• $(\Omega,\mathcal A,\operatorname P)$ be a probability space
  • $(X_t)_{t\ge0}$ be a Feller process on $(\Omega,\mathcal A,\operatorname P)$
  • $(h_d)_{d\in\mathbb N}\subseteq(0,\infty)$ with $$h_d\xrightarrow{n\to\infty}0$$
  • $\left(Y^{(d)}_n\right)_{n\in\mathbb N_0}$ be a time-homogeneous Markov chain on $(\Omega,\mathcal A,\operatorname P)$ and $$X^{(d)}_t:=Y^{(d)}_{\lfloor\frac t{h_d}\rfloor}\;\;\;\text{for }t\ge0$$ for $d\in\mathbb N$
  • $N$ be a Poisson process on $(\Omega,\mathcal A,\operatorname P)$ with parameter $1$ independent of $Y^{(d)}$ for all $d\in\mathbb N$ and $$N^{(d)}_t:=N_{\frac t{h_d}}\;\;\;\text{for }t\ge0$$ as well as $$\tilde X^{(d)}_t:=Y^{(d)}_{N^{(d)}_t}\;\;\;\text{for }t\ge0$$ for $d\in\mathbb N$

Note that $N^{(d)}$ is a Poisson process with parameter $h_d^{-1}$ for all $d\in\mathbb N$.

How can we show that (in probability with respect to the Skorohod topology) $X^{(d)}\xrightarrow{d\to\infty}X$ iff $\tilde X^{(d)}\xrightarrow{d\to\infty}X$?

In the book of Kallenberg, the author is mentioning that the claim follows from the following two theorems:


I don't get how we need to apply them. Clearly, for fixed $t\ge0$, we can consider $$\frac1d\sum_{i=1}^d\left(N^{(i)}_t-N^{(i-1)}_t\right)$$ with $N^{(0)}_t:=0$. However, while independent, the $N^{(i)}_t-N^{(i-1)}$ are not identically distributed ...

If it's hard to prove in the general setting, it's okay for me to assume $h_d^{-1}=d$ for all $d\in\mathbb N$. In that case, the strong law of large numbers yields $$\sup_{t\in[0,\:T]}\left|\frac1d N^{(d)}_t-t\right|\xrightarrow{d\to\infty}0\;\;\;\text{almost surely for all }T>0\tag1.$$ Now, let $\tau^{(d)}_0:=0$, $$\tau_n^{(d)}:=\inf\left\{t>\tau^{(d)}_{n-1}:N^{(d)}_t-N^{(d)}_{\tau^{(d)}_{n-1}}>0\right\}\;\;\;\text{for }d\in\mathbb N$$ and $$\lambda^{(d)}_t:=\sum_{n=0}^\infty1_{\left[\frac nd,\:\frac{n+1}d\right)}(t)\left(\tau^{(d)}_n+(dt-n)\left(\tau^{(d)}_{n+1}-\tau^{(d)}_n\right)\right)\;\;\;\text{for }t\ge0$$ for $d\in\mathbb N$. Moreoverr, let $T>0$ and $\rho_T$ denote the metric inducing the Skorohod $J_1$-topology on the space of càdlàg functions $[0,T]\to\mathbb R$. We should obtain $$\rho_T\left(X^{(d)},\tilde X^{(d)}\right)\le\sup_{t\in[0,\:T]}\left|\lambda^{(d)}_t-t\right|+\sup_{t\in[0,\:T]}\left|X^{(d)}_t-\tilde X^{(d)}_{\lambda^{(d)}_t}\right|\tag2,$$ where the last term should be $0$. So, if we could show that the first term converges in probability to $0$ as $d\to\infty$, we should be able to conclude (since $T$ was arbitrary).


Update: The consequences of Theorem 1, below, have been more neatly stated in Corollary 1 and Lemma 1. To Lemma 1, I am adding an extra-assumption: the limit process $X$ has continuous sample-paths, almost surely.

I will build from what you proposed and I may repeat some passages to make sure everything is in place.

Theorem 1. $\mathbb{P}\left(\rho_{T}\left(\overline{X}^{(d)},X^{(d)}\right)>\epsilon\right)\overset{d\rightarrow \infty}\longrightarrow 0$, for any $\epsilon>0$.

