Expectation of an Itô integral

Question

I'm interested in computing the following expectation: $$\mathbb{E}\left[W_T\cdot\int_0^T f(s)\mathrm{d}W_s\right].$$ Here $\{W_t\}_{t\ge 0}$ is a standard $\mathbb{R}$-valued Brownian motion and $f:[0,\infty)\rightarrow\mathbb{R}$ is a $C^1$-function. Does someone know how to compute this?

Answer 1

I'm also studying a course where things like this appear, so I'll present an attempt but corrections are welcome. I know stochastic calculus doesn't always play by the rules.

I get that: $$\mathbb{E}[W_T \cdot \int_0^Tf(s)dW_s] = \int_0^Tf(s)ds$$

If $\mathbf{P} = \{t_0 = 0, t_1,t_2,\dots,t_n = T\} $ is a tagged partition, $||\mathbf{P}|| = \max_{k=1,\dots,n}{\{t_k -t_{k-1}\}}$ is the 'mesh' of the partition and $t_k^* \in (t_k, t_{k+1})$ are the tags then:

$$\mathbb{E}[W_T \cdot \int_0^Tf(s)dW_s] = \mathbb{E}[W_T \cdot \lim_{||\mathbf{P}|| \rightarrow0}\sum_{k=0}^{n-1}f(t_k^*)(W_{t_{k+1}}-W_{t_k})] = \lim_{||\mathbf{P}|| \rightarrow0}\mathbb{E}[\sum_{k=0}^{n-1}(W_{t_{k+1}}-W_{t_k}) \cdot \sum_{k=0}^{n-1}f(t_k^*)(W_{t_{k+1}}-W_{t_k})] = \lim_{||\mathbf{P}|| \rightarrow0}\mathbb{E}[\sum_{k=0}^{n-1}f(t_k^*)(W_{t_{k+1}}-W_{t_k})^2]$$

This line follows from linearity of the expectation (which allows us to take expectations termwise) and the independent increments property (whereby the expectation of a product of disjoint increments is zero).

$$= \lim_{||\mathbf{P}|| \rightarrow0}\sum_{k=0}^{n-1}f(t_k^*)\mathbb{E}[(W_{t_{k+1}}-W_{t_k})^2] = \lim_{||\mathbf{P}|| \rightarrow0}\sum_{k=0}^{n-1}f(t_k^*)(t_{k+1}-t_k) = \int_0^Tf(s)ds$$.