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How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Weak Laws and Series of Independent Random Variables

1 · Prerequisites

2 · Summary

Weak laws arise first from variance control and then from truncating large summands. For independent series, first-crossing inequalities turn tail control into almost-sure convergence; symmetrization gives the necessity half of the three-series theorem. Kronecker summation connects convergent random series to normalized strong laws. The final results characterize deterministic weak-law centering and show that probability and almost-sure convergence agree for independent partial sums.

All variables are finite real measurable functions. Sums start at one with an empty initial sum of zero, truncation retains equality at its positive cutoff, and all almost-sure conclusions use one measurable probability-one event.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Identical distribution and IID families

Definition

Let (Xi)iI be random elements with the same measurable target (E,E). They are identically distributed if P(XiB)=P(XjB) for all i,jI and BE, that is, their laws in Law or distribution of a random element agree. They are independent and identically distributed (IID) if, in addition, the whole family is independent in Independent random elements. Independence means mutual independence, not merely pairwise independence. No moment assumption is part of either definition. The empty family satisfies these universal conditions vacuously.

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Partial sums, row sums and sample means

Definition

For real random variables X1,X2, on one probability space, define S0=0,Sn=k=1nXk,Xn=Sn/n(n1). For a triangular array with finite row length rn0, write Sn=k=1rnXn,k. An empty row has sum zero. These are finite sums in Finite sums and finite products, by recursion, with its index shifted by one. Each sum and each sample mean is a real random variable by Arithmetic and lattice operations preserve measurability whenever they are defined and Random elements and real random variables. No independence, common law, or integrability is implicit.

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

Chebyshev weak law for uncorrelated arrays

Statement

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.

Facts & Assumptions

[F1]

Variance and covariance identities for random variables: Let X,Y be square-integrable real random variables on one probability space. Then Var(X)=E[X2]E[X]2, Cov(X,Y)=E[XY]E[X]E[Y]. Moreover, covariance is symmetric and bilinear on finite linear combinations. On finite full-power-set probability spaces these formulas reduce to the published finite identities.

[F2]

Chebyshev's inequality for random variables: If X is a square-integrable real random variable and a>0, then P(XE[X]a)Var(X)a2.

[F3]

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.

[F4]

Lp convergence for random variables: Let 1p. For real random variables whose classes lie in Lp(P) as defined by def-l-p-space-as-a-quotient-by-null-functions, write XnX in Lp when [Xn][X]Lp(P)0. For p<, this norm is [Xn][X]Lp(P)=(EXnXp)1/p; for p=, it is the essential-supremum norm. Thus the assertion concerns almost-everywhere equivalence classes, not chosen representatives.

[F5]

Linearity, monotonicity, and the modulus bound for expectation: Let X,Y be integrable real or complex random variables on one probability space. 1. For scalars a,b, E[aX+bY]=aE[X]+bE[Y]. 2. If X and Y are real-valued and XY almost surely, then E[X]E[Y]. 3. E[X]E[X].

Proof

Given: The objects and hypotheses of the statement.

1.1

Put Zn=bn1k(Xn,kEXn,k)=(SnESn)/bn. Finite linearity gives EZn=0. Square integrability of the finite sum follows from (k=1rzk)2rkzk2 for r1; the empty sum is zero.

F5givenalgebra
2.1

Covariance bilinearity and the zero off-diagonal covariances give EZn2=bn2kVar(Xn,k)=vn. This includes a singleton row and zero variances without division by a variance. Thus the L2 norm tends to zero.

F1F4step 1.1algebra
3.1

For every ε>0, Chebyshev gives P(Znε)vn/ε20. This also bounds the strict event defining convergence in probability.

F2F3step 2.1
CorollaryStatement: AI-adaptedProof: AI-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

IID finite-variance weak law

Statement

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.

Facts & Assumptions

[F1]

Identical distribution and IID families: Let (Xi)iI be random elements with the same measurable target (E,E). They are identically distributed if P(XiB)=P(XjB) for all i,jI and BE, that is, their laws in def-law-or-distribution-of-a-random-element agree. They are independent and identically distributed (IID) if, in addition, the whole family is independent in def-independent-random-elements. Independence means mutual independence, not merely pairwise independence. No moment assumption is part of either definition. The empty family satisfies these universal conditions vacuously.

[F2]

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.

[F3]

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.

Proof

Given: The objects and hypotheses of the statement.

1.1

IID gives common mean and variance. For distinct indices, factorization of the integrable variables gives E(XiXj)=μ2, hence zero covariance. Thus the first n variables form an uncorrelated row.

F1F3given
2.1

Apply the row result with rn=n and bn=n. Its variance sum is nσ2, so it gives the displayed identity, both convergences, and the probability bound. This calculation holds for n=1 and for σ2=0.

F2step 1.1algebra
DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-09-07Open item page →

Zero truncation at a positive level

Definition

For a real random variable X and a deterministic level A>0, its zero truncation is X(A)=X1{XA}. The threshold event is measurable because X is measurable and [A,A] is Borel; its indicator and the product are measurable by Arithmetic and lattice operations preserve measurability whenever they are defined. Thus X(A) is a real random variable as in Random elements and real random variables. It equals X at both cutoff endpoints and is zero outside the interval. Since X(A)A, for every 0<p< its absolute pth moment is at most ApP(Ω)=Ap. This is not clipping to the endpoints.

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

Khinchin weak law for integrable IID variables

Statement

If (Xk)k1 are IID real random variables and EX1<, then, with Sn=k=1nXk and μ=EX1, ESn/nμ0. Consequently Sn/nμ in probability.

Facts & Assumptions

[F1]

Zero truncation at a positive level: For a real random variable X and a deterministic level A>0, its zero truncation is X(A)=X1{XA}. The threshold event is measurable because X is measurable and [A,A] is Borel; its indicator and the product are measurable by thm-arithmetic-and-lattice-operations-preserve-measurability. Thus X(A) is a real random variable as in def-random-element-and-real-random-variable. It equals X at both cutoff endpoints and is zero outside the interval. Since X(A)A, for every 0<p< its absolute pth moment is at most ApP(Ω)=Ap. This is not clipping to the endpoints.

[F2]

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.

[F3]

Measurable coordinatewise functions preserve independence: Let (Xi)iI be an independent family of random elements Xi:(Ω,F,P)(Si,Σi). For each i, let gi:(Si,Σi)(Ti,Ti) be measurable. Then the family (giXi)iI is independent.

[F4]

Dominated convergence: Let f and (fn) be measurable complex-valued functions such that fnf almost everywhere and fng almost everywhere for a single nonnegative measurable function g with gdμ<+. Then fL1(μ), fnfdμ0, and hence fndμfdμ.

[F5]

Markov's inequality for random variables: If X:Ω[0,+] is a nonnegative random variable on a probability space and a>0, then P(Xa)E[X]a.

[F6]

Finite-measure Lr includes into Lp for p<r: Let (X,A,μ) be a measure space with μ(X)<. 1. If 1p<r< and fLr(μ), then fLp(μ) and fpμ(X)1/p1/rfr. 2. If 1p< and fL(μ), then fLp(μ) and fpμ(X)1/pf.

[F7]

Linearity, monotonicity, and the modulus bound for expectation: Let X,Y be integrable real or complex random variables on one probability space. 1. For scalars a,b, E[aX+bY]=aE[X]+bE[Y]. 2. If X and Y are real-valued and XY almost surely, then E[X]E[Y]. 3. E[X]E[X].

Proof

Given: The objects and hypotheses of the statement.

1.1

Fix A>0 and set Yk=Xk(A), Rk=XkYk. The Yk are bounded IID variables: measurable transformations preserve independence and their laws remain equal by inverse images. The finite-variance result gives n1k(YkEYk)2=Var(Y1)/n.