Theorem 1 will imply Corollary 1 and Lemma 1 further ahead.

Preliminary Remarks. I am assuming that $D_{\left[0,T\right]}$ is endowed with the (incomplete) metric $\rho_{T}(X,Y)=\inf_{\lambda\in \Lambda_T}\left\{\left|\left|\lambda-{\sf id}\right|\right|\vee \left|\left|X\circ\lambda-Y\right|\right|\right\}$, where

$\Lambda_T\overset{\Delta}=\left\{\lambda\,:\,\left[0,T\right]\rightarrow \left[0,T\right]\,:\,\lambda\mbox{ is bijective, continuous and }\lambda(0)=0,\,\lambda(T)=T\right\}$;

${\sf id}$ is the identity map from $\left[0,T\right]$ onto itself; and we have defined

$\left|\left|X\right|\right|=\sup_{t\in\left[0,T\right]} \left|X(t)\right|$

as the $\sup$ norm on the interval $\left[0,T\right]$.

Note that for a particular $\lambda\in\Lambda_T$, we have $\rho_T(X,Y)\leq\left|\left|\lambda-{\sf id}\right|\right|+\left|\left|X\circ\lambda-Y\right|\right|$ (as pointed out in your equation (2)).

Proof to Theorem 1. We need to slightly correct your $\lambda^{(d)}_t$ so it belongs to $\Lambda_T$ (almost surely), since in your case $\lambda_T^{(d)}\neq T$ almost surely (and we need $\lambda(0)=0$ and $\lambda(T)=T$). Define $n^{\star}\overset{\Delta}=\min\left\{n\in\mathbb{N}_0\,:\,\tau_{n+1}^{(d)}> T\right\}$, -- note that $n^{\star}(\omega)=N_{T}^{(d)}(\omega)$ for all $\omega\in\Omega$, -- and let us redefine your $\lambda_t^{(d)}$ rather as


Now we have that $\lambda^{(d)}_t\in \Lambda_T$ for all $d$, almost surely. Note, in particular, that $\lambda^{(d)}_T=T$.

We have that

$\rho_T(\overline{X}^{(d)},X^{(d)})\leq\left|\left|\lambda^{(d)}-{\sf id}\right|\right|+\left|\left|\overline{X}^{(d)}\circ\lambda^{(d)}-X^{(d)}\right|\right|=\left|\left|\lambda^{(d)}-{\sf id}\right|\right|.\tag{1}$

Note that without the correction on $\lambda^{(d)}_t$ the second term on the left-hand side of the identity above would not be zero.

Now, we observe that

$\left|\left|\lambda^{(d)}-{\sf id}\right|\right|=\frac{1}{d}\sup_{k\in\left\{0,1,\ldots,N_{T}^{(d)}\right\}}\left|\tau_k-k\right|=\sup_{t\in\left[0,\tau_{n^{\star}}\right]}\frac{1}{d}\left|N_{t}^{(d)}-td\right|\leq \sup_{t\in\left[0,T\right]}\frac{1}{d}\left|N_{t}^{(d)}-td\right|$,

where for the first identity above: i) without loss of optimality, we can restrict attention to the jump moments plus the final moment $T$; (ii) in the final moment $T$, $\lambda_T^{(d)}-T=0$, thus we can restrict attention to the jumps within the interval $\left[0,T\right]$ and ignore the moment $T$. Observe that without the correction on $\lambda_t^{(d)}$ the first identity would not hold true (and the devised upper-bound above would not follow).

Note that $N^{(d)}_t-td$ is a martingale with $N_{T}^{(d)}-Td\in L_2$ and from Doob's inequality

$\mathbb{P}\left(\sup_{t\in\left[0,T\right]}\left|\lambda_{t}^{(d)}-t\right|>\epsilon\right)\leq\mathbb{P}\left(\sup_{t\in\left[0,T\right]}\frac{1}{d}\left|N_{t}^{(d)}-td\right|>\epsilon\right)\leq \frac{E\left[\left(N_{T}^{(d)}-Td\right)^2\right]}{d^2\epsilon^2}=\frac{Td}{d^2 \epsilon^2}=\frac{T}{d\epsilon}\overset{d\rightarrow \infty}\longrightarrow 0$.

