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I have the following definitions.

Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probabilised space. Let's take two stochastic processes $(X(t))_{i\in I}$ and $(Y(t))_{i\in I}$ with values in a separable Banach space $E$ endowed with its borelian sigma-algebra $\mathcal{E}=\mathcal{B}(E)$, where $I$ is an arbitrary set of indices.

  • $Y$ is a modification of $X$ if for all $t\in I$, $\mathbb{P}(X(t)=Y(t))=1$.
  • $X$ and $Y$ are equivalent if for all $n\in\mathbb{N}^*$, for all $(t_1,\dots,t_n)\in I^n$, the random variables $(X(t_1),\dots,X(t_n))$ and $(Y(t_1),\dots,Y(t_n))$ with values in $E^n$ follow the same distribution.
  • $Y$ and $Y$ are indistinguishable if there exists $\Omega_0\subset\Omega$ with $\mathbb{P}(\Omega_0)=1$ such that for all $\omega\in\Omega_0$, for all $t\in I$, $X(t)=Y(t)$, or equivalently $\mathbb{P}(\sup\limits_{t\in I}\vert X(t)-Y(t)\vert>0)=0$.

I have shown that if two stochastic processes are indistinguishable, then they are modification of each other. But I am not able to prove that if they are modification of each other, then they must be equivalent.

From the definition, I can only consider the different probabilies for each $t_1,\dots,t_n$,but how do I assemble this to make sure that the laws are the same ?

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1 Answer 1

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Let $A=\bigcap_i \{X(t_i)=Y(t_i)\}$ Then $P(A)=1$ and $(X_{t_1},X_{t_2},\dots,X_{t_n})=(Y_{t_1},Y_{t_2},\dots,Y_{t_n})$ on $A$. Hence $P((X_{t_1},X_{t_2},\dots,X_{t_n}) \in C)=P((Y_{t_1},Y_{t_2},\dots,Y_{t_n}) \in C)$ for any Borel set $C$ in $ E^{n}$.

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  • $\begingroup$ Hi Kavi, thanks for the answer. Shouldn't it be any borel set in $E^n$ ? $\endgroup$
    – Flewer47
    Nov 4, 2020 at 10:52
  • $\begingroup$ @Flewer47 Thanks. Answer corrected. $\endgroup$ Nov 4, 2020 at 11:33
  • $\begingroup$ Why is $\mathbb{P}(A)=1$ ? It is an uncountable intersection of sets, each being of probability 1, but the fact that it is uncountable is a bit problematic, isn't it ? $\endgroup$
    – Flewer47
    Nov 4, 2020 at 13:45
  • $\begingroup$ It is a finite intersection. $\endgroup$ Nov 4, 2020 at 13:50
  • $\begingroup$ Indeed, you consider $n\in\mathbb{N}^*$, times $t_1,\dots,t_n$ and then do your calculations. Got it, thanks ! $\endgroup$
    – Flewer47
    Nov 5, 2020 at 12:46

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