F1F3F2given
2.1

On a probability space the L1 norm is at most the L2 norm. Finite linearity, the triangle inequality and the modulus bound give En1k(RkERk)2ER1. Therefore ESn/nμVar(Y1)/n+2E(X11{X1>A}).

F6F7step 1.1algebra
3.1

For each fixed A let n in this bound. Then let A run through positive integers tending to infinity. The residual is dominated by the integrable X1 and tends pointwise to zero, so dominated convergence makes the remaining bound tend to zero. Finally Markov applied to Sn/nμ proves convergence in probability. No division by a moment occurs, so constant or zero variables are included.

F4F5step 2.1
DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Almost-sure convergence of a random series

Definition

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 Almost-sure convergence of real random variables. With S0=0 from Partial sums, row sums and sample means, its convergence event is C=r1N1jiN{SjSi<1/r}. This is exactly the real Cauchy condition, with the indexing of A series converges iff for every ε>0 there is N with am+1++an<ε for all n>mN 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, 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 Sequential suprema, infima, limsup, liminf, and pointwise limits of measurable functions are measurable and Arithmetic and lattice operations preserve measurability whenever they are defined 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.

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

Kolmogorov maximal inequality

Statement

Let X1,,Xn be independent centered square-integrable real random variables, n1, and Sk=j=1kXj. For every λ>0, P(max1knSkλ)Var(Sn)λ2=j=1nVar(Xj)λ2. Thus controlling the whole finite maximum costs no larger bound than controlling the final sum by Chebyshev.

Facts & Assumptions

[F1]

Disjoint groups of an independent sigma-algebra family remain independent: Let (Fi)iI be an independent family of sigma-algebras on a probability space, and let J0,,Jm1I be pairwise disjoint index sets. For each r<m, define Gr:=σ(iJrFi). Then the sigma-algebras G0,,Gm1 are independent.

[F2]

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.

[F3]

Variance and covariance identities for random variables: Let X,Y be square-integrable real random variables on one probability space. Then Var(X)=E[X2]E[X]2, Cov(X,Y)=E[XY]E[X]E[Y]. Moreover, covariance is symmetric and bilinear on finite linear combinations. On finite full-power-set probability spaces these formulas reduce to the published finite identities.

[F4]

Arithmetic and lattice operations preserve measurability whenever they are defined: Let (X,A) be a measurable space and let f,g:XR be measurable. Then: 1. cf is measurable for every real scalar c; 2. max(f,g), min(f,g), f, f+, and f are measurable; 3. if f+g is pointwise defined, then f+g is measurable; 4. with the convention of rem-zero-times-infinity-convention-for-pointwise-products, the pointwise product fg is measurable.

Proof

Given: The objects and hypotheses of the statement.

1.1

Let Ak={Sj<λ (j<k), Skλ}. These measurable events are disjoint, and their union is the event in the statement. Measurability follows by finite arithmetic.

F4given
2.1

Grouping shows that Sk1Ak and SnSk are independent. Both are integrable (their squares have finite expectation), and the latter has mean zero. Factorization therefore gives E[Sk1Ak(SnSk)]=0. For k=n the tail is zero and the identity still holds.

F1F2step 1.1
3.1

Expand the square on Ak: E[Sn21Ak]=E[Sk21Ak]+E[(SnSk)21Ak]λ2P(Ak). Sum over the disjoint events. Centering gives ESn2=Var(Sn), and covariance bilinearity with factorization cancels every off-diagonal covariance. This proves the bound even when the variance is zero or n=1.

F3F2step 1.1step 2.1algebra
TheoremStatement: AI-adaptedProof: AI-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Kolmogorov convergence criterion

Statement

For independent centered square-integrable real random variables (Xn)n1, if n1Var(Xn)<, then n1Xn converges almost surely and in L2 to the same finite real random variable.

Facts & Assumptions

[F1]

Kolmogorov maximal inequality: Let X1,,Xn be independent centered square-integrable real random variables, n1, and Sk=j=1kXj. For every λ>0, P(max1knSkλ)Var(Sn)λ2=j=1nVar(Xj)λ2. Thus controlling the whole finite maximum costs no larger bound than controlling the final sum by Chebyshev.

[F2]

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.

[F3]

A series converges iff for every ε>0 there is N with am+1++an<ε for all n>mN: Let (ak) be a sequence of reals, with partial sums sn=k<nak (def-series). Then ak converges if and only if for every real ε>0 there is NN such that k=m+1nak<ε for all n>mN. The block k=m+1nak is the finite sum am+1++an of def-finite-sum, and it equals sn+1sm+1. This is the Cauchy criterion transported from sequences to series. Its value is that it decides convergence without producing, or even naming, the sum.

[F4]

Continuity from below for measures: Let (En)nN be an increasing sequence of measurable sets for a measure μ, so EnEn+1. Then μ(nNEn)=supnNμ(En). No finiteness hypothesis is required.

[F5]

Continuity from above when one set has finite measure: Let (En)nN be a decreasing sequence of measurable sets for a measure μ. If μ(En0)<+ for some n0, then μ(nNEn)=infnNμ(En).

[F6]

Variance and covariance identities for random variables: Let X,Y be square-integrable real random variables on one probability space. Then Var(X)=E[X2]E[X]2, Cov(X,Y)=E[XY]E[X]E[Y]. Moreover, covariance is symmetric and bilinear on finite linear combinations. On finite full-power-set probability spaces these formulas reduce to the published finite identities.

[F7]

Riesz-Fischer completeness of Lp for 1p: Let (X,A,μ) be a measure space and let 1p. Then Lp(μ), with the norm of thm-the-l-p-norm-descends-to-the-quotient-and-makes-l-p-a-normed-space, is complete. Equivalently, the metric induced by that norm is a complete metric in the sense of def-complete-metric-space. Moreover, if a sequence in Lp(μ) converges in norm, then some subsequence admits measurable representatives converging almost everywhere in the sense of def-convergence-almost-everywhere-relative-to-a-measure.

[F8]

Lp convergence implies convergence in probability: Let 1p<. If XnX in Lp, then XnX in probability.

[F9]

Almost-sure convergence implies convergence in probability: If XnX almost surely, then XnX in probability.

[F10]

Limits in probability are unique almost surely: If XnX and XnY in probability, then X=Y almost surely.

Proof

Given: The objects and hypotheses of the statement.

1.1

Write S0=0, Sn=k=1nXk, and vm=k>mVar(Xk). Applying the maximal inequality to each block Xm+1,,XN and then continuity from below gives P(supjmSjSm>t)vm/t2 for t>0. The strict supremum event is the increasing union of finite strict maximum events, each bounded by the corresponding non-strict estimate.

F1F4given
1.2

For n>m, the same variance expansion used in the maximal inequality gives ESnSm2=k=m+1nVar(Xk)vm0. Hence the classes of Sn are Cauchy in L2; completeness gives an L2 limit class with a finite measurable representative T0. Set T=ReT0. This is a finite measurable real variable, and SnTSnT0 pointwise, so SnT in L2 even if completeness was formulated over complex scalars.

F1F6F7given
2.1

Let wm=supi,jmSiSj. Its strict level events are countable unions of measurable events and decrease with m. Since wm2supjmSjSm, continuity from above gives P(m{wm>2/r})=0 for every integer r1. Outside the union of these null events, for each r some m has wm2/r; this is the real Cauchy condition. Completeness supplies a finite limit, extended measurably by zero as in the series definition.

F5F3F2step 1.1
3.1

The L2 convergence gives convergence in probability to T, and the almost-sure convergence gives convergence in probability to the limit S from the Cauchy event. Uniqueness gives S=T almost surely. These arguments allow all variances to vanish and finite tails to be identically zero.

