Central Limit Theorems — Examples

The normal law assigns zero mass to each singleton: its bounded density integrates to zero on a Lebesgue-null singleton by [F3]–[F4]. The boundary of [a,b] is contained in the two endpoints, so it is a continuity set. Apply [F1] and the continuity-set implication of [F2] to get the displayed limit. Open, closed or half-open choices of the two fixed standardized endpoints have the same limit. If a=b the closed singleton has limiting probability zero; if a>b the event is empty.

Take n=100,p=1/2 and the event $45\le B_{100}\le 55$. Its mean is 50 and standard deviation is $\sqrt{100\mathrm{/}4}=5$. Raw standardization gives endpoints -1 and 1, and the corresponding normal probability is $\mathrm{\Phi }(1)-\mathrm{\Phi }(-1)\approx 0.68268949$. The half-unit cell endpoints $89\mathrm{/}2$ and $111\mathrm{/}2$ give corrected standardized endpoints -1.1 and 1.1, and the corrected normal probability is $\mathrm{\Phi }(1.1)-\mathrm{\Phi }(-1.1)\approx 0.72866788$. These decimal evaluations are of the displayed normal integrals. The theorem does not bound either finite-n approximation error or prove that the correction always improves it. AC is inherited through [F1] and the normal-density and null-set construction.

The nonnegative Borel density ${\mathbf{1}}_{[0,1]}$ defines a measure by [F1]. On [0,1] the primitives x, $x^2\mathrm{/}2$ and $x^3\mathrm{/}3$ have derivatives 1,x and $x^2$ by [F4]. Those derivatives are continuous and integrable by [F9], so [F3] and [F5] give integrals 1,1/2 and 1/3 respectively. Thus the measure is a probability. Apply [F2] separately to the globally nonnegative functions $x^{+}=\max \{x,0\}$ and $x^{-}=\max \{-x,0\}$. Their products with the density are respectively $x{\mathbf{1}}_{[0,1]}$ and zero, so [F10] gives $\mathbb{E}U=1\mathrm{/}2-0=1\mathrm{/}2$. Applying [F2] to the nonnegative function $x^2$ gives $\mathbb{E}{U}^2=1\mathrm{/}3$, hence $\mathrm{Var}(U)=1\mathrm{/}3-1\mathrm{/}4=1\mathrm{/}12$. The density-supported integrands are bounded, so no tail or improper integral is involved.

If copies need realization, [F7] lets AC supply the DC and countable choice in [F6]; the resulting coordinate variables have exactly this density law and are iid. For given iid U_k the same moment computation applies. [F8] gives $({\sum }_k{U}_k-n\mathrm{/}2)\mathrm{/}\sqrt{n\mathrm{/}12}\Rightarrow N(0,1)$, and $1\mathrm{/}\sqrt{n\mathrm{/}12}=\sqrt{12\mathrm{/}n}$ proves the displayed form. The variance is strictly positive and n>=1, so the normalization is defined. AC is used in the integral bridge, copy construction when needed, and CLT supplier.

Use [F2]–[F3] on the law assigning mass 1/2 to each sign, and index its coordinates by $n(n-1)\mathrm{/}2+k$ for $1\le k\le n$. These indices are distinct across the array and exhaust the positive integers. Thus the required signs exist and are independent. Direct two-point integration gives $\mathbb{E}{X}_{n,k}=(k-k)\mathrm{/}2=0$, $\mathbb{E}{X}_{n,k}^2=k^2$ and $\mathbb{E}\mathrm{\mid }{X}_{n,k}{\mathrm{\mid }}^3=k^3$. Different k have disjoint supports $\{-k,k\}$, so their laws differ.

