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Let $X_1,X_2,\dots$ be real independent and identically distributed random variables with expected value $\mu =\mathbb{E}[X_1]$ and finite, strictly positive variance ${\sigma }^2=\mathbb{E}[(X_1-\mu {)}^2]$. Write $S_n={\sum }_{j=1}^n{X}_j$ and ${\overline{X}}_n=S_n/n$ for their sum and sample mean. The central limit theorem statesPrecisely, for every real $z$,where $\Phi$ is the standard normal distribution function. The conclusion is convergence in distribution to the standard normal distribution. If the variance is zero, every observation equals $\mu$ almost surely; the displayed normalization is then undefined and there is no fluctuating error to describe.

The striking feature is that the observations need not have a normal distribution. They may be binary, bounded, or strongly skewed. Their detailed probability distribution disappears from this limit except through the expected value and variance. The subtraction removes predictable growth: $\mathbb{E}[S_n]=n\mu$. The division implements variance normalization: variance additivity for independent random variables gives $\mathrm{Var}(S_n)=n{\sigma }^2$, so every $Z_n$ has expected value zero and variance one. For nonzero $\mu$, an uncentered sum drifts; without rescaling its spread grows.

For a concrete example, let each $X_j$ have a Bernoulli distribution with success probability $p=0.2$. Then $K_n=S_n$ has a binomial distribution, with expected value $np$ and variance $np(1-p)$. Its exact probability mass function and its normalized count areHere $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ is a binomial coefficient. Individual trials are conspicuously skewed, yet their centered, rescaled counts approach the same standard normal distribution as sums of fair trials.

At $n=100$, $K_n$ has expected value $20$ and standard deviation $4$. The exact binomial distribution gives $\mathbb{P}(16\le K_n\le 24)=0.740141\dots$. A normal approximation with the continuity correction uses the interval $15.5$ to $24.5$:The small discrepancy is an approximation error, not an equality supplied by the central limit theorem.

The strong law of large numbers and weak law of large numbers locate the average: ${\overline{X}}_n\to \mu$, respectively by almost sure convergence and convergence in probability. The central limit theorem magnifies that shrinking error and describes its shape. In terms of the sample mean, the resulting normal approximation is $N(\mu ,{\sigma }^2/n)$, with standard error $\sigma /\sqrt{n}$. This explains why four times as many independent observations halve the standard error. When $\sigma$ is known, ${\overline{X}}_n\pm 1.96\sigma /\sqrt{n}$ has approximately $95\%$ coverage for $\mu$ at large $n$, giving a confidence interval. This coverage refers to repeated samples, not a guarantee for an individual interval.

A useful application is the Monte Carlo method: estimate a success probability by the proportion of successful simulated trials. Nicholas Metropolis and S. Ulam's original paper, "The Monte Carlo Method" (1949), illustrates this with games of solitaire and estimation of a region's volume by randomly selected points, and discusses why estimates require probabilistic error control. The strong law of large numbers justifies convergence of such proportions for independent repetitions; the central limit theorem supplies their large-sample error scale when $0<p<1$.^[ref]

There are three useful limits to this picture:

A historical original-paper reference is J. W. Lindeberg's "Eine neue Herleitung des Exponentialgesetzes in der Wahrscheinlichkeitsrechnung" (1922), published in Mathematische Zeitschrift 15, pp. 211–225. The Lindeberg condition and Lindeberg-Feller central limit theorem below describe a version for sums whose summands need not share a probability distribution.^[ref]

A brief proof outline uses the characteristic function of $Y=(X_1-\mu )/\sigma$. Its expected value is zero and its variance is one, so the finite second moment givesThe Taylor expansion of the complex exponential, with its remainder controlled by $Y^2$, and the dominated convergence theorem justify this step without a third-moment assumption. Independence then gives the characteristic function of $Z_n$:The limit is the characteristic function of the standard normal distribution; the Lévy continuity theorem turns this into convergence in distribution.

