Context
There are two different Wiener processes $W_t$ and $V_t$. It's known that they are independent. Additionally, we are given with third Wiener process $B_t$ that is given by the formula $$B_t = aW_t+bV_t, \quad \quad a^2 + b^2= 1.$$
Problem
Find the limit in $L^2$ of $$S_n = \sum_{i=1}^n\left[B_{it/n} - B_{(i-1)t/n}\right]\left[V_{it/n} - V_{(i-1)t/n}\right]$$
as $n$ tends to infinity.
My ideas
I assume that this is the type of task where we need to calculate the expected value and the variance. As the latter tends to $0$ (it should), we can say that the desired limit is the expected value. The issue is that it's very overextended work to calculate the $E(S_n)$.
Let $W_{it/n} = X_i$ and $V_{it/n} = Y_i$. We have
$$\Bbb E(S_n) = \Bbb E\sum_{i=1}^n [aX_i + bY_i - aX_{i-1} - bY_{i-1}][Y_{i} - Y_{i-1}]$$ which can be written as $$\sum \Bbb E\bigg( aX_iY_i + bY_i^2 - aX_{i-1}Y_i - bY_{i-1}Y_i - aX_iY_{i-1} - bY_iY_{i-1} + aX_{i-1}Y_{i-1} + bY_{i-1}^2\bigg).$$ Next calculations confuse me (what is $\Bbb E(X_i Y_{i-1})$?) Is it zero? And the main question how to calculate the variance?
If I'm not mistaken, $\Bbb E(S_n) = nb \to \infty$, so we don't need variance. Am I right?
Let $P_i=W_{it/n}-W_{(i-1)t/n}$ and $Q_i=V_{it/n}-V_{(i-1)t/n}$. This means that $P_i,Q_i\sim{\sf N}(0,t/n)$, so\begin{align}\Bbb E[S_n]&=\sum_{i=1}^n\Bbb E\left[(aP_i+bQ_i)Q_i\right]=\sum_{i=1}^na\Bbb E[P_i]\Bbb E[Q_i]+b\Bbb E[Q_i^2]\\&=b\sum_{i=1}^n\Bbb V[Q_i]+\Bbb E[Q_i]^2=b\sum_{i=1}^n\frac tn\\&=bt.\end{align} Similarly, since $P_i$ is independent of $P_{i-1}$ and likewise for $Q_i$, we have \begin{align}\Bbb V[S_n]&=\sum_{i=1}^n\Bbb V\left[(aP_i+bQ_i)Q_i\right]\\&=\sum_{i=1}^n\Bbb E\left[(aP_i+bQ_i)^2Q_i^2\right]-\Bbb E[(aP_i+bQ_i)Q_i]^2\\&=\sum_{i=1}^na^2\Bbb E[P_i^2]\Bbb E[Q_i^2]+2ab\Bbb E[P_i]\Bbb E[Q_i^3]+b^2\Bbb E[Q_i^4]-\left(\frac{bt}n\right)^2\\&=\sum_{i=1}^na^2\left(\frac tn\right)^2+b^2\cdot3\left(\frac tn\right)^2-b^2\left(\frac tn\right)^2\\&=\frac{(a^2+2b^2)t^2}n\stackrel{n\to\infty}\to0.\end{align}