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Weak Laws and Series of Independent Random Variables — Examples

1 · Prerequisites

2 · Summary

These examples distinguish IID hypotheses, variance control, tail control, and absolute convergence. Bernoulli sample means illustrate the weak laws; fair-sign series show the sharp square-summability threshold. Three constructions isolate the three-series conditions, while rare large jumps separate almost-sure convergence from summability of untruncated variances. Cauchy and logarithmic tails mark the boundary for deterministic weak-law centering.

Sequence-existence constructions retain the countable-choice and dependent-choice hypotheses of the countable product results.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-adaptedVerification: AI-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Bernoulli sample frequencies

Example

Assume countable choice and dependent choice. For IID Bernoulli(p) variables (Xk)k1, where 0p1, let Sn=k=1nXk. Then Sn/np in probability and P(Sn/npε)p(1p)nε2(ε>0). The endpoint laws p=0,1 are included.

Facts & Assumptions

[F1]

IID finite-variance weak law: Let (Xk)k1 be IID square-integrable real random variables, with μ=EX1 and σ2=Var(X1). For Sn=k=1nXk, ESn/nμ2=σ2/n, and Sn/nμ in L2 and in probability. Also P(Sn/nμε)σ2/(nε2) for ε>0.

[F2]

Bernoulli random variables and binomial random variables as sums of independent Bernoulli trials: For p[0,1], a Bernoulli(p) random variable takes the value 1 with probability p and 0 with probability 1p. For nN, a binomial(n,p) random variable is a sum X=i<nXi of n mutually independent Bernoulli(p) variables. When n=0, this is the constant zero random variable.

[F3]

A Bernoulli(p) variable has mean p and variance p(1p); a binomial(n,p) variable has mean np and variance np(1p): If X is Bernoulli(p), then E[X]=p and Var(X)=p(1p). If S is binomial(n,p), then E[S]=np,Var(S)=np(1p). These formulas include p=0, p=1, and n=0.

[F4]

Countably many independent copies of a prescribed law exist: Assume countable choice and dependent choice. Every probability measure ν on (S,Σ) is the common law of a countable independent family of S-valued random elements.

Verification

Given: The construction and assumptions above.

1.1

The Bernoulli law puts masses p and 1p at 1 and 0. Under countable choice and dependent choice, the countable-copy result constructs a common-space IID sequence with this law. Its mean is p and variance p(1p), including both endpoints.

F2F4F3given
2.1

The finite-variance IID weak law and its probability bound apply with μ=p and σ2=p(1p), giving the displayed estimate and convergence. If p=0 or p=1, all the variables equal p on the intersection of their countably many probability-one events, so every sample mean equals p there and the error probability is zero, even at n=1.

F1step 1.1algebra
ExampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

A nonidentical Bernoulli weak law

Example

Assume countable choice and dependent choice. Let independent Xk be Bernoulli(pk) with pk=1/4 for odd k and pk=3/4 for even k. Then for Sn=k=1nXk, ESnESnn2=316n,Sn/n1/2in probability. Thus a common distribution is not required.

Facts & Assumptions

[F1]

Chebyshev weak law for uncorrelated arrays: For each n1, let Xn,1,,Xn,rn be square-integrable real random variables on one probability space, pairwise uncorrelated within the row, where rn0 is finite. Set Sn=k=1rnXn,k and let bn>0 be deterministic. If vn:=bn2k=1rnVar(Xn,k)0, then (SnESn)/bn0 in L2 and in probability. More precisely, its second moment is vn, and its probability of absolute value at least ε>0 is at most vn/ε2. No independence between rows is required.

[F2]

Coordinate random elements of a countable product are independent: Under the measure of thm-countable-product-of-probability-spaces, the coordinate maps Xn(x)=xn have laws μn and are independent.

[F3]

Assuming countable and dependent choice, countable products of arbitrary probability spaces: Assume countable choice and dependent choice. For probability spaces (En,En,μn)nN there is a unique probability measure μ on CN such that, for every finite F, its F-coordinate marginal is nFμn.

[F4]

A Bernoulli(p) variable has mean p and variance p(1p); a binomial(n,p) variable has mean np and variance np(1p): If X is Bernoulli(p), then E[X]=p and Var(X)=p(1p). If S is binomial(n,p), then E[S]=np,Var(S)=np(1p). These formulas include p=0, p=1, and n=0.