F8F9F10step 2.1step 1.2
CorollaryStatement: AI-adaptedProof: AI-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Kolmogorov two-series sufficiency

Statement

Let (Xn)n1 be independent square-integrable real random variables. If n1EXn converges in R and n1Var(Xn)<, then n1Xn converges almost surely and in L2.

Facts & Assumptions

[F1]

Kolmogorov convergence criterion: For independent centered square-integrable real random variables (Xn)n1, if n1Var(Xn)<, then n1Xn converges almost surely and in L2 to the same finite real random variable.

[F2]

Measurable coordinatewise functions preserve independence: Let (Xi)iI be an independent family of random elements Xi:(Ω,F,P)(Si,Σi). For each i, let gi:(Si,Σi)(Ti,Ti) be measurable. Then the family (giXi)iI is independent.

[F3]

Variance and covariance identities for random variables: Let X,Y be square-integrable real random variables on one probability space. Then Var(X)=E[X2]E[X]2, Cov(X,Y)=E[XY]E[X]E[Y]. Moreover, covariance is symmetric and bilinear on finite linear combinations. On finite full-power-set probability spaces these formulas reduce to the published finite identities.

Proof

Given: The objects and hypotheses of the statement.

1.1

Set Yn=XnEXn. These are independent centered square-integrable variables, and Var(Yn)=Var(Xn) by covariance bilinearity. The convergence criterion supplies an almost-sure and L2 limit Y for their series.

F2F3F1given
2.1

Let a=nEXn. The identity k=1nXk=k=1nYk+k=1nEXk gives pointwise convergence to Y+a on the same event. Its L2 error is at most the centered L2 error plus k=1nEXka, which tends to zero. This includes zero variances and conditionally convergent deterministic means.

step 1.1givenalgebra
DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Symmetric real random variables

Definition

A real random variable X is symmetric if its law as defined in Law or distribution of a random element equals the law of X. Equivalently, P(XB)=P(XB) for every Borel BR, where B={b:bB}. No existence of an expectation is assumed in this definition. In particular atoms, including an atom at zero, are allowed.

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

Independent-copy symmetrization of random series

Statement

Given an independent sequence (Xn)n1 on (Ω,F,P), form the product probability space (Ω2,FF,PP). Write Un(ω,ω)=Xn(ω), Vn(ω,ω)=Xn(ω), and Zn=UnVn. Then (Un) and (Vn) are independent copies of the whole sequence, and the Zn are independent symmetric real random variables. Almost-sure convergence of nXn implies almost-sure convergence of nZn. If XnA almost surely for every n, with 0A<, then Zn2A almost surely, EZn=0, and Var(Zn)=2Var(Xn).

Facts & Assumptions

[F1]

Independent random elements: Let Xi:(Ω,F,P)(Si,Σi) be random elements on a common probability space. For each i, write σ(Xi):={Xi1(B):BΣi}F. The family (Xi)iI is independent when the sigma-algebras (σ(Xi))iI are independent in the sense of def-independent-sigma-algebras-and-events. When Σi is a Borel sigma-algebra, this agrees with the notation σ(Xi) from def-sigma-algebra-generated-by-a-function.

[F2]

Symmetric real random variables: A real random variable X is symmetric if its law as defined in def-law-or-distribution-of-a-random-element equals the law of X. Equivalently, P(XB)=P(XB) for every Borel BR, where B={b:bB}. No existence of an expectation is assumed in this definition. In particular atoms, including an atom at zero, are allowed.

[F3]

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.

[F4]

For sigma-finite factors, the product measure exists, has the rectangle formula, is sigma-finite, and is unique: Let (X,A,μ) and (Y,B,ν) be sigma-finite measure spaces. Then: 1. the set function E(μ×ν)(E) of def-product-measure-on-sigma-finite-spaces is a measure on AB; 2. for measurable rectangles, (μ×ν)(A×B)=μ(A)ν(B); 3. the measure μ×ν is sigma-finite; and 4. it is the unique measure on AB with the rectangle formula.

[F5]

Disjoint groups of an independent sigma-algebra family remain independent: Let (Fi)iI be an independent family of sigma-algebras on a probability space, and let J0,,Jm1I be pairwise disjoint index sets. For each r<m, define Gr:=σ(iJrFi). Then the sigma-algebras G0,,Gm1 are independent.

[F6]

Measurable coordinatewise functions preserve independence: Let (Xi)iI be an independent family of random elements Xi:(Ω,F,P)(Si,Σi). For each i, let gi:(Si,Σi)(Ti,Ti) be measurable. Then the family (giXi)iI is independent.

[F7]

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.

[F8]

Variance and covariance identities for random variables: Let X,Y be square-integrable real random variables on one probability space. Then Var(X)=E[X2]E[X]2, Cov(X,Y)=E[XY]E[X]E[Y]. Moreover, covariance is symmetric and bilinear on finite linear combinations. On finite full-power-set probability spaces these formulas reduce to the published finite identities.

Proof

Given: The objects and hypotheses of the statement.

1.1

Probability measures are finite and hence sigma-finite, so the two-factor product theorem applies. Its rectangle formula gives independent coordinate sigma-algebras with the original marginals. For any finite list of restrictions on the Ui and Vi, the rectangle formula followed by independence of the original Xi factors its probability into all the individual probabilities. Thus the combined family is independent.

F4F1given
2.1

Group each pair (Ui,Vi): distinct pair sigma-algebras are independent, so their measurable differences are independent. The pair law is the product of two equal marginal laws, invariant under exchanging coordinates (first on rectangles, then by product-measure uniqueness). Its difference therefore has the same law as its negative.

F5F6F4F2step 1.1
2.2

If C is the original probability-one convergence event, then C×C has probability one. On it the partial sums of Zn are differences of two convergent real sequences, hence converge finitely. This is almost-sure series convergence.

F3F4step 1.1algebra
3.1

Under the boundedness hypothesis, Zn2A and EUn=EVn, so EZn=0. Factorization gives E(UnVn)=(EXn)2. Expanding the square gives EZn2=2EXn22(EXn)2=2Var(Xn). This includes A=0 and deterministic laws.

F7F8step 1.1algebra
LemmaStatement: AI-adaptedProof: AI-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Bounded centered convergent series have summable variances

Statement

Let (Xn)n1 be independent centered real random variables with XnC almost surely for one finite constant C0. If nXn converges almost surely, then nVar(Xn)<. The bound is two-sided and uniform in n.

Facts & Assumptions

[F1]

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.

[F2]

Disjoint groups of an independent sigma-algebra family remain independent: Let (Fi)iI be an independent family of sigma-algebras on a probability space, and let J0,,Jm1I be pairwise disjoint index sets. For each r<m, define Gr:=σ(iJrFi). Then the sigma-algebras G0,,Gm1 are independent.

[F3]

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.

[F4]

Variance and covariance identities for random variables: Let X,Y be square-integrable real random variables on one probability space. Then Var(X)=E[X2]E[X]2, Cov(X,Y)=E[XY]E[X]E[Y]. Moreover, covariance is symmetric and bilinear on finite linear combinations. On finite full-power-set probability spaces these formulas reduce to the published finite identities.

[F5]

Continuity from below for measures: Let (En)nN be an increasing sequence of measurable sets for a measure μ, so EnEn+1. Then μ(nNEn)=supnNμ(En). No finiteness hypothesis is required.

[F6]

Linearity, monotonicity, and the modulus bound for expectation: Let X,Y be integrable real or complex random variables on one probability space. 1. For scalars a,b, E[aX+bY]=aE[X]+bE[Y]. 2. If X and Y are real-valued and XY almost surely, then E[X]E[Y]. 3. E[X]E[X].

Proof

Given: The objects and hypotheses of the statement.

