Computing Covariance of Linear Brownian Motion Transformations

Summary

The example computes the covariance between a linear transformation of Brownian motion at later times and Brownian motion at an earlier time, with constants included in both expressions. It uses the identity that covariance equals the expected product minus the product of expectations. Expanding the product shows that additive constants and terms with zero expectation do not contribute to the result.

The remaining terms are evaluated using the standard Brownian covariance rule: the covariance of two Brownian values is the minimum of their time arguments. Since the question specifies that the earlier time is below both later sampling times, this rule gives the stated answer, a product of the earlier time with the difference between the coefficient time and the other sampling time. The response illustrates a direct calculation rather than a more general covariance technique. It assumes standard Brownian motion and the time ordering supplied in the question.

Key ideas

• Covariance can be calculated as the expected product minus the product of the expectations.
• Additive constants do not affect covariance.
• For standard Brownian motion, covariance at two times equals the smaller time.
• Use the supplied time ordering to evaluate each Brownian covariance term.
• The example applies to the stated linear transformation and does not develop a general method beyond it.

Tags

# Covariance of two Brownian Motions


# Covariance of two Brownian Motions












During revision, I came across the following question in a past paper:

Suppose $(B_t, t\geq0)$ is a standard Brownian motion. Compute for $0<s<t$ the covariance $$cov(tB_{3t}-B_{2t}+5, B_s-1).$$

Now, the answers simply state that the solution is $ts-s$. However, the only notes we have been given are that: $$cov(B_t,B_s) = min\{t,s\},$$ for which the proof involves taking iterated expectations. Do I apply the same method for solving this, or are there any better / more intuitive methods for finding the covariances between transformations of a standard Brownian motion?

## Answer by R. Rayl (score 6, accepted)

https://quant.stackexchange.com/a/63910

Since $\text{Cov}(X, Y) = E(XY) - EX EY$, we have

\begin{align} \text{Cov}(tB_{3t} - B_{2t} + 5, B_s - 1) &= E[tB_{3t}B_s - tB_{3t} - B_{2t}B_s + B_{2t} + 5B_s - 5] - (5)(-1) \\ &= tE[B_{3t}B_s] - E[B_{2t}B_s] \\ &= ts - s \end{align} where the first equaltiy is just mutliplying out the product, the second equality comes from discarding zero expectation terms, and the third equality comes from the relationship: \begin{equation} \text{Cov}(B_s, B_t) = \text{min}\{s, t\} \end{equation} that you correctly wrote out.

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