Find parameters so that random variables (connected to Brownian movement) are independent.

Question

$W_t\sim\mathcal{N}(0,t)$ is Brownian movement, find values of parameters $a, b$ for which $aW_1-W_2$ and $W_3+bW_5$ are independent.

I don't even know where to start, so any hint is highly appreciated, though that's my ''solution''.

Do I have to find $\mathrm{Cov}(aW_1-W_2,W_3+bW_5)=0$?

If so then

$\mathrm{Cov}(aW_1-W_2,W_3+bW_5)$ $=a\mathrm{Cov}(W_1,W_3)+ab\mathrm{Cov}(W_1,W_5)+\mathrm{Cov}(-W_2,W_3)+b\mathrm{Cov}(-W_2,W_5)$ $=a+ab-2-2b$ Hence $a=2,\ b=-1$

Is this the right solution? I would be grateful if you could point out every mistake.

Thanks in advance!

Answer 1

The point is that jointly normal random variables are independent iff they are uncorrelated. So yes, you find the covariance, and yes, it is $a + ab - 2 - 2b$. And since $a + ab - 2 - 2b = (a-2)(1+b)$, the answer is $a=2$ or $b=-1$ (the way you wrote it, it looks like you want $a=2$ and $b=-1$, but that is overkill).

Answer 2

Yes indeed you have to find a,b such that

$$cov(aW_1-W_2,W_3+bW_5)=0$$ Because you know that $(W_1,W_2,W_3,W_5)$ follows a normal distribution. Hence $(aW_1-W_2,W_3+bW_5)$ also follows a normal distribution. Now, we know that for normal distributed variables , independence holds if and only if the covariance between them is 0.