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I quote Kuo (2006).

Definition A stochastic process $\tilde{X}(t)$ is called a version of $X(t)$ if $\mathbb{P}\{\tilde{X}(t)=X(t)\}=1$ for each $t$.

Definition A stochastic process $\tilde{X}(t)$ is called a realization of $X(t)$ if there exists $\Omega_0$ such that $\mathbb{P}(\Omega_0)=1$ and for each $\omega\in\Omega_0, \tilde{X}(t,\omega)=X(t,\omega)$ for all $t$.

Then, I read that:

Obviously, a realization is also a version, but not vice versa.

and that is pretty clear to me.

What I cannot understand is the following statement:

But if the time parameter set is countable, then these two concepts are equivalent.

Why is the immediately above statament true?

Strictly_increasing
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1 Answers1

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That's because a countable union of null sets is a null set, i.e., if the (time) parameter set $T$ is countable, then $$ \mathsf{P}\!\left(\bigcup_{t\in T}\{\tilde{X}_t\ne X_t\}\right)\le \sum_{t\in T}\mathsf{P}\!\left(\tilde{X}_t\ne X_t\right)=0. $$

  • So, if I am not mistaken, I could rewrite your passage as: $$\mathsf{P}\bigg(\Omega_0^C\bigg)=0$$ where $$\Omega_0^c={\omega: \tilde{X}_t(\omega)\ne X_t(\omega)}$$ Hence $$\mathsf{P}\bigg(\Omega_0\bigg)=1$$ However, I cannot understand why a finite union of null sets is not a null set. Could you please help me understand such a fact? [continue] @d.k.o. – Strictly_increasing Oct 02 '20 at 09:41
  • [continue] That is, couldn't I state, for $T$ finite set such that $T={t_0,\ldots,t_n}$ that: $$\mathsf{P}!\left(\bigcup_{i=t_0}^{t_n}{\tilde{X}i\ne X_i}\right)\le \sum{i=t_0}^{t_n}\mathsf{P}!\left(\tilde{X}_i\ne X_i\right)=0.$$? @d.k.o. – Strictly_increasing Oct 02 '20 at 09:43
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    What exactly are you asking about? A finite union of null sets is obviously a null set. –  Oct 02 '20 at 09:56
  • Of course. So, why with $$\mathsf{P}!\left(\bigcup_{i=t_0}^{t_n}{\tilde{X}i\ne X_i}\right)\le \sum{i=t_0}^{t_n}\mathsf{P}!\left(\tilde{X}_i\ne X_i\right)=0.$$ am I not allowed to say that a version is also a realization for finite set $T$? (as written in the question, on my reference I read instead that for finite set $T$ a version is not necessarily a realization) @d.k.o. – Strictly_increasing Oct 02 '20 at 09:58
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    You're surely allowed to say that when $T$ is a finite set. –  Oct 02 '20 at 10:19
  • So, why on my reference I read "Obviously, a realization is also a version, but not vice versa. But if the time parameter set is countable, then these two concepts are equivalent." (meaning that if time set $T$ is finite a version is not necessarily a realization)? That's the question @d.k.o. – Strictly_increasing Oct 02 '20 at 10:21
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    A countable set is either a finite set or a countably infinite set. –  Oct 02 '20 at 10:58
  • Sorry, probably I am not getting to the point in a clear way. That a realization is also a version (in any case, both for $T$ finite set and for $T$ countable set is clear, as you have just shown in a perfect way). The fact I cannot understand is why on my reference it is said that $$\text{for finite set } T\text{, a version is not necessarily a realization}$$Why that? @d.k.o. – Strictly_increasing Oct 02 '20 at 10:59
  • Where exactly is that quote (in Kuo's book)? –  Oct 02 '20 at 11:03
  • Sorry, my very silly mistake (I did not have definition of a countable set clear in mind). What it should be meant there is that $$\text{for uncountable set } T\text{, a version is not necessarily a realization}$$Original statement is "Obviously, a realization is also a version, but not vice versa. But if the time parameter set is countable, then these two concepts are equivalent" (ch. 3, p. 30). [continue] – Strictly_increasing Oct 02 '20 at 11:08
  • [continue] So, at this point, if I am not mistaken, for uncountable set $T$ a version is not necessarily a realization since I cannot state that $$\mathsf{P}!\left(\bigcup_{t\in T}{\tilde{X}t\ne X_t}\right)\le \sum{t\in T}\mathsf{P}!\left(\tilde{X}_t\ne X_t\right)=0\tag{1}$$Finally, why cannot I state so (that is, $(1)$)? Thank you a lot for your patience and time. @d.k.o. – Strictly_increasing Oct 02 '20 at 11:08
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    Probably it is easier to look at a counterexample. Look at the example in this quesiton –  Oct 02 '20 at 11:15
  • Thank you a lot! I have also found such a question https://math.stackexchange.com/questions/2910175/why-is-an-uncountable-union-of-null-sets-not-necessarily-a-null-set. Accepted answer clears out any doubt @d.k.o. – Strictly_increasing Oct 02 '20 at 11:17