The distribution of the sum of two standard Brownian motions, B(u) + B(v), is characterized by a mean of 0 and a variance of u + v. This conclusion is derived from the properties of standard Brownian motion, where B(u) and B(v) are independent for u, v ≥ 0. The variance of the sum of independent random variables is indeed the sum of their variances, confirming that Var(B(u) + B(v)) = Var(B(u)) + Var(B(v)). Therefore, for specific values, such as Var(B(1) + B(1)), the result is 2.
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BrownianMan said: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?
BrownianMan said:Aren't B(u) and B(v) independent? If so, then the variance of their sum should be the sum of their variance.