Brownian motions

Question

If $A_t$ and $B_t$ are independent Brownian motions and $a$ is a number, can we say that $(\sin a )A_t+ (\cos a) B_t$ is also a Brownian motion?

Answer 1

If $A_t$ and $B_t$ are independent Brownian motions, then $\sin(a)A_t+\cos(a)B_t$ is a centered Gaussian process (due to independence, $\sin(a)A_t+\cos(a)B_t$ has law $N(0,(\sin(a)^2+\cos(a)^2)t)=N(0,t)$ at fixed $t$) and the covariance of the process is

$$K(s,t)=\operatorname{Cov}(\sin(a)A_s+\cos(a)B_s,\sin(a)A_t+\cos(a)B_t)=\sin(a)^2s\wedge t+\cos(a)^2s\wedge t+0=s\wedge t$$

where the $s\wedge t$ (minimum of $s$ and $t$) comes from the fact that $A_t$ and $B_t$ are Brownian motions and the $0$ comes from the fact that $A_t$ and $B_s$ are independent.

We have shown that $\sin(a)A_t+\cos(a)B_t$ is a centered Gaussian process with covariance $K(s,t)=s\wedge t$. Obviously the trajectories are still continuous since multiplication by a constant preserves continuity and the sum of two continuous functions is continuous. This characterizes Brownian motion, so yes, this is definitely once again a Brownian motion.