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

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)?

## 2 Answers

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 $$q_n for all $$n$$. The stochastic processes given by: $$X^n(t,\omega)= I_{s are all predictable. Hence any limit is also. But the limit is $$X(t,\omega)=I(s which isn't left-continuous.

• Thank you this answers my first question Commented Jul 12, 2023 at 11:54

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 $$(*)$$.