Suppose $B_t$ is a Brownian motion. As I understand, $B_2-B_1$ is independent of $B_3-B_2$ from properties of Brownian motion. Does it also mean that $B_1$ and $B_2$ are also independent? Can I use this independence to find the joint density of $B_1+B_2+B_3$ as each Brownian process is a normal process of mean 0 and variance t, it should be trivial.
I've another related question. To find the expectation over a Brownian process, can I integrate my stochastic process over the normal density function for Brownian motion (mean 0 and variance t)? I hope this makes sense.
Best Answer
$B_1$ and $B_2$ are not independent.
Since $B_1$ and $B_2-B_1$ are independent, $$0=\mathrm{Cov}(B_1,B_2-B_1)=E[B_1 (B_2-B_1)]=E[B_1 B_2]-E[B_1^2]=E[B_1 B_2] - 1$$ So, $$\mathrm{Cov}(B_1,B_2)=E[B_1 B_2]=1$$
You can use this result and the fact that linear combinations of normal variables are normal to calculate the distribution of $B_1+B_2+B_3$.
I'm not sure what exactly you meant by "expectation over a Brownian process". Can you please give us a little background of the problem you are trying to solve? Your method of taking an expectation with respect to $N(0,\sqrt{t})$ is fine when you want the expected value of some function of $B_t$.
There is also a theory of stochastic integration with respect to stochastic processes which might be of use to you.