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?

  • 1
    $\begingroup$ Hi: It's easiest to take the covariance of each piece seperately. So, the covariance of your expression is equal to $cov(t B_{3t}, B_{s}) + cov(-B_{2t}, B_{s})$. $\endgroup$
    – mark leeds
    May 11 at 14:36

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.

  • $\begingroup$ Thanks! This is much simpler than I initially thought. Would the same approach apply for two separate processes, say $B_{t_1}$ and $W_{t_2}$ in $cov(B_{t_1}, W_{t_2})$? $\endgroup$
    – Kris
    May 11 at 23:14
  • 1
    $\begingroup$ @Kris In that case, you need to be aware of whether $B_t$ and $W_s$ are independent or correlated. If they are independent, Cov is zero of course. If they are correlated, then you can write $B_t = \rho W_t + \sqrt{1-\rho^2}\tilde{W}_t$ (where $\tilde{W}_t$ is a third process independent of both $B$ and $W$) and you can now apply similar arguments as before. $\endgroup$
    – R. Rayl
    May 12 at 7:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.