4

I have an intuitive stochastic process as follows, but not sure how to construct it rigorously on some probability space.

Consider the unit interval $[0,1]$, each point $x\in [0,1]$ is associated with an independent Bernoulli($0.5$) random variable. Then each realization can be viewed as a function $f$ from $[0,1]$ to $\{0,1\}$. The level set $\{x: f(x) = 1\}$ is then a random set on $[0,1]$. I want to argue the set is almost surely (Lebesgue)-measurable, and almost-surely has measure $0.5$.

However, the above construction does not seem to be rigorous. Therefore, I want to know if there exists a probability space $\Omega$ that admits the a stochastic process $f(t,\omega) \in \{0,1\}$, such that $1.$ all the finite dimensional distribution is a product of independent Bernoulli. $2.$ For almost every $\omega$, the set $\{x: f(x,\omega) = 1\}$ is measurable, and has probability $0.5$.

It seems the first condition can be guaranteed using the Kolmogorov's extension theorem. But I have not idea how to guarantee the second condition.

shong
  • 187
  • 3
    I disagree with your intuition that such a (non-rigorously defined) process should be such that almost surely the set ${x : f(x) = 1}$ is Lebesgue-measurable. Lebesgue-measurable subsets of $[0,1]$ have to do with the actual interval $[0,1]$ (and the structure of $\mathbb{R}$), whereas for your situation, $[0,1]$ just represents an uncountable set of points that have no relation between one another. – mathworker21 Jul 14 '22 at 02:20
  • Such white noise processes are often used in a Gaussian setting. So I would look up how they do this for Gaussian processes. Bernoulli should then be a simple consequence (i.e. Gaussian >0 <0) – g g Jul 16 '22 at 09:35
  • @mathworker21 Thanks! I agree that the set ${x: f(x) = 1}$ is not necessarily measurable just from the definition of the process. I am interested in whether it is possible to construct a rigorous process on certain state space such that the random set is almost surely measurable. – shong Jul 17 '22 at 17:19
  • @mathworker21 I agree with your intuition, see the answer below. – Yuval Peres Jul 18 '22 at 17:58

1 Answers1

3

It seems this is impossible. Let $\mu$ denote Lebesgue measure on $[0,1]$. If $A(\omega)=\{x: f(x,\omega) = 1\}$ is measurable then $\mu\bigl(A(\omega) \cap [A(\omega)+t]\bigr) \to \mu(A)=1/2$ as $t \to 0$. But $$E\Bigl(\mu\bigl(A(\omega) \cap [A(\omega)+t]\bigr)\Bigr) $$ $$=\int_\Omega \int_0^1 f(x,\omega) f(x-t,\omega) \,dx \,dP $$ $$= \int_0^1 \int_\Omega f(x,\omega) f(x-t,\omega) \,dP \, dx=1/4\,,$$ and this is a contradiction in view of Lebesgue's bounded convergence theorem.

Yuval Peres
  • 22,782