# Linear combination of two Brownian Motions

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

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

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.