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Let $(\Omega,\mathcal F, \mathbb P)$ be a probability space on which a Brownian motion $W$ is defined, and $\mathcal U$ be the set of progressively measurable (w.r.t. the Brownian filtration) processes $p=(p_t)_{t\ge 0}$ taking values in $\mathbb R_+$ s.t.

$$\mathbb E\left[\int_0^T p_t dt\right]<\infty,\quad \forall T\ge 0. $$

For each $n\ge 1$, denote by $\mathcal U_n\subset \mathcal U$ the subset of $p=(p_t)_{t\ge 0}$ taking values in $[1/n,\infty)$. Define, for $p\in\mathcal U$, $t\in [0,1]$ and $x\in [0,1]$,

\begin{eqnarray} J(t,x,p):=\mathbb E\left[\int_t^{\min(1,\tau^{p,t,x})}\big(1+\log(p_s)\big)ds\right] \end{eqnarray}

where $\tau^{p,t,x}:=\{s\ge t: X^{p,t,x}_s\notin (0,1)\}$ and $dX^{p,t,x}_s=\sqrt{2p_s}dW_s$ for all $s\ge t$ with $X^{p,t,x}_t:=x$. Can we prove the existence of some $N$ s.t. for any $p\in\mathcal U$, there exists $q\in\mathcal U_N$ satisfying $J(t,x,p)\le J(t,x,q)$?

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  • $\begingroup$ there are some notation issues to fix first, eg. in the integral you have $\log p_{s}$ but I think you meant $\log p_{u}$. Also, later you again use the same variable $u$ for both integrating and to denote constant control. $\endgroup$ Commented Feb 16, 2023 at 19:06
  • $\begingroup$ For Q1, How about we replace the supremum over $p\in \mathcal{U}_{n}$ by over $p\in \mathcal{U}$ by writing $1_{[1/n,n]}p$ inside the expected value? Then it is more of a question of applying Dominated convergence theorem in the expected value and integral. $\endgroup$ Commented Feb 16, 2023 at 19:14
  • $\begingroup$ For Q2, it seems unlikely that there is a deterministic such $N$ because the stopping time $\tau^{p,t,x}$ shows up in the integral it varies randomly based on $n$. It would mean that the maximizer $p_{*}$ is almost surely within a deterministic interval $[0,M]$ but that is never case for random objects because it suffices to take some realization $\omega$ where all the $p$ are large. $\endgroup$ Commented Feb 16, 2023 at 19:41
  • $\begingroup$ @ThomasKojar Thanks for pointing out the typo. It's corrected $\endgroup$
    – Fawen90
    Commented Feb 17, 2023 at 9:28

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