Let $\mathcal{A}\subseteq\mathbb R$ be a compact interval, $T\in\mathbb R_+$ be a finite horizon, and $g:\mathbb R\to\mathbb R_+$ be a continuous function with $g\leq 1+|\cdot|$. Consider an optimal control problem where, observing a standard Brownian motion $\{B_t\}_t$, I choose an $\mathcal A$-valued control $\{a_t\}_t$ to control the drift. I want to maximize $\mathbb E g(X_T)$, where $X_t:=B_t + \int_0^t a_s \text{ d}s$.

I have proven the following:

1) The optimal value function $v:[0,T]\times\mathbb R\to\mathbb R_+$ (taking a time $t$ and current level of $X_t$ to an optimal continuation value) is finite-valued.

2) There is a continuous cutoff function $\kappa:[0,T]\to\mathbb R$ such that, for every $t\in[0,T)$, the function $v(t,\cdot)$ is strictly increasing on $[\kappa(t),\infty)$ and strictly decreasing on $(-\infty,\kappa(t)]$.

I want to show that there is an optimal control where I choose $\max \mathcal A$ whenever $X_t > \kappa(t)$ and $\min \mathcal A$ whenever $X_t < \kappa(t)$. Ideally, I would show that any other optimal control almost surely agrees with this one at almost every time.

This seems, intuitively, like it must be true, but I don't see an easy proof (or know a reference), especially given that I've made no differentiability assumptions.

  • $\begingroup$ "The optimal value function $v$ ... is finite-valued" you certainly need some assumptions about the growth of $g$. $\endgroup$
    – zhoraster
    Apr 15, 2020 at 13:52
  • $\begingroup$ Yes, sorry for being unclear. I meant that I had an instance of this problem in which I had already proven it was finite-valued. In my instance, $g\leq 1+|\cdot|$, so that finite-valuedness is easy. I'll edit to add this feature to the problem. $\endgroup$ Apr 15, 2020 at 15:05


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