Suppose I have two independent Brownian motions $B^1\_t, B^2\_t$ and $\mathbb F\_t$ be the natural filtration generated by them. Let $T > 0$ be a fixed finite number. Let $q\_t$ be a $[-1,1]$ valued $\mathbb{F}\_t$ martingale that the analyst controls. 


Let $\mathcal Q$ be the set of $[-1,1]$ valued $\mathbb F\_t$ martingales. The control problem is: $$\sup\_{q \in \mathcal Q} E\_{0,q\_0} f(q\_T, B^1\_T)$$


where $q\_0$ is the value of $q$ at time $0$. $f(\cdot, \cdot)$ is linear in each argument. As we can see, the objective function does not depend on $B^2\_t$. 


I would conjecture that, given the linearity of $f(\cdot, \cdot)$ in each of its argument and no dependence on $B^2\_t$, that we can restrict attention to $[-1,1]$ valued $\sigma(B^1\_t)$ martingales where $\sigma(B^1\_t)$ is a coarser filtration generated only by $B^1\_t$. 


Does that sound like a reasonable conjecture? And how can I argue this if true? 



Conditioning on an irrelevant variable in a martingale control problem
This is true. For any $q \in \mathcal{Q}$, you may define $\tilde{q}$ as the *optional projection* of $q$ onto the filtration generated by $B^1$. This $\tilde{q}$ remains a martingale and achieves the same value thanks to linearity of $f$. More generally, using Jensen’s inequality, your conjecture is true as long as $f$ is concave in $q$.


This notion of *optional projection* is well established in stochastic analysis. Essentially, it gives you a process $\tilde{q}$ adapted to the $B^1$ filtration, which I’ll denote $F^1$, satisfying $\tilde{q}\_t = E[q\_t | F^1\_t]$ almost surely for each $t$. A google search should lead you easily to references on how it is defined and the theorem stating that the optional projection of a martingale remains a martingale.