Let $X_1,X_2,\dots$ be independent and identically distributed random variables taking $-1$ and $1$ with equal probabilities, the Rademacher distribution. The two opposite complex exponentials give ${\phi }_X(t)=(e^{it}+e^{-it})/2=\cos t$. Their expected value is zero and their variance is one. For $S_n={\sum }_{j=1}^n{X}_j$, the characteristic functions of its sample mean and its variance normalization are respectivelyFor each fixed $t$, the bases are positive for all sufficiently large $n$. The Taylor expansion $\log \cos u=-u^2/2+O(u^4)$ at zero then givesExponentiating proves the two displayed pointwise limits. The first limit is the characteristic function of the constant zero; the second is that of the standard normal distribution. The Lévy continuity theorem gives the corresponding convergence in distribution statements. Averaging has variance $1/n$, whereas dividing the sum by $\sqrt{n}$ preserves variance one. This sign calculation illustrates why the law of large numbers and the central limit theorem use different scales; it is not a proof for arbitrary probability distributions.

Average $1-\cos (uy)$ over $0\le u\le 1/\lambda$. Multiplication by $\lambda$ gives $1-\sin (y/\lambda )/(y/\lambda )$, with value zero at $y=0$. This is at least $1-\sin 1$ when $|y|\ge \lambda$. Tonelli's theorem then gives the displayed bound for a probability measure with characteristic function $\varphi$. The bound converts control of the characteristic function near zero into tightness of probability measures and is useful in proving the Lévy continuity theorem.

Every characteristic function is uniformly continuous and satisfies $\phi (0)=1$ and $|\phi (t)|\le 1$. Continuity follows from the dominated convergence theorem applied to $e^{itX}$.

More explicitly, if $t_n\to t$, then $e^{i{t}_{n}X}\to e^{itX}$ pointwise and $|e^{i{t}_{n}X}|=1$. Apply the dominated convergence theorem to the real part and imaginary part separately. There is no finite moment hypothesis. For uniform continuity, the stronger estimatefollows by factoring out $e^{itX}$, whose absolute value is one. The integrand tends pointwise to zero and is bounded by two, so the dominated convergence theorem applies again. The bound is independent of $t$, which supplies uniform continuity on the entire real line.^[ref]

If the first absolute moment is finite, $|e^{iu}-1|\le |u|$ additionally gives $|{\phi }_X(t+h)-{\phi }_X(t)|\le |h|\mathbb{E}[|X|]$, a Lipschitz continuity estimate. The earlier argument matters because continuity survives even when this expected value is infinite. For convergence in distribution, the separate Lévy continuity theorem concerns limits of whole sequences of characteristic functions; the argument here establishes continuity of one characteristic function.

The position is the sum of $k$ independent symmetric signs. Its characteristic function after division by $\sqrt{k}$ is $\cos (u/\sqrt{k}{)}^k\to e^{-u^2/2}$. The Lévy continuity theorem gives the normal distribution limit. Scaling the same skeleton to radius $2^{-n}$ and using Brownian exit-skeleton clock convergence identifies its other limit as $B_{\mathbb{E}{\sigma }_1}$, forcing $\mathbb{E}{\sigma }_1=1$.

Let $\varphi (u)=\int e^{iuy}\mu (dy)$. Since $1-\cos (uy)\ge 0$, the Tonelli theorem giveswith the fraction defined to be one at zero. For $|x|\ge 1$, $\sin x/x\le \sin 1$. To verify the constant, the function is even; on $[1,\pi ]$ it decreases because $(x\cos x-\sin x{)}^{\prime }=-x\sin x<0$ on $(0,\pi )$ and $x\cos x-\sin x$ starts from zero. On $[\pi ,\infty )$, $\sin x/x\le 1/\pi <\sin 1$. Consequently the integrand is at least $1-\sin 1$ on $|y|\ge \lambda$, proving the characteristic-function tail boundFor pointwise convergence ${\varphi }_n\to \varphi$, where $\varphi$ is a characteristic function, the dominated convergence theorem applies on each finite integration interval, since the integrands lie between zero and two. HenceContinuity of $\varphi$ at zero and $\varphi (0)=1$ make the right side tend to zero as $\lambda \to \infty$. Given $\epsilon >0$, choose $\lambda$ so it is below $\epsilon /2$. The tail probability is then below $\epsilon$ for all sufficiently large $n$. Enlarge the compact interval to control the finitely many remaining measures, each of which has vanishing tails. This proves uniform tightness of the entire sequence, not just of its tail. It is the characteristic-function tightness bound principle with the sharper displayed constant.