1.1

Put S0=0 and Sn=k=1nXk. Almost every convergent path is bounded. The measurable events {supnSnl} for positive integers l increase to a probability-one event. Continuity from below supplies one integer l with probability δ>0. Set F0=Ω and Fn={maxknSkl}; thus P(Fn)δ.

F1F5given
2.1

The past event Fn1 and Sn1 are independent of Xn by grouping. All moments below are finite by boundedness. Expanding and factoring the cross term and the square term gives E[Sn21Fn1]=E[Sn121Fn1]+Var(Xn)P(Fn1). This also holds at n=1, where the past sum is zero.

F2F3F4step 1.1
3.1

On Fn1Fn, the triangle inequality and XnC give Snl+C almost surely. Split the expectation in the previous identity over Fn and Fn1Fn. Monotonicity yields δVar(Xn)E[Sn21Fn]E[Sn121Fn1]+(l+C)2P(Fn1Fn).

F6step 2.1algebra
4.1

Sum from 1 to N. The expectation differences telescope, the exit events are disjoint, and SN2l2 on FN. Thus δn=1NVar(Xn)l2+(l+C)2 for every N. The increasing nonnegative partial sums are bounded, so their series is finite. No division by C or a variance is used, and C=0 is included.

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

Necessity of the truncated mean and variance conditions

Statement

Let independent real random variables (Xn)n1 have nXn convergent almost surely. For every fixed A>0, set Yn=Xn(A)=Xn1{XnA}. Then nP(Xn>A)<,nVar(Yn)<, and the real numerical series nEYn converges.

Facts & Assumptions

[F1]

Independent-copy symmetrization of random series: Given an independent sequence (Xn)n1 on (Ω,F,P), form the product probability space (Ω2,FF,PP). Write Un(ω,ω)=Xn(ω), Vn(ω,ω)=Xn(ω), and Zn=UnVn. Then (Un) and (Vn) are independent copies of the whole sequence, and the Zn are independent symmetric real random variables. Almost-sure convergence of nXn implies almost-sure convergence of nZn. If XnA almost surely for every n, with 0A<, then Zn2A almost surely, EZn=0, and Var(Zn)=2Var(Xn).

[F2]

Bounded centered convergent series have summable variances: Let (Xn)n1 be independent centered real random variables with XnC almost surely for one finite constant C0. If nXn converges almost surely, then nVar(Xn)<. The bound is two-sided and uniform in n.

[F3]

Kolmogorov convergence criterion: For independent centered square-integrable real random variables (Xn)n1, if n1Var(Xn)<, then n1Xn converges almost surely and in L2 to the same finite real random variable.

[F4]

Zero truncation at a positive level: For a real random variable X and a deterministic level A>0, its zero truncation is X(A)=X1{XA}. The threshold event is measurable because X is measurable and [A,A] is Borel; its indicator and the product are measurable by thm-arithmetic-and-lattice-operations-preserve-measurability. Thus X(A) is a real random variable as in def-random-element-and-real-random-variable. It equals X at both cutoff endpoints and is zero outside the interval. Since X(A)A, for every 0<p< its absolute pth moment is at most ApP(Ω)=Ap. This is not clipping to the endpoints.

[F5]

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.

[F6]

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

[F7]

Measurable coordinatewise functions preserve independence: Let (Xi)iI be an independent family of random elements Xi:(Ω,F,P)(Si,Σi). For each i, let gi:(Si,Σi)(Ti,Ti) be measurable. Then the family (giXi)iI is independent.

[F8]

A series converges iff each of its tail series converges, and the sum splits as sN plus the N-th tail: Let (ak) be a sequence of reals with partial sums sn=k<nak, let NN, and let tj:=i<jaN+i be the partial sums of the N-th tail series kNak (def-series). Then: 1. tj=sj+NsN for every jN; 2. ak converges if and only if its N-th tail series converges, and in that case k=0ak  =  sN  +  k=Nak; 3. hence the following are equivalent: ak converges; every tail series of ak converges; some tail series of ak converges. In words: convergence of a series is a property of its terms from any index on, and changing finitely many terms changes the sum but not the fact of convergence.

Proof

Given: The objects and hypotheses of the statement.

1.1

Convergence of partial sums implies Xn0 on its probability-one event. Therefore {Xn>A} occurs only finitely often almost surely. These events are independent, by measurable transformations. If their probability sum were infinite, the second Borel–Cantelli lemma would instead make their infinitely-often event have probability one. Hence the sum is finite.

F7F6given
2.1

The Yn are independent and bounded by A. By the first Borel–Cantelli lemma, Yn=Xn eventually almost surely. Finite-change invariance, with the real-series indices shifted by one, gives almost-sure convergence of nYn.

F4F7F5F8step 1.1
3.1

Symmetrize this bounded sequence on the two-factor product. The differences Zn are independent, centered, bounded by 2A, and their series converges almost surely. The bounded-centered lemma gives nVar(Zn)<. Since Var(Zn)=2Var(Yn), the variance sum for Yn is finite.

F1F2step 2.1
4.1

The independent centered variables YnEYn now satisfy the convergence criterion. On the intersection of its probability-one event with that from the truncations, subtract the two convergent partial sums: their difference is the deterministic sequence k=1nEYk. That sequence therefore converges in R. A probability-one event is nonempty, and this argument selects only one path to establish a deterministic conclusion. All truncated expectations are finite, including when Yn=0.

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

Kolmogorov three-series theorem

Statement

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.

Facts & Assumptions

[F1]

Zero truncation at a positive level: For a real random variable X and a deterministic level A>0, its zero truncation is X(A)=X1{XA}. The threshold event is measurable because X is measurable and [A,A] is Borel; its indicator and the product are measurable by thm-arithmetic-and-lattice-operations-preserve-measurability. Thus X(A) is a real random variable as in def-random-element-and-real-random-variable. It equals X at both cutoff endpoints and is zero outside the interval. Since X(A)A, for every 0<p< its absolute pth moment is at most ApP(Ω)=Ap. This is not clipping to the endpoints.

[F2]

Necessity of the truncated mean and variance conditions: Let independent real random variables (Xn)n1 have nXn convergent almost surely. For every fixed A>0, set Yn=Xn(A)=Xn1{XnA}. Then nP(Xn>A)<,nVar(Yn)<, and the real numerical series nEYn converges.

[F3]

Kolmogorov two-series sufficiency: Let (Xn)n1 be independent square-integrable real random variables. If n1EXn converges in R and n1Var(Xn)<, then n1Xn converges almost surely and in L2.

[F4]

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.

[F5]

Measurable coordinatewise functions preserve independence: Let (Xi)iI be an independent family of random elements Xi:(Ω,F,P)(Si,Σi). For each i, let gi:(Si,Σi)(Ti,Ti) be measurable. Then the family (giXi)iI is independent.

[F6]

A series converges iff each of its tail series converges, and the sum splits as sN plus the N-th tail: Let (ak) be a sequence of reals with partial sums sn=k<nak, let NN, and let tj:=i<jaN+i be the partial sums of the N-th tail series kNak (def-series). Then: 1. tj=sj+NsN for every jN; 2. ak converges if and only if its N-th tail series converges, and in that case k=0ak  =  sN  +  k=Nak; 3. hence the following are equivalent: ak converges; every tail series of ak converges; some tail series of ak converges. In words: convergence of a series is a property of its terms from any index on, and changing finitely many terms changes the sum but not the fact of convergence.

Proof

Given: The objects and hypotheses of the statement.

1.1

If the original series converges almost surely, the necessity lemma gives all three numerical conditions for this arbitrary fixed A>0. In particular all truncated means and variances used here are finite.