[F5]

Expectations factor over finite products of independent random variables: Let n1, let X0,,Xn1 be independent real random variables on a common probability space, and let gi:RR be Borel measurable for each i<n. 1. If every gi is nonnegative, then E[i<ngi(Xi)]=i<nE[gi(Xi)] in [0,+]. 2. If every gi(Xi) is integrable, then i<ngi(Xi) is integrable and the same factorization holds in R.

Verification

Given: The construction and assumptions above.

1.1

Under countable choice and dependent choice, take the countable product of the two-point Bernoulli probability spaces with the prescribed pk (shift the product index by one). Its coordinates are independent with the required laws. Each has mean pk and variance 3/16. Independence gives zero mixed centered moments, hence zero off-diagonal covariances.

F3F2F4givenF5
2.1

The row weak law with the first n entries and normalizer n gives centered second moment n(3/16)/n2=3/(16n) and convergence in probability to zero. Meanwhile ESn/n=1/2 for even n and 1/21/(4n) for odd n, including n=1. For any ε>0 the deterministic error is eventually below ε/2, so the probability of Sn/n1/2>ε is at most the probability that the centered average exceeds ε/2 in absolute value, which tends to zero.

F1step 1.1algebra
ExampleConstruction: AI-adaptedVerification: AI-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Rademacher-series threshold

Example

Assume countable choice and dependent choice, and let (ϵn)n1 be independent fair signs taking exactly the values 1 and 1. For real α, the series n1ϵnnα converges almost surely exactly when α>1/2; if α1/2 it diverges almost surely. Its absolute series converges exactly when α>1.

Facts & Assumptions

[F1]

Kolmogorov three-series theorem: Let (Xn)n1 be independent real random variables and fix A>0. Put Yn=Xn1{XnA}. Then nXn converges almost surely if and only if all three conditions hold: nP(Xn>A)<,nEYn converges in R,nVar(Yn)<. The conditions hold for some A>0 if and only if they hold for every A>0. No moment assumption is imposed on the untruncated variables.

[F2]

Almost-sure convergence of an independent series is a zero-one event: Let (Xn)nN be an independent sequence of real random variables. Then the event {n=0Xn converges} has probability 0 or 1.

[F3]

Countably many independent copies of a prescribed law exist: Assume countable choice and dependent choice. Every probability measure ν on (S,Σ) is the common law of a countable independent family of S-valued random elements.

[F4]

The p-series for a real exponent p converges exactly when p is greater than one: For every real p, k11kp convergesp>1.

Verification

Given: The construction and assumptions above.

1.1

Under countable choice and dependent choice, construct IID copies of the probability law on the two-point space {1,1}, each point having mass 1/2. For α>0, the summands have magnitude nα1. At cutoff A=1, the tail probabilities and truncated means are zero, and the truncated variances are n2α.

F3givenalgebra
2.1

The three-series theorem and the real p-series test therefore give almost-sure convergence for α>1/2 and rule out probability-one convergence for 0<α1/2. In the latter range the convergence event is a tail event of an independent sequence, so its zero-one law forces its probability to be zero. In particular the boundary α=1/2 has the divergent harmonic variance series.

F1F4F2step 1.1
3.1

If α0, the magnitudes nα1 do not tend to zero at any point, so the partial sums cannot converge. Finally at every point the absolute series equals nnα, which converges exactly for α>1 by the p-series test. This also checks the absolute boundary α=1 and the term-test boundary α=0.

F4step 1.1algebra
ExampleConstruction: AI-adaptedVerification: AI-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Almost-sure conditional convergence

Example

Assume countable choice and dependent choice. On a probability space carrying independent fair signs ϵn{1,1}, the random harmonic series n1ϵn/n converges almost surely, but n1ϵn/n diverges at every sample point. By comparison the deterministic harmonic series diverges and its alternating version converges.

Facts & Assumptions

[F1]

Kolmogorov three-series theorem: Let (Xn)n1 be independent real random variables and fix A>0. Put Yn=Xn1{XnA}. Then nXn converges almost surely if and only if all three conditions hold: nP(Xn>A)<,nEYn converges in R,nVar(Yn)<. The conditions hold for some A>0 if and only if they hold for every A>0. No moment assumption is imposed on the untruncated variables.