For the integrable mean-zero summands, differentiation at zero is justified byThe dominated convergence theorem gives ${\varphi }^{\prime }(0)=i\mathbb{E}{X}_1=0$. DefineIt is continuous away from zero by continuity of the characteristic function, and continuous at zero by the zero derivative. Since $|\varphi (u)|\le 1$, the finite geometric series yields, for $n\ge 1$,and the case $n=0$ has both sides zero. Thus the requested uniform-in-$n$ estimate isThis is the characteristic-function derivative under an absolute first moment estimate.

Independence makes the characteristic function of $S_n/n=(X_1+\cdots +X_n)/n$ equal to $\varphi (u/n{)}^n$. Therefore $|1-\varphi (u/n{)}^n|\le |u|\psi (u/n)$. Applying the already proved characteristic-function tail bound to its distribution, for every $\epsilon >0$,The convergence follows because ${\sup }_{0\le u\le 1/\epsilon }\psi (u/n)\to 0$, and the remaining integral of $u$ is finite. Hence $\boxed{S_n/n\to 0\text{ in probability}}$, the mean-zero weak law of large numbers. This deduction uses the tail bound directly and does not presuppose the Lévy continuity theorem.

Write $m_l=\lfloor N{t}_l\rfloor -\lfloor N{t}_{l-1}\rfloor$ and $\varphi (v)=\mathbb{E}{e}^{iv\xi }$. Since $\mathbb{E}\xi =0$ and $\mathbb{E}{\xi }^2=1$, the characteristic function expansion at zero is $\varphi (v)=1-v^2/2+o(v^2)$. Indeed, $|e^{ix}-1-ix|\le x^2/2$ for real $x$, and $(e^{iv\xi }-1-iv\xi )/v^2\to -{\xi }^2/2$. The dominated convergence theorem, with dominating variable ${\xi }^2/2$, proves the expansion after taking expectations. Initially omit the endpoint interpolation terms. The integer-time increments involve disjoint sets of independent steps, so their joint characteristic function isHere $m_l/N\to t_l-t_{l-1}$. Taking the logarithm near $1$ givesThis proves the claimed limit for the integer-time approximation.

At a fixed time $t$, the omitted endpoint term is $R_N(t)=N^{-1/2}(Nt-\lfloor Nt\rfloor ){\xi }_{\lfloor Nt\rfloor +1}$, and $\mathbb{E}|R_N(t){|}^2\le 1/N$. A fixed linear combination of the errors in the $k$ increments therefore tends to zero in $L^1$, by the triangle inequality and Cauchy-Schwarz inequality, even when two endpoint terms share a step. Since $|e^{ix}-e^{iy}|\le |x-y|$, restoring interpolation changes the joint characteristic function by a quantity tending to zero. HenceThe limit is the joint characteristic function of independent centered normal increments with the indicated variances. By Lévy continuity theorem, the increment vectors converge in distribution to Brownian increment vectors. The value vector is obtained by the fixed invertible triangular map of cumulative sums; thus the continuous mapping theorem gives convergence of $(S_{t_1}^N,\dots ,S_{t_k}^N)$ to $(B_{t_1},\dots ,B_{t_k})$. This proves every required finite-dimensional distribution limit. The finite fourth moment is not needed for this part; it will give tightness in part (b).

For probability measures on a metric space, weak convergence of probability measures ${\mu }_n\Rightarrow \mu$ means $\int fd{\mu }_n\to \int fd\mu$ for every bounded continuous real function $f$. On the real line it is equivalently convergence of the cumulative distribution functions at every continuity point of the limiting cumulative distribution function. Bounded continuous upper and lower approximations to interval indicators give one direction; a partition at continuity points, uniform approximation on a compact interval, and control of the tails give the reverse direction.

For a probability measure $\mu$ on $\mathbb{R}$, its characteristic function is ${\phi }_{\mu }(t)=\int e^{itx}\mu (dx)$. The Lévy continuity theorem states: ${\mu }_n\Rightarrow \mu$ implies ${\phi }_{{\mu }_n}(t)\to {\phi }_{\mu }(t)$ for every real $t$; conversely, if these characteristic functions have a pointwise limit $\phi$ which is continuous at zero, there is a unique probability measure $\mu$ with characteristic function $\phi$, and ${\mu }_n\Rightarrow \mu$. In particular, for a specified limiting probability measure, weak convergence is equivalent to pointwise convergence of its characteristic functions.