F2F1given
1.2

Conversely suppose the three conditions hold. The bounded truncations are independent square-integrable variables. Two-series sufficiency makes nYn converge almost surely. The first Borel–Cantelli lemma makes Xn=Yn eventually almost surely; finite-change invariance then gives convergence of nXn. This also covers zero truncations and finite exceptional sets.

F1F5F3F4F6given
2.1

Conditions at one positive cutoff give convergence by the preceding direction; convergence gives the conditions at every positive cutoff by the first direction. Conditions at every positive cutoff give them at, for example, A=1. This is an equivalence between deterministic numerical conditions, and needs no intersection over uncountably many cutoff-dependent events.

step 1.1step 1.2algebra
LemmaStatement: AI-adaptedProof: AI-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Kronecker summation lemma

Statement

Let (xn)n1 be real and let 0<bn be deterministic, nondecreasing, and tend to infinity. If n1xn/bn converges in R, then 1bnk=1nxk0. Repeated values of bn are allowed.

Facts & Assumptions

[F1]

Abel summation by parts: with An=k<nak one has k<nakbk=Anbn1k<n1Ak+1(bk+1bk) for every n1: Let (ak) and (bk) be sequences of reals and let An  :=  k<nak(nN) be the partial sums of ak (def-series, def-finite-sum), so that A0=0 and ak=Ak+1Ak for every k. Then for every natural number n1 k<nakbk  =  Anbn1    k<n1Ak+1(bk+1bk). Both sides are finite sums in the sense of def-finite-sum; at n=1 the right-hand sum is empty and the identity reads a0b0=A1b0. The hypothesis n1 is what makes the statement legitimate, not merely convenient: the index n1 occurs on the right, and n1 is a natural number exactly when n1. At n=0 there is nothing to state, both the left-hand side and A0 being 0.

Proof

Given: The objects and hypotheses of the statement.

1.1

Put t0=b0=0 and tn=k=1nxk/bkt. Abel summation, shifted from its zero-based indices, gives bn1k=1nxk=tnk=1n(bkbk1)tk1/bn. One can verify the same identity by substituting xk=bk(tktk1) and telescoping; for n=1 it reads x1/b1=t1.

F1givenalgebra
2.1

The weights wn,k=(bkbk1)/bn are nonnegative and sum to one. For ε>0, take M such that tjt<ε for jM. With K=max0j<Mtjt, the weighted average differs from t by at most KbM/bn+ε for n>M. The first term tends to zero; no division by K is required, even if it vanishes. Repeated normalizers merely give zero weights.

step 1.1givenalgebra
3.1

As ε is arbitrary the weighted average tends to t. Subtracting it from tnt in the finite identity proves the assertion, including the zero sequence.

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

Strong law under summable normalized variances

Statement

Let (Xn)n1 be independent square-integrable real random variables. Let 0<bn be deterministic and nondecreasing with bn. If n1Var(Xn)bn2<, then 1bnk=1n(XkEXk)0almost surely. In particular, for IID centered square-integrable variables and any ε>0, Sn/[n(logn)1/2+ε]0 almost surely (the displayed normalization is used for n2).

Facts & Assumptions

[F1]

Kolmogorov convergence criterion: For independent centered square-integrable real random variables (Xn)n1, if n1Var(Xn)<, then n1Xn converges almost surely and in L2 to the same finite real random variable.

[F2]

Kronecker summation lemma: Let (xn)n1 be real and let 0<bn be deterministic, nondecreasing, and tend to infinity. If n1xn/bn converges in R, then 1bnk=1nxk0. Repeated values of bn are allowed.

[F3]

Measurable coordinatewise functions preserve independence: Let (Xi)iI be an independent family of random elements Xi:(Ω,F,P)(Si,Σi). For each i, let gi:(Si,Σi)(Ti,Ti) be measurable. Then the family (giXi)iI is independent.

[F4]

Variance and covariance identities for random variables: Let X,Y be square-integrable real random variables on one probability space. Then Var(X)=E[X2]E[X]2, Cov(X,Y)=E[XY]E[X]E[Y]. Moreover, covariance is symmetric and bilinear on finite linear combinations. On finite full-power-set probability spaces these formulas reduce to the published finite identities.

[F5]

The integral test: for f0 nonincreasing on [0,), kf(k) converges if and only if the sequence (0Nf)N is bounded, with 0Nfk<Nf(k)f(0)+0Nf: For nonnegative nonincreasing f on [0,), the series k0f(k) converges if and only if the proper integrals 0Nf are bounded above as integers N vary.

Proof

Given: The objects and hypotheses of the statement.

1.1

The variables Yn=(XnEXn)/bn are independent, centered, and square-integrable, with variances Var(Xn)/bn2. Measurable transformations give independence, and covariance bilinearity gives the variance identity. The convergence criterion makes nYn converge on one probability-one event.

F3F4F1given
2.1

On each path in that event apply Kronecker to xn=XnEXn and the given bn. The deterministic conclusion is precisely the asserted normalized convergence. Zero variances and repeated positive normalizers present no exception.

F2step 1.1
3.1

For the rate assertion, set bn=n(logn)1/2+ε for n2 and choose b1=b2. These are positive and nondecreasing. The variance sum from n=2 is σ2n2[n(logn)1+2ε]1. Apply the zero-based integral test to f(t)=[(t+2)(log(t+2))1+2ε]1 on [0,): it is nonnegative and decreasing, and substitution u=log(t+2) bounds its integrals by log2u12εdu=(log2)2ε/(2ε). The first variance term is finite. The result already proved therefore gives the rate, also when σ2=0.

F5step 1.1step 2.1algebra
CorollaryStatement: AI-adaptedProof: AI-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Strong law for independent nonidentical variables

Statement

For independent square-integrable real random variables (Xn)n1, the condition n1Var(Xn)/n2< implies SnESnn0almost surely,Sn=k=1nXk. The variables need not have a common law or a common mean.

Facts & Assumptions

[F1]

Strong law under summable normalized variances: Let (Xn)n1 be independent square-integrable real random variables. Let 0<bn be deterministic and nondecreasing with bn. If n1Var(Xn)bn2<, then 1bnk=1n(XkEXk)0almost surely. In particular, for IID centered square-integrable variables and any ε>0, Sn/[n(logn)1/2+ε]0 almost surely (the displayed normalization is used for n2).

Proof

Given: The objects and hypotheses of the statement.

1.1

Take bn=n for n1. This sequence is positive, increasing, and tends to infinity, and the required normalized variance series is exactly the one in the hypothesis.

givenalgebra
2.1

The normalized-variance strong law gives n1k=1n(XkEXk)0 almost surely. Finite linearity identifies the numerator with SnESn. This includes zero variance and deterministic variables and requires no relation between different means.

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One-sided maximal inequality for symmetric independent sums

Statement

For independent symmetric real random variables X1,,Xn, n1, let Sk=j=1kXj. For every real a, P(max1knSk>a)2P(Sn>a). Consequently for every t>0, P(max1knSk>t)2P(Sn>t). No moment assumptions are needed.

Facts & Assumptions

[F1]

Symmetric real random variables: A real random variable X is symmetric if its law as defined in def-law-or-distribution-of-a-random-element equals the law of X. Equivalently, P(XB)=P(XB) for every Borel BR, where B={b:bB}. No existence of an expectation is assumed in this definition. In particular atoms, including an atom at zero, are allowed.

[F2]

Disjoint groups of an independent sigma-algebra family remain independent: Let (Fi)iI be an independent family of sigma-algebras on a probability space, and let J0,,Jm1I be pairwise disjoint index sets. For each r<m, define Gr:=σ(iJrFi). Then the sigma-algebras G0,,Gm1 are independent.

[F3]

Measurable coordinatewise functions preserve independence: Let (Xi)iI be an independent family of random elements Xi:(Ω,F,P)(Si,Σi). For each i, let gi:(Si,Σi)(Ti,Ti) be measurable. Then the family (giXi)iI is independent.

