Krdytkyu

Let

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


• 
  $left(Y^{(n)}_kright)_{kinmathbb N_0}$ be a time-homogeneous Markov chain on $(Omega,mathcal A,operatorname P)$ and $$X^{(n)}_t=Y^{(n)}_{lfloor ntrfloor};;;text{for all }tin[0,1]$$
  


• 
  $(N_t)_{tge0}$ be a Poisson process on $(Omega,mathcal A,operatorname P)$ with intensity $1$ independent of $Y^{(n)}$ for all $ninmathbb N$ and $$Z^{(n)}_t:=begin{cases}Y^{(n)}_{N_{nt}}&text{for }tin[0,1)\ Z^{(n)}_{1-}end{cases}$$ for $ninmathbb N$
  

Now, let $tau_0:=0$, $$tau_k:=infleft{t>tau_{k-1}:Delta N_t=1right};;;text{for }kinmathbb N$$ as well as $$tau^{(n)}_k:=frac{tau_k}n;;;text{for }kinmathbb N$$ and $$lambda^{(n)}_t:=sum_{k=0}^infty 1_{left[frac kn,:frac{k+1}nright)}(t)left(tau^{(n)}_k+(nt-k)left(tau^{(n)}_{k+1}-tau^{(n)}_kright)right);;;text{for }tge0$$ for $ninmathbb N$.

How can we show that $$sup_{tge0}left|lambda^{(n)}_t-tright|=sup_{tin[0,:1)}left|frac{N_{nt}}n-tright|tag1$$ for all $ninmathbb N$? (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 $N_{nt}=lfloor nfrac{N_{nt}}nrfloor$ for all $tge0$ and $ninmathbb N$, 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 $kinmathbb N$, are meant), but I don't know why.

However, we may observe the following: Let $tinleft[frac kn,frac{k+1}nright)$. Then, there is an $alphain[0,1)$ with $$t=frac{k+alpha}n$$ and hence $$lambda^{(n)}_t-t=left(tau^{(n)}_k-frac knright)+left(tau^{(n)}_{k+1}-tau^{(n)}_kright)alpha.tag2$$ Since this is linear in $alpha$, the supremum of $(2)$ is attained for $alpha=0$ or $alpha=1$. Thus, we should obtain $$sup_{tge0}left|lambda^{(n)}_t-tright|=sup_{kinmathbb N_0}left|tau^{(n)}_k-frac knright|=frac1nsup_{kinmathbb N_0}left|tau_k-kright|tag3.$$ Maybe we need to use $k=N_{tau_k}$ almost surely such that $$sup_{tge0}left|lambda^{(n)}_t-tright|=sup_{kinmathbb N_0}left|frac{N_{ntau^{(n)}_k}}n-tau^{(n)}_kright|tag4;;;text{almost surely}.$$ Now, clearly, $left{N_{ntau^{(n)}_k(omega)}(omega):kinmathbb N_0right}=left{N_{tau_k(omega)}(omega):kinmathbb N_0right}=left{N_t(omega):tge0right}$, but this still doesn't yield $(1)$. However, it yields $$left[tau^{(n)}_k,tau^{(n)}_{k+1}right)ni tmapstofrac{N_{nt}}n-t=k-t$$ which clearly attains its supremum at $t=tau^{(n)}_k$ or $t=tau^{(n)}_{k+1}$.

We may note that $left(frac{N_{nt}}n-tright)_{tge0}$ is a martingale, but I'm not sure if we need this fact here.