The forward implication follows by testing cosine and sine. The dominated convergence theorem shows that every characteristic function is continuous at zero and has value one there. For the converse, first establish tightness of probability measures. By Fubini's theorem, for $h>0$,The integrand on the right is at least $1/2$ when $|x|\ge 2/h$. This proves the characteristic-function tightness boundThe integrals converge for each $h$ by dominated convergence, and continuity of $\phi$ at zero makes the limit as small as desired by choosing $h$ small. Thus all sufficiently late measures have a common tail bound; the finitely many earlier measures also have small tails once the radius is increased. This is uniform tightness of probability measures.

We give the required compactness argument explicitly. For any subsequence, choose a further diagonal subsequence whose cumulative distribution functions $F_n$ converge at every rational $q$, to $f(q)$. DefineThis function is nondecreasing and right-continuous: given a rational $q>x$ with $f(q)$ close to $F(x)$, the same bound holds for $F(y)$ when $x<y<q$. Tightness gives $F(-\infty )=0$ and $F(+\infty )=1$, so $F$ is a cumulative distribution function of a probability measure $\nu$. For rationals $r<x<s$, $F_n(r)\le F_n(x)\le F_n(s)$. The limits bracket $F(x-)$ and $F(x)$; at a continuity point of $F$ they coincide. Thus $F_n(x)\to F(x)$ at all such points. To obtain weak convergence, choose a compact interval with small tails and partition it using continuity points of $F$. A bounded continuous function is uniformly continuous there, so step-function approximations and convergence of interval probabilities prove convergence of its integral. This proves Helly selection for distribution functions and supplies a weak subsequential limit $\nu$.

Its characteristic function is $\phi$ by the already proved forward implication. We also prove uniqueness rather than assume it. If two probability measures $\nu ,\rho$ have the same characteristic function, convolve each with a centered normal distribution of variance ${\epsilon }^2>0$. Their densities are, by the Gaussian integral and Fubini's theorem,and likewise for $\rho$. The integrable Gaussian factor justifies exchanging the integrals, so these densities agree. For every bounded continuous $f$, letting $\epsilon \downarrow 0$ in $\int \mathbb{E}[f(x+\epsilon Z)]\nu (dx)$, with $Z$ a standard normal random variable, gives $\int fd\nu$ by bounded convergence theorem. Hence the two measures have equal integrals against all these test functions, and approximating interval indicators shows $\nu =\rho$. This proves the uniqueness theorem for characteristic functions.

Every subsequence therefore has a further subsequence weakly converging to the same unique $\mu$. If the original sequence failed weak convergence, some bounded continuous test function would have a subsequence of integrals staying a fixed distance from its integral under $\mu$, contradicting that further convergence. This completes both directions of the theorem. Continuity at zero is essential: the laws $N(0,n)$ have pointwise limiting characteristic function zero away from zero and one at zero, and are not tight.

The force from one location is defined almost surely, since the uniform distribution assigns probability zero to the origin. Symmetry of that uniform distribution makes its characteristic function real:Near zero the integrand is bounded by $2$. For large $x$ it is at most $m^2{t}^2/(2x^4)$. The monotone convergence theorem therefore gives the finite integralIndependence of the locations gives the characteristic function of the total force as $q_n(t{)}^n$. For fixed $t$, $I_n(t)$ is bounded, so Taylor expansion of the logarithm givesThe logarithm is legitimate for all sufficiently large $n$.

For $t\ne 0$, substitute $u=m|t|/x^2$. The cosine integral for a symmetric stable exponent, evaluated directly in part (b), givesAt $t=0$ both sides vanish. The limiting characteristic function is continuous at zero, so the Lévy continuity theorem proves weak convergence of random variables to the symmetric stable distribution of index $1/2$ withThe coefficient reflects the spatial density $n/(2n)=1/2$. Symmetry was used in the characteristic function, without assuming that the singular force has a finite mean.

Interpret the signed inverse-power force as $m\mathrm{sign}(x)/|x{|}^p$, so noninteger $p$ is meaningful on either side of the origin. Write $\alpha =1/p\in (0,2)$. Exactly as in part (a), the one-star characteristic function is $1-I_{n,p}(t)/n$, whereThe integrand is bounded near zero and is at most $m^2{t}^2/(2x^{2p})$ at infinity. The condition $2p>1$ thus makes its integral finite. Independence and the logarithm calculation in part (a) give the limiting characteristic function $\exp (-I_p(t))$, withHere we substituted $u=m|t|/x^p$. The integral for $J_{\alpha }$ is finite because its integrand is $O(u^{1-\alpha })$ at zero and $O(u^{-1-\alpha })$ at infinity.

