0

Let $T$ be a parameter set (countable or uncountable), and we have the probability space $(\Omega,\mathcal{F},P)$ with a collection of r.v. $\{X_t:t\in T\}$, where $X_t$ are $(E,\mathcal{E})$-valued r.v..

How can we prove that $X^{-1}(\mathcal{E}^T)\subset \mathcal{F} \Leftrightarrow X_t^{-1}(\mathcal{E})\subset \mathcal{F}$?

I think I understand why $X$ can be viewed as $(E^T,\mathcal{E}^T)$-valued r.v.

Also, I think that if we could generate $\mathcal{E}^T$ from $\mathcal{E}$ it would be easier... but not exactly sure how.

An old man in the sea.
  • 5,422

1 Answers1

2

$\mathcal E^T$ is by definition the smallest $\sigma$-algebra on $E^T$ such that for every $t\in T$ the projection $\pi_t: E^T\to E$ is measurable.

So we have:$$\mathcal E^T=\sigma\left(\bigcup_{t\in T}\pi_t^{-1}(\mathcal E)\right)$$

Then: $$X^{-1}(\mathcal E^T)=X^{-1}\left(\sigma\left(\bigcup_{t\in T}\pi_t^{-1}(\mathcal E)\right)\right)=\sigma\left(X^{-1}\left(\bigcup_{t\in T}\pi_t^{-1}(\mathcal E)\right)\right)=\sigma\left(\bigcup_{t\in T}X^{-1}\left(\pi_t^{-1}(\mathcal E)\right)\right)=$$$$\sigma\left(\bigcup_{t\in T}X_t^{-1}(\mathcal E)\right)\tag1$$

The second equality needs a proof which can be found here.

Equality $(1)$ makes it easy to prove that: $$X^{-1}(\mathcal E^T)\subseteq\mathcal F\iff\forall t\in T\;[ X_t^{-1}(\mathcal E)\subseteq\mathcal F]$$

drhab
  • 153,781
  • Many thanks for your answer/help. May I interest you also in this question? https://stats.stackexchange.com/questions/369591/definition-of-stochastic-process-as-probability-measure-in-a-prob-space – An old man in the sea. Oct 01 '18 at 13:31