- EdisonCruise
**Posts:**118**Joined:**

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
**Posts:**118**Joined:**

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).

(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.

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