$\operatorname{var}[\frac{b}{a} B(a-b)-b B(b)]$ with $b\leq a$ and $b\geq 0$; $B=$brownian motion

Question

I want to calculate:

$\operatorname{var}[\frac{b}{a} B(a-b)-b B(b)]$ with $b\leq a$ and $b\geq 0$; $B=$brownian motion.

I started like this: $(\frac{b}{a})^2 \operatorname{var}[ B(a-b)]+-b ^2 \operatorname{var}[ B(b)]+ 2 \operatorname{Cov}(B(a-b), B(b))$

Can someone help me here or comment on my approach?

Answer 1

It is not enterely correct. You should find:

$$\left(\frac{b}{a}\right)^2 \operatorname{var}[ B(a-b)]+(-b) ^2 \operatorname{var}[ B(b)]+ 2 \frac{b}{a}(-b)\operatorname{Cov}(B(a-b), B(b))$$

because $\operatorname{var}(aX+bY) = a^2\operatorname{var}(X) + b^2\operatorname{var}(Y) + 2ab\operatorname{Cov}(X,Y)$

You have now to use classical properties of the covariance in a Browian motion: $$\operatorname{Cov}(B(t),B(s)) = \min(s,t).$$ You're almost done ;)