Skorohod space on $[0,1]$

cadlag functions $x: [0,1] \to \mathbb{R}$
Define $T\subset [0,1]$, $w_x(T): = \sup_{s,t \in T} |x(s)-x(t)|$
Modulus of continuity $w_x(\delta)= \sup_{0\le t \le 1-\delta}w_x([t,t+\delta])$

Properities of cadlag functions

Consider cadlag modulus $w_x’(\delta)= \inf_{{t_i}} \max_{1\le i \le v}w_x([t_{i-1},t_i))$ The infimum over all $\delta$-sparese sets ${t_i}$, ie. $\min_{1\le i \le v} (t_i -t_{i-1}) > \delta$.

characterisation: $\lim_{\delta \to 0}w_x’(\delta)=0$ this implies:
at most countably many discontinuities
$x$ is bounded
$x$ can be uniformly approximated by simple functions constant over intervals

Skorohod topology finite time

Class of strictly increasing, continuous mappings from $[0,1]$ onto itself: $\Lambda$. In particular, $\lambda(0)=0, \lambda(1)=1$. Metric on $D$, $d(x,y)= \inf_{\lambda \in \Lambda} \{ \|\lambda -I \|_{\infty[0,1]} \vee \| x - y\lambda\|_{\infty [0,1]} \}$

Convergence in Skorohod topology iff $\lim_n x_n(\lambda_n t)= x(t)$ uniformly in $t$ and $\lim_n \lambda_n t =t$ uniformly in $t$.
Uniform convergence implies Skorohod convergence
Skorohod convergence implies $x_n(t) \to x(t)$ for all $t$ continuity point of $x$.
If $x$ is continuous on $[0,1]$, then Skorohod convergence implies uniform convergence.
However, $D$ is not complete under $d$. It is under $d^{\circ}(x,y)=\inf_{\lambda\in\Lambda} \{ \| \lambda {\|^{\circ}}_{\infty} \vee \| x- y\lambda\|_{\infty} \}$, $\|\lambda\|^{\circ}=\sup_{s<t}|\log \frac{\lambda t-\lambda s}{t-s} |$
$D$ is separable and complete.
Compactness: Theorem 12.3 analogue of Arzela-Asocli theorem, Theorem 12.4 via ${w}^{''}_x(\delta)=\sup_{t_1 \le t \le t_2, t_2 -t_1 \le \delta} \{|x(t)-x(t_1)| \wedge |x(t_2)-x(t)|\}$ (Billingsley)
Finite dimensional sets

Weak convergence and tightness in $D$

Compare with the continuous function version Theorem 7.3 via Arzeal-Ascoli.

Skorohod topology $[0,\infty)$

$D[0,\infty)$ cadlag functions on $[0,\infty)$. Introduce $D_t$ cadlag functions on $[0,t]$, $d_t^{\circ}(x,y)$ which is analogous to $d^{\circ}$ but for functions on $[0,t]$, similarly for $|x |m=\sup{s\le t}|x(s)|$.

Define $g_m(t)=\begin{cases} 1 \quad &\text{if } t\le m-1,\\ m-t \quad &\text{if } m-1\le t \le m,\\ 0 \quad &\text{if } t \ge m.\end{cases}$ For $x\in D_\infty$, $x^m\in D_\infty$, $x^m(t)=g_m(t)x(t), t\ge 0.$ $d^{\circ}_\infty(x,y)= \sum_{m=1}^{\infty} 2^{-m} (1 \wedge d_m^{\circ}(x^m, y^m))$ Convergence in $d_\infty^{\circ}(x_n,x) \to 0$ in $D_{\infty}$ iff there exists $\lambda_n \in \Gamma_\infty$, $\sup_{t<\infty}|\lambda_n t -t| \to 0$ and for each $m$ $\sup_{t\le m}|x_n(\lambda_n t)- x(t)|\to 0$.

and iff for each continuity point $t$ of $x$, $d_t^{\circ}(x_n,x)\to 0$.

Define $\phi_m x$ as $x^m$ restricted to $[0,m]$. $D_{\infty}$ is separable and complete. $D_{\infty}$ isometric to the product metric on $D_1 \times D_2 \times …$

Compactness

relative compact in $D_{\infty}$ iff $\phi_m A$ is relatively compact in $D_m$ for each $m$.

We have

\[w^{'}_{m}(x,\delta)=\inf \sup_{1 \le i \le v} w(x, [t_{i-1},t_i))\]

, where infimum extends over all decompositions $[t_{i-1},t_i)$, $1\le i \le v$ of $[0,m)$, such that it is $\delta$-sparse for $1\le i < v$ but does not require $t_v - t_{v-1}>\delta$.

Theorem 16.5

$A$ is relatively compactness iff for every $m$, $\sup_{x \in A} \| x \|_{m} < \infty$ and $\lim_{\delta \to 0} \sup_{x \in A} w^{'}_m (x, \delta) = 0$.

The tightness is characterised by the sequence ${P_n}$ is tight if and only if for each $m$,

\[\lim_{a\to\infty} \limsup_{n} P_n (x: \| x\|_m \ge a) = 0\]

with $\lVert x\rVert_{m} = \sup_{s\le m} |x(s)|$ . for each $m$ and $\epsilon$

$\lim_{\delta} \limsup_{n} P_n(x: w^{'}_m(x,\delta)\ge \epsilon)=0$.

This translate to the relevant tightness sufficeint condition for measures on $D_\infty$. $\lim_{a \to \infty}\limsup_n P[|X^n|_m \ge a]=0$

and for each $\epsilon, \eta,m$, there exists a $\delta_0$ and $n_0$ such that if $\delta<\delta_0$, $n\ge n_0$, and $\tau$ a discrete $X^n$ stopping time with $\tau\le m$, then

\[P(|X^n_{\tau+\delta} - X^n_\tau| \ge \epsilon |)\le \eta\]

Proof idea is to show the second condition implies

\[\lim_{\delta}\limsup_n P[w^{'}_m (X^n,\delta)\ge \epsilon]=0.\]

Reference: Convergence of Probability Measures, Patrick Billingsley Chapter 12, 13, 16

Finite dimensional sets Weak convergence Tightness criterion