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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?

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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 $n\in\mathbb N$.

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