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I found proofs in books (for example Oksendal's) where there is a process $S_t$ which satisfies $\mathrm E[S_t^2]=0$ for all $t$. Based on this, a conclusion is made that

$$ \mathbb P[S_t = 0 \textrm{ for all } t \in \mathbb Q \cap [0,T]]=1 $$ for some $T>0$ and where $\mathbb Q$ is the set of rational numbers. Then, using the continuity of $S_t$ a further conclusion is made that $$ \mathbb P[S_t = 0 \textrm{ for all } t \in [0,T]] = 1 $$

My question is, why can one conclude the first probability but not the second from $\mathbb E[S_t^2]=0$? Also, probably related, how do I show these conclusions rigorously?

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    $\begingroup$ Regarding your first question: $E[S_t^2]=0\Rightarrow P(S_t=0)=1$ is a standard result for random-variables, but you cannot conclude $\forall t\leq T:P(S_t=0)=1\Rightarrow P(\forall t\leq T:S_t=0)$. (Think for example of a process on $[0,1]$ with a jump at a random point $U\in\mathcal{U}([0,1])$.) But at least, now you can argue with $\sigma$-continuity arguments, because you are dealing with a countable union. This is not possible for the second probability. $\endgroup$
    – Graf Zahl
    Nov 10, 2020 at 11:10
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    $\begingroup$ The second probability should follow by the fact, that your process is $\omega$-wise continuous, so if for allmost-all $\omega$ the realizations are constant on a dense subset of $[0,T]$, they must be constant on $[0,T]$. $\endgroup$
    – Graf Zahl
    Nov 10, 2020 at 11:16
  • $\begingroup$ Thanks! For the first part, can you give some more references on those "$\sigma$-continuity arguments"? $\endgroup$
    – Tohiko
    Nov 10, 2020 at 11:23
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    $\begingroup$ Sorry, I think $\sigma$-additivity makes more sense. Let $(A_n)$ be a countable family of measurable sets with probability 1. You can argue $1\geq P(\cap_n A_n)=1-P(\cup_n A_n^c)\geq 1- \sum_nP(A_n^c)=1$. $\endgroup$
    – Graf Zahl
    Nov 10, 2020 at 11:30

1 Answer 1

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For a single $t$, we have $E(S_t^2)=0$ implies $P(S_t\neq 0)=0$. Now if $t_1,t_2,\ldots$ are any countable set of times, the probability that $S_{t_i}\neq 0$ for some $i$ is at most the sum of the individual probabilities, i.e. $0$.

But we cannot in general make the same conclusion for uncountable sets of times; for example if $S_t$ is the indicator function of a Poisson point process then we have $E(S_t^2=0)$ for any given $t$, yet almost surely there are infinitely many values of $t$ with $S_t=1$.

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