From the bound (1), we have $\rho_T(\overline{X}^{(d)}(\omega),X^{(d)}(\omega))>\epsilon \Rightarrow \left|\left|\lambda^{(d)}(\omega)-{\sf id}\right|\right|>\epsilon$ and thus,

$\mathbb{P}\left(\rho_T(\overline{X}^{(d)},X^{(d)})>\epsilon\right)\leq \mathbb{P}\left(\left|\left|\lambda^{(d)}-{\sf id}\right|\right|>\epsilon\right)\overset{d\rightarrow\infty}\longrightarrow 0$. $\tag*{$\blacksquare$}$

Corollary 1.[Convergence in probability] For every $T>0$, we have

$\mathbb{P}\left(\rho_T\left(X^{(d)},X\right)>\epsilon\right)\longrightarrow 0$ for all $\epsilon>0$ $\Leftrightarrow \mathbb{P}\left(\rho_T\left(\overline{X}^{(d)},X\right)>\epsilon\right)\longrightarrow 0$ for all $\epsilon>0$, i.e., $X^{(d)}\rightarrow X$ in probability w.r.t. $\left(\rho_T, D_{\left[0,T\right]}\right)$ if and only if $\overline{X}^{(d)}\rightarrow X$ in probability w.r.t. $\left(\rho_T, D_{\left[0,T\right]}\right)$.

Proof to Corollary 1. Obvious from Theorem 1. $\tag*{$\blacksquare$}$

In what follows, $\rho^{o}_T$ is a metric that is topologically equivalent to $\rho_T$, i.e., it induces the same (Skorokhod) topology on $D_{\left[0,T\right)}$, except that the metric space $\left(\rho^{o}_T,D_{\left[0,T\right)}\right)$ is complete. $\rho^{o}$ is a metric built upon $\left\{\rho^{o}_T\right\}_{T=1}^{\infty}$ and inducing the Skorokhod topology on $D_{\left[0,\infty\right)}$, with $\left(\rho^{o},D_{\left[0,\infty\right)}\right)$ complete. Their explicit characterizations can be abstracted in what follows, but can be found in equations (16.4) for $\rho^{o}$ and (12.16) for $\rho^{o}_{T}$ of Patrick Billingsley "Convergence of Probability Measures".

Lemma 1.[Weak convergence] If $\mathbb{P}\left(X\in\mathcal{C}_{\left[0,\infty\right)}\right)=1$, then $X^{(d)}\longrightarrow X$ weakly w.r.t the Skorokhod topology in $D_{\left[\left.0,\infty\right)\right.}$ if and only if $\overline{X}^{(d)}\longrightarrow X$ weakly w.r.t. the Skorokhod topology in $D_{\left[\left.0,\infty\right)\right.}$.

Proof to Lemma 1. Let $X^{(d)}\longrightarrow X$ weakly in $D_{\left[\left.0,\infty\right)\right.}$. Then, in view of the Skorokhod Representation Theorem, Theorem 6.7 in Billingsley, we have $\widetilde{X}^{(d)}\equiv X^{(d)}$ and $\widetilde{X}\equiv X$, where $\equiv$ stands for equal in distribution, so that $\rho^{o}(\widetilde{X}^{(d)}(\omega),\widetilde{X}(\omega))\rightarrow 0$, for all $\omega\in \Omega$. Note that $\mathbb{P}\left(\widetilde{X}\in\mathcal{C}_{\left[0,\infty\right)}\right)=\mathbb{P}\left(X\in\mathcal{C}_{\left[0,\infty\right)}\right)=1$ and from Theorem 16.2, Billingsley, we have that $\rho^{o}_{T}(\widetilde{X}^{(d)},\widetilde{X})\rightarrow 0$ for all $T>0$, almost surely. This further implies that $X^{(d)}\longrightarrow X$ weakly with respect to $D_{\left[0,T\right]}$. Now we resort to Theorem 4.28 referred to in the question and to Theorem 1. Let $\epsilon,\delta>0$ and choose $d$ large enough so that $\mathbb{P}\left(\rho_{T}\left(X^{(d)},\overline{X}^{(d)}\right)\leq\epsilon\right)\geq 1-\delta$, then $E\left[\rho_{T}\left(X^{(d)},\overline{X}^{(d)}\right)\wedge 1\right]\leq \epsilon+\delta$, and thus we have $\limsup_{d\rightarrow\infty} E\left[\rho_{T}\left(X^{(d)},\overline{X}^{(d)}\right)\wedge 1\right]=0$. This implies that $\overline{X}^{(d)}\longrightarrow X$ weakly with respect to $D_{\left[0,T\right]}$ in light of Theorem 4.28. This convergence holds for all $T$. With the same Skorokhod representation+Theorem 16.2, we can conclude that $\overline{X}^{(d)}\longrightarrow X$ converges weakly with respect to the Skorokhod topology in $D_{\left[0,\infty\right)}$. $\tag*{$\blacksquare$}$

