Itō Integral multiplied by Riemann Integral

Question

I was wondering whats the result of an Itō integral multiplied by a Riemann Integral. For example, what is $$\left(\int_0^T f(u)\ \mathsf dW_u\right)\left(\int_0^T g(v)\ \mathsf dv\right)$$ where $W$ is a standard Wiener process. Since $\mathsf dW_t\mathsf dt=0$, does that mean the multiplication of the two integrals is also zero? Is there a formal way of proving the result? Thanks in advance for any help.

Answer 1

Why it is not $0$:

The expression $\text{d}W_t\, \text{d}t = 0$ is valid when it is being jointly integrated with respect to the same variable $t$, thus, symbolically $$\int_{t=0}^T f(t) \text{d}W_t\, \text{d}t = 0.$$ In your case you have two different integrals with respect to different integration variables $u$ and $v$, so you can't say that $\text{d}W_u\, \text{d}v$ is null. You can interpret it as a "double differential" in a similar fashion than $\text{d}x\, \text{d}y$ in double integrals.

What is it then?

It is a good exercise to prove that an integral of the form $\int_0^T f(s)dW_s$ is normally distributed with zero mean and variance $\int_0^T |f(s)|^2ds $.

(Hint: write the Riemann sums and interpret them as a sum of independent normally distributed random variables.)

Thus, the factor to the left is a random variable. On the other hand, assuming that $g$ is a real valued function (non-random), the factor to the right is a number, call it $\sigma$. The product of those two then is a normally distributed random variable of mean $0$ and variance $\sigma^2\int_0^T |f(s)|^2ds$.