Markov Chain holding time

Question

Let $X$ be a continuous-time Markov chain.

How does one justify

$P(X(s)=x,0\leq s\leq t\mid X(0)=x)=\lim_{n\to\infty}P(X(kt/n)=x,k=0,1,\dots,n\mid X(0)=x)$

without prior knowledge of the distribution of the holding times?

Answer 1

I will assume in addition that $X = X(t)_{t\geq 0}$ has right continuous sample paths since otherwise, the set $\{X(s) = x, 0 \leq s \leq t \}$ would not be even measurable. With right-continuity at hand, we can write

$$\{X(s) = x, 0 \leq s \leq t \} = \bigcap_{q \in \mathbb Q \cap [0,t]}\{X(q) =x \} $$ Let

$$ A_n := \bigcap_{i=0}^{2^n} \left\{ X \left(\frac{ti}{2^n} \right) = x \right\}.$$

By construction,

$$ A_{n+1} \subset A_n, \forall n \geq 0 \quad \text{and} \quad \bigcap_{n=0}^\infty A_n = \bigcap_{q \in \mathbb Q \cap [0,t]}\{X(q) =x \}. $$

Hence, by continuity of (probability) measures, (denote $\mathbb P^x:= P(\cdot|X(0)=x)$) one has

$$ \mathbb P^x(X(s) = x, 0\leq s \leq t) =\lim_{n\rightarrow \infty} \mathbb P^x(A_n). $$