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Starting with the Bellman equation for the optimal stopping problem: $$F(x,t)=max\{\Omega(x,t), \pi(x,t)+(1+\rho dt)^{-1} E[F(x+dx, t+dt)|x]\}$$ In the continuation region where the second term is the greatest they get the following equation after expanding by Ito's Lemma: $$\frac{1}{2}b^2(x,t)F_{xx}(x,t)+a(x,t)F_x(x,t)+F_t(x,t)-\rho F(x,t)+\pi(x,t)=0 $$ (13)

$F(x,t)$ is the value of an investment. $\pi(x,t)$ is the profit flow, $\rho$ is the discount rate and $\Omega(x,t)$ is the value achieved if stopping. $x$ is assumed to follow an Ito process: $$dx=a(x,t)dt+b(x,t)dz$$ I dont understand how (13) is reached. The closest i get is: $$F(x,t)=\pi(x,t)+(1+\rho dt)^{-1}E[F(x+dx, t+dt)|x]$$ Multiplying with $(1+\rho dt)$ $$F(x,t)+\rho F(x,t)dt=\pi(x,t)(1+\rho dt)+E[F(x+dx, t+dt)|x]$$ $$\rho F(x,t)dt=\pi(x,t)(1+\rho dt)+E[F(x+dx, t+dt)|x]-F(x,t)$$ $$\rho F(x,t)dt=\pi(x,t)(1+\rho dt)+E[F(x+dx, t+dt)-F(x,t)|x]$$ $$\rho F(x,t)dt=\pi(x,t)(1+\rho dt)+E[dF|x]$$ Expanding dF with Itos lemma: $$dF=[F_t(x,t)+a(x,t)F_x(x,t)+\frac{1}{2}b(x,t)^2F_{xx}(x,t)]dt+b(x,t)F_x(x,t)dz$$ Now I can continue. $$\rho F(x,t)dt=\pi(x,t)(1+\rho dt)+[F_t(x,t)+a(x,t)F_x(x,t)+\frac{1}{2}b(x,t)^2F_{xx}(x,t)]dt$$ Divide by $dt$ $$\rho F(x,t)=\pi(x,t)\frac{1+\rho dt}{dt}+F_t(x,t)+a(x,t)F_x(x,t)+\frac{1}{2}b(x,t)^2F_{xx}(x,t)$$ Then rearranging I may have. $$\frac{1}{2}b(x,t)^2F_{xx}(x,t)+a(x,t)F_x(x,t)+F_t(x,t)-\rho F(x,t)+\pi(x,t)\frac{1+\rho dt}{dt}=0$$ To have the same equation as in the book i can think of two adjustment/assumptions. $\rho dt = 0$ and $\pi(x,t)$ needs to be replaced with $\pi(x,t)dt$ in the first Bellman equation. Is this how equation (13) is achieved or am I missing something?

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