predictable $\Rightarrow$ left continous? progressive $\Rightarrow$ adapted and left continous?

Question

Assume we have a real valued continous time stochastic process $X:=(X_t)_{t\in [0,T]}$ $(T>0)$ defined on a complete, filtered probability space $(\Omega, \mathscr{F},\mathrm{P},(\mathscr{F}_t)_{t\in [0,T]})$.

• $X$ is said to be predictable if $X_t$ is measurable wrt. $\sigma(Y : Y\text{ is left continous and adapted})$ for all $t\in[0,T]$
• $X$ is said to be progressive or progressively measurable if $[0,t]\times \Omega \ni (s,\omega)\mapsto X_s(\omega)\in \mathbb{R}$ is $\mathscr{B}([0,t])\otimes \mathscr{F}_t$ -$\mathscr{B}$- measurable for all $t\in [0,T]$.
• $X$ is said to be joint measurable if $[0,T]\times \Omega \ni (s,\omega)\mapsto X_s(\omega)\in \mathbb{R}$ is $\mathscr{B}([0,T])\otimes \mathscr{F}_t$ -$\mathscr{B}$- measurable.

Question 1: Is a predictable process also left continous?

Because then we would have (in the sense of being a version):

predictable $\Leftrightarrow$ adapted and left continous $\Rightarrow$ progressively measurable $\Leftrightarrow$ joint measurable and adapted

Question 2: Are there any conditions such that

progressively measurable $\Rightarrow$ adapted and left continous

holds (in the sense of being a version)?

Answer 1

I suppose the answer to my second question is actually pretty trivial:

poisson process

Because the process $N:= (N_t)_{t\geq 0} $ given by

$N_t:=\#\lbrace S_1, S_2, \ldots \rbrace\cap [0,t]$

where $ S_1, S_2-S_1, S_3-S_2 \ldots$ are i.i.d. exponentiell distributed with parameter $\lambda>0$ , is a poisson process. It is progressive wrt. its augmented filtration, since all paths are right continous and the process is adapted.

If $N$ would have a left continous version $\tilde{N}$, then $\tilde{N}\overset{fidi}{=}N$ i.e. their finite dimensional distributions are the same, since $\mathbb{P}(N_t=\tilde{N}_t)=1$ for alle $t\geq 0$.

Define the random variables (they don't have to be stopping times)

$\tau:=\mathrm{inf}[t\geq 0 | N_t \geq 1]$ and $\tilde{\tau}:=\mathrm{inf}[t\geq 0 | \tilde{N}_t \geq 1]$.

There measurable since for arbitrary $\alpha>0$ $[N_\alpha\geq 1]=[\tau < \alpha]$ and respectively the same holds for $\tilde{\tau}$.

Since $N$ and $\tilde{N}$ equal in their finite dimensional distributions we have

$N_\tau\overset{d}{=} N_{\tilde{\tau}}$ $(*)$.

But $\mathbb{P}(N_\tau =1)=1$ and $\mathbb{P}(\tilde{N}_\tilde{\tau}=1)=0$, since $\tilde{N}_t$ is left continous (Note that $\tilde{N}_t$ can't be right continous at the same time, since then it would be continous and would reach any value in $\mathbb{R}$, which makes no sense since $\tilde{N}_t$ is Poisson distributed for all $t>0$.)

This contradicts $(*)$.

Answer 2

I don't think that a predictable process is always left-continuous. My reasoning depends on the following observation:

Let $q_n$ be a sequence of increasing numbers with limit $q$ and such that $q_n<q$ for all $n$. The stochastic processes given by: $$X^n(t,\omega)= I_{s<t\leq q_n}$$ are all predictable. Hence any limit is also. But the limit is $$X(t,\omega)=I(s<t<q)$$ which isn't left-continuous.