[F4]

Independent random elements have product joint law: Let n1, and let Xi:(Ω,F,P)(Si,Σi) for i<n be independent random elements. Define X=(X0,,Xn1):Ωi<nSi. Then X is a random element of (i<nSi,i<nΣi), and its law is the finite product of the marginal laws: PX=i<nPXi.

Proof

Given: The objects and hypotheses of the statement.

1.1

Let Ak={Sja (j<k), Sk>a}. They partition the crossing event. The unused tail Rk=SnSk is independent of the past by grouping. Its law is symmetric: the independent marginal laws are unchanged when each remaining variable is negated, so their sum has the same law as its negative. Hence P(Rk0)1/2, including Rn=0.

F2F3F1givenF4
2.1

On Ak{Rk0} one has Sn>a. Independence gives P(Ak{Rk0})P(Ak)/2. These events are disjoint over k, so summing proves the one-sided assertion. The argument works unchanged at a=0 and at negative a.

step 1.1algebra
3.1

Apply that assertion to (Xj) and (Xj). The event of a strict absolute crossing of t is contained in the union of a positive and a negative crossing. Their final events {Sn>t} and {Sn>t} are disjoint for t>0, giving the displayed two-sided bound. Atoms at thresholds do not enter the strict events.

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

Tail comparisons under independent-copy symmetrization

Statement

Let X be an independent copy of a real random variable X. For every t>0, P(XX>t)2P(X>t/2). There exists a finite M0 with P(XM)1/2; for every such M, P(XX>t)12P(X>t+M).

Facts & Assumptions

[F1]

Independent-copy symmetrization of random series: Given an independent sequence (Xn)n1 on (Ω,F,P), form the product probability space (Ω2,FF,PP). Write Un(ω,ω)=Xn(ω), Vn(ω,ω)=Xn(ω), and Zn=UnVn. Then (Un) and (Vn) are independent copies of the whole sequence, and the Zn are independent symmetric real random variables. Almost-sure convergence of nXn implies almost-sure convergence of nZn. If XnA almost surely for every n, with 0A<, then Zn2A almost surely, EZn=0, and Var(Zn)=2Var(Xn).

[F2]

Continuity from below for measures: Let (En)nN be an increasing sequence of measurable sets for a measure μ, so EnEn+1. Then μ(nNEn)=supnNμ(En). No finiteness hypothesis is required.

Proof

Given: The objects and hypotheses of the statement.

1.1

The triangle inequality gives {XX>t}{X>t/2}{X>t/2}. The union bound and equality of the two marginal laws give the upper estimate. Independent copies can be realized on the two-factor product described by symmetrization.

F1givenalgebra
2.1

The intervals [m,m] increase to R as positive integers m increase. Continuity from below gives P(Xm)1, so there is a suitable finite M. For any such M, the event {X>t+M, XM} implies XX>t. Independence makes its probability P(X>t+M)P(XM), giving the lower bound. This includes M=0 when allowed by the law.

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Truncation weak law for independent arrays

Statement

For each n1 let Xn,1,,Xn,rn be independent real random variables on one probability space, with finite rn0. Let deterministic bn>0 tend to infinity and set Yn,k=Xn,k1{Xn,kbn}. If k=1rnP(Xn,k>bn)0,bn2k=1rnVar(Yn,k)0, then k=1rnXn,kk=1rnEYn,kbn0in probability. No independence between rows is required.

Facts & Assumptions

[F1]

Zero truncation at a positive level: For a real random variable X and a deterministic level A>0, its zero truncation is X(A)=X1{XA}. The threshold event is measurable because X is measurable and [A,A] is Borel; its indicator and the product are measurable by thm-arithmetic-and-lattice-operations-preserve-measurability. Thus X(A) is a real random variable as in def-random-element-and-real-random-variable. It equals X at both cutoff endpoints and is zero outside the interval. Since X(A)A, for every 0<p< its absolute pth moment is at most ApP(Ω)=Ap. This is not clipping to the endpoints.

[F2]

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.

[F3]

Measurable coordinatewise functions preserve independence: Let (Xi)iI be an independent family of random elements Xi:(Ω,F,P)(Si,Σi). For each i, let gi:(Si,Σi)(Ti,Ti) be measurable. Then the family (giXi)iI is independent.

[F4]

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.

[F5]

Finite and countable subadditivity of measures: Let μ be a measure and let (Ek)kN be measurable. Then μ(kNEk)k=0μ(Ek). For every mN one also has μ(k<mEk)k<mμ(Ek), including m=0, where both sides are 0.

[F6]

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.

Proof

Given: The objects and hypotheses of the statement.

1.1

Each truncated row is independent by measurable transformations and bounded by bn. Its distinct centered mixed moments vanish: independence factors expectations of bounded products, so the row is uncorrelated. The row weak law therefore makes bn1k(Yn,kEYn,k) tend to zero in probability.

F1F3F2givenF6
2.1

Let En=k=1rn{Xn,k>bn}. Outside En the original and truncated sums agree. Hence for ε>0 the probability of the claimed error exceeding ε is bounded by kP(Xn,k>bn) plus the corresponding centered truncated probability. Both tend to zero. For empty rows all sums and the union are zero or empty, so the argument includes them; equality at the cutoff is retained.

F5F4step 1.1givenalgebra
LemmaStatement: AI-adaptedProof: AI-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Vanishing tail control bounds truncated second moments

Statement

Let X be a real random variable with nP(X>n)0 as positive integers n. Then xP(X>x)0 for real x, and E[X21{Xn}]n0. Moreover EXp< for every 0<p<1.

Facts & Assumptions

[F1]

Zero truncation at a positive level: For a real random variable X and a deterministic level A>0, its zero truncation is X(A)=X1{XA}. The threshold event is measurable because X is measurable and [A,A] is Borel; its indicator and the product are measurable by thm-arithmetic-and-lattice-operations-preserve-measurability. Thus X(A) is a real random variable as in def-random-element-and-real-random-variable. It equals X at both cutoff endpoints and is zero outside the interval. Since X(A)A, for every 0<p< its absolute pth moment is at most ApP(Ω)=Ap. This is not clipping to the endpoints.

[F2]

For 0 < p < infinity, the layer-cake formula computes the integral of |f|^p from the distribution function: Let (X,A,μ) be a measure space, let f:XC be measurable, and let 0<p<. Then Xfpdμ=p0tp1μ({f>t})dt=p0tp1Af(t)dt, where either side may be +.

Proof

Given: The objects and hypotheses of the statement.

1.1

For x1, put m=x. Monotonicity of the tail gives xP(X>x)(m+1)P(X>m), which tends to zero. Consequently h(t)=tP(X>t) is bounded on [0,) and tends to zero.

givenalgebra
2.1

Apply layer cake with exponent 2 to X1{Xn}. Its tail is at most that of X for t<n and is zero for tn. Thus its second moment is at most 20nh(t)dt. If h(t)ε for tM, division by n bounds this by 2n10Mh(t)dt+2ε for nM. Let n then ε0. No moment assumption on the untruncated square was used.

F1F2step 1.1algebra
3.1

For 0<p<1, layer cake gives EXp=p0tp1P(X>t)dt. On (0,1) this is at most p01tp1dt=1. If hK, the remaining integral is at most pK1tp2dt=pK/(1p)<. This also covers X=0 and bounded laws. Neither endpoint p=0 nor p=1 is asserted.

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

Largest-summand bound for independent symmetric variables

Statement

Let Y1,,Yn be independent symmetric real random variables, n1, and Sn=k=1nYk. For t>0, P(Snt)12P(maxknYkt). The same bound holds when both inequalities inside the probabilities are strict. If the Yk are IID and p=P(Y1>t), then P(Sn>t)12(1(1p)n)12(1enp).

