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?

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- April 14th 2013, 09:11 AMBrownianManSum of two Brownian Motions.
*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? - April 14th 2013, 11:20 AMSpringFan25Re: Sum of two Brownian Motions.
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?