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 $[s,t]∩[s′,t′] =ø$


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

  • $\begingroup$ Variance of a sum of two independent r.v's is the sum of the varinaces. $\endgroup$ May 22, 2023 at 11:33
  • $\begingroup$ So how do I compute it? $\endgroup$
    – RavenBoy7
    May 22, 2023 at 11:35
  • $\begingroup$ Do you know about characteristic functions? If so, you can easily verify this property of normals $\endgroup$
    – kibble
    May 23, 2023 at 22:45

1 Answer 1


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.

  • $\begingroup$ So what is the variance? Is it t-s? $\endgroup$
    – RavenBoy7
    May 22, 2023 at 11:36
  • $\begingroup$ Yes, that is correct. $\endgroup$ May 22, 2023 at 12:51

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .