Lagmental Vicfred

Brownian Covariance Is the Minimum of Two Times by Vicfred

tags: mathematics, stochastic processes, order statistics, information theory

Independent Gaussian increments force the covariance of Brownian motion to equal the shared elapsed time. I want the notation, the mechanism, and the failure mode visible at the same time.

Objects and notation

Continuous-time processes are described by finite-dimensional distributions plus path regularity. A Poisson process \(N_t\) has independent increments, while Brownian motion \(B_t\) has Gaussian increments.

$$ B_0=0,\qquad B_t-B_s\sim N(0,t-s) $$

I read the first line as input and the second as output. The symbols \(\forall\) and \(\exists\) are not interchangeable, and neither may be upgraded silently to \(\Longleftrightarrow\).

$$ \operatorname{Cov}(B_s,B_t)=\min(s,t) $$

Push the symbols

The middle display is intentionally dense: it is where signs, bounds, multiplicities, or normalising factors are most likely to be lost.

$$ 0\le s\le t:\qquad\mathbf E[B_sB_t]=\mathbf E[B_s(B_s+B_t-B_s)]=s+0=s $$

Structural reading

A good test for understanding is to change the presentation while keeping the invariant fixed. The aligned form makes that comparison unusually easy.

$$ \begin{aligned} \mathsf{D}\;&:\quad B_0=0,\qquad B_t-B_s\sim N(0,t-s),\\[5pt] \mathsf{C}\;&:\quad \operatorname{Cov}(B_s,B_t)=\min(s,t). \end{aligned} $$

A hypothesis worth keeping

Matching means and variances does not identify a distribution. Independence, increment laws, and sample-path properties are separate ingredients.

$$ \boxed{\begin{gathered} \text{compact conclusion}\\[-2pt] \operatorname{Cov}(B_s,B_t)=\min(s,t) \end{gathered}} $$

The result is compact enough to reuse without pretending that the caveat has disappeared. The worked line remains the quickest consistency check.

This article was posted on Sat 01 December 2018. Facts and circumstances may have changed since publication.
Please contact me before jumping to conclusions if something seems wrong or unclear.