Let $(W_t)_t$ be a Brownian motion.
Prove that, when $n\to\infty,$ $$\sum_{k=0}^{2^n-1}W_{(k+1/2)2^{-n}}~~(W_{(k+1)2^{-n}}~~-W_{k2^{-n}}~)$$ converges in $L^2$ to $\int_0^1 W_tdW_t+1/2$
I tried to compute $$E\left[\left(\sum_{k=0}^{2^n-1}W_{(k+1/2)2^{-n}}~~(W_{(k+1)2^{-n}}-W_{k2^{-n}})-\int_0^1 W_tdW_t-1/2\right)^2\right],$$ where I am having troubles finding $$E\left[W_{(k+1/2)2^{-n}}~~(W_{(k+1)2^{-n}}~~-W_{k2^{-n}}~)\int_0^1 W_tdW_t\right]$$ and $$E\left[\left(\sum_{k=0}^{2^n-1}W_{(k+1/2)2^{-n}}~~(W_{(k+1)2^{-n}}~~-W_{k2^{-n}})\right)^2\right].$$
Any ideas how to treat this sort of problem ? (We also didn't cover yet Ito's formula)
By the definition of the Ito integral, $$\sum_{k=0}^{2^n-1}W_{k 2^{-n}}~~(W_{(k+1)2^{-n}}~~-W_{k2^{-n}}~)$$ converges in $L^2$ to $\int_0^1 W_tdW_t$.
It remains to calculate the $L^2$ limit as $n \to \infty$ of $$\sum_{k=0}^{2^n-1}A_k(A_k+B_k)\,,$$ where $$A_k=W_{(k+1/2)2^{-n}}- W_{k2^{-n}} \quad \text{and} \quad B_k= ~W_{(k+1)2^{-n}}~~-W_{(k+1/2)2^{-n}} \,.$$
Now $S_A(n):= \sum_{k=0}^{2^n-1}A_k^2$ has the same law as the $n$th dyadic approximation to the quadratic variation of Brownian motion over an interval of length $1/2$, so it must converge in $L^2$ to $1/2$. More directly, $S_A(n)$ has mean $1/2$ and variance of order $2^{-n}$, so it converges in $L^2$ to $1/2$.
The remaining sum $\sum_{k=0}^{2^n-1}A_k B_k$ has mean zero and variance $2^{-n-2}$, so it tends in $L^2$ to $0$.