Hitting Times: Continuous-Time Chains and associated Jump chain

Question

Consider $A\subseteq I$ , with $I$ countable.

Define:

$(1)\quad \mathcal{D}^A:\Omega\rightarrow I \text{ , given by}\quad \mathcal{D}^A(\omega):=\inf\{t\geq 0:X_t(\omega)\in A \} \text{ (Hitting time for continuous Markov chain)}$

$(2)\quad \mathcal{H}^A:\Omega\rightarrow I \text{ , given by}\quad \mathcal{H}^A(\omega):=\inf\{ n\geq 0:Y_n(\omega)\in A \}$ (Hitting time for the jump chain)

In Chapter 3 of J. Norris's book Markov Chains, the author states that: $\{ \mathcal{D}^A<\infty \}=\{\mathcal{H}^A<\infty \}$

The following is my attempt at proving this result.

Suppose by absurd that $\{ \mathcal{D}^A<\infty \}\not\subset\{\mathcal{H}^A<\infty \}\iff \exists \text{ }\tilde{\omega}\in\Omega:\tilde{\omega}\in\{ \mathcal{D}^A<\infty \} \text{ }\&\text{ } \tilde{\omega}\not\in\{ \mathcal{H}^A<\infty \} .$

Then , follow that:

$(i)\quad \exists\text{ }\tilde{t}\geq 0:\mathcal{D}^A(\tilde{\omega})=\tilde{t},\text{ i.e , }\tilde{t}=\inf{\{t\geq 0:\tilde{X}_t:=X_t(\tilde{\omega})\in A\}}; $

$(ii)\quad \forall n\geq 0 ,\tilde{Y}_n:=Y_n(\tilde{\omega}):=X_{J_n}(\tilde{\omega})\not\in A \text{ , with }J_n\text{ jump time}$

Particularly $\tilde{Y}_0:=\tilde{X}_0\not\in A $ (Remeber $J_0:=0$), then $\tilde{X}_t\neq \tilde{X}_0$.

However, $\tilde{t}$ is the first time at which $\tilde{X}_t\in A$, So it is the first time at which they differ and therefore $\tilde{t}=J_1:=\inf{\{t\geq 0:\tilde{X}_t\neq\tilde{X_0} \}}.$

Then $\tilde{Y}_1:=\tilde{X}_{J_1}=\tilde{X}_{\tilde{t}}\in A$

Which is in contradiction with item (ii).

I believe the other inclusion would be similar.

My questions are:

Is the way I approached this correct?

Is there a more straightforward way to demonstrate this?

Answer 1

Your proof uses the equality $\mathcal{D}^A(\tilde{\omega}) = \tilde{t} = J_1(\tilde{\omega})$, which is not necessarily true. For instance, it is possible that $Y_0 \notin A$, $Y_1 \notin A$, and $Y_2 \in A$. A small modification fixes this issue: by the right-continuity of continuous-time Markov chains, $\mathcal{D}^A(\tilde{\omega}) = \tilde{t} = J_m(\tilde{\omega})$ for some $m \geq 0$.

I also have two recommendations. First, following the generally good practice of attempting a direct proof before attempting a proof by contradiction [https://math.stackexchange.com/q/319896], I find a direct proof of $\{\mathcal{D}^A < \infty\} \subseteq \{\mathcal{H}^A < \infty\}$ more straightforward: if $\mathcal{D}^A(\omega) < \infty$, then $\mathcal{D}^A(\omega) = J_m(\omega)$ for some $m \geq 0$, which implies that $\mathcal{H}^A(\omega) = \inf\{n \geq 0 : X_{J_n} \in A\} \leq m < \infty$.

Second, I personally found the extra tildes (as in $\tilde{X}_{\tilde{t}}$) a bit hard to read. In probability theory, we often omit any indication of the $\tilde{\omega}$'s in an expression like $J_1(\tilde{\omega}) = \inf\{t \geq 0 : X_t(\tilde{\omega}) \neq X_0(\tilde{\omega})\}$.

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Here is a very compact proof of the stated result. By the definition of the infimum and the right-continuity of continuous-time Markov chains, $$ \{\mathcal{D}^A < \infty\} = \bigcup_{t \geq 0} \{X_t \in A\} = \bigcup_{n \geq 0} \{X_{J_n} \in A\} = \{\mathcal{H}^A < \infty\}. $$ (One can read "$\bigcup_{t \geq 0}$" as "there exists $t \geq 0$ such that".)