Distribution of Sum of Two Brownian Motions

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How do we find the distribution of the sum of two Brownian Motions?

The questions was asked here: Distribution of Brownian motion, and was answered with

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 $B(v)$.

How do we know we can write this? Where do I look this up / learn it? I'm missing the basics.

Many thanks.

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1 Answer

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The idea is natural here: if we have two independent random variables with normal distribution, we know that the sum is normally distributed. We also know that the increments of Brownian motion are independent, hence it is logical to try to express $B(u+v)+B(v)$ as an increment plus an other term. The other terms is two times an increment, which remains independent of the first one, because $u+v>v>0$.

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