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Let $\theta \in \{0,1\}$ be an unknown state of the world. Let $P_0 := Prob(\theta = 1)$ at time $0$.

Let $G_t$ and $B_t$ be two Poisson processes with stochastic intensity $\lambda_g e_t \theta$ and $\lambda_b (1-e_t) (1-\theta)$. Let $\mathcal{F_t} := \sigma(G_s,B_s: s\in [0,t])$.

Let $P_t := Prob(\theta = 1 | \mathcal{F}_t)$.

The agent's control variable is $e_t$, which is $\mathcal{F}_t$ adapted.

Notice that at most one of $G_t$ or $B_t$ will generate any arrival. Moreover, if $G_t = 1 \Rightarrow P_t = 1$, $B_t = 1 \Rightarrow P_t = 0$. Otherwise, $$ dP_t = - [\lambda_g e_t - \lambda_b (1-e_t)] P_t (1-P_t) dt$$

Let $(p_1,p_2)$ be an interval such that $0 < p_1 < p_2 < 1$ and $s(p)$ be a continuous, bounded function on $[p_1,p_2]$.

\begin{align*} \tau := \inf \{t: P_t \notin(p_1,p_2)\} \end{align*}

The agent solves the following problem:

\begin{align*} V(p) := \sup_{e_t\in[0,1]} \mathbb{E} \left[ \int_0^\tau \exp\left(\int_0^t -s(P_u) du\right) \exp(-t) dt + e^{-\tau} V(P_\tau) \big| P_0 = p\right] \end{align*}

I would guess that for a smooth, well-behaved (Lipschitz) $s(p)$, $V$ would be $C^1$. But only argument to this effect that I can think of would need the viscosity approach. Is there a shorter way of doing this?

Thanks.

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