epsilon$. On $\mathrm{\mid }x\mathrm{\mid }>\epsilon$ we have $t^2{x}^2>16$, and $1-\cos (tx)\le 2$ yields $d_t(x)\ge t^2{x}^2\mathrm{/}2-2\ge t^2{x}^2\mathrm{/}4$. On the complementary set d_t is nonnegative. Integrating and summing gives ${\sum }_k\mathbb{E}[X_{n,k}^2{\mathbf{1}}_{\{\mathrm{\mid }{X}_{n,k}\mathrm{\mid }>\epsilon \}}]\le 4D_n(t)\mathrm{/}{t}^2\to 0$. This is precisely [F11], for every positive epsilon. AC is inherited from the target normal-law construction; the proof uses no Helly selection, uniqueness inversion or backward application of sufficiency. Zero entries and t=0 in the earlier steps are harmless, but the final chosen t is nonzero.

Fix $\epsilon >0$. On $\mathrm{\mid }{X}_{n,k}\mathrm{\mid }>\epsilon s_n>0$, monotonicity of the positive real power gives $\mathrm{\mid }{X}_{n,k}{\mathrm{\mid }}^{\delta }>(\epsilon s_n{)}^{\delta }$. Multiplying by $X_{n,k}^2\mathrm{/}(\epsilon s_n{)}^{\delta }$ gives $X_{n,k}^2\le \mathrm{\mid }{X}_{n,k}{\mathrm{\mid }}^{2+\delta }\mathrm{/}(\epsilon s_n{)}^{\delta }$ there. Off that event the truncated square is zero and the right side is nonnegative, including X=0. [F3] therefore bounds $L_n(\epsilon )\le {\epsilon }^{-\delta }{s}_n^{-(2+\delta )}{\sum }_k\mathbb{E}\mathrm{\mid }{X}_{n,k}{\mathrm{\mid }}^{2+\delta }$. The exponent simplification uses [F4]. For positive delta, monotonicity of log and exp in the defining formula [F5] gives the asserted power monotonicity.

The right side tends to zero for this fixed positive epsilon, so Lindeberg holds for every epsilon. All hypotheses of [F2] are now satisfied: finite second moments, centered independent rows and positive total standard deviations were given. Apply it to obtain the stated limit. The positive delta is fixed across all rows; delta=0 is excluded because the displayed normalized second-moment sum would be one. AC is inherited from [F2].

For any square-integrable vector X, covariance entries ${\mathrm{\Sigma }}_{ij}=\mathbb{E}[(X_i-m_i)(X_j-m_j)]$ exist because $2\mathrm{\mid }ab\mathrm{\mid }\le a^2+b^2$. They are symmetric. Finite linearity gives $u^T\mathrm{\Sigma }u=\mathbb{E}({\sum }_i{u}_i(X_i-m_i){)}^2\ge 0$, so covariance matrices are positive semidefinite. Conversely let a symmetric positive semidefinite Sigma be given and take its nonnegative symmetric square root A by [F2], so $A{A}^T=A^2=\mathrm{\Sigma }$.

By [F3]–[F4], AC realizes d independent standard normals $Z_1,\dots ,Z_d$, each of mean zero and second moment one by [F1]. Define $X=m+AZ$. Coordinate linear combinations are measurable, so X is a Borel random vector and its pushforward is a probability by [F9]. For any u and real t, [F5] and [F1] give $\mathbb{E}{e}^{itu\cdot X}=e^{itu\cdot m}{\prod }_j{e}^{-t^2(A^{T}u{)}_j^2\mathrm{/}2}=e^{itu\cdot m-t^2{u}^T\mathrm{\Sigma }u\mathrm{/}2}$. This equals the characteristic function of $N(u\cdot m,u^T\mathrm{\Sigma }u)$ by [F1]; [F6] gives equality of scalar laws. Thus the required projection condition holds, including u=0 and all null directions.

Finite linearity gives $\mathbb{E}X=m$. By [F7], $\mathbb{E}{Z}_i{Z}_j=0$ for i different from j, and [F1] gives $\mathbb{E}{Z}_i^2=1$. Hence the covariance of AZ is $AI{A}^T=\mathrm{\Sigma }$, with every product integrable by the bound in step 1.1. If Sigma=0 then A=0 and the law is the point mass at m; no inverse or density is required even when only some directions are null. For d=1 the construction agrees with the scalar affine normal. If dimension zero is admitted, use the unique law on the singleton empty tuple instead. AC is spent in the standard-normal construction, independent-copy realization and scalar uniqueness supplier.

For fixed t, [F1] makes $t\cdot X$ scalar normal of mean $t\cdot m$ and variance $t^T\mathrm{\Sigma }t$. Evaluate its characteristic function from [F2] at scalar frequency one. This gives the displayed formula. If t=0 both sides are one; if $t^T\mathrm{\Sigma }t=0$ the scalar law is the point mass at $t\cdot m$ and the same formula applies.

Let Y have another Borel probability law with the same displayed vector transform. For every u and scalar s, ${\phi }_{u\cdot Y}(s)={\mathrm{\Phi }}_Y(su)={\mathrm{\Phi }}_X(su)={\phi }_{u\cdot X}(s)$. Scalar uniqueness [F3] identifies each pair of projection laws. Then [F4] identifies the vector laws. Thus the construction in [F1] is independent of any realization choices. No determinant or inverse of Sigma is used. AC is inherited from [F1]–[F4]; in dimension zero the single possible law makes the assertion immediate.