Can you factor out a known Brownian Motion during integration?

Question

Say I have a problem which is of the form, where $W(t)$ is a Brownian Motion and $f(a,b)$ is just some function where we will plug in terms $u, W(u)$. For this example, let's say that $f(a,b) = b-a^2b$. If I have the following form:

$$ \int^{t}_0 W(t)f\big(u,W(u)\big)dW(u), $$

would I be able to take out the $W(t)$ and treat it as if it were a constant so that we have the form

$$ W(t)\int^{t}_0 f\big(u,W(u)\big)dW(u)? $$

If not how would I go about solving such a problem?

Answer 1

The problem that I see with the first integral is that it might not be well-defined. Following Øksendal, Definition 3.1.4, the integrand, let us call it $F(s,\omega)$ with $s:=u$, must satisfy,

2. $F(s,\omega)$ is $\mathcal{F}_s$-adapted, i.e., $\omega\to F(s,\omega)$ must be $\mathcal{F}_s$ measurable for all $s$, where $\mathcal{F}_s$ is the $\sigma$-algebra generated by the Brownian motion up to $s$, Definition 3.1.2 Øksendal.

Clearly, this does not seem to be the case for your integral, as it depends on $W(t)$ with $t\ge s$ for all $s$.