covariance of the integral of brownian motion over disjoint intervals

Question

Let $W_i = \int_{t_{i-1}}^{t_i} B(t) \,dt$ be the integral of Brownian motion over the time $t \in [t_{i-1},t_i]$. I'm reading a paper which says that $W_i$ and $W_j$ are independent Gaussian random variables. I can see why they're both Gaussian but it is not clear to me that they are independent.

If I look at $E[W_i W_j]$ for $j>i$, clearly the intervals $[t_{i-1},t_i]$ and $[t_{j-1},t_j]$ are disjoint. So $E[W_i W_j] = \int_{t_{i-1}}^{t_i}\int_{t_{j-1}}^{t_j} E[B(s)B(t)] \,ds \,dt$.

$E[B(s)B(t)] = s$ considering that $i < j$ and so the interval $[t_{j-1},t_j]$ has larger values than $[t_{i-1},t_i]$. Now, I don't see the double integral being zero!

I am missing out on something here?

Edit: paper here:

Answer 1

In this paper, $W_t$ isn't a Brownian motion. It's a "white noise excitation" or more simply just a "white noise", and it's integral $$B_t=\int_0^t W_s ds$$ is the corresponding Brownian motion.

So, the paper is making the fairly uncontroversial claim that independent increments of a Brownian motion are independent Gaussians.

Note that there are various ways of making the meaning of the integral above precise. One common one is to take the white noise $W_s$ to be defined by its integral (or rather by the integrals $\int W_t$ and $\int W_t^2$ over intervals or something similar) and side-step the question of what the white noise function $W_t$ is exactly.

Answer 2

You're not missing anything; those are not independent. In fact, as your post suggests, they're positively correlated which makes quite a bit of intuitive sense. Is it possible that there's a term missing that you cut out? If we instead defined $$W_i' = \int_{t_{i-1}}^{t_i} B(t) \,dt - (t_i - t_{i-1}) B_{t_{i-1}}$$ then you'd be looking at independent Gaussians. Perhaps this is what the paper was saying (or what they meant to say)?