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)?
Best Answer
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.