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If $X_t$ is an $\mathbb{R}$- valued stochastic process with continuous paths, show that the following two conditions are equivalent:

(i) for all $f\in C^2(\mathbb{R})$ the process $$f(X_t) − f(X_0) −\int_0^t Af(X_s)ds$$ is a martingale,

(ii) for all $f\in C^{1,2}([0,\infty)\times\mathbb{R})$ the process $$f(t,X_t)−f(0,X_0)−\int_{0}^t(\partial_t f(X_s)+A f(X_s))ds$$ is a martingale.

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Hi, I can't remember the exactly the argument but essentially the goal of the game is to reduce the case (ii) to case (i) (and vice versa) by using the right transform of the transition Kernel and to see that infinitesimal generator of this now a homogenous Markov process is $\partial_t+ A$. I wouldn't be surprised if this was done in the book of Oksendal on SDEs. –  TheBridge Mar 30 '11 at 14:40
    
How did you get that $X_t$ is Markov? It doesn't follow from the statement of the problem. –  Ilya Apr 6 '11 at 19:16
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I'm trying to give a partial answer.

(ii) $\Rightarrow$ (i) is trivial.

(i) $\Rightarrow$ (ii). I cannot prove it completely, but the following argument should apply to the case $f(t,x)=g(t)h(x)$, where the variables are separable.

Assume (i). $M_t:=h(X_t) − h(X_0) −\int_0^t Ah(X_s)ds$ is a continuous martingale, with $M_0=0$. Apply the integration by parts formula for stochastic integrals, $$\int_0^t g(s)dM_s=g(t)M_t-\int_0^tM_sg'(s)ds.$$ The left-hand side is a martingale. One can show that the right-hand side equals $$g(t)h(X_t)-g(0)h(X_0)-\int_0^t [g'(s)h(X_s)+g(s)Ah(X_s)]ds,$$ because $$\int_0^tg'(s)\int_0^s Ah(X_u)duds=g(t)\int_0^tAh(X_s)ds-\int_0^tg(s)Ah(X_s)ds.$$ Thus, the spacial case is proved. To get the general result, use some kind of extension argument?

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