I'm tasked with showing that the Ito integral with respect to a standard Wiener process
$$
\int^t_0 g(W_s,s)dW_s = \lim \limits_{n \to \infty} \sum^{n-1}_{k=0} f(W_{t_i},t_i)(W_{t_i + 1}-W_{t_i})
$$
where $f$ is a simple process, $t_i = \frac{it}{n}$, $0=t_0 < t_1 < t_2 < … < t_{n-1} < t_n = t$, and $n \in \mathbb{N}$ has an expectation of 0
$$
\mathbb{E}\bigg[\int^t_0 g(W_s,s)dW_s\bigg]=0
$$
I assume that this should follow the same lines as this post; however, I am unsure how the 2 fit together. Is it that $C_s$ is directly equivalent to $g_s$ here?
Best Answer
Here's what i reckon you want to do. Write $h(t) = \int_0^t g(W_s, s)dW_s $. Find $\mathbb{E}[h(t)-h(s) + h(s) | \mathscr{F}_s]$.
You can say that $\mathbb{E}[h(t)-h(s) | \mathscr{F}_s] =0$.
Why is it equal to $0$? It is nothing but the integral between $s$ and $t$, and it does not mind about the past. So you can turn it into an unconditional expectation and use the mean value property to say it is zero.
and $ \mathbb{E}[h(s) | \mathscr{F}_s] = h(s)$ as this is deterministic.