I've been trying to figure this out for a long time and I cannot seem to wrap my head around it:

a) We have an index set $\mathcal{I}=[0,1]$, and for each $i\in\mathcal{I}$, $x_i$ is drawn i.i.d. from $\{0,1\}^3$ according to some distribution $F_x$. We have a measurable map $a_i(x_i):\{0,1\}^3\to A$ where $A$ is a finite set. We define the set $Q=\{i\in\mathcal{I}:a_i(x_i)=a\}$ for some $a\in A$ at a given realization of the random process $(x_i)_{i\in\mathcal{I}}$. Essentially then, $Q$ is a random set that depends on the draws $(x_i)_{i\in\mathcal{I}}$. The question i'm trying to address is whether or not I can be sure that the process induces a probability distribution over the family of the sets of the form of $Q$.

b) Pretty related, if $s_\alpha:S\to\mathbb{R}$ is a random variable, and $\mathcal{A}$ is an uncountable index set, how can we make sure that a given realization of the process $(s_\alpha)_{\alpha\in\mathcal{A}}$ generate a $Y=\{s:s=s_\alpha, i\in\mathcal{A}\}$ that is a measurable set and also induces a probability distribution on it?

If the family indices for these two questions were countable, I know how to do it, using the product topology by having the measurability of each of the coordinates with the sigma algebra generated by cylinders, which I would have due to the sets being formed by realizations of random variables and mapping them through measurable functions, however, when I introduce the uncountability factor in the family of realizations, I want to be sure that the measurability is still preserved. I have seen this in the context of Brownian Motion and more generally, continuous stochastic processes, but I'm not sure if it translates perfectly in what I have, given that I don't necessarily have a filtration.

Many thanks

  • $\begingroup$ What do you mean by measurability of a random set? $\endgroup$ – d.k.o. Feb 15 at 3:30
  • $\begingroup$ In this case, every realization $(x_i)_{i\in\mathcal{I}}$ generates a set $Q$ defined as above, so ex ante, before such realization, the set $Q$ depends on the underlying sampling space. What I'm interested in knowing is when that realization generates a set that is measurable, for instance, with respect to the Lebesgue Measure. I'm afraid that there could exist some realizations of the process $(x_i)_{i\in\mathcal{I}}$ that could generate a set $Q$ that is actually not measurable. $\endgroup$ – Zeky Murra Feb 15 at 11:55
  • $\begingroup$ Intuitively for me, if each $x_i$ is a random variable, each of them maps a Borel set back into the sigma algebra generated by $x_i$, so if that is true for each $i$, I think that every set on $\sigma(\cup_{i\in\mathcal{I}} \mathcal{B}(\mathbb{R}))$ should map back into $\sigma(x_i:i\in\mathcal{I})$. If $\mathcal{I}$ is a countable set, then i know that $\sigma(\cup_{i\in\mathcal{I}} \mathcal{B}(\mathbb{R}))=\prod_{i\in\mathcal{I}}\mathcal{B}(\mathbb{R})$, and then I don't have a problem. I'm just struggling to extend this notion to an uncountable index set. $\endgroup$ – Zeky Murra Feb 15 at 12:02
  • $\begingroup$ Also what do you mean by the distribution over the family of sets of the form $Q$? Are you interested in the distribution of $Q$ for particular value of $a$? $\endgroup$ – d.k.o. Feb 15 at 19:11
  • $\begingroup$ Yes, Indeed, I'm interested, on computing the probability of a specific set $Q$ happening. $\endgroup$ – Zeky Murra Feb 17 at 15:31

Let $(\Omega,\mathcal{F},\mathsf{P})$ be the underlying probability space. Then $Q(\omega)$ need not be (Borel/Lebesgue) measurable for all $\omega\in \Omega$. A trivial example would be $A=\{0,1\}$ and $a_i=1\{i\in N\}$, where $N$ is a non-measurable subset of $[0,1]$. In this case each $a_i$ is measurable (a constant function) and $Q=N$ for $a=1$.

Let $\xi_i=a_i\circ x_i$. Then $(\xi_i:i\in \mathcal{I})$ is a stochastic process consisting of mutually independent random variables. You are interested in the sections of $\Xi=\{(i,\omega):\xi_i(\omega)=a\}$ for some $a\in A$. A sufficient condition ensuring that $Q(\omega)=\Xi^{\omega}$ is measurable is the (joint) measurability of $\xi$ w.r.t. $\mathcal{B}\otimes \mathcal{F}$, where $\mathcal{B}$ is a $\sigma$-algebra on $\mathcal{I}$. Unfortunately, a process like $\xi$ is not necessarily measurable (see , for example, Section 19.5 in Stoyanov, Counterexamples in Probability).

By the way, when $\mathcal{I}$ is countable, each $\Xi^{\omega}$ is automatically measurable ($\because \mathcal{B}=2^{\mathcal{I}}$).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.