0
$\begingroup$

Suppose I have a discrete-time Markov Chain $(X_n)_{n \geq 0}$ with the hitting time $H_x:= \inf \{n \geq 0 \colon X_n = x\}$ for some $x \in E$, where $E$ is a countable state space.

Consider now

$E[1(H_x < \infty) f(X_n)]$ and suppose that we can write $f(X_n)= g(X_n) \circ \theta_m$ where $\circ \theta_m$ denotes the composition with a shift operator (for whatever reason), then we have

$$E[1(H_x < \infty) f(X_n)]= \sum_{m =1}^{\infty} E[1(H_x = m) f(X_n)] = \sum_{m=1}^{\infty} E[1(H_x=m) g(X_n) \circ \theta_m].$$

When I want to apply the Markov property to it, is it the simple or the strong one, as I have that $H_x=m$, so a stopping time takes on this value, but we also have a deterministic value. I hope my question is clear! Thank you very much.

$\endgroup$
1
$\begingroup$

Having broken things down according to the value of $H_x$, you can use the simple Markov property on each term in the sum on the far right — the event $\{H_x=m\}$ is $\mathcal{F}_m$-measurable because $H_x$ is a stopping time .

Alternatively, you could write $f(X_n)$ as $g(X_n)\circ\theta_{H_x}$ on $\{H_x<\infty\}$ and use the strong Markov property at the stopping time $H_x$.

$\endgroup$
1
  • $\begingroup$ Thanks, this is exactly what I was looking for. As the conditioning on $\{H_x=m\}$ has other reasons than applying the Markov Property, I must go the first way, though! $\endgroup$
    – user136457
    Jul 11 '15 at 17:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.