"Сross сubic variation" of two Brownian motions and interpretation of the simulation result

Question

Consider two independent 1-dimensional Brownian motions $W_{t},B_{t}$, with an equidistant partition of the interval $[0,T]$, and $n\Delta≡T$.

How to calculate the expression below? Can we rewrite the limit of the summation when $\Delta$ goes to zero as an Ito integral? $$ S=\lim_{\Delta\to 0} \sum_{i=0}^{n-1}(\frac{1}{\Delta})⋅(W_{(i+1)\Delta}-W_{i\Delta})^{2}⋅(B_{(i+1)\Delta}-B_{i\Delta}) $$

NOTE:

I have done many numerical experiments by Monte-Carlo simulation, I find on interval $[0,1]$ the summation converge to a random variable (I guess it is normal) with mean 0 and standard deviation 1.73 (approximately) by the function "normfit()" in MATLAB.

However, I can not guess the final expression of $S$ because of the "weird" 1.73.

Answer 1

As Bjørn already mentioned, the CLT shows that the limit is indeed normal with variance $3 T$. A more interesting case is that of $$ I = \lim_{\Delta \to 0} \sum_{i=0}^n f(W_{i\Delta},B_{i\Delta})\Delta^{-1} \delta W_i^2 \delta B_i\;. $$ (Here $\delta W_i = W_{(i+1)\Delta} - W_{i\Delta}$.) Here, one would expect the limit to be given by $$ I = \int_0^T f(W_s,B_s)\,dB_s + \sqrt 2\int_0^T f(W_s,B_s)\,d\tilde B_s\;, $$ where $\tilde B$ is a Brownian motion independent of both $W$ and $B$, at least if $f$ is sufficiently regular (say uniformly Hölder continuous). The reason is that you can again use the CLT to show that, with $f = 1$, the triple $(I, B_T, W_T)$ converges to $(B_T + \sqrt 2 \tilde B_T, B_T, W_T)$. From there it's a simple approximation argument.