To determine the constant rather than leave it as an unevaluated integral, use the Gamma function representationThe integrand multiplied by $1-\cos u$ is nonnegative, so the Tonelli theorem permits exchanging the integrals. For $v>0$ the inner integral isConsequently, substituting $z=v^2$ and using the Euler beta function and Gamma reflection formula,Since $\Gamma (1+\alpha )=\alpha \Gamma (\alpha )$, this givesEvery factor in $c_p$ is positive in the stated range. The Lévy continuity theorem gives weak convergence of random variables because this characteristic function is continuous at zero. The limit is strictly stable: if $Z_1,Z_2$ are independent copies of it, the characteristic function of $a{Z}_1+b{Z}_2$ iswhich is the characteristic function of $(|a{|}^{\alpha }+|b{|}^{\alpha }{)}^{1/\alpha }{Z}_1$. This proves the asserted index, as well as identifying the symmetric stable distribution. In particular $p=2$ gives $c_2=\sqrt{\pi m/2}$, while $p=1$ gives $c_1=\pi m/2$, the centered Cauchy distribution with that scale.

Condition on the number $M$ of stars. The independence assumptions give $\mathbb{E}[e^{it{F}_n}\mid M]=q_n(t{)}^M$, with the inverse-square one-star characteristic function $q_n$ from part (a). Averaging over the Poisson distribution therefore gives the compound Poisson distribution formula directly:Since $I_n(t)\to \sqrt{\pi m/2}|t{|}^{1/2}$, the Lévy continuity theorem gives exactly the same weak convergence of random variables as in part (a):The random number of stars introduces no additional limiting term: its effect is already contained in the exact exponential formula. This is the symmetric inverse-power Poisson field on a line at spatial intensity $1/2$ and exponent $p=2$.

Write the constant limit as $c$. First, convergence in distribution to a constant implies convergence in probability: the Portmanteau theorem applied to the closed set $\{y:|y-c|\ge \epsilon \}$ givesNow compare characteristic functions without imposing independence. For fixed $t\in \mathbb{R}$,The second bound uses $|e^{iu}-e^{iv}|\le \min \{2,|u-v|\}$. By convergence in probability, the limiting upper bound is $|t|\epsilon$; then let $\epsilon \downarrow 0$. Meanwhile $X_n$ has weak convergence of random variables to $X$, so $\mathbb{E}{e}^{it{X}_n}\to \mathbb{E}{e}^{itX}$. Thus the characteristic function of the sum tends to $e^{itc}\mathbb{E}{e}^{itX}$, the characteristic function of $X+c$, which is continuous at zero. The Lévy continuity theorem provesThis is the additive case of the Slutsky theorem, with its proof supplied. The constant limiting marginal removes the dependence obstruction in part (b).

Let ${\xi }_j=B_{{\sigma }_1^j}-B_{{\sigma }_1^{j-1}}$. Continuity makes each displacement equal to $+1$ or $-1$. Reflection symmetry of a fresh Brownian motion gives probability one half to each sign, and the Strong Markov property makes these signs independent across successive exits. Thus $B_{{\sigma }_1^n}={\sum }_{j=1}^n{\xi }_j$.

For the normalized position, the characteristic function issince $\log \cos z=-z^2/2+O(z^4)$ near zero. By the Lévy continuity theorem, this provesThis is the Gaussian limit of a Brownian exit skeleton.

Compare along the subsequence $n=2^{2m}$ with ${\tau }_m={\sigma }_{2^{-m}}^{2^{2m}}$. Joint Brownian scaling of the successive exits givesThe right-hand side converges weakly to $N(0,1)$. The preceding part and path continuity give $B_{{\tau }_m}\to B_C$ almost surely on the original Brownian probability space. Hence its weak limit is also the law $N(0,C)$. Uniqueness of the weak limit forces $C=1$, so finally $\mathbb{E}{\sigma }_a=a^2$. Independence of the position and its random clock was not assumed in this comparison.

For the central limit theorem, let $X_1,X_2,\dots$ be independent and identically distributed random variables with expectation $\mu$ and finite positive variance ${\sigma }^2$. ThenThus the centered, normalized sum has convergence in distribution to the standard normal distribution. The positive-variance assumption is needed for this normalization; if ${\sigma }^2=0$, each summand equals $\mu$ almost surely and the centered sum is identically zero.

