# Covariance of $cov(5W_7+6W_9,W_7)$ where $W_t$ is a standard Brownian motion

I'm having trouble deducing the value for the problem in the title.

Here is what I have done so far. (Given a standard Brownian motion (BM) $W_t, t\geq0$ with $W_0 = 0$ and $\sigma^2=1$)

The standard covariance formula is given by $cov(W_x,W_y) = E(W_xW_y) - E(W_x)E(W_y)$ and because BM is also a Markov process $E(W_x)E(W_y) = 0$ so the covariance formula can be simplified to $cov(W_x,W_y)=E(W_xW_y)$

BM is also a Gaussian process so $cov(W_iW_j) = E(W_i,W_j) = min(W_i,W_j)$ where $i,j\in[0,\infty]$ so e.g. as $\sigma^2=1$, $cov(W_7,W_9) = 7$

Going back to my original question of finding $cov(5W_7+6W_9,W_7)$, is the following computation correct?

$cov(5W_7+6W_9,W_7) = cov(5W_7,W_7) + cov(6W_9,W_7)$

$=E(5W_7,W_7) + E(6W_9,W_7)$

$=5E(W_7,W_7) + 6E(E_9,W_7)$

$=5*7 + 6*7$

$=77$

• Your title says $cov(5W_7+6W_9,W_3)$ but your calculations use $cov(5W_7+6W_9,W_7)$. – Henry Sep 14 '14 at 16:42
• Good point. I've amended the title to match. thanks – Paul P M Sep 14 '14 at 17:14
• Unless I missed something, that calculation looks okay now – Glen_b Sep 14 '14 at 17:47
• Thanks @Glen_b, I just wasn't sure and couldn't find any sources that went through this kind of problem even though it is really just a combination of simpler problems ... still so much to learn. – Paul P M Sep 15 '14 at 0:16

Another way of getting the same result is to note that $W_9-W_7$ is independent of $W_7$ so $$cov(5W_7+6W_9,W_7) = cov(11W_7+6(W_9-W_7),W_7)$$ $$= 11 var(W_7)+6cov(W_9-W_7,W_7) =11\times 7 +6\times 0.$$
• Thanks. btw, which rule did you use to get from $cov(5W_7+6W_9,W_7)$ to $cov(11W_7+2(W_9-W_7),W_7)$ ? – Paul P M Sep 15 '14 at 0:33