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I am working on exercise 1.3 in the book of Bass "Stochastic Processes". $X_t$ is $F_t$ adapted and $F_t$ is a complete, right-continuous filtration. First I proved that $$X^{(n)}_t(\omega) = \sum_{k=0}^\infty X_{k/2^n}(\omega)1_{[k/2^n, (k+1)2^n}(t)$$ is progressively measurable where I am viewing $X$ as a map $$X:\Omega \times [0,s] \to \mathbb{R}$$

Now, assuming $X$ is a.s. left continuous I should be able to use this to show that $X$ is progressively measurable. Let $N \subset \Omega$ be the nullset on which $X$ is not left continuous.

$$X^{-1}(B) = (X^{-1}(B) \cap (N\times[0,s])) \cup (X^{-1}(B) \cap (\Omega \backslash N\times[0,s]))$$ The second term in the union is measurable since on this set $X = \limsup_{n\to \infty} X^{(n)}$.

But the first term is not clear to me. $N$ is a nullset and thus every subset of $N$ is in $F_s$. But here I have some arbitrary subset of $N \times [0,s]$ and I don't see why this should be in $F_s \otimes \mathcal{B}([0,s])$.

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(The exercise in Bass assumes that $X$ has left-continuous paths -- no a.s.!) Proceeding as you outline, assuming only that $X$ is a.s. left continuous, the process $\limsup_{n\to\infty}X^{(n)}$ is progressive and $X_t(\omega)=Y_t(\omega)$ for all $t\ge 0$ and all $\omega\in\Omega\setminus N$. And this is the best you can do: $X$ is indistinguishable from a progressively measurable process.

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  • $\begingroup$ Ah ok thank you. I got confused since he defines cadlag on the pages before s.t. it wasn't clear to me if left continuous means the same thing as a.s. left continuous in his book. I know this is off-topic, but since you know the book apparently: Is it a good choice for deepening my stochastic processes knowledge compared with shreve/karatzas or Steele? $\endgroup$
    – fubal
    Dec 6, 2017 at 19:03

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