Here is a characteristic function proof using only the finite second moment. Put $U=(X_1-\mu )/\sigma$, so $\mathbb{E}U=0$ and $\mathbb{E}{U}^2=1$. The elementary integral remainder identity gives $|e^{iz}-1-iz|\le z^2/2$ for real $z$. ConsequentlyBy the dominated convergence theorem, the characteristic function satisfies ${\phi }_U(t)=1-t^2/2+o(t^2)$ as $t\to 0$. For a fixed real $t$, independence therefore givesThe allowed complex-number limit makes this tend to $e^{-t^2/2}$, the characteristic function of the standard normal distribution. It is continuous at zero, so the Lévy continuity theorem proves the asserted convergence in distribution. No third moment or boundedness of the summands was used.

On the real line, Prohorov's theorem states that a sequence of probability measures is uniformly tight if and only if every subsequence has a further subsequence converging weakly to a probability measure. Here uniformly tight means that, for every $\epsilon >0$, some compact interval contains mass at least $1-\epsilon$ for every measure in the sequence. This is relative sequential compactness for weak convergence of probability measures, not merely the existence of one convergent subsequence.

We first prove the characteristic-function tail bound. Put $c_0=1-\sin 1>0$. For $y\ne 0$,At $y=0$ the expression is zero by continuity. If $|y|\ge \lambda$, the supplied elementary bound, with $t=|y|/\lambda$, makes it at least $c_0$. The integrand is nonnegative, so Tonelli's theorem yieldsThus one valid universal constant isThe estimate controls spatial tails using only the characteristic function near zero.

Apply it to each ${\mu }_n$. For fixed $\lambda$, pointwise convergence of the characteristic functions and the bound $|{\varphi }_n|\le 1$ allow dominated convergence on the finite integration interval, givingContinuity of a characteristic function and $\varphi (0)=1$ make the last expression tend to zero as $\lambda \to \infty$. Given $\epsilon >0$, choose $\lambda$ making it less than $\epsilon /2$; then for all sufficiently large $n$, the preceding tail is less than $\epsilon$. Enlarge the compact interval to handle the finitely many remaining probability measures. This proves uniform tightness of the whole sequence.

By Prohorov's theorem, every subsequence has a further weakly convergent subsequence, say with limit $\nu$. The real and imaginary parts of $e^{iuy}$ are bounded continuous functions, so along that subsequenceThe uniqueness theorem for characteristic functions implies $\nu =\mu$. All subsequential weak limits are therefore $\mu$. If the original sequence failed to converge weakly to $\mu$, a bounded continuous test function and a subsequence would keep its integrals a fixed distance from the limiting integral; a further weakly convergent subsequence would contradict that. HenceThis supplies the required proof of the Lévy continuity theorem in the stated case.

For the final request, the printed $Q$ is not defined; the intended function is $\varphi$, the characteristic function of a summand. We prove that ${\varphi }^{\prime }(0)=ia$ without assuming integrability. Write $Y_n=S_n/n$. Independence gives $\mathbb{E}{e}^{iu{Y}_n}=\varphi (u/n{)}^n$. Moreover convergence in probability to $a$ implies uniform convergence of these characteristic functions on each bounded frequency interval: for $|u|\le T$ and $\delta >0$,First let $n\to \infty$, then $\delta \downarrow 0$.

The potential logarithm-branch ambiguity needs to be resolved, not discarded. By continuity of $\varphi$ near zero, its principal complex logarithm is defined at $u/n$ for all $|u|\le 1$ and all large $n$, with value zero at $u=0$. Alsouniformly on $[-1,1]$, so $\mathrm{Log}{\Psi }_n(u)\to 0$ uniformly there. The two continuous logarithms differ by an integer multiple of $2\pi i$:That integer is continuous in $u$ and zero at zero, hence identically zero. ConsequentlyFor arbitrary nonzero $h\to 0$, take $n=\lfloor 1/|h|\rfloor$ and $u=nh$. Eventually $1/2\le |u|\le 1$, so dividing the preceding error by $u$ shows $\mathrm{Log}\varphi (h)/h\to ia$. Exponentiating at zero proves the derivative forced by the weak law:This is the weak law forces a characteristic-function derivative argument. Its estimates use bounded exponentials, not $\mathbb{E}|X_1|$. If $X_1$ is integrable, dominated convergence permits differentiation inside the expectation and gives ${\varphi }^{\prime }(0)=i\mathbb{E}{X}_1$, so then $a=\mathbb{E}{X}_1$.

