SDE multiplied by previsible process

Question

I'm reading "Financial Calculus" of Baxter and Rennie, and have a question regarding some substitution in SDE. Let's suppose that $\Phi_t$ and $\Psi_t$ are previsible processes. We have SDE: $$dS_t=\sigma S_t dW_t + rS_t dt $$ where $\sigma$ and $r$ are constants. Suppose then, that $$dV_t=\Phi_tdS_t + \Psi_tdB_t$$ where $dB_t=rB_t dt$. Substituting first SDE into second one, we have $$dV_t=(\sigma S_t \Phi_t)dW_t + (rS_t\Phi_t + r\Psi_tB_t)dt$$ Although it's very clear in terms of calculations, I am just wondering why we are allowed to do that? I mean, why we can simply write $$\Phi_tdS_t=\Phi_t\sigma S_t dW_t + \Phi_trS_t dt$$ when $\Phi_t$ is not a constant? We use shortened notation above, but I am reffering to the notation with integrals e.g. (let's omit the second part of $V_t$ equation) $$V_t=V_0+\int_0^u\Phi_tdS_u$$ Thank you in advance for your help.

Answer 1

This substitution is valid by associativity of the stochastic integral.

Theorem (Associativity of the Stochastic integral): Let $X$ be a semimartingale and let $Z$ be an $X$-integrable process. Suppose that $Y = \int Z\,dX$. Then, $Y$ is a semimartingale and a predictable process $\alpha$ is $Y$-integrable if and only if $\alpha Z$ is $X$-integrable, in which case $$\displaystyle \int \alpha \,dY = \int \alpha Z \,dX. $$

For example, if $$S_t = S_0 + \int_0^t \sigma S_s \,dW_s + \int_0^t r S_s\, ds$$ then associativity of the stochastic integral tells us that for $X_t = \int_0^t \sigma S_s \,dW_s$ and $Y_s = \int_0^t r S_s \,ds$, we have \begin{align*} \int_0^t \Phi_s \,dS_s =& \int_0^t \Phi_s \,dX_s + \int_0^t \Phi_s \, dY_s \\=& \int_0^t \Phi_s \cdot (\sigma S_s) \, dW_s + \int_0^t \Phi_s \cdot (r S_s) \,ds \end{align*} which, in differential notation, is exactly $\Phi_t dS_t = \sigma S_t \Phi_t dW_t + r S_t \Phi_t dt$.