Facts & Assumptions

[F1]

Symmetric real random variables: A real random variable X is symmetric if its law as defined in def-law-or-distribution-of-a-random-element equals the law of X. Equivalently, P(XB)=P(XB) for every Borel BR, where B={b:bB}. No existence of an expectation is assumed in this definition. In particular atoms, including an atom at zero, are allowed.

[F2]

Independent random elements have product joint law: Let n1, and let Xi:(Ω,F,P)(Si,Σi) for i<n be independent random elements. Define X=(X0,,Xn1):Ωi<nSi. Then X is a random element of (i<nSi,i<nΣi), and its law is the finite product of the marginal laws: PX=i<nPXi.

[F3]

Arithmetic and lattice operations preserve measurability whenever they are defined: Let (X,A) be a measurable space and let f,g:XR be measurable. Then: 1. cf is measurable for every real scalar c; 2. max(f,g), min(f,g), f, f+, and f are measurable; 3. if f+g is pointwise defined, then f+g is measurable; 4. with the convention of rem-zero-times-infinity-convention-for-pointwise-products, the pointwise product fg is measurable.

Proof

Given: The objects and hypotheses of the statement.

1.1

On Rn let Bj be the Borel set where j is the least index attaining the largest coordinate magnitude. The sets Bj partition the space and are invariant under flipping the sign of coordinate j. The product joint law and symmetry of each marginal make that sign flip measure preserving. Ties and zero coordinates are included by the least-index rule.

F2F1F3given
2.1

Fix j, and write R=kjyk. Since 2yj=(R+yj)(Ryj)R+yj+Ryj, at least one of R+yj,Ryj is at least yj. On Bj{yjt} the two indicators of a final magnitude at least t, before and after the sign flip, therefore sum to at least one. Integrate using flip invariance to obtain 2P(Bj{Snt})P(Bj{Yjt}). The identical argument on yj>t uses strict final events and proves their version directly.

step 1.1algebra
3.1

Summing over j gives both bounds. Under IID, independence gives P(maxkYk>t)=1(1p)n. Finally 1pep for 0p1, so (1p)nenp. This includes p=0, p=1, and n=1.

F2step 2.1algebra
TheoremStatement: AI-adaptedProof: AI-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Exact tail criterion for a truncated-centered IID weak law

Statement

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.

Facts & Assumptions

[F1]

Identical distribution and IID families: Let (Xi)iI be random elements with the same measurable target (E,E). They are identically distributed if P(XiB)=P(XjB) for all i,jI and BE, that is, their laws in def-law-or-distribution-of-a-random-element agree. They are independent and identically distributed (IID) if, in addition, the whole family is independent in def-independent-random-elements. Independence means mutual independence, not merely pairwise independence. No moment assumption is part of either definition. The empty family satisfies these universal conditions vacuously.

[F2]

Independent-copy symmetrization of random series: Given an independent sequence (Xn)n1 on (Ω,F,P), form the product probability space (Ω2,FF,PP). Write Un(ω,ω)=Xn(ω), Vn(ω,ω)=Xn(ω), and Zn=UnVn. Then (Un) and (Vn) are independent copies of the whole sequence, and the Zn are independent symmetric real random variables. Almost-sure convergence of nXn implies almost-sure convergence of nZn. If XnA almost surely for every n, with 0A<, then Zn2A almost surely, EZn=0, and Var(Zn)=2Var(Xn).

[F3]

Tail comparisons under independent-copy symmetrization: Let X be an independent copy of a real random variable X. For every t>0, P(XX>t)2P(X>t/2). There exists a finite M0 with P(XM)1/2; for every such M, P(XX>t)12P(X>t+M).

[F4]

One-sided maximal inequality for symmetric independent sums: For independent symmetric real random variables X1,,Xn, n1, let Sk=j=1kXj. For every real a, P(max1knSk>a)2P(Sn>a). Consequently for every t>0, P(max1knSk>t)2P(Sn>t). No moment assumptions are needed.

[F5]

Truncation weak law for independent arrays: For each n1 let Xn,1,,Xn,rn be independent real random variables on one probability space, with finite rn0. Let deterministic bn>0 tend to infinity and set Yn,k=Xn,k1{Xn,kbn}. If k=1rnP(Xn,k>bn)0,bn2k=1rnVar(Yn,k)0, then k=1rnXn,kk=1rnEYn,kbn0in probability. No independence between rows is required.

[F6]

Vanishing tail control bounds truncated second moments: Let X be a real random variable with nP(X>n)0 as positive integers n. Then xP(X>x)0 for real x, and E[X21{Xn}]n0. Moreover EXp< for every 0<p<1.

[F7]

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.

[F8]

Finite and countable subadditivity of measures: Let μ be a measure and let (Ek)kN be measurable. Then μ(kNEk)k=0μ(Ek). For every mN one also has μ(k<mEk)k<mμ(Ek), including m=0, where both sides are 0.

[F9]

Largest-summand bound for independent symmetric variables: Let Y1,,Yn be independent symmetric real random variables, n1, and Sn=k=1nYk. For t>0, P(Snt)12P(maxknYkt). The same bound holds when both inequalities inside the probabilities are strict. If the Yk are IID and p=P(Y1>t), then P(Sn>t)12(1(1p)n)12(1enp).

Proof

Given: The objects and hypotheses of the statement.

1.1

Assume the tail condition and take row Xn,k=Xk for 1kn, with bn=n. The sum of row tail probabilities is nP(X1>n)0. The normalized truncated variance sum is at most n1E[X121{X1n}]0 by the second-moment lemma. The truncated array law yields the claimed convergence with the explicit finite μn.

F6F5F1given
1.2

For necessity suppose constants μn give the convergence. On the two-factor product take an independent copy Xk and let Zk=XkXk, Tn=k=1nZk. The Zk are IID and symmetric. The triangle and union bounds give P(Tn>εn)2P(Snnμn>εn/2)0. The deterministic center cancels exactly.

F2F8F7given
2.1

For pn=P(Z1>εn), the largest-summand bound gives P(Tn>εn)(1enpn)/2. Since the left side tends to zero and npn0, necessarily npn0; otherwise a positive lower bound along a subsequence would keep the right side away from zero.

F9step 1.2algebra
3.1

Choose finite M0 with P(X1M)1/2. The symmetrization lower bound with t=n/2 gives P(Z1>n/2)P(X1>n/2+M)/2P(X1>n)/2 for n2M. Multiply by n and use the previous step with ε=1/2. This proves necessity, including atomic or deterministic laws.

F3step 2.1algebra
4.1

An alternative check of the maximal step uses Zk=TkTk1 with T0=0, whence maxkZk2maxkTk. The symmetric maximal inequality gives P(maxkZk>2εn)2P(Tn>εn)0. Independence then gives 1(1P(Z1>2εn))n0, hence again nP(Z1>2εn)0 by 1pep. Both checks retain strict events and allow atoms.

F4step 1.2algebra
LemmaStatement: AI-adaptedProof: AI-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Levy maximal bound from uniform tail bounds

Statement

Let X1,,Xn be independent real random variables, n1, with Sk=j=1kXj. Let l>0 and 0δ<1. If P(j=inXjl/2)δ(1in), then P(maxknSkl)δ1δ. No centering or moment assumption is required.

Facts & Assumptions

[F1]

Disjoint groups of an independent sigma-algebra family remain independent: Let (Fi)iI be an independent family of sigma-algebras on a probability space, and let J0,,Jm1I be pairwise disjoint index sets. For each r<m, define Gr:=σ(iJrFi). Then the sigma-algebras G0,,Gm1 are independent.

