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Dear all,

[Bauer, Probability Theory, Exercise 2 of Chapter 39] -->

http://books.google.de/books?id=w76IHsPHybcC&pg=PA339#v=onepage&q&f=false

gives the following characterisation for the continuity of sample paths of a $\mathbb{R}^d$ valued stochastic process:


A stochastic process $(\Omega,\mathcal{A},P,(X_t)_{t\geq0})$with state space $\mathbb{R}^d$has a continuous modification $(X'_t)_{t\geq0}$if and only if there is a countable dense subset $S$ of $\mathbb{R}_+$with the following properties: - For every pair of positive realnumbers $\eta,k$ $$\lim_{\delta\downarrow 0}P(\bigcup_{s,t\in S,s,t\leq k,|s-t|\leq \delta}\{|X_s-X_t|\geq\eta\})=0$$ - For every $t\in\mathbb{R}_+$there is a sequence $(s_n)$ in $S$ converging to $t$ such that the sequence $(X_{s_n})$converges stochastically to $X_t$.


My question now is: Does anyone know a reference for an analogous statement for processes with a general complete and separable metric state space $(E,d)$?

Of course, one then has to substitute $d(X_t,X_s)$ for $|X_s-X_t|$ in the first condition; but apart from that I do not see any reason why the result should not carry over to this more general setting. But I did not find any reference in the literature yet!

Thanks for answers! Kind regards

lpdbw

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    $\begingroup$ I don't think this question is really research-level. I think if you read the proof (which Google Books will not show me) you will find that it goes through for a general complete metric space. You might find a more explicit statement in another textbook, but this fact seems sufficiently simple that one wouldn't really need to give a reference. Note that the first property says that, almost surely, $X_s$ is uniformly continuous on $S \cap [0,k]$ for every $k$. Hence it has a unique continuous extension to all of $\mathbb{R}$. $\endgroup$ Commented Jul 8, 2012 at 17:29
  • $\begingroup$ And the second property guarantees that this extension is a modification of the original process. $\endgroup$ Commented Jul 8, 2012 at 17:29
  • $\begingroup$ lpdbw, having created a new account, wrote the following in an answer: `Thanks for your answer. In fact, I did't claim this was a "research level" problem. All I wanted was a convenient reference suitable for citation since I could not find one.' $\endgroup$
    – S. Carnahan
    Commented Jul 9, 2012 at 4:34

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