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:
$\displaystyle B_u + B_v = \left[2B_u \right] + \left[B_v - B_u \right]$
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.
$\displaystyle 2B_v$ has variance 4v
$\displaystyle B_v - B_u$ has variance (v-u).
So what can you say about the variance of the sum?
And hence the distribution of the sum?