ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14

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Linear combinations of Brownian values are Gaussian

Example

Assume the Axiom of Choice. Let $B$ be a standard Brownian motion. For every finite list $t_{1}, \dots, t_{n} \geq 0$ and $a_{1}, \dots, a_{n} \in R$ ,

$\sum_{j = 1}^{n} a_{j} B_{t_{j}} \sim N ⁣ (0, \sum_{i, j = 1}^{n} a_{i} a_{j} \min (t_{i}, t_{j})) .$

This includes repeated and zero times, zero coefficients, variance zero, and the empty sum when $n = 0$ .

Facts & Assumptions

Given: AC, a standard Brownian motion $B$ , and finite time and coefficient lists as in the example.

[F1]

Brownian motion is a centered Gaussian process with covariance kernel $K (s, t) = \min (s, t)$ . Brownian motion, Gaussian process.

[F2]

A finite evaluation vector of a Gaussian process has a possibly singular multivariate normal law, and every scalar projection of $N_{n} (m, Σ)$ has law $N (u \cdot m, u^{T} Σ u)$ ; the parameters are its mean and variance. Gaussian process, Multivariate normal law, including singular covariance.

[F3]

Covariance is bilinear on finite linear combinations. Covariance is symmetric and bilinear in finite linear combinations.

[F4]

The minimum kernel is positive semidefinite for every finite, repeated, or zero time list, including the empty list. Positive semidefiniteness of the Brownian covariance kernel.

[F5]

AC is inherited through the Gaussian and Brownian normal-law interfaces. The Axiom of Choice.

Verification

technique · direct

1.1

First suppose $n \geq 1$ and write $X = (B_{t_{1}}, \dots, B_{t_{n}})$ and $a = (a_{1}, \dots, a_{n})$ . By [F1]--[F2], $X$ is multivariate normal with mean vector zero and covariance matrix $K_{i j} = \min (t_{i}, t_{j})$ , even if some coordinates repeat or are deterministic. Its projection $a \cdot X$ therefore has law $N ⁣ (0, a^{T} K a) = N ⁣ (0, \sum_{i, j = 1}^{n} a_{i} a_{j} \min (t_{i}, t_{j})) .$

F1F2algebra

2.1

Independently, covariance bilinearity computes $Var ⁣ (\sum_{j = 1}^{n} a_{j} B_{t_{j}}) = \sum_{i, j = 1}^{n} a_{i} a_{j} Cov (B_{t_{i}}, B_{t_{j}}) = \sum_{i, j = 1}^{n} a_{i} a_{j} \min (t_{i}, t_{j}),$ confirming that the second parameter in step 1.1 is the actual variance. It is nonnegative by [F4], including when cancellations make it zero; in that case [F2] interprets the law as the point mass $N (0, 0)$ .

step 1.1F1F2F3F4

3.1

If $n = 0$ , the sum and the double sum are both empty and equal zero, so the random variable is the constant zero and has law $N (0, 0)$ by [F2]. Zero coefficients, $t_{j} = 0$ , and repeated times require no deletion and are already covered by the possibly singular matrix in steps 1.1--2.1. AC is used only through [F1]--[F2]; the finite algebra and the positive-semidefinite calculation make no further choice.

step 1.1step 2.1F2F4F5∎

Source notes

Yoshida, Lemma 6.1.3 and equation (6.5), printed pp. 174--175, identify Brownian finite collections as mean-zero Gaussian variables with covariance $\min (s, t)$ in dimension one. The displayed projection and singular-case calculation are supplied explicitly above.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

Nobuaki Yoshida, Probability Theory, Section 6.1 (standard reference, not scraped)