# Steps of a Markov chain subordinated to a Poisson process

Let

• $$(\Omega,\mathcal A,\operatorname P)$$ be a probability space
• $$\left(Y^{(n)}_k\right)_{k\in\mathbb N_0}$$ be a time-homogeneous Markov chain on $$(\Omega,\mathcal A,\operatorname P)$$
• $$D([0,1]):=\left\{f:[0,1]\to\mathbb R\mid f\text{ is càdlàg and left-continuous at }1\right\}$$
• $$X^{(n)}$$ be a $$D([0,1])$$-valued random variable on $$(\Omega,\mathcal A,\operatorname P)$$ with $$X^{(n)}_t=Y^{(n)}_{\lfloor nt\rfloor}\;\;\;\text{for all }t\in[0,1)$$
• $$(N_t)_{t\ge0}$$ be a Poisson process on $$(\Omega,\mathcal A,\operatorname P)$$ with intensity $$1$$ independent of $$Y^{(n)}$$ for all $$n\in\mathbb N$$ and $$Z^{(n)}_t:=\begin{cases}Y^{(n)}_{N_{nt}}&\text{for }t\in[0,1)\\ Z^{(n)}_{1-}\end{cases}$$

I've read that first $$(n-1)\wedge N_{n-}$$ steps of $$X^{(n)}$$ and $$Z^{(n)}$$ coincide, but appear at different times ($$\frac kn$$ vs the $$k$$th jump time of $$\left(N_{nt}\right)_{t\ge0}$$). What's exactly meant and how can we prove it rigorously?

I've got a vague intuition what's meant: Assuming that $$N$$ is right-continuous (in practice, we can always find a right-continuous modification), $$N$$ is almost surely nondecreasing and makes almost surely jumps of size $$1$$. So, $$t\mapsto N_{nt}$$ somehow behaves like $$t\mapsto\lfloor nt\rfloor$$, but the time-scale is stretched. How can we formulate this rigorously?

What's confusing me most is that it's written that only the first $$(n-1)\wedge N_{n-}$$ steps coincide. Where does the $$N_{n-}=\lim_{t\to n-}N_t$$ come from? Shouldn't the first $$n-1$$ steps coincide?

## 1 Answer

It's easy to see that $$X^{(n)}_{[0,\:1)}=\left\{Y^{(n)}_0,\ldots,Y^{(n)}_{n-1}\right\}$$ and $$Z^{(n)}_{[0,\:1)}=\left\{Y^{(n)}_0,\ldots,Y^{(n)}_{N_{n-}}\right\}$$ for all $$n\in\mathbb N$$. In order to make these processes coincide, we can apply a random time change $$\lambda^{(n)}_t:=\sum_{k=0}^\infty 1_{\left[\frac kn,\:\frac{k+1}n\right)}(t)\left(\tau^{(n)}_k+(nt-k)\left(\tau^{(n)}_{k+1}-\tau^{(n)}_k\right)\right)\;\;\;\text{for }t\ge0$$ with $$\tau_0:=0$$, $$\tau_k:=\inf\left\{t>\tau_{k-1}:\Delta N_t=1\right\}\;\;\;\text{for }k\in\mathbb N$$ and $$\tau^{(n)}_k:=\frac{\tau_k}n\;\;\;\text{for }k\in\mathbb N$$ for $$n\in\mathbb N$$.