ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedaudited 2026-09-22

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A deterministic step integrand

Example

Assume the Axiom of Choice and the standing hypothesis (H) of Elementary predictable Brownian integrands. Let $h = \sum_{k = 0}^{m - 1} a_{k} 1_{(t_{k}, t_{k + 1}]}$ be a deterministic step function on $[0, T]$ , with real coefficients $a_{k}$ and partition $0 = t_{0} < \dots < t_{m} = T$ . Then $\int_{0}^{t} h_{s} d B_{s} = \sum_{k = 0}^{m - 1} a_{k} (B_{t \land t_{k + 1}} - B_{t \land t_{k}}) (0 \leq t \leq T),$ and at the terminal time the integral has law $N (0, \sum_{k} a_{k}^{2} (t_{k + 1} - t_{k}))$ ; in particular it has mean $0$ and variance $\int_{0}^{T} h^{2} d s$ .

Facts & Assumptions

Given: AC, the standing hypothesis (H), a deterministic step function $h = \sum_{k} a_{k} 1_{(t_{k}, t_{k + 1}]}$ on $[0, T]$ and $t \in [0, T]$ .

[F1]

A deterministic step function is an elementary predictable integrand with coefficients $a_{k} \in F_{t_{k}}$ (constants), and its elementary integral is the finite sum $\sum_{k} a_{k} (B_{t \land t_{k + 1}} - B_{t \land t_{k}})$ ; the general integral agrees with the elementary one on this subspace. Elementary predictable Brownian integrands Ito integral of an elementary predictable process Ito integral for square-integrable predictable processes

[F2]

For deterministic $h \in L^{2} [0, T]$ the integral is centered normal with variance $\int_{0}^{T} h^{2} d s$ . Deterministic Ito integrals are Gaussian Standard normal and normal laws

[F3]

AC is declared for the ambient interfaces. The Axiom of Choice

Verification

technique · direct

1.1

Substituting the deterministic coefficients into the definition gives the displayed finite sum for every $t$ , and at $t = T$ it is $\sum_{k} a_{k} (B_{t_{k + 1}} - B_{t_{k}})$ , a linear combination of the independent increments over the partition intervals.

F1given

2.1

The squared $L^{2} [0, T]$ norm of $h$ is $\int_{0}^{T} h^{2} d s = \sum_{k} a_{k}^{2} (t_{k + 1} - t_{k})$ , so by [F2] the law of the terminal integral is $N (0, \sum_{k} a_{k}^{2} (t_{k + 1} - t_{k}))$ , with mean $0$ and that variance.

F2step 1.1

3.1

The cases are covered: a single-interval step ( $m = 1$ , $a_{0} = a$ ) gives $a (B_{T} - B_{0}) = a B_{T}$ of law $N (0, a^{2} T)$ ; the degenerate case $a_{k} = 0$ for some $k$ contributes zero variance on that block; and $h = 0$ gives the Dirac law $N (0, 0)$ at $0$ . AC enters only through [F3].

F2F3step 2.1given∎

Source notes

Lawler, Section 3.2.2, defines the integral of a simple process exactly as this finite sum; the distribution statement is the deterministic-step instance of the deterministic-integrand corollary.

Depends on

Used by

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Dependency tree · two levels

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Sources

Gregory F. Lawler, Stochastic Calculus: An Introduction with Applications, Section 3.2.2 (standard reference, not scraped)