Use Borel-Cantelli to show that$\sum\limits_{m \leq 2^{n}} \Delta _{m,n}^{2} \to t$-a.s. by computing $E(\sum\limits_{m \leq 2^{n}} \Delta _{m,n}^{2} - t)^{2}$
where $ \Delta _{m,n}=B_{\frac{mt}{2^{n}}}-B_{\frac{(m-1)t}{2^{n}}}$ where $(B_{t})_{t \geq 0}$ is a brownian motion
I am struggling to compute to compute $E(\sum\limits_{m \leq 2^{n}} \Delta _{m,n}^{2} - t)^{2}$:
$E(\sum\limits_{m \leq 2^{n}} \Delta _{m,n}^{2} - t)^{2}=E((\sum\limits_{m \leq 2^{n}} \Delta _{m,n}^{2})^{2}-2t\sum\limits_{m \leq 2^{n}} \Delta _{m,n}^{2}+t^2)=E((\sum\limits_{m \leq 2^{n}} \Delta _{m,n}^{2})^{2})-2tE(\sum\limits_{m \leq 2^{n}} \Delta _{m,n}^{2})+t^2$
I do not know how to continue with the first term
First notice that $$\mathbb E\left(\sum\limits_{1\leq m \leq 2^{n}} \Delta _{m,n}^{2} - t\right)^{2}= \sum_{m_1=1}^{2^n}\sum_{m_2=1}^{2^n}\mathbb E\left[\left(\Delta _{m_1,n}^{2} - t2^{-n}\right)\left(\Delta _{m_2,n}^{2} - t2^{-n}\right)\right].$$ If $m_1\neq m_2$, the term $\mathbb E\left[\left(\Delta _{m_1,n}^{2} - t2^{-n}\right)\left(\Delta _{m_2,n}^{2} - t2^{-n}\right)\right]$ is zero due to the independence of increments of a Brownian motion. We thus obtain $$\mathbb E\left(\sum\limits_{1\leq m \leq 2^{n}} \Delta _{m,n}^{2} - t\right)^{2}= \sum_{m =1}^{2^n} \mathbb E\left[\left(\Delta _{m ,n}^{2} - t2^{-n}\right)^2\right].$$ To conclude, observe that $\Delta_{m,n}$ has a centered normal distribution with variance $t2^{-n}$.