Expectation of Product of Ito Integrals wrt Independent Brownian Motions

Question

Let $W_1(t)$, $W_2(t)$, $W_3(t)$ be independent Brownian motions and $f$, $g$ smooth functions. I want to know if the following is true: $$ \mathbb{E}\left[ \left( \int\limits_0^t f(W_3(s),s) \,dW_1(s) \right) \left( \int\limits_0^t g(W_3(s),s) \,dW_2(s) \right) \right] = 0 $$ I think it is, because, as an argument from intuition: $$ \begin{align} \mathbb{E}&\left[ \left( \int\limits_0^t f(W_3(s),s) \,dW_1(s) \right) \left( \int\limits_0^t g(W_3(s),s) \,dW_2(s) \right) \right]\\ &= \mathbb{E}\left[ \left( \lim_{n\rightarrow\infty}\sum\limits_{i=0}^{n-1} f(W_3(t_i),t_i) [W_1(t_{i+1})-W_1(t_{i})] \right) \left( \lim_{n\rightarrow\infty}\sum\limits_{j=0}^{n-1} g(W_3(t_j),t_j) [W_2(t_{j+1})-W_2(t_{j})] \right) \right]\\ &= \lim_{n\rightarrow\infty} \sum\limits_{i=0}^{n-1} \sum\limits_{j=0}^{n-1} \mathbb{E}\left[ f(W_3(t_i),t_i) g(W_3(t_j),t_j) \right] \underbrace{ \mathbb{E}\left[ W_1(t_{i+1})-W_1(t_{i}) \right] }_0 \underbrace{ \mathbb{E}\left[ W_2(t_{j+1})-W_2(t_{j}) \right] }_0\\ &=0 \end{align} $$ since increments of $W$ are Gaussian. But I am not sure if there are some stochastic analysis concepts I am missing. Thanks!

Answer 1

As Did mentioned, $[W_1,W_2](t)=0$. Let $\{t_i\}_{i=1}^{n}$ be a partition of $[0,t]$. We want to consider $$A_n=\sum_{i=0}^{n-1}(\,W_1(t_{i+1})-W_1(t_{i})\,)(\,W_2(t_{i+1})-W_2(t_{i})\,)$$ Using independent of $W_1$ and $W_2$ , thus $\mathbb{E}[A_n]=0$. Since increment of Wiener process are independent, the variance of sum is sum of variance , and we have $$\operatorname{Var}(A_n)=\sum_{i=0}^{n-1}\mathbb{E}[(\,W_1(t_{i+1})-W_1(t_{i})\,)^2]\,\,\mathbb{E}[(\,W_2(t_{i+1})-W_2(t_{i}))^2\,]$$ $$\operatorname{Var}(A_n)=\sum_{i=0}^{n-1}(t_{i+1}-t_i)^2 $$ $$\operatorname{Var}(A_n)\le\sum_{i=0}^{n-1}(t_{i+1}-t_i)\max\{t_{i+1}-t_i\}_{i=0}^{n-1}$$ $$\operatorname{Var}(A_n)\le\max\{t_{i+1}-t_i\}_{i=0}^{n-1}\sum_{i=0}^{n-1}(t_{i+1}-t_i)$$ $$\operatorname{Var}(A_n)\le\max\{t_{i+1}-t_i\}_{i=0}^{n-1}\,\,t$$ therefore $\operatorname{Var}(A_n)=\mathbb{E}[A_n^2]\to 0$ as $\delta_n=\max\{t_{i+1}-t_i\}_{i=0}^{n-1}\to 0$. This implies that $A_n\to 0$, in probability. Now set $$X_t=\int\limits_0^t f(W_3(s),s) \,dW_1(s) $$ in other words $$dX_t=f(W_3(t),t) \,dW_1(t)\quad,\quad X_0=0$$ and $$Y_t=\int\limits_0^t g(W_3(s),s) \,dW_2(s) $$ in other words $$dY_t=g(W_3(t),t) \,dW_2(t)\quad,\quad Y_0=0$$ By application of Ito's lemma $$d(X_tY_t)=Y_t dX_t+X_tdY_t+\underbrace{d[X_t,Y_t]}_{0}$$ as a result $$X_tY_t=\underbrace{X_0Y_0}_{0}+\int_{0}^{t}Y_sdX_s+\int_{0}^{t}X_sdY_s$$ finally, we have $$\mathbb{E}\left[X_tY_t\right]=\mathbb{E}\left[\int_{0}^{t}Y_sdX_s\right]+\mathbb{E}\left[\int_{0}^{t}X_sdY_s\right]=0$$