Let $W_1(t)$ and $W_2(t)$ be two independent Brownian motions. Define the new process $X(t) =(W_1(t) - W_2(t))/√2$. Is $X(t)$ then another Brownian motion? I.e check that
1.$X(0) = 0$.
2.$X(t)−X(s)$ is independent from $X(t′)−X(s′)$ whenever $[s,t]∩[s′,t′] =ø$
3.$X(t)−X(s)∼N(0,t−s)$.
4.$X(t)$ is continuous.
I have managed to show properties 1,2 and 4, as well as the fact that the mean is 0. I am left to show that the variance of $X(t)-X(s)=t-s$
How can I show this? Is $X(t)$ a Brownian motion? Any help is appreciated
If $A$ and $B$ are two independent random variables, then $\mathrm{Law}(A+B)=\mathrm{Law}(A)*\mathrm{Law}(B)$. Use this with $A=(W_1(t)-W_1(s))/\sqrt{2}$ and $B=-(W_2(t)-W_2(s))/\sqrt{2}$ along with the fact that the convolution of two normal distributions is normal with variance given by the sum of the squares of the individual variances.