What is the limit of sum of split Wiener process? I wish to find the limit in $L^2$ for $n\rightarrow\infty$ of
$A_n = \sum_{i=1}^nW^a(it/n)-W^a((i-1)t/n))(W^b(it/n)-W^b((i-1)t/n))$
How can I go about that?
 A: I assume that $W^a,W^b$ are independent $1$ dimensional BMs.
Let $Y_n = \sum_{i=1}^n \left(W^a(\frac{it}n) - W^a\left(\frac{(i-1)t}{n}\right)\right)\left(W^b(\frac{it}n) - W^b\left(\frac{(i-1)t}{n}\right)\right)$. Note that for different values of $i$, the random variables inside the bracket are independent by the independence of increments of a BM, along with the independent blocks theorem.
Therefore, $E[Y_n^2] = \sum_{i=1}^n E\left[\left(W^a(\frac{it}n) - W^a\left(\frac{(i-1)t}{n}\right)\right)^2\left(W^b(\frac{it}n) - W^b\left(\frac{(i-1)t}{n}\right)\right)^2\right]$
again by independence, the inner expectation breaks :
$$
E\left[\left(W^a\left(\frac{it}n\right) - W^a\left(\frac{(i-1)t}{n}\right)\right)^2\left(W^b\left(\frac{it}n\right) - W^b\left(\frac{(i-1)t}{n}\right)\right)^2\right]\\ = E\left[\left(W^a\left(\frac{it}n\right) - W^a\left(\frac{(i-1)t}{n}\right)\right)^2\right]E\left[\left(W^b\left(\frac{it}n\right) - W^b\left(\frac{(i-1)t}{n}\right)\right)^2\right]  \\ = \frac 1{n} \cdot \frac 1n = \frac 1{n^2}
$$
and when we take the sum, we just get $E[Y_n^2] = \frac 1n$. Hence it is clear that $Y_n \to 0$ in $L^2$.

Note that if $W^a$ and $W^b$ are correlated random variables and jointly Gaussian (note : if uncorrelated and jointly Gaussian, then they are independent by a well known property of Gaussian random variables), it is still possible to perform this kind of calculation, by creating BMs using appropriate functionals of $W^a$ and $W^b$.
A: I'm sorry for not doing this in TeX-form but it is easier to understand in handwritten form. 
Fill free to ask if you don't understand smth in my solution.