At least n/2 integers k in the row satisfy $k\ge n\mathrm{/}2$, so $s_n^2\ge (n\mathrm{/}2)(n\mathrm{/}2{)}^2=n^3\mathrm{/}8$. Also ${\sum }_{k=1}^n{k}^3\le n{n}^3=n^4$. Hence $s_n^{-3}{\sum }_k\mathbb{E}\mathrm{\mid }{X}_{n,k}{\mathrm{\mid }}^3\le 8^{3\mathrm{/}2}\mathrm{/}\sqrt{n}\to 0$. These bounds remain valid for n=1. The rows are centered, independent and have finite moments and positive s_n, so [F1] with delta=1 proves the assertion. AC is used by the copy construction and inherited in [F1].

Each $0<p_{n,k}<1\mathrm{/}2$ defines a two-point probability. Index the pairs by $n(n-1)\mathrm{/}2+k$ and use [F1]–[F3] to construct all coordinates independently. By [F4], the centered entry has mean zero and variance $p_{n,k}(1-p_{n,k})$. Its values are $-p_{n,k}$ and $1-p_{n,k}$, both of absolute value less than one. Distinct p have distinct negative support points with positive mass, hence distinct centered laws. Normalizing every entry in a fixed row by the same positive s_n also preserves this distinction.

Since $1-p_{n,k}>1\mathrm{/}2$, $s_n^2\ge \frac{1}{2}{\sum }_{k=1}^{n}k\mathrm{/}(2n+2)=n\mathrm{/}8$. Here ${\sum }_{k=1}^{n}k=n(n+1)\mathrm{/}2$, obtained by pairing k with n+1-k and adding the n equal pair sums. Thus s_n is positive and tends to infinity. For any fixed epsilon>0, eventually $\epsilon s_n>1$, so every event $\mathrm{\mid }{X}_{n,k}\mathrm{\mid }>\epsilon s_n$ is empty. The Lindeberg sum is then exactly zero. All second moments are finite, so [F5] gives the claimed limit. AC is used only through the stated product construction and CLT suppliers; no cross-row independence is needed by the theorem.

The vector has mean (0,0), second norm moment $2\mathbb{E}{Y}_k^2=2{\sigma }^2$, and each covariance entry equals $\mathbb{E}{Y}_k^2={\sigma }^2$. The two columns of Sigma agree and are nonzero, so its rank is one and determinant zero. Its eigenvectors (1,1) and (1,-1) have eigenvalues $2{\sigma }^2$ and zero. [F1] therefore gives convergence to $N_2(0,\mathrm{\Sigma })$.

For a scalar $Z\sim N(0,{\sigma }^2)$, every projection of (Z,Z) is $(u_1+u_2)Z$. By [F4]–[F6], this has law $N(0,{\sigma }^2(u_1+u_2{)}^2)$, including a negative coefficient and coefficient zero. Since $u^T\mathrm{\Sigma }u={\sigma }^2(u_1+u_2{)}^2$, [F2]–[F3] identify (Z,Z) with the target law. It is supported on the diagonal, and the (1,-1) projection is identically zero both before and after the limit. If the optional case sigma=0 is allowed, all centered Y_k vanish almost surely and the example reduces to the point mass (0,0). AC is inherited from the Gaussian and multivariate CLT suppliers.

First construct the symmetric Laplace density $l(x)=e^{-\mathrm{\mid }x\mathrm{\mid }}\mathrm{/}2$. On [0,R] the primitive of $e^{-x}$ is $-e^{-x}$, so [F1]–[F3], [F5] and [F20] give integral $1-e^{-R}$; the other half interval gives the same value by the primitive $e^x$. The bridge [F2] and MCT [F6] yield $\int l=1$. The elementary exponential series bound $e^R\ge 1+R$ makes $e^{-R}\to 0$. Thus [F8] defines a probability with density l. For every bounded measurable complex $h=u+iv$, apply [F9] to $u^{+},u^{-},v^{+},v^{-}$; these four functions are bounded by $\mathrm{\mid }h\mathrm{\mid }$, so their products with $l$ have finite integral. The real/imaginary definition [F21] then gives $\int hd(ld\lambda )=\int hld\lambda$. In particular the characteristic function of the constructed law is its density integral. For real t, componentwise differentiation using [F3]–[F5] gives the primitive $e^{(-1+it)x}\mathrm{/}(-1+it)$ on the positive half-line and $e^{(1+it)x}\mathrm{/}(1+it)$ on the negative half-line. Their nonzero denominators have modulus at least one. Taking integer R limits with DCT majorant $e^{-\mathrm{\mid }x\mathrm{\mid }}$ and [F19] gives ${\phi }_l(t)=\frac{1}{2}((1-it{)}^{-1}+(1+it{)}^{-1})=1\mathrm{/}(1+t^2)$.

