I need to prove that $$ \lim_n \sum_{i=0}^{2^n -1} W(\frac{i}{2^n}) ( W(\frac{i+1}{2^n}) - W(\frac{i}{2^n}) ) \rightarrow \frac{1}{2}W_1^2 - \frac{1}{2}$$ in $L^2$.
where $W$ is a standard Wiener process.
So I started computing
$$E[( \sum_{i=0}^{2^n -1} W(\frac{i}{2^n}) ( W(\frac{i+1}{2^n}) - W(\frac{i}{2^n}) ) + \frac{1}{2}(1-W_1^2) )^2]$$
and I split have that, after expanding the square:
$$E[( \sum_{i=0}^{2^n -1} W(\frac{i}{2^n}) ( W(\frac{i+1}{2^n}) - W(\frac{i}{2^n}) ))^2]=\frac{1}{2}$$
by independence the Gaussian distribution of independent increments.
Then, the double-product term $$\frac{1}{2} E[(1-W_1^2) \cdot \sum_{i=0}^{2^n -1} W(\frac{i}{2^n}) ( W(\frac{i+1}{2^n}) - W(\frac{i}{2^n}) )) ] = 0 $$ since it's a standard Wiener process and by independence of increments
The last term to compute is
$$ \frac{1}{4} E[(1-W_1^2)^2)] = \frac{1}{4} E[1+W_1^4 - 2 W_1^2 ] = \frac{1}{4} (1+3-2)=\frac{1}{2} $$
using that $E[W_1^2] = 1$ and $E[W_1^4] = 3$.
But in this way the sum is $1$ and not $0$... so what am I missing?
You expectation numbers are incorrect, you could see that in your proof, your solution is independent of $n$...
First notice that
$$ W_1^2= \sum_{i=0}^{2^n -1} W(\frac{i+1}{2^n})^2- W(\frac{i}{2^n})^2$$ $$ W_1^2= \sum_{i=0}^{2^n -1}\left( W(\frac{i+1}{2^n})+ W(\frac{i}{2^n})\right)\left( W(\frac{i+1}{2^n})- W(\frac{i}{2^n})\right)$$
Therefore, by factorizing, we have $$\sum_{i=0}^{2^n -1} W(\frac{i}{2^n}) ( W(\frac{i+1}{2^n}) - W(\frac{i}{2^n}) ) + \frac{1}{2}(1-W_1^2) =\frac{1}{2}\left[1-\sum_{i=0}^{2^n -1} ( W(\frac{i+1}{2^n}) - W(\frac{i}{2^n}) )^2\right]$$
You want to calculate
$$E\left[\frac{1}{4}\left[1-\sum_{i=0}^{2^n -1} ( W(\frac{i+1}{2^n}) - W(\frac{i}{2^n}) )^2\right]^2\right]$$
The variance of a Brownian motion increment is known, we have
$$E\left[\sum_{i=0}^{2^n -1} ( W(\frac{i+1}{2^n}) - W(\frac{i}{2^n}) )^2\right] =\sum_{i=0}^{2^n -1}\frac{1}{2^n }=1$$
Finally,
$$E\left[\left[\sum_{i=0}^{2^n -1} ( W(\frac{i+1}{2^n}) - W(\frac{i}{2^n}) )^2\right]^2\right]=E\left[\sum_{i,j=0}^{2^n -1} ( W(\frac{i+1}{2^n}) - W(\frac{i}{2^n}) )^2 (W(\frac{j+1}{2^n}) - W(\frac{j}{2^n}) )^2\right]$$
or $$\sum_{i,j=0, i \ne j}^{2^n -1} E\left[( W(\frac{i+1}{2^n}) - W(\frac{i}{2^n}) )^2 (W(\frac{j+1}{2^n}) - W(\frac{j}{2^n}) )^2\right]+\sum_{i=0}^{2^n -1} E\left[( W(\frac{i+1}{2^n}) - W(\frac{i}{2^n}) )^4\right]$$
Increments are independent , and you know the 4th moment of a Brownian motion, you can conclude when you make $n$ goes to infinity.