How to compute $E[B_t(\int_0^t f(s)dB_s)]=E[\int_0^tdB_s\int_0^t f(s)dB_s]=E[\int_0^t\int_0^t f(s)dB_sdB_s] $?

Question

For standard $1$-dimensional Brownian motion $B_t$, if for a function $f$ we have $$ E\left[\int_0^t f(s)dB_s\right]=0 $$ what is the expectation of following and its distribution? $$ B_t\left(\int_0^t f(s)dB_s\right) $$

I try to rewrite it $$ E\left[B_t\int_0^t f(s)dB_s\right]=E\left[\int_0^tdB_s\int_0^t f(s)dB_s\right]=E\left[\int_0^t\int_0^t f(s)dB_sdB_s\right] $$ I am stuck here.

There is a useful formula: $$ \int_0^t f(s)dB_s=f(t)B_t-\int_0^tB_sdf(s) $$

Answer 1

As a consequence of the Itô isometry, the covariance of $B_t$ and $\int_0^t f(s) dB_s$ is given by:

$$\mathbb{E} \left( B_t \int_0^t f(s) dB_s \right) = \mathbb{E} \left( \int_0^t dB_s \int_0^t f(s) dB_s \right) = \int_0^t f(s) d\langle B, B \rangle_s = \int_0^t f(s) ds$$

Being the $L^2$ limit of Gaussian processes, $Y_t = \int_0^t f(s) dB_s$ is a Gaussian process itself, and by Itô's isometry, $Y_t \sim N\left(0, \int_0^t f(s)^2 ds \right)$. Thus, $(B_t, Y_t)$ is a jointly normal random variable, with mean zero and covariance given by: $$\Sigma = \begin{pmatrix} s& \int_0^t f(s)ds \\ \int_0^t f(s)ds & \int_0^t f(s)^2ds \end{pmatrix} $$

It follows that $B_t Y_t$ follows a product-normal distribution.