Does this martingale have right-continuous (or cadlag) sample paths?

Question

Let $(X_t)$ be an irreducible continuous-time Markov chain $(X_t)_{t \ge 0}$ with respect to its canonical filtration $(\mathcal G_t)_{t \ge 0}$. Suppose

• $(\mathcal G_t)_{t \ge 0}$ is completed and right-continuous.

• The state space $V$ is finite end endowed with discrete topology.

Let $f, \phi$ be functions from $V$ to $\mathbb R_+$ and $a \in (0,1)$. Due to the finiteness of $V$ and Dynkin’s formula, the stochastic process $(M_t)_{t \ge 0}$ defined by $$M_t = f\left(X_{t}\right) a^{t} -f(x)-\int_{0}^{t} \phi\left(X_{s}\right) a^{s}\mathrm{d} s$$ is a $\mathcal G_t$-martingale under $\mathbb P_x$ for all $x \in V$.

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I would like to ask if the martingale $(M_t)_{t \ge 0}$ defined above has right-continuous (or cadlag) sample paths.

Thank you so much for your help!

Answer 1

For every submartingale $M$ holds, that it has a cadlag version if and only if it $t\mapsto E(M_t)$ is right-continuous. Since $E(M_t)=0$ for martingales, $M$ has a cadlag version.

I'm not sure whether this sufficient to answer your question.

Answer 2

The paths of the Markov chain $(X_t)$ can be taken to be a right-continuous step functions. In this case the paths of $(M_t)$ are cadlag.