$\mathbb{E}(\xi_i^2)<\infty$, $\xi_i \in \mathcal{F}_{t_i}$, $B(t)$is a brownian motion, how to prove $\xi_i^2 (B(t_{i+1})-B(t_i))^2$ is integrable？

Question

$\mathbb{E}(\xi_i^2)<\infty$,

$\xi_i \in \mathcal{F}_{t_i}$,

$B(t)$is a brownian motion,

how to prove $\xi_i^2 (B(t_{i+1})-B(t_i))^2$ is integrable？

where 0< $t_1$ < $t_2$ <......<$t_{n-1}$ <$t_{n}$ < t ,

my tries: $\mathbb{E}(\xi_i^2 (B(t_{i+1})-B(t_i))^2=\mathbb{E}(\mathbb{E}(\xi_i^2 (B(t_{i+1})-B(t_i))^2|\mathcal{F}_{t_i}))$=$\mathbb{E}(\xi_i^2) (t_{i+1}-t_i)$,

then $\sum_{i=1}^\infty \mathbb{E}(\xi_i^2) (t_{i+1}-t_i)=t\mathbb{E}(\xi_i^2 )$

RHS<$\infty$,

can I say the $\xi_i^2 (B(t_{i+1})-B(t_i))^2$ is integrable? or there is a regular routine to prove the variable is integrable?

Answer 1

If $X$ and $Y$ are integrable and independent then $XY$ is integrable and $XY=EXEY$. All you need is to observe that $B(t_{i+1})-B(t_i)$ is independent of $\mathcal F_{t_i}$ which makes $\xi_i^{2}$ and $(B(t_{i+1})-B(t_i))^{2}$ independent.