Prove that $\mathbb{E} \left[ \int_0^t f_s \, dW_s \right] = 0$, for any continuous stochastic process $f_s$

Question

I am currently trying to prove that $$ \mathbb{E} \left[ \int_0^t f_s \, dW_s \right] = 0 $$ where $f_s = f(W_s)$ is some continuous bounded stochastic process.

My attempt to do this is as follows: $$ \mathbb{E} \left[ \int_0^t f_s \, dW_s \right] = \mathbb{E} \left[ \lim_{n \rightarrow \infty} \sum_{i=0}^{n-1} f_{s_i} \cdot \Delta W_{s_i} \right] $$ where $\Delta W_{s_i} := W_{s_{i+1}} - W_{s_i}$. This is then equal to $$ \lim_{n \rightarrow \infty} \sum_{i=0}^{n-1} \mathbb{E} \left[f_{s_i} \cdot \Delta W_{s_i} \right] $$ which I believe is then equal to $$ \lim_{n \rightarrow \infty} \left( n \cdot \mathbb{E} \left[f_{s_i} \right] \cdot \mathbb{E} \left[ \Delta W_{s_i} \right] \right) $$ Thus, since $\mathbb{E} \left[ \Delta W_{s_i} \right] = 0$, we have our result.

So my question is as follows:

1. Is this a correct proof (i.e. Have I made any false assumptions)?
2. Could this have been done better?

Answer 1

You are on the right track. Intuitively, your proposal works quite well, especially if you do not require mathematical rigor. If you require so, two comments are:

(1) Specific reasons are necessary for the commutativity of $\mathbb{E}$ and $\lim_{n\to\infty}$, and

(2) Conditional expectation could be adopted to clarify the expectation of each term of the sum.

Here is a sketch of proof.

Recall the definition of Ito's integral $$ Y_t=\int_0^tf_s{\rm d}W_s:=\lim_{n\to\infty}\sum_{j=0}^{n-1}f_{s_j}\left(W_{s_{j+1}}-W_{s_j}\right), $$ where $0=s_0<s_1<\cdots<s_n=t$ forms a partition of $\left[0,t\right]$. Note that the limit hereinabove is in the sense of the existence of some square-integrable $Y_t$ that satisfies $$ \lim_{n\to\infty}\mathbb{E}\left(Y_t-\sum_{j=0}^{n-1}f_{s_j}^n\left(W_{s_{j+1}}-W_{s_j}\right)\right)^2=0, $$ where $\left\{f_t^n\right\}_{n\in\mathbb{N}}$ is a sequence of random simple function that approximates $f_t$, i.e., $f_t^n\to f_t$ almost surely (guaranteed by the boundedness of $f_t$, or more generally, square-integrability of $f_t$). Denote $$ Y_t^n=\sum_{j=0}^{n-1}f_{s_j}^n\left(W_{s_{j+1}}-W_{s_j}\right). $$

With these understandings, our target could be figured out by $$ \mathbb{E}Y_t=\mathbb{E}Y_t^n+\mathbb{E}\left(Y_t-Y_t^n\right). $$ Thanks to the vanishing of the last term, i.e. (using Cauchy-Schwarz inequality and the definition of $Y_t$), $$ \left|\mathbb{E}\left(Y_t-Y_t^n\right)\right|\le\mathbb{E}\left|Y_t-Y_t^n\right|\le\sqrt{\mathbb{E}\left(Y_t-Y_t^n\right)^2}\to 0, $$ it suffices to figure out the limit of $\mathbb{E}Y_t^n$.

$Y_t^n$ is a finite sum, for which \begin{align} \mathbb{E}Y_t^n&=\mathbb{E}\left(\sum_{j=0}^{n-1}f_{s_j}^n\left(W_{s_{j+1}}-W_{s_j}\right)\right)\\ &=\sum_{j=0}^{n-1}\mathbb{E}\left(f_{s_j}^n\left(W_{s_{j+1}}-W_{s_j}\right)\right)\\ &=\sum_{j=0}^{n-1}\mathbb{E}\left(\mathbb{E}\left(f_{s_j}^n\left(W_{s_{j+1}}-W_{s_j}\right)|\mathcal{F}_{s_j}\right)\right)\\ &=\sum_{j=0}^{n-1}\mathbb{E}\left(f_{s_j}^n\mathbb{E}\left(W_{s_{j+1}}-W_{s_j}|\mathcal{F}_{s_j}\right)\right)\\ &=\sum_{j=0}^{n-1}\mathbb{E}\left(f_{s_j}^n\cdot 0\right)\\ &=0. \end{align} Therefore, $\mathbb{E}Y_t^n\to 0$, for which $\mathbb{E}Y_t=0$ follows immediately as per the reasoning from above.

You may refer to here for more detailed steps with regards to this topic.