ExampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-22

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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.
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Expected dyadic quadratic variation

Example

Let $B$ be a standard Brownian motion, fix $T > 0$ and for $n \geq 1$ let $π_{n}$ be the dyadic partition of $[0, T]$ with points $k T / 2^{n}$ . Then the expected quadratic sum over the $2^{n}$ equal subintervals is exactly the elapsed time, $E \sum_{k = 1}^{2^{n}} (B_{k T / 2^{n}} - B_{(k - 1) T / 2^{n}})^{2} = T,$ for every $n \geq 1$ ; no independence is needed for this mean computation.

Facts & Assumptions

Given: AC, a standard Brownian motion $B$ , $T > 0$ and $n \geq 1$ with $h = T / 2^{n}$ and $Δ_{k} = B_{k h} - B_{(k - 1) h}$ .

[F1]

The terminal dyadic quadratic sum in this example is the finite sum $\sum_{k = 1}^{2^{n}} Δ_{k}^{2}$ .

[F2]

Each increment over an interval of length $h$ has law $N (0, h)$ and $E (Δ B)^{2} = h$ . Brownian motion Gaussian even moments for Brownian increments

[F3]

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

Verification

technique · direct

1.1

For each $k$ the increment $Δ_{k}$ has the law $N (0, h)$ by [F2], so $E Δ_{k}^{2} = h$ .

givenF2

2.1

By [F1] the terminal dyadic sum is $\sum_{k = 1}^{2^{n}} Δ_{k}^{2}$ , and linearity of expectation with [step 1.1] gives $E \sum_{k} Δ_{k}^{2} = \sum_{k} h = 2^{n} \cdot T / 2^{n} = T$ .

step 1.1F1

3.1

The cases are covered: $T > 0$ and $n \geq 1$ give $h > 0$ and $2^{n}$ summands, including the degenerate case $n = 1$ ; the mean computation uses only the marginal law of each increment, not independence or any joint distribution; and AC enters only through [F3].

step 2.1F3given∎

Source notes

Lawler, Section 2.8, computes the mean of the squared-increment sums as the total elapsed time. The example isolates the mean computation, which uses only the variance of the increments.

Depends on

Used by

Variance of dyadic quadratic variation Example

Dependency tree · two levels

17 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Gregory F. Lawler, Stochastic Calculus: An Introduction with Applications, Section 2.8 (standard reference, not scraped)