ExampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedaudited 2026-09-22

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.

Brownian density and Gaussian convolution

Example

Assume the Axiom of Choice and let $p_{t} (x, y) = (2 π t)^{- 1 / 2} e^{- (y - x)^{2} / (2 t)}$ be the Brownian transition The Brownian transition semigroup. Then for all $s, t > 0$ and $x, z \in R$ , $\int_{R} p_{s} (x, y) p_{t} (y, z) d y = p_{s + t} (x, z),$ and the integral is evaluated below by completing the square, giving the constant $\sqrt{2 π s t / (s + t)}$ .

Facts & Assumptions

Given: AC, $s, t > 0$ , $x, z \in R$ and the kernel $p$ .

[F1]

$p_{r} (x, y) = (2 π r)^{- 1 / 2} e^{- (y - x)^{2} / (2 r)}$ , and the semigroup identity holds for these kernels. The Brownian transition semigroup The Brownian kernels form a semigroup

[F2]

$\int_{R} e^{- u^{2}} d u = \sqrt{π}$ , and affine substitutions on compact intervals with the continuous integrand $e^{- u^{2}}$ pass to the improper limit by monotone convergence. The Gaussian integral $\int_{- \infty}^{\infty} e^{- x^{2}} d x = \sqrt{π}$ Substitution: if $φ$ is differentiable on $[c, d]$ with $φ^{'}$ integrable and $f$ is continuous on an interval containing $φ ([c, d])$ , then $\int_{φ (c)}^{φ (d)} f = \int_{c}^{d} (f \circ φ) φ^{'}$ Monotone convergence for the integral

[F3]

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

Verification

technique · direct

1.1F1algebra

Put $A = \frac{1}{2 s} + \frac{1}{2 t} = \frac{s + t}{2 s t}$ , $m = \frac{t x + s z}{s + t}$ and $C = \frac{(z - x)^{2}}{2 (s + t)}$ . Expanding squares gives $\frac{(y - x)^{2}}{2 s} + \frac{(z - y)^{2}}{2 t} = A (y - m)^{2} + C$ : the coefficient of $y^{2}$ is $A$ , the coefficient of $- 2 y$ is $2 A m = \frac{z}{t} + \frac{x}{s}$ , and subtracting the square leaves the constant $\frac{x^{2}}{2 s} + \frac{z^{2}}{2 t} - \frac{(z / t + x / s)^{2}}{4 A}$ , which equals $\frac{(z - x)^{2}}{2 (s + t)}$ .

2.1step 1.1F1

Consequently $p_{s} (x, y) p_{t} (y, z) = \frac{1}{2 π \sqrt{s t}} e^{- C} e^{- A (y - m)^{2}}$ for every $y$ , a nonnegative continuous function of $y$ .

2.2F2step 1.1

For $L > 0$ , the affine substitution $u = \sqrt{A} (y - m)$ on $[m - L / \sqrt{A}, m + L / \sqrt{A}]$ and [F2] give $\int_{m - L / \sqrt{A}}^{m + L / \sqrt{A}} e^{- A (y - m)^{2}} d y = \frac{1}{\sqrt{A}} \int_{- L}^{L} e^{- u^{2}} d u$ ; letting $L \to \infty$ with [F2] gives $\int_{R} e^{- A (y - m)^{2}} d y = \frac{\sqrt{π}}{\sqrt{A}} = \sqrt{\frac{2 π s t}{s + t}}$ .

3.1F1step 2.1step 2.2

Multiplying by the constant of step 2.1, $\int_{R} p_{s} (x, y) p_{t} (y, z) d y = \frac{1}{2 π \sqrt{s t}} \sqrt{\frac{2 π s t}{s + t}} e^{- C} = \frac{1}{\sqrt{2 π (s + t)}} e^{- (z - x)^{2} / (2 (s + t))} = p_{s + t} (x, z)$ , which is the displayed identity; the semigroup identity for the operators follows from it as in the cited lemma.

4.1F1F3givenstep 3.1∎

The degenerate cases are excluded or harmless as stated: $s, t > 0$ keeps $A$ finite and positive and all square roots real, the case $s = t$ and $x = z$ are included, and the cases $s = 0$ or $t = 0$ belong to the identity operator convention of the semigroup rather than to this convolution. AC is used only through [F3].

Source notes

Lawler, Section 2.6, computes the Gaussian convolution by completing the square; the example records the exact constant produced by the one-dimensional Gaussian integral.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

46 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 2.6 (standard reference, not scraped)