[F2]

Countably many independent copies of a prescribed law exist: Assume countable choice and dependent choice. Every probability measure ν on (S,Σ) is the common law of a countable independent family of S-valued random elements.

[F3]

The p-series for a real exponent p converges exactly when p is greater than one: For every real p, k11kp convergesp>1.

[F4]

The alternating series test: if (bk) is nonincreasing with bk0 then k(1)kbk converges, the sum lies between any two consecutive partial sums, and the error after n terms is at most bn: Let (εk) be the alternating sequence of lem-alternating-sequence, that is the unique sequence of reals with ε0=1 and εk+1=εk, which is what is usually written εk=(1)k; let e and o be its even and odd index maps, so that εej=1, εoj=1, and every natural number is ej for exactly one j or oj for exactly one j. Let (bk) be a sequence of reals that is nonincreasing (def-monotone-sequence) and converges to 0 (def-real-limit); then bk0 for every k. Write tn:=k<nεkbk for the partial sums (def-series). Then: 1. the series εkbk converges; write L for its sum; 2. tejLtoj for every jN, and for every nN the sum L lies between the two consecutive partial sums tn and tn+1; 3. Ltnbn for every nN. Claim 3 is the error bound: the partial sum tn, which uses the n terms ε0b0,,εn1bn1, differs from the sum by at most the first term omitted. Only claim 1 is a corollary of thm-dirichlet-test. Claims 2 and 3 are not: they come from the interlacing of the even-index and odd-index partial sums, and that argument is carried out below rather than smuggled into the Dirichlet estimate, which produces no bracketing at all.

Verification

Given: The construction and assumptions above.

1.1

Under countable choice and dependent choice, use the countable-copy theorem for the fair law on {1,1}. At cutoff A=1, all summands Xn=ϵn/n are retained, including the first one. Their means are zero and their variances are 1/n2, whose sum is finite. Three-series therefore gives almost-sure convergence.

F2F3F1givenalgebra
2.1

At every point ϵn/n=1/n, and the harmonic p-series diverges. Thus on the probability-one convergence event the convergence is conditional. The same p-series test gives deterministic harmonic divergence. Apply the zero-based alternating-series test with bk=1/(k+1) to obtain convergence of n1(1)n1/n; bk decreases to zero and is nonnegative.

F3F4step 1.1algebra
ExampleConstruction: AI-generatedVerification: AI-generatedaudited 2026-09-07Open item page →

The three series impose separate conditions

Example

Assume countable choice and dependent choice. At cutoff A=1, each of the following independent-sequence constructions violates exactly one of the three-series conditions:

  1. For n2, let Xn=2 with probability 1/n and Xn=0 otherwise; set X1=0. Only the large-jump probability series diverges.
  2. Let Xn=1/n deterministically. Only the truncated mean series diverges.
  3. Let Xn=ϵn/n for independent fair signs. Only the truncated variance series diverges.

None of these series converges almost surely.

Facts & Assumptions

[F1]

Kolmogorov three-series theorem: Let (Xn)n1 be independent real random variables and fix A>0. Put Yn=Xn1{XnA}. Then nXn converges almost surely if and only if all three conditions hold: nP(Xn>A)<,nEYn converges in R,nVar(Yn)<. The conditions hold for some A>0 if and only if they hold for every A>0. No moment assumption is imposed on the untruncated variables.

[F2]

Second Borel-Cantelli lemma under pairwise independence: Let (An)nN be pairwise independent events with n=0P(An)=+. Then P(An i.o.)=1.

[F3]

The p-series for a real exponent p converges exactly when p is greater than one: For every real p, k11kp convergesp>1.

[F4]

Coordinate random elements of a countable product are independent: Under the measure of thm-countable-product-of-probability-spaces, the coordinate maps Xn(x)=xn have laws μn and are independent.

[F5]

Assuming countable and dependent choice, countable products of arbitrary probability spaces: Assume countable choice and dependent choice. For probability spaces (En,En,μn)nN there is a unique probability measure μ on CN such that, for every finite F, its F-coordinate marginal is nFμn.

Verification

Given: The construction and assumptions above.