By [F10]–[F11], ${\int }_{-R}^R(1+t^2{)}^{-1}dt=\arctan R-\arctan (-R)<\pi$. The integrands are continuous, so the bridge and MCT make ${\phi }_l$ Lebesgue integrable. [F12] supplies the continuous density $f(y)=(2\pi {)}^{-1}\int e^{-ity}\mathrm{/}(1+t^2)dt$ for the Laplace law. It equals l everywhere: both are continuous and have the same integral on every interval, whereas any nonzero value of f-l would, by continuity, have one strict sign bounded away from zero on a nonempty interval, contradicting that integral equality and [F13]. At y=0 this identity gives $\int (1+t^2{)}^{-1}dt=\pi$. Therefore [F8] makes c a probability density. At y=-u the same identity, multiplied by two, gives $\int e^{iux}c(x)dx=e^{-\mathrm{\mid }u\mathrm{\mid }}$. No contour integral is used.

For R>=1, ${\int }_1^R{x}^{2}c(x)dx\ge (R-1)\mathrm{/}(2\pi )$ because $x^2\mathrm{/}(1+x^2)\ge 1\mathrm{/}2$ for x>=1. Nonnegative integral monotonicity and [F13] give this lower bound, tending to infinity; [F9] identifies the second moment as infinite. By [F14]–[F15], AC realizes iid copies of the c law. By [F16], their normalized-sum characteristic function is $(e^{-\mathrm{\mid }t\mathrm{\mid }\mathrm{/}\sqrt{n}}{)}^n=e^{-\sqrt{n}\mathrm{\mid }t\mathrm{\mid }}$. It tends to zero for t different from zero and stays one at zero.

If those row laws had any weak limit probability, [F17] would make its characteristic function equal to the pointwise limit just calculated. That function is discontinuous at zero, contradicting [F18]. Thus there is no weak probability limit, and in particular no normal limit. AC was used through the Lebesgue bridge, density inversion and independent-copy construction; the direct moment lower bound and discontinuity argument make no additional selections.

On the standard-normal probability space supplied by [F7], let Z be its coordinate and define $X_{n,1}=Z$, $X_{n,k}=0$ for $2\le k\le n$. By [F1], every entry is centered and the total variance is one. Each row is independent: if an event for a zero coordinate excludes zero, both the intersection probability and the product are zero; otherwise all those events are the full space and the identity reduces to the event for Z. Thus the row sum is exactly Z for every n, its law is N(0,1), and the maximum summand variance equals one. Independence between rows is not required.

For each fixed epsilon>0, [F2] gives the Lindeberg quantity $\mathbb{E}[Z^2{\mathbf{1}}_{\{\mathrm{\mid }Z\mathrm{\mid }>\epsilon \}}]$, independent of n. On $[\epsilon +1,\epsilon +2]$, the integrand against the normal density is at least $(\epsilon +1{)}^2{e}^{-(\epsilon +2{)}^2\mathrm{/}2}\mathrm{/}\sqrt{2\pi }>0$, by [F3] and [F5]. Integrating this bound over the length-one interval with [F4]–[F6] proves strict positivity. Hence the Lindeberg limit is not zero for any epsilon>0, despite exact normality of all row sums. The example works for n=1 as well. AC is inherited solely through construction of the normal law and its Lebesgue density; one Z is reused and no independent sequence across rows is constructed.