  • $\begingroup$ I will check this answer soon. At the moment, I'm a bit struggling with the usual definitions. If we consider càdlàg functions $[0,t]\to E$, where $(E,d)$ is a metric space, is the following the (incomplete) metric which induces the so-called $J_1$-topology?: Let $\Lambda:=\left\{\lambda:[0,t]\to[0,t]\mid\lambda\text{ is bijective and continuous with }\lambda(0)=0\right\}$ (I guess $λ(0)=0$ is to ensure that $λ$ is increasing (and not decreasing)). Then the $J_1$-metric should be $$\sigma(x,y):=\inf_{λ\in\Lambda}\left(\sup_{s\le t}|λ(s)-s|+\sup_{s\le t}d(x(s),(y\circλ)(s))\right),$$ right? $\endgroup$ – 0xbadf00d Mar 4 '19 at 17:41
  • $\begingroup$ For the metric, I am resorting to Patrick Billingsley's "Convergence of Probability Measures" definition. What you wrote is consistent with the definition I am using, but for the extra condition $\lambda(t)=t$. This is relevant. One (maybe superficial) reason why this is relevant is that the metric is devised to measure distances between points in $D_{\left[0,T\right]}$. If you relax the constraints $\lambda(0)=0$ and $\lambda(t)=t$, you may be running outside the domain of one of the functions. The goal of the abscissa-wiggling function $\lambda$ is (to be continued) $\endgroup$ – Augusto Santos Mar 4 '19 at 21:22
  • $\begingroup$ to lip the following undesired fact. Let $f(t)=1_{t\in\left[\left.1,\infty\right)\right.}$ and $f_n(t)=1_{t\in\left[\left.1+1/n,\infty\right)\right.}$. These are cadlag functions. $f_n$ does not converge to $f$ w.r.t. the $\sup$-norm, but it converges with respect to the above Skorokhod metric. This is because of the extra wiggling $\lambda$. Another reason why it is important to insist on $\lambda(T)=T$ (this is more related to your problem), it is because, if otherwise, the second term in your equation (2) would be no longer zero (and we would have a serious issue). $\endgroup$ – Augusto Santos Mar 4 '19 at 21:33
  • $\begingroup$ I've taken a look into Billingsley's book, but it seems like he's only considering real-valued càdlàg functions. Moreover, he's defining $\Lambda$ to be the set of increasing continuous functions mapping $[0,1]$ onto itself. That should be consistent with my definition. Any continuous function from a closed interval to another closed interval is bijective if and only if it is strictly monotone. That's why in my definition I need to explicitly require $\lambda(0)=0$ (since this yields that $\lambda$ is (strictly) increasing instead of decreasing). Am I missing something? $\endgroup$ – 0xbadf00d Mar 5 '19 at 13:12
  • $\begingroup$ That is all correct. In your previous comment, unless I missed something, you missed reinforcing that $\lambda(t)=t$. I just tried to explain why this latter is important. I will also update the answer with the definition of $\Lambda$ as soon as possible. $\endgroup$ – Augusto Santos Mar 5 '19 at 13:35

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