[F2]

Arithmetic and lattice operations preserve measurability whenever they are defined: Let (X,A) be a measurable space and let f,g:XR be measurable. Then: 1. cf is measurable for every real scalar c; 2. max(f,g), min(f,g), f, f+, and f are measurable; 3. if f+g is pointwise defined, then f+g is measurable; 4. with the convention of rem-zero-times-infinity-convention-for-pointwise-products, the pointwise product fg is measurable.

Proof

Given: The objects and hypotheses of the statement.

1.1

Let Ak={Sj<l (j<k), Skl} and A=kAk. These are measurable disjoint first-crossing events. For k<n, Ak is independent of the remaining tail Rk=SnSk by grouping. For k=n, Rn=0, so its probability of magnitude at least l/2 is zero.

F2F1given
2.1

On Ak{Snl/2}, the triangle inequality gives Rkl/2. Thus P(A{Snl/2})kP(Ak)P(Rkl/2)δP(A). The complementary part has probability at most P(Sn>l/2)δ, using the hypothesis at i=1. Hence (1δ)P(A)δ, and division by the positive 1δ proves the assertion. This includes δ=0 and n=1.

step 1.1givenalgebra
LemmaStatement: AI-adaptedProof: AI-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Cauchy sequences in probability have a measurable limit

Statement

Let (Yn)n1 be real random variables on one probability space. Suppose that for every ε,η>0 there is N such that P(YnYm>ε)<η(n,mN). Then there is a finite measurable real random variable Y such that YnY in probability.

Facts & Assumptions

[F1]

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.

[F2]

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.

[F3]

A series converges iff for every ε>0 there is N with am+1++an<ε for all n>mN: Let (ak) be a sequence of reals, with partial sums sn=k<nak (def-series). Then ak converges if and only if for every real ε>0 there is NN such that k=m+1nak<ε for all n>mN. The block k=m+1nak is the finite sum am+1++an of def-finite-sum, and it equals sn+1sm+1. This is the Cauchy criterion transported from sequences to series. Its value is that it decides convergence without producing, or even naming, the sum.

[F4]

Sequential suprema, infima, limsup, liminf, and pointwise limits of measurable functions are measurable: Let (X,A) be a measurable space and let fn:XR be measurable for every nN. Then the functions supnfn,infnfn,lim supnfn,lim infnfn are measurable. The set {x:limnfn(x) exists in R} is measurable. In particular, if fnf pointwise, then f is measurable.

[F5]

Almost-sure convergence implies convergence in probability: If XnX almost surely, then XnX in probability.

[F6]

Finite and countable subadditivity of measures: Let μ be a measure and let (Ek)kN be measurable. Then μ(kNEk)k=0μ(Ek). For every mN one also has μ(k<mEk)k<mμ(Ek), including m=0, where both sides are 0.

Proof

Given: The objects and hypotheses of the statement.

1.1

Set n0=1. For each k1, choose recursively the least integer nk>nk1 for which all pairs of indices at least nk have probability less than 2k of separation exceeding 2k. Such an integer exists by the hypothesis. Thus, for every k1, P(Ynk+1Ynk>2k)<2k. The first Borel–Cantelli lemma gives a measurable probability-one event where these inequalities fail only finitely often.

F2given
2.1

On that event the series of absolute successive differences is finite: its finite initial part is finite because all values are real, and its remaining part is bounded by a geometric series. Therefore the subsequence is Cauchy and has a finite real limit. Its finite convergence event is measurable: intersect the measurable extended-limit event with {supkYnk<}, a countable union of countable intersections. Define Y to be this limit there and zero outside. The corresponding restricted sequence converges everywhere, so measurable limits give a real random variable.

F3F4step 1.1algebra
3.1

This subsequence converges almost surely and hence in probability to Y. Fix ε,η>0, choose N so late-pair errors at ε/2 are below η/2, and then choose k with nkN and P(YnkY>ε/2)<η/2. For all nN, the triangle and union bounds give P(YnY>ε)<η. This proves convergence of the full sequence, including constant sequences.

F5F6F1step 2.1givenalgebra
TheoremStatement: AI-adaptedProof: AI-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Convergence in probability and almost surely agree for independent series

Statement

For partial sums Sn=k=1nXk of independent real random variables (Xn)n1 on one probability space, the following are equivalent: (Sn) is Cauchy in probability; (Sn) converges in probability to a finite real random variable; (Sn) converges almost surely to a finite real random variable. The probability and almost-sure limits agree almost surely.

Facts & Assumptions

[F1]

Levy maximal bound from uniform tail bounds: Let X1,,Xn be independent real random variables, n1, with Sk=j=1kXj. Let l>0 and 0δ<1. If P(j=inXjl/2)δ(1in), then P(maxknSkl)δ1δ. No centering or moment assumption is required.

[F2]

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.

[F3]

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.

[F4]

Continuity from below for measures: Let (En)nN be an increasing sequence of measurable sets for a measure μ, so EnEn+1. Then μ(nNEn)=supnNμ(En). No finiteness hypothesis is required.

[F5]

Continuity from above when one set has finite measure: Let (En)nN be a decreasing sequence of measurable sets for a measure μ. If μ(En0)<+ for some n0, then μ(nNEn)=infnNμ(En).

[F6]

A series converges iff for every ε>0 there is N with am+1++an<ε for all n>mN: Let (ak) be a sequence of reals, with partial sums sn=k<nak (def-series). Then ak converges if and only if for every real ε>0 there is NN such that k=m+1nak<ε for all n>mN. The block k=m+1nak is the finite sum am+1++an of def-finite-sum, and it equals sn+1sm+1. This is the Cauchy criterion transported from sequences to series. Its value is that it decides convergence without producing, or even naming, the sum.

[F7]

Almost-sure convergence implies convergence in probability: If XnX almost surely, then XnX in probability.

[F8]

Limits in probability are unique almost surely: If XnX and XnY in probability, then X=Y almost surely.

[F9]

Cauchy sequences in probability have a measurable limit: Let (Yn)n1 be real random variables on one probability space. Suppose that for every ε,η>0 there is N such that P(YnYm>ε)<η(n,mN). Then there is a finite measurable real random variable Y such that YnY in probability.

Proof

Given: The objects and hypotheses of the statement.

1.1

Convergence in probability implies the Cauchy condition: for ε>0, the event SnSm>ε is contained in {SnS>ε/2}{SmS>ε/2}, whose probabilities are uniformly small for sufficiently large n,m. Conversely the Cauchy-in-probability completeness lemma gives a measurable finite probability limit.

F3F9givenalgebra
2.1

Assume the Cauchy condition. Fix t>0 and 0<δ<1. For all sufficiently large m and all N>m, every tail of the finite block Xm+1,,XN is an increment SNSi1 with both indices sufficiently large. Its probability of magnitude at least t/2 is at most δ: use the Cauchy condition at the strictly smaller tolerance t/4. The tail maximal lemma gives P(maxm<jNSjSmt)δ/(1δ).

F1step 1.1given
3.1

Pass to the infinite strict supremum by continuity from below. Let wm=supi,jmSiSj. Since wm2supjmSjSm, the previous estimate bounds P(wm>2t) by δ/(1δ) for sufficiently large m. These events decrease with m. Continuity from above and arbitrariness of δ show P(m{wm>2t})=0. Take t=1/r for all positive integers r. Outside a single null set the partial sums are real Cauchy, hence converge finitely; their limit extended by zero is measurable as in the series definition.

F4F5F6F2step 2.1
4.1

Almost-sure convergence implies convergence in probability. Its probability limit and any probability limit from the first step coincide almost surely by uniqueness. Thus all three conditions are equivalent. No moment hypothesis was introduced, and zero or deterministic increments cause no exception.

F7F8step 1.1step 3.1

5 · Examples, counterexamples and false statements

None yet.

Sources