Alphabeta Math
ExampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-22
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Variance of dyadic quadratic variation

Example

With the notation of Expected dyadic quadratic variation, the dyadic quadratic sum Qn=k=12n(BkhB(k1)h)2 over [0,T] has Var(Qn)=2T22n, so QnT in L2 as n.

Facts & Assumptions

Given: AC, a standard Brownian motion B, T>0, n1 with h=T/2n and Δk=BkhB(k1)h.

[F1]

The increments over disjoint intervals are independent with laws N(0,h), and E(ΔB)2=h, E(ΔB)4=3h2 for an increment of length h. Brownian motion Gaussian even moments for Brownian increments

[F2]

The mean of the dyadic sum is EQn=T. Expected dyadic quadratic variation

[F3]

AC is the ambient assumption of the Brownian interfaces. The Axiom of Choice

Verification

technique · direct
1.1

For each k, [F1] gives Var(Δk2)=EΔk4(EΔk2)2=3h2h2=2h2.

givenF1
2.1

The variables Δk2 are functions of increments over disjoint intervals, hence independent by [F1], so the variance of the sum is the sum of the variances: Var(Qn)=k=12n2h2=2n2(T/2n)2=2T2/2n.

step 1.1F1
3.1

Since EQn=T by [F2], E(QnT)2=Var(Qn)=2T2/2n0, which is the L2 convergence QnT.

step 2.1F2
4.1

The cases are covered: the independence of the squared increments is the only place where the joint law is used; the value n=1 is included and gives Var(Q1)=T2; the limit is taken as n with T fixed and positive; and AC enters only through [F3].

step 2.1F3given

Source notes

Lawler, Theorem 2.8.1, obtains the variance of the quadratic sums from the fourth Gaussian moment and the independence of the increments, giving the mean-square convergence used in the dyadic quadratic-variation theorem.

Depends on

Used by

Nothing in the library uses this result yet.

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Sources