What is the limit of sum of split Wiener process?

Question

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?

Answer 1

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.

Answer 2

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 $L^2$.

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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 $W^b$.