Absolute integrability is genuinely unnecessary. For an explicit weak law without an absolute first moment, take a symmetric variable withwith mass $1-1/e$ at zero and no mass with $0<|X|<e$. This is a valid law: the specified tail decreases from $1/e$ to zero, and its derivative defines the continuous part beyond $e$. Its absolute first moment is infinite because ${\int }_e^{\infty }dx/(x\log x)=\infty$. For independent copies, set $Y_{n,j}=X_j{1}_{\{|X_j|\le n\}}$. Symmetry gives $\mathbb{E}{Y}_{n,j}=0$, and integration of the tail givesFor the last estimate, split the integral at $\sqrt{n}$; the initial part is $O(\sqrt{n})$ and the remainder is at most $2n/\log n$, apart from a fixed initial constant. Chebyshev's inequality now gives $n^{-1}{\sum }_{j\le n}{Y}_{n,j}\to 0$ in probability. The probability that any truncation changes a summand is at most $n\mathbb{P}(|X|>n)=1/\log n\to 0$, so $S_n/n\to 0$ in probability although $\mathbb{E}|X|=\infty$. The derivative is zero, but the usual expectation of $X$ does not exist.

If either the positive or negative part is integrable, the finite weak-law limit does force full integrability. For example, if $\mathbb{E}{X}^{-}<\infty$, then $X\wedge A$ is integrable and its sample averages have the weak law of large numbers limit $\mathbb{E}(X\wedge A)$. Since they are bounded above by the original averages, $\mathbb{E}(X\wedge A)\le a$ for every $A$. Letting $A\to \infty$ shows $\mathbb{E}{X}^{+}<\infty$. The symmetric argument handles the other one-sided hypothesis. The nonintegrable example necessarily has both one-sided expectations infinite.

Fix times $t_1,\dots ,t_d\in [0,1]$, and setThese are iid bounded centred random vectors, andTheir covariance matrix isWe prove the required central limit theorem calculation explicitly. For $u\in {\mathbb{R}}^d$, the scalar $u\cdot V_1$ is bounded, has mean zero and second moment $u^{T}Cu$. The Taylor expansion of the exponential, with its uniformly bounded cubic remainder, givesIndependence implies that the characteristic function of the sum vector is the $n$th power of this expression, so it tends to $\exp (-u^{T}Cu/2)$. This is the characteristic function of a centred multivariate normal distribution with covariance $C$. The Lévy continuity theorem states that convergence of characteristic functions to a characteristic function continuous at zero gives convergence in distribution; apply it here.

By part (ii), the corresponding Brownian bridge vector has exactly this covariance and mean. ThereforeRepeated times or the endpoint times may make $C$ singular; the characteristic-function argument still applies. This proves every finite-dimensional distribution convergence requested, the uniform case of the Finite-dimensional Brownian bridge limit of an empirical distribution process.

Suppose $X_n\to X$ in probability. For a fixed real $t$ and any $\delta >0$, split according to $|X_n-X|\le \delta$. The supplied exponential inequality and the bound $|e^{it{X}_n}-e^{itX}|\le 2$ giveLet $n\to \infty$, then $\delta \downarrow 0$. Thus the characteristic functions converge pointwise to ${\phi }_X$. The Lévy continuity theorem states that pointwise convergence of characteristic functions to a characteristic function continuous at zero implies convergence in distribution. Every characteristic function is continuous at zero, so

Weak convergence $X_k\Rightarrow X$ means that $\mathbb{E}f(X_k)\to \mathbb{E}f(X)$ for every bounded continuous real function $f$ on ${\mathbb{R}}^d$; it is a statement about their distributions, not their coupling.

The central limit theorem says that for independent identically distributed real variables of finite mean $\mu$ and variance ${\sigma }^2$, $\sqrt{n}({\bar{X}}_n-\mu )\Rightarrow N(0,{\sigma }^2)$. For $\sigma >0$ this is equivalent to convergence of $(\sum X_j-n\mu )/(\sigma \sqrt{n})$ to the standard normal distribution. For $\sigma =0$, every variable equals $\mu$ almost surely and the limit is the point mass at zero.