Expectation of an Ito integral involving arbitrary function in integrand

Question

Suppose, we have a have stochastic integral $\int ^t_0 I(t) dB(t)$ and $\int ^t_0 \frac{S(t)}{I(t)} dB(t)$, where $I(t)$ and $\frac{S(t)}{I(t)}$ are continuous bounded functions and they are actually Susceptible and infected populations of an epidemic model. I have to prove that the expectation of both the integrals are zero.

My thought: I write $\mathbb{E}\int ^t_0 I(t) dB(t)= \mathbb{E} ~\Sigma_{j} I(t_j) \{B_{t_{i+1}}-B_{t_{i}}\}= \Sigma_{j} \mathbb{E}\bigg\{I(t_j) \{B_{t_{i+1}}-B_{t_{i}}\}\bigg\}=\Sigma_{j} \mathbb{E}\big\{I(t_j)\big\} \mathbb{E}\big\{B_{t_{i+1}}-B_{t_{i}}\big\}$. Now since Brownian motion has mean zero, so $\mathbb{E}\int ^t_0 I(t) dB(t)=0$. Is it right? Please help to prove it more elaborately.

Answer 1

The integral $I = \int_a^bf(s)dB_s$ is the is the mean square limit of the partial sums $S_n = \sum_{i=0}^{n - 1}f(t_i)(B_{t_{i+1}} - B_{t_i}).$ Then $$E[S_n] = \sum_{i=0}^{n - 1}E[f(t_i)]E[(B_{t_{i+1}} - B_{t_i})]$$

We used the fact that $f(t_i)$ is independent of $B_{t_{i+1}} - B_{t_i}$ to get that expectation. Now we use the fact that the increments of Brownian motion are normally distributed with mean zero and variance $t_{i+1} - t_i$, $$B_{t_{i+1}} - B_t \sim N(0, t_{i+1} - t_i).$$

This gives

$$E[S_n] = \sum_{i=0}^{n - 1}f(t_i)E[(B_{t_{i+1}} - B_{t_i})] = 0$$

and since $I = \int_a^bf(s)dB_s = \lim_{n \to \infty} S_n$. We have that $$E[I] = \lim_{n \to \infty} E[S_n] = 0$$

Answer 2

I was thinking in the way @user1 and thanks for his answer. But now I am thinking in the following way:

Since the integrated $I(t)$ is a non-anticipative continuous function and also $\mathbb{E} \bigg\{\int_0^ t |I(t)|^2 dt\bigg\} < \infty $, so we can write $$\int_0^t I(t)~ dB(t)= \lim _{\Delta t \rightarrow 0} \Sigma_{j=1}^N~ I(t_{j-1})~(B(t_j)-B(t_{j-1})).$$ Now $\mathbb{E} \bigg\{\int_0^t I(t)~ dB(t)\bigg\}=\lim _{\Delta t \rightarrow 0} \Sigma_{j=1}^N~ \mathbb{E}(I(t_{j-1}))~\mathbb{E}((B(t_j)-B(t_{j-1})))$. Now as increments of Brownian motion are normally distributed with mean zero, variance $t_j−t_{j-1}$ and the function is non-anticipative, so we can say that $\mathbb{E} \bigg\{\int_0^tI(t)~dB(t)\bigg\}=0.$

Can I do it this way? Please give some advice. Thank you.