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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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    How did you get that $X_t$ is Markov? It doesn't follow from the statement of the problem.2011-04-06

1 Answers 1

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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?