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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?

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    $\begingroup$ 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
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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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  • $\begingroup$ 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
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    $\begingroup$ @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

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