# Distribution of Brownian motion

How would I go about finding the distribution of $B(u) + B(u+v)$ where $u+v > u$?

I know that both $B(u)$ and $B(u+v)$ are normal random variables. The sum of two normal random variables is also normal with parameters equal to the sum of means and variances. However, this only applies for independent normal variables. How can I make $B(u) + B(u+v)$ independent? Or am I completely on the wrong track?

Thanks for the help.

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Beware !!! The sum of two gaussian variables is not always Gaussian. The correct statement here is any linear combination of coordinates of a Gaussian Vector is gaussian. Regards – TheBridge Apr 4 '11 at 9:17

We can write $B(u+v)+B(u) = (B(u+v)-B(u))+2B(u)$. But $B(u+v)-B(u)$ is independent of $B(u)$ and should have the same distribution as $B(v)$.
So $E[B(u+v) + B(u)]$ should be $E[B(v)]+E[2B(u)] =E[B(v)]+2E[B(u)]$, and $var[B(u+v) + B(u)]$ should be $var[B(v)]+var[2B(u)] =var[B(v)]+4var[B(u)]$