Concerning Proposition 1.1.32 of Jacod and Shiryaev

Question

I am trying to understand the proof of Proposition 1.1.32 in Jacod and Shiryaev. I summarize the problem for the sake of brevity. Part of the proof is contained in this answer to my previous question.

Suppose $X$ is a cadlag process. Define, for each $n$, the sequence of stopping time $\{S(n,p),p\in\mathbb{N}\}$ as $S(n,0)=0$ and

$$ S(n,p+1)= \inf\left\{t>S(n,p)\mid \left|X_t-X_{S(n,p)}\right|>2^{-n}\right\} $$

It can be proved that $S(n,p)\rightarrow\infty$. Consider the event $$ A(n,p)\doteq\{S(n,p)<\infty,\Delta X_{S(n,p)}\neq 0\}=\{S(n,p)<\infty,X_{S(n,p)}-X_{S(n,p)-}\neq 0\} $$ and the stopping time $$ T(n,p)\doteq\begin{cases} S(n,p) & \textrm{ on }A_{n,p}\\ +\infty & \textrm{ on }A_{n,p}^c \end{cases} $$ The authors say that since $S(n,p)\rightarrow\infty$ it is clear that $$ \left\{(\omega,t)\in\Omega\times[0,\infty)\mid\Delta X_t(\omega)\neq 0\right\}=\bigcup_{n,p}\left\{(\omega,t)\in\Omega\times[0,\infty)\mid t=T(n,p)(\omega)\right\},\quad(1) $$ For me the inclusion $\supseteq$ is ok: if $(\omega,t)$ belongs to the right-hand side of equation $(1)$ then $\exists n,p$ such that $t=T(n,p)(\omega)$ and, since $t<\infty$, then it must also be that $\omega\in A_{n,p}$ and so $\Delta X_{S(n,p)}\neq 0$, whence $(\omega,t)$ belongs to the left-hand side of equation $(1)$.

The problem is to verify the inclusion $\subseteq$ in identity $(1)$. If $(\omega,t)$ belongs to the left-hand side, I can say that $\Delta X_t(\omega)=X_t(\omega)-X_{t-}(\omega)\neq 0$, whence there must be $n\in\mathbb{N}$ such that $\left|\Delta X_t(\omega)\right|>2^{-n}$. Now I need to prove that $(\omega,t)$ belongs to the right-hand side. For that, I have to find the $p$ (the $n$ is already found). My idea is to take $p$ such that

$$ p^{\star}\doteq\min\{p\in\mathbb{N}\mid S(n,p)\geq t\} $$

which exists being $\{p\in\mathbb{N}\mid S(n,p)\geq t\}\neq \emptyset$ (a consequence of $S(n,p)\rightarrow+\infty$). But I miss how to prove that $$ \Delta X_{S(n,p^{\star})} = X_{S(n,p^{\star})}-X_{S(n,p^{\star})-}\neq 0 $$

Answer 1

You want to show that $$ \{(\omega,t) \in \Omega \times [0,\infty) : \Delta X_t(\omega) \neq 0\} = \bigcup_{n,p}\{(\omega,t) \in \Omega \times [0,\infty) : t = T(n,p)(\omega)\} $$

For you the inclusion $\supseteq$ seems ok but I'll write it again for completeness. If $(\omega,t) \in RHS$ then there is an $n,p \in \mathbb N$ such that $t = T(n,p)(\omega)$. Since $t<\infty$, $T(n,p)(\omega)<\infty$. By definition of $T$, this means that $\omega \in A_{n,p}$ and $t=T(n,p) = S(n,p)$. Now, as $\omega \in A_{n,p}$, $\Delta X_{S(n,p)}(\omega) \neq 0$, the same as $\Delta X_t(\omega) \neq 0$. Thus, $(\omega,t) \in LHS$.

Suppose that $(\omega,t) \in LHS$. Then, we know that $t<\infty$ and $\Delta X_t(\omega) \neq 0$. We want to show that $t = T(n,p)(\omega)$ for some $n,p$. This is equivalent to showing that $t = S(n,p)(\omega)$ for some $n,p$, which seems apparently difficult given the definition of $S(n,p)$.

Let us revisit the definition of $S(n,p)$. With $n$ fixed, we are tracking using the index $p$, those times when we are at least $2^{-n}$ away from the previous tracked point. For each fixed $n$, we have $S(n,p) \to \infty$ (on the entirety of $\Omega$, by removing the null set if necessary).

Now, suppose that $t \neq S(N,P)(\omega)$ for some fixed value of $N$ and all values of $P$. By definition, $t \neq 0$, since $S(N,0)=0$ for all $N \geq 1$. Hence, we will assume that $t \in (0,\infty)$.

Because $S(N,p)(\omega) \to \infty$, there exists $P'(n,\omega)$ such that $S(N,P')(\omega) < t < S(N,P'+1)(\omega)$. By definition of $$ S(n,p)(\omega) = \inf\{t > S(n,p-1)(\omega) : |X_{t}(\omega) - X_{S(n,p-1)}(\omega)| > 2^{-n}\} $$ this means that $$ |X_s(\omega) - X_{S(N,P')}(\omega)| \leq 2^{-N} \ \ \forall S(N,P')(\omega) \leq s \leq t $$ Now we use the triangle inequality. Indeed, for all $S(N,P')(\omega) \leq s \leq t$, we have by the above that $$ |X_s(\omega) - X_t(\omega)| \leq |X_s(\omega) - X_{S(N,P')}(\omega)| + |X_t(\omega) - X_{S(N,P')}(\omega)| \leq 2^{-N}+2^{-N} \leq 2^{1-N} $$ As the left limit $\Delta X_t(\omega) = \lim_{s \to t^-} (X_t(\omega)-X_s(\omega))$ exists, the above implies that $\Delta X_t(\omega) \leq 2^{1-N}$.

We have proved the following in the above few lines :

If , for some $N>1$, it is true that $t \neq S(N,P)(\omega)$ for all $P \geq 0$, then $|\Delta X_t(\omega)| \leq 2^{1-N}$.

Taking the contrapositive of this statement,

If , for some $N>1$, it is true that $|\Delta X_t| > 2^{1-N}$, then it must happen that $t = S(N,P)$ for some $P \geq 0$.

Now we return to our setup. For $(\omega,t) \in LHS$, $t<\infty$, $\Delta X_t(\omega)\neq 0$, we know that $|\Delta X_t(\Omega)| > 2^{1-N}$ for some large $N$. Using the blockquote above, there is a $P$ such that $t=S(N,P)(\omega)$ for some $P\geq 0$. Now, as $t<\infty$, $S(N,P)<\infty$ and as $\Delta X_t(\omega)\neq 0$, $\Delta X_{S(N,P)}(\omega) \neq 0$. We have fulfilled the conditions for $\omega\in A(N,P)$, whence $t = S(N,P)= T(N,P)$. Hence, $(\omega,T) \in RHS$, as desired.