B(t) is a standard Brownian Motion. u and v are both => 0. What is the distribution of B(u) + B(v)?
The mean is 0.
For the variance I get Var(B(u)+B(v)) = u+v. Is this right?
Assuming v>u:
Consider the right hand side.
By the basic properties of brownian motions, The two bracketed terms are independent and normally distributed. Their sum will be normally distributed as well.
The expected value is 0, so all you need is the total variance.
has variance 4v
has variance (v-u).
So what can you say about the variance of the sum?
And hence the distribution of the sum?