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Let $X$ be a RCLL Markov Process with generator $A$. Then I know that

$$ M^f = f(X)-f(X_0)-\int Af(X_s)ds $$

is a martingal for every $f\in \mathcal{D}_A$. If we suppose that $Af=0$, we see that $f(X)-f(X_0)$ is a martingale. Further I know that from the Markov property

$$E[f(X_{t+h})|\mathcal{F}_t]=P_hf(X_t)$$

Where $(P_t)$ is the transition semigroup. Let $X_0=x$ be the starting point. If we assume that $f(X)-f(x)$ is a martingale, then

$$P_tf(x)-f(x)=E[f(X_t)-f(x)]=0$$

Why is this equation true? I wanted to use the property above with conditional distribution, without success.

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2 Answers 2

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If $Af=0$, then, for every $t\geqslant0$, $\mathrm e^{tA}f=f+\displaystyle\sum_{n\geqslant0}\frac{t^{n+1}}{(n+1)!}A^n(Af)=f$. By definition of $A$, $\mathrm e^{tA}=P_t$. Hence $P_tf=f$.

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Why does from $P_tf=f$ follows $P_tf(x)=E[f(X_t)]$. We slightly covered Markov process in our class, so sorry for my ignorance of this. –  user20869 Jul 27 '12 at 10:05
    
?? How do you define $P_tf(x)$? –  Did Jul 27 '12 at 10:13

$P_t$ is defined to be the transition semigroup, thus the definition is given by $P_t f(x) = \mathbb{E}[f(X_t) \, | \, X_0 = x]$.

There seems to be some confusion, as what you know is that $$ M^f := f(X_t) - f(X_0) - \int_0^t Af(X_s) ds $$ is a martingale for every $f \in \mathcal{D}_A$, and not that $f(X) - f(X_0) - \int_0^t Af(X_s) ds$ is a martingale. The expression $f(X)$ does not make any sense.

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