1.1

Under countable choice and dependent choice, the countable product of the stated finite probability spaces constructs the first and third independent sequences; deterministic coordinates construct the second. In the first construction the zero truncations at 1 are all zero, so their mean and variance series vanish, but n2P(Xn>1)=n21/n=. The second Borel–Cantelli lemma gives infinitely many terms equal to 2 almost surely; hence the terms fail to tend to zero.

F5F4F3F2given
1.2

For the second construction, every term is retained at A=1, its variance is zero, and there are no large jumps. Its truncated mean series is n1/n=. Thus exactly the mean condition fails and its deterministic partial sums diverge.

F3givenalgebra
2.1

For the third construction every term, including n=1, is retained, there are no large jumps, and the means vanish. Its variance series is n1/n=. The three-series theorem rules out almost-sure convergence. Each construction therefore isolates exactly the claimed failed condition.

F3F1givenalgebra
CounterexampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

A macroscopic row term defeats averaging

Statement refuted

The assertion that independent centered rows automatically satisfy a weak law with normalization n is false. Assume countable choice and dependent choice. A witness is the row of length n defined by Xn,1=nϵn and Xn,k=0 for 2kn, where the ϵn are independent fair signs. With Sn=k=1nXn,k, Sn/n does not converge in probability to zero.

Facts & Assumptions

[F1]

Chebyshev weak law for uncorrelated arrays: For each n1, let Xn,1,,Xn,rn be square-integrable real random variables on one probability space, pairwise uncorrelated within the row, where rn0 is finite. Set Sn=k=1rnXn,k and let bn>0 be deterministic. If vn:=bn2k=1rnVar(Xn,k)0, then (SnESn)/bn0 in L2 and in probability. More precisely, its second moment is vn, and its probability of absolute value at least ε>0 is at most vn/ε2. No independence between rows is required.

[F2]

Convergence in probability: For real random variables (Xn) and X on one probability space, write XnX in probability when, for every ε>0, P(XnX>ε)0. This is precisely def-convergence-in-measure for the probability measure.

[F3]

Countably many independent copies of a prescribed law exist: Assume countable choice and dependent choice. Every probability measure ν on (S,Σ) is the common law of a countable independent family of S-valued random elements.

Counterexample

Given: The construction and assumptions above.

1.1

Under countable choice and dependent choice take IID fair signs using the countable-copy theorem. A row consisting of one random entry and constants is independent: any finite intersection of coordinate events reduces to the one nonconstant event or is empty. Each entry is centered and square-integrable. Its variance sum is n2, so the row weak law has normalized variance 1, not a quantity tending to zero.

F3F1givenalgebra
2.1

The sum is exactly Sn=nϵn. Thus P(Sn/n>1/2)=1 for every n1, contradicting the defining requirement for convergence in probability to zero. At n=1 the row has only its one random entry. This is an array witness and asserts no failure of the IID integrable weak law.

F2step 1.1algebra
CounterexampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Summable untruncated variances are not necessary

Statement refuted

It is false that almost-sure convergence of a series of independent centered square-integrable variables forces summability of their untruncated variances. Assume countable choice and dependent choice. Let X1=0 and for n2 take independent Xn with P(Xn=n)=P(Xn=n)=12n2,P(Xn=0)=11n2. Then nXn converges absolutely almost surely, although EXn=0 and Var(Xn)=1 for every n2.

Facts & Assumptions

[F1]

First Borel-Cantelli lemma for events: Let (An)nN be events in a probability space. If n=0P(An)<+, then P(An i.o.)=0. No independence hypothesis is needed.

[F2]

Coordinate random elements of a countable product are independent: Under the measure of thm-countable-product-of-probability-spaces, the coordinate maps Xn(x)=xn have laws μn and are independent.

[F3]

Assuming countable and dependent choice, countable products of arbitrary probability spaces: Assume countable choice and dependent choice. For probability spaces (En,En,μn)nN there is a unique probability measure μ on CN such that, for every finite F, its F-coordinate marginal is nFμn.

[F4]

The p-series for a real exponent p converges exactly when p is greater than one: For every real p, k11kp convergesp>1.

