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Can we construct an i.i.d. family of Rademacher random variables $(X_t)_{t\in\mathbb{R}}$ defined on some probability space $(\Omega,\mathcal{F},\mathbb{P})$ (so, in particular, $\forall t \in \mathbb{R}, \forall i\in\{-1,+1\}, \mathbb{P}[X_t=i]=\frac{1}{2}$) such that for $\mathbb{P}$-a.s. $\omega \in \Omega$ we have that $$\mathbb{R} \to \{-1,+1\}, t\mapsto X_t(\omega)$$ is measurable? If so, and $-\infty<a<b<+\infty$, what could be said of the random variable $$\omega\mapsto \frac{1}{b-a}\int_{[a,b]}X_t(\omega)\operatorname{d}t?$$ Does it satisfy some kind of strong law of large numbers? Intuitively - and here my intuition could be very wrong - this integral "should be" just a series of infinitesimal i.i.d. random variables of zero mean...

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    $\begingroup$ Maybe you can appeal to an appropriate ergodic theorem? $\endgroup$ May 21, 2021 at 13:32
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    $\begingroup$ How do you define independence for an uncountable collection of random variables? $\endgroup$
    – user159517
    May 21, 2021 at 14:46
  • $\begingroup$ Nice question: in fact it could be part of having a negative of positive answer. What I had in mind was the classical "finite" definition, i.e. that for every two finite disjoint subsets of the real numbers, say $A$ and $B$, we have that the $\sigma$-algebras generated by $(X_t)_{t\in A}$ and $(X_t)_{t\in B}$ are $\mathbb{P}$-independent. But probably at this point we would require a stronger definition in order to obtain something sensible, maybe something like if $A$ and $B$ are two disjoint intervals of the real numbers then $(X_t)_{t\in A}$ and $(X_t)_{t\in B}$ are $\mathbb{P}$-independent $\endgroup$
    – Bob
    May 21, 2021 at 15:01
  • $\begingroup$ Could you clarify what kind of behavior you're interested in for that integral? E.g. thinking of $|b-a| \rightarrow 0$, $|b-a| \rightarrow \infty$, or just a general question of what its distribution is for arbitrary $a,b$? $\endgroup$ May 21, 2021 at 16:14
  • $\begingroup$ I'm mainly interested in the behaviour for fixed $-\infty<a<b<+\infty$ $\endgroup$
    – Bob
    May 21, 2021 at 16:22

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I will show that joint measurability is impossible. That is, you cannot cook up such a probability space with $(\omega,t) \mapsto X_{t}(\omega)$ jointly measurable (that is, as a map from $\Omega \times \mathbb{R}$ with product of $\mathcal{F}$ and Lebesgue to $\mathbb{R}$.) This seems to be a stronger assumption than what you asked for. At the same time, it is somewhat awkward to say "$t \mapsto X_{t}(\omega)$ is measurable for $\mathbb{P}$-a.e. $\omega \in \Omega$" since it's not clear that "is measurable" will be in $\mathcal{F}$ or how to build such a $\Omega$.

Given $a,b \in \mathbb{Q}$, define $Y_{a,b} = \int_{a}^{b} X_{s} \, ds$. Notice that Fubini's Theorem implies \begin{equation*} \mathbb{E}(Y_{a,b}^{2}) = \int_{a}^{b} \int_{a}^{b} \mathbb{E}(X_{s} X_{\xi}) \, ds \, d \xi = 0. \end{equation*} Therefore, $Y_{a,b} = 0$ $\mathbb{P}$-a.s. $\mathbb{Q} \times \mathbb{Q}$ is countable so, in fact, $\{Y_{a,b} \, \mid \, a,b \in \mathbb{Q}\} = \{0\}$ almost surely. Since $t \mapsto X_{t}$ is measurable in this event, we are left to conclude that $X \equiv 0$ almost surely. (A bounded, measurable function whose integral vanishes in every interval with rational endpoints equals zero almost surely.) This contradicts your specification that $\{X_{t}\}_{t \in \mathbb{R}} \subseteq \{-1,1\}$.

The argument is taken directly from Chapter 1 of Revuz and Yor (see Section 3).

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  • $\begingroup$ Can you explain why should it be awkward to say that $t↦X_t(ω)$ is measurable for $\mathbb{P}$-a.e. $\omega \in \Omega$? If I'm not wrong, here I'm just asking that the trajectories of the process are $\mathbb{P}-a.e.$ measurable (a milder condition than the continuity of the trajectories, that is used for example in brownian motion). If I'm not wrong the joint measurability is such a strong requirement, that the reference to which you point to establishes that gaussian processes satisfying such condition don't exist... $\endgroup$
    – Bob
    May 22, 2021 at 10:14
  • $\begingroup$ "If I'm not wrong the joint measurability is such a strong requirement, that the reference to which you point to establishes that gaussian processes satisfying such condition don't exist..." Brownian motion is Gaussian and jointly measurable (suitably defined). Revuz and Yor show that there is no way of getting a jointly measurable Gaussian process with each time independent of the other (i.e. your example, but with Gaussians replacing Bernoulli r.v.) $\endgroup$
    – user711689
    May 22, 2021 at 17:12
  • $\begingroup$ "Can you explain why should it be awkward to say that..." To check that $(X_{t})_{t \in \mathbb{R}}$ is measurable requires checking the behavior at uncountably many times. Uncountable intersections of events aren't necessarily events. $\endgroup$
    – user711689
    May 22, 2021 at 17:20

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