 EdisonCruise
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### How to derive Hamilton-Jacobi-Bellman Equation from supreme of expectation?

thanks Alan
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### How to derive Hamilton-Jacobi-Bellman Equation from supreme of expectation?

Here is a partial development.Define $g(t,V_t,X_t) = E_t[above]$, which can be rewritten $M_t = E_t[M_T]$, defining $M_t \equiv g(t,V_t,X_t)$. This shows $M_t$ is a martingale, which means its Ito's rule expansion has no "dt" term. That expansionis $dg = (g_t + A g) dt + ...$, so $g_t + A g = 0$, where $A g$ is the same operator on g as the operator on v in the "sup" in (2). Applying the sup to the construction left to you. p.s. This type of thing is textbook. See, for example, Merton's 'Continuous-time Finance', pg 128
Last edited by Alan on January 13th, 2015, 11:00 pm, edited 1 time in total. EdisonCruise
Topic Author
Posts: 97
Joined: September 15th, 2012, 4:22 am

### How to derive Hamilton-Jacobi-Bellman Equation from supreme of expectation?

Thank you, Alan. I know what you mean. I agree that Equation (2) can be derived from g(t, Vt, Xt)=sup[E( g(T, VT, XT) ) ].But I am still not sure how to get g(t, Vt, Xt) from Equation (1). There is an exponetial function and a parameter alpha in Equation (1), which finally disppear in Equaiton (2). Alan
Posts: 9788
Joined: December 19th, 2001, 4:01 am
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### How to derive Hamilton-Jacobi-Bellman Equation from supreme of expectation?

(2) still follows if (1) said: $v(t,V_t,X_t) = E_t[\phi(T,V_T,X_T)]$, assuming the expectation exists. In other words, the specific form of $\phi(\cdot)$ plays no role in the argument to get (2). It just serves as a terminal condition (like an initial condition) under which (2) must be solved. Everything I wrote still applies to this -- more general -- problem.
Last edited by Alan on January 14th, 2015, 11:00 pm, edited 1 time in total.  