[F5]

Almost-sure convergence of a random series: For real random variables (Xn)n1, the series n1Xn converges almost surely if its partial sums Sn converge to a finite real limit on an event of probability one, as in def-almost-sure-convergence-of-random-variables. With S0=0 from def-partial-sums-and-sample-means, its convergence event is C=r1N1jiN{SjSi<1/r}. This is exactly the real Cauchy condition, with the indexing of thm-series-cauchy-criterion shifted by one. Measurable arithmetic makes every event in this countable expression measurable. For any fixed m, the union over N may be restricted to Nm; then each difference uses only Xm+1,Xm+2,. Thus C is in the tail sigma-algebra, without assuming independence. Under independence, cor-almost-sure-convergence-of-an-independent-series-is-a-zero-one-event gives P(C){0,1}. Set S=limnSn on C and S=0 off C. The functions 1CSn converge everywhere to S, so thm-sequential-suprema-infima-limsup-liminf-and-pointwise-limits-are-measurable and thm-arithmetic-and-lattice-operations-preserve-measurability make S measurable. For Borel sets Bn, the event {XnBn infinitely often}=mnm{XnBn} is likewise tail measurable. Changing finitely many summands adds an eventually constant finite difference to Sn; divided by deterministic cn>0 tending to infinity that difference tends to zero, so the normalized limsup is unchanged. The sign of the unnormalized limsup need not be unchanged: the all-zero sequence has limsup zero, while changing its first term to 1 makes the limsup of partial sums equal to 1.

Counterexample

Given: The construction and assumptions above.

1.1

Under countable choice and dependent choice take the countable product of these finite probability spaces. The specified masses are nonnegative and sum to one; its independent coordinates have the desired laws. For n2, direct finite expectation gives EXn=0 and EXn2=n2(1/n2)=1, hence variance 1. The first coordinate is zero.

F3F2givenalgebra
2.1

The sum n2P(Xn0)=n2n2 is finite. The first Borel–Cantelli lemma gives only finitely many nonzero terms almost surely. On that event the absolute sum is a finite sum of finite numbers, hence finite, and the original partial sums converge. But their untruncated variance sum is n21=. The example has no uniform bound on all summands.

F4F1F5step 1.1algebra
CounterexampleConstruction: AI-adaptedVerification: AI-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Cauchy averages admit no deterministic weak centering

Statement refuted

Assume countable choice and dependent choice. For IID real variables with density f(x)=1/[π(1+x2)] on R, there is no deterministic real sequence (μn) for which Sn/nμn0 in probability, where Sn=k=1nXk. Thus IID alone cannot guarantee a weak law even with varying deterministic centering.

Facts & Assumptions

[F1]

Exact tail criterion for a truncated-centered IID weak law: For IID real random variables (Xn)n1 and Sn=k=1nXk, there exist deterministic real constants (μn) with Sn/nμn0 in probability if and only if nP(X1>n)0. When this condition holds, μn=E[X11{X1n}] works. Neither existence of an untruncated mean nor convergence of (μn) is asserted.

[F2]

Countably many independent copies of a prescribed law exist: Assume countable choice and dependent choice. Every probability measure ν on (S,Σ) is the common law of a countable independent family of S-valued random elements.

[F3]

Law or distribution of a random element: Let X:(Ω,F,P)(S,Σ) be a random element. Its law or distribution is the set function PX:Σ[0,+],PX(B):=P(X1(B)). Thus the law of X records the probability of each measurable target set by pulling it back to an event in the original probability space.

[F4]

Probability laws correspond to distribution functions: Assume the Axiom of Countable Choice. 1. Let X be a real random variable, let PX be its law, and let FX(x)=P(Xx). Then FX is nondecreasing and right-continuous, satisfies limxFX(x)=0,limx+FX(x)=1, and obeys PX((a,b])=FX(b)FX(a)(a<b). 2. Conversely, if F:RR is nondecreasing and right-continuous with limxF(x)=0,limx+F(x)=1, then there is a unique Borel probability measure μ on R such that μ((a,b])=F(b)F(a)(a<b), equivalently F(x)=μ((,x])(xR).

Counterexample

Given: The construction and assumptions above.

1.1

The nonnegative density has total integral [arctanx/π]=1 and F(x)=1/2+arctan(x)/π is nondecreasing and continuous with limits zero and one. The distribution-function correspondence therefore supplies its Borel probability law; the fundamental theorem of calculus identifies its density as f. Under countable choice and dependent choice, construct IID copies with that law. Symmetry of the density gives P(X1>n)=(2/π)n(1+x2)1dx.

