Expectation of product of brownian motion and stochastic integral

Question

Let $f:[0,\infty)\to\mathbb{R}$ be a deterministic continuous function and $B$ a Brownian motion with $B_0=0$. I need to prove that $$\mathbb{E}\left(B_t\int_0^tf(s)dB_s\right)=\int_0^tf(s)ds.$$ I really am not sure where to start with this. I have considered trying to use limit definition of the stochastic integral by an approximation of $f$ as a simple function, but then I am unsure about taking the limit out of the expectation and if this would even really take me anywhere.

Answer 1

Hint: Compute $d(X_tY_t)$ where $X_t = B_t$ and $Y_t = \int_0^t f(s)dB_s$.

Answer 2

Another option could be that of using Itô's isometry.

Notice that

$$\mathbb E\left[\bigg|\int_0^{\infty} f(s)dB(s)+\int_0^{\infty} g(s)dB(s)\bigg|^2\right]=\mathbb E\left[\bigg|\int_0^{\infty} [f(s)+g(s)] dB(s)\bigg|^2\right]=\int_0^{\infty}[f(s)+g(s)]^2 ds,$$ and this implies that $$\mathbb E\left[\left(\int_0^{\infty} f(s)dB(s)\right)\left(\int_0^{\infty} g(s)dB(s)\right)\right]=\int_0^{\infty} f(s)g(s)ds.$$

Now notice that $B_t=\int_0^{\infty} \mathbf{1}_{[0,t]}(s)dB(s)$ and then the result follows.