Conditional expectation of controlled diffusion process with respect to a trajectory of the control Suppose $a_t$ is the solution to an SDE controlled by the process $b_t$ (both processes are defined on the same probability space)
\begin{align*}
    a_t &= a_0 + \int_0^t f_a(a_s, b_s)ds + \int_0^t \sigma_a(a_s, b_s) dW_s \\
    %b_t &=b_0 + \int_0^t f_b(a_s, b_s)ds + \int_0^t \sigma_b dW^b_s
\end{align*}
Let $b_{0\leq s\leq t}$ denote a trajectory of the process $b$ up to time $t$ and consider the conditional expectation
\begin{align*}
  \textbf a_t :=  \mathbb E[a_t \mid b_{0\leq s\leq t}]
\end{align*}
Question: given a trajectory $b_{\leq t}$ does $\textbf a_t$ solve the SDE
\begin{align*}
   \textbf a_t &= \textbf a_0 + \int_0^t f_a(\textbf a_s, b_s)ds
\end{align*}
?
What if $f_a$ is linear or when $\sigma_a$ is constant?
This sounds plausible by taking the conditional expectation of the initial SDE and using the fact that Ito stochastic integrals are martingales so should vanish in expectation, however, I am not able to prove it rigorously. Maybe with a generalization of Ito's lemma for stochastic processes?
Any help or pointers to references (books, papers) that could help me solve this would be really appreciated.
 A: If $b_t$ and $W_t$ are independent, we should have formally
$$\mathbb{E}\left[\int_0^t\sigma_a(a_s,b_s) dW_s| b_{0\leq \tau \leq t}\right] =\mathbb{E}\left[\lim_{n \to \infty} \sum_{k}^{n} \sigma_a(a_{\frac{kt}{n}},b_{\frac{tk}{n}}) (W_{\frac{t(k+1)}{n}}-W_{\frac{tk}{n}})| b_{0\leq \tau \leq t}\right] \\= \lim_{n \to \infty} \sum_{k}^{n}\mathbb{E}\left[ \sigma_a(a_{\frac{kt}{n}},b_{\frac{tk}{n}}) (W_{\frac{t(k+1)}{n}}-W_{\frac{tk}{n}})| b_{0\leq \tau \leq t}\right] $$
Now if I am not mistaken, without any additional assumptions this will be hard to tackle. If we have that $\sigma_a$ is constant, or only depends on $b$, then the conditional expectation will be $0$.
In this case, we are left with
$$\mathbb{E}\left[\int_0^t f_a(a_s,b_s)ds|b_{0\leq \tau \leq t}\right]= \int_0^t \mathbb{E}[f_a(a_s,b_s)|b_{0\leq \tau \leq t}] ds.$$
Again, in general we will not have that
$$ \mathbb{E}[f_a(a_s,b_s)|b_{0\leq \tau \leq t}] = f_a(\mathbb{E}[a_s|b_{0\leq \tau \leq t}],b_s),$$
so the assertion should not hold in general.
If $f$ is linear, we can however make this move and then indeed $ \textbf{a}$ solves the desired equation (still assuming that also $\sigma$ is constant).
