Random time change for a Poisson process and convergence with respect to the Skorohod topology Let


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*$(\Omega,\mathcal A,\operatorname P)$ be a probability space

*$\left(Y^{(n)}_k\right)_{k\in\mathbb N_0}$ be a time-homogeneous Markov chain on $(\Omega,\mathcal A,\operatorname P)$ and $$X^{(n)}_t=Y^{(n)}_{\lfloor nt\rfloor}\;\;\;\text{for all }t\in[0,1]$$

*$(N_t)_{t\ge0}$ be a Poisson process on $(\Omega,\mathcal A,\operatorname P)$ with intensity $1$ independent of $Y^{(n)}$ for all $n\in\mathbb N$ and $$Z^{(n)}_t:=\begin{cases}Y^{(n)}_{N_{nt}}&\text{for }t\in[0,1)\\ Z^{(n)}_{1-}\end{cases}$$ for $n\in\mathbb N$
Now, let $\tau_0:=0$, $$\tau_k:=\inf\left\{t>\tau_{k-1}:\Delta N_t=1\right\}\;\;\;\text{for }k\in\mathbb N$$ as well as $$\tau^{(n)}_k:=\frac{\tau_k}n\;\;\;\text{for }k\in\mathbb N$$ and $$\lambda^{(n)}_t:=\sum_{k=0}^\infty 1_{\left[\frac kn,\:\frac{k+1}n\right)}(t)\left(\tau^{(n)}_k+(nt-k)\left(\tau^{(n)}_{k+1}-\tau^{(n)}_k\right)\right)\;\;\;\text{for }t\ge0$$ for $n\in\mathbb N$.

How can we show that $$\sup_{t\ge0}\left|\lambda^{(n)}_t-t\right|=\sup_{t\in[0,\:1)}\left|\frac{N_{nt}}n-t\right|\tag1$$ for all $n\in\mathbb N$? (I'm unsure whether there isn't an error in the domain of one of the suprema. I could imagine that the left should be taken only over $[0,1)$ or the right actually over $[0,\infty)$.)

I've read that we obtain claim by noting that $N_{nt}=\lfloor n\frac{N_{nt}}n\rfloor$ for all $t\ge0$ and $n\in\mathbb N$, but while that's trivially true, I don't see how we can make use of it.
Moreover, I've read that the supremum is attained at one of the jump times (I guess the jump times of $X^{(n)}$, i.e. $\frac kn$ for $k\in\mathbb N$, are meant), but I don't know why.
However, we may observe the following: Let $t\in\left[\frac kn,\frac{k+1}n\right)$. Then, there is an $\alpha\in[0,1)$ with $$t=\frac{k+\alpha}n$$ and hence $$\lambda^{(n)}_t-t=\left(\tau^{(n)}_k-\frac kn\right)+\left(\tau^{(n)}_{k+1}-\tau^{(n)}_k\right)\alpha.\tag2$$ Since this is linear in $\alpha$, the supremum of $(2)$ is attained for $\alpha=0$ or $\alpha=1$. Thus, we should obtain $$\sup_{t\ge0}\left|\lambda^{(n)}_t-t\right|=\sup_{k\in\mathbb N_0}\left|\tau^{(n)}_k-\frac kn\right|=\frac1n\sup_{k\in\mathbb N_0}\left|\tau_k-k\right|\tag3.$$ Maybe we need to use $k=N_{\tau_k}$ almost surely such that $$\sup_{t\ge0}\left|\lambda^{(n)}_t-t\right|=\sup_{k\in\mathbb N_0}\left|\frac{N_{n\tau^{(n)}_k}}n-\tau^{(n)}_k\right|\tag4\;\;\;\text{almost surely}.$$ Now, clearly, $\left\{N_{n\tau^{(n)}_k(\omega)}(\omega):k\in\mathbb N_0\right\}=\left\{N_{\tau_k(\omega)}(\omega):k\in\mathbb N_0\right\}=\left\{N_t(\omega):t\ge0\right\}$, but this still doesn't yield $(1)$. However, it yields $$\left[\tau^{(n)}_k,\tau^{(n)}_{k+1}\right)\ni t\mapsto\frac{N_{nt}}n-t=k-t$$ which clearly attains its supremum at $t=\tau^{(n)}_k$ or $t=\tau^{(n)}_{k+1}$.
We may note that $\left(\frac{N_{nt}}n-t\right)_{t\ge0}$ is a martingale, but I'm not sure if we need this fact here.
 A: New edit. An update has been introduced to highlight that equation (1) as stated does not hold unless the domain of the $\sup$ in both sides of the equation is $\mathbb{R}_{+}$.

Maybe I am missing something, but you have actually done everything. We have,
$\sup_{t\geq 0}\left|\lambda^{(n)}_t-t\right|=\frac{1}{n}\sup_{k\in\mathbb{N}_0} \left|\tau_k-k\right|$ (this is your equation (3))
$\sup_{t\geq 0} \left|\frac{N_{nt}}{n}-t\right|=\sup_{t\geq 0} \left|\frac{N_{nt}-nt}{n}\right|=\frac{1}{n}\sup_{t\geq 0} \left|N_{nt}-nt\right|=\frac{1}{n}\sup_{k\in\mathbb{N}_0} \left|N_{\tau_k}-\tau_k\right|$ (as before, without loss of optimality, we can restrict the domain to the jump moments of the Poisson process $N_t$, hence the last equality.)
As you pointed $N_{\tau_k}=k$, and the identity $\sup_{t\geq 0} \left|\frac{N_{nt}}{n}-t\right|=\sup_{t\geq 0}\left|\lambda^{(n)}_t-t\right|$ holds.
Remark. Identity (1) does not hold with the domains $t\geq 0$ on the left hand side and $t\in \left[\left.0,1\right)\right.$ on the right hand side. In fact, in this case $\sup_{t\in \left[\left.0,1\right)\right.} \left|\frac{N_{nt}}{n}-t\right|=\frac{1}{n}\sup_{k\in\left\{0,1,\ldots,N_n\right\}}\left|\tau_k-k\right|\vee \left|n-N_n\right|$ and we have that $\sup_{k\in\left\{0,1,\ldots,N_n\right\}}\left|\tau_k-k\right|\vee \left|n-N_n\right|<\sup_{k\in\mathbb{N}_0}\left|\tau_k-k\right|$ with positive probability, in particular, the strict inequality holds for all realizations $\omega\in\Omega$ so that $\tau_1(\omega)>n+\epsilon$ with $\epsilon>0$ -- which holds with positive probability.
