I am wondering if $\int_0^tW(s)ds$ is independent of $\int_t^TW(s)ds$, where $W$ is a standard brownian motion/wiener process, and for $0 \leq t \leq T$
Writing them as limits of Lebesgue Integrals, with $t_n = t$ and $t_l = T$ $\int_0^tW(s)ds = \sum_{i=1}^nW_{i} \cdot (t_{i+1} - t_i)$ and $\int_t^TW(s)ds = \sum_{i=n+1}^lW_{i} \cdot (t_{i+1} - t_i)$. In this sense, they seem to be independent to me, but I know that $\int_t^TW(s)ds$ is not independent of $W_t$ (or the filtration $\mathcal F^W_t$) since $W_s$ is a martingale and so $\mathbb E\left[ \int_t^TW(s)ds | W_t\right] = W_t \cdot (T-t)$ This was making me wonder if $\int_0^tW(s)ds$ had 'enough information' to affect the value of $\int_t^TW(s)ds$.
I was also wondering for the general case of time integrals, if $\int_0^tf(s,W(s))ds$ and $\int_t^Tf(s,W(s))ds$, but it looks like the answer should be the same as for the case of the brownian motion.
Thanks for the help!
Let $T=2t$. Then $$ E \left[ \left( \int_0^t W_s \, d s \right)\left(\int_t^{2t} W_s \, d s\right) \right] = E \left[ \left( \int_0^t W_s \, d s \right)\left(\int_0^t W_{t+s }\, d s\right) \right] = E \left[ \int_0^t \int_0^t W_s W_{t+u}\, d s \, d u \right] = \int_0^t \int_0^t E (W_s W_{t+u}) \, d s \, d u = \int_0^t \int_0^t E (W_s (W_{t+u} - W_s)) \, d s \, d u + \int_0^t \int_0^t E (W_s^2) \, d s \, d u = \int_0^t \int_0^t s \, d s \, d u = \frac{t^3}{2} \neq 0.$$