F3F2givenalgebraF4
2.1

For xn1, x2/(1+n2)(1+x2)1x2. Integrating and multiplying by n gives 2/[π(1+n2)]nP(X1>n)2/π. Thus this tail quantity tends to 2/π, not zero. Necessity in the truncated-centering criterion rules out every deterministic centering sequence.

F1step 1.1algebra
ExampleConstruction: AI-adaptedVerification: AI-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

An infinite-mean law requiring diverging centering

Example

Assume countable choice and dependent choice. Let Xe have survival function P(X>x)=1/(xlogx) for xe, with an atom of mass 11/e at e. For IID copies (Xk), the untruncated mean is infinite but Sn/nμn0in probability,μn=e+loglogn1logn(ne). Here μn=E[X1{Xn}]; for integer n<e set μn=0. More generally, the survival family 1/[x(logx)α] for xe, α0, has infinite second moment for every α, finite first moment exactly for α>1, and admits deterministic weak-law centering exactly for α>0.

Facts & Assumptions

[F1]

Exact tail criterion for a truncated-centered IID weak law: For IID real random variables (Xn)n1 and Sn=k=1nXk, there exist deterministic real constants (μn) with Sn/nμn0 in probability if and only if nP(X1>n)0. When this condition holds, μn=E[X11{X1n}] works. Neither existence of an untruncated mean nor convergence of (μn) is asserted.

[F2]

Countably many independent copies of a prescribed law exist: Assume countable choice and dependent choice. Every probability measure ν on (S,Σ) is the common law of a countable independent family of S-valued random elements.

[F3]

Layer-cake formulas for random variables: Let (Ω,F,P) be a probability space. 1. If X:Ω[0,+] is measurable, then E[X]=0P(X>t)dt, where the right-hand side may be +. 2. If X is an integrable real random variable, then E[X]=0P(X>t)dt0P(X<t)dt.

[F4]

Probability laws correspond to distribution functions: Assume the Axiom of Countable Choice. 1. Let X be a real random variable, let PX be its law, and let FX(x)=P(Xx). Then FX is nondecreasing and right-continuous, satisfies limxFX(x)=0,limx+FX(x)=1, and obeys PX((a,b])=FX(b)FX(a)(a<b). 2. Conversely, if F:RR is nondecreasing and right-continuous with limxF(x)=0,limx+F(x)=1, then there is a unique Borel probability measure μ on R such that μ((a,b])=F(b)F(a)(a<b), equivalently F(x)=μ((,x])(xR).

Verification

Given: The construction and assumptions above.

1.1

Define F(x)=0 for x<e and F(x)=11/[x(logx)α] for xe, where α0. It is nondecreasing and right-continuous with limits zero and one at the two infinities. Its jump at e is 11/e. The distribution-function theorem constructs its Borel probability law (using countable choice); under countable choice and dependent choice the countable-copy result constructs IID variables with it.

F4F2givenalgebra
2.1

Layer cake gives EX=e+e[x(logx)α]1dx=e+1uαdu. This is finite exactly for α>1, when it equals e+1/(α1). Applying layer cake to X2 and substituting t=x2 gives EX2=e2+2e(logx)αdx=: eventually (logx)αx, so the last integrand dominates 1/x. The comparison follows from uαeu for large u, for example by an exponential-series term of integer degree greater than α.

F3step 1.1algebra
2.2

For ne the tail quantity is nP(X>n)=(logn)α. It tends to zero exactly for α>0, whereas for α=0 it equals one. Both directions of the truncated-centering criterion therefore give exactly the asserted centering range, even in the infinite-mean cases 0<α1.

F1step 1.1algebra
3.1

For α=1 use the pointwise identity min(X,n)=X1{Xn}+n1{X>n}. Layer cake for the bounded minimum gives μn=0nP(X>t)dtnP(X>n)=e+loglogn1/logn for real ne. At n=e this is e1, exactly the contribution of the atom; below e the zero truncation is zero. These finite centers work by the preceding step, although EX= and μnloglogn.

F3step 2.1step 2.2algebra

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