Characterizing Brownian Motion with Gaussian Covariance and Continuity

Summary

The document asks why a continuous Gaussian process with zero mean and covariance equal to the minimum of its two time arguments qualifies as Brownian motion. It compares this characterization with a definition requiring a zero initial value, independent increments relative to the process’s past, normally distributed increments with variance equal to elapsed time, and almost surely continuous paths.

The question is foundational probability theory that can support stochastic modeling in quantitative finance. The document itself supplies the theorem conditions and asks for a proof, but gives no proof or financial application. Establishing the equivalence relies on the properties of Gaussian vectors: the covariance structure yields the increment distributions and makes disjoint increments independent; continuity and the initial condition complete the characterization. No model calibration, trading evidence, or market-specific caveat is discussed.

Key ideas

• The process is assumed Gaussian, centered, and to have covariance given by the minimum of the time arguments.
• Almost sure path continuity is part of the stated characterization.
• The comparison definition specifies a zero start and normally distributed increments.
• For Gaussian processes, zero covariance between disjoint increments implies their independence.
• The document poses the proof question but does not provide the proof.

Tags

# Equivalent definition of brownian motion


# Equivalent definition of brownian motion












I'm having a question about this characterization of Brownian Motion :

Theorem : If a process : $\big( X_t \big)_{t\geq 0}$ satisfies these conditions,

- $\big( X_t \big)_{t\geq 0}$ is a Gaussian process,

- For all $t,s\geq 0$ : $$\mathbb{E}(X_t)=0\quad \text{and} \quad \mathbb{C}ov(X_s,X_t) = \min(s,t)$$

- $\big( X_t \big)_{t\geq 0}$ is a continuous process (their paths are a.s. continuous)

then it's a Brownian motion.

The definition was given as this :

Definition : Let a process : $\big( X_t \big)_{t\geq 0}$ satisfies these conditions,

- $X_0 \underset{a.s.}{=} 0$

- For all $t>s\geq 0$ : the increment $X_t-X_s$ is independent of the $\sigma$-Algebra : $$ \mathcal{F}^X_s = \sigma\bigg( X_u, \ u\leq s \bigg) $$

- For all $t > s\geq 0$ : $$X_t - X_s \sim \mathcal{N}(0,t-s)$$

- $\big( X_t \big)_{t\geq 0}$ is a continuous process (their paths are a.s. continuous)

then we call it a Brownian motion.

I'm looking for a proof of the theorem

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