2
$\begingroup$

Given a probability space $\Omega$, a measurable space $S$ and a set $T$, a stochastic process $X: \Omega \times T \to S$ is defined as a $T$-indexed family of random variables $\{X_t: \Omega \to S, t \in T\}$.

It can be viewed as a functional-valued mapping $\Omega \to S^T$ with the product sigma algebra on $S^T$.

I was wondering how to show that this functional-valued mapping is measurable wrt the product sigma algebra on $S^T$?

Thanks and regards!

$\endgroup$
3
$\begingroup$

The product sigma-algebra on $S^T$ is generated by the projections

$$\Pi_t: S^T \to S, f \mapsto f(t) \qquad (t \in T)$$

i.e. it's the smallest sigma-algebra such that projections $\Pi_t$ ($t \in T$) are measurable. This means that a mapping $X: \Omega \to S^T$ is measurable (with respect to the product sigma-algebra on $S^T$) iff the mappings

$$\Pi_t \circ X: \Omega \to S, \omega \mapsto X(\omega)(t) = X_t(\omega)$$

are measurable for all $t \in T$.

Let $Y$ a set, $(Y_i,\mathcal{A}_i)$ measure spaces ($i \in I$) and $f_i: Y \to Y_i$ arbritary mappings ($i \in I$). Denote by $\sigma(f_i,i \in I)$ the $\sigma$-algebra generated by the mappings $f_i$. Moreover, let $(\Omega,\mathcal{A})$ a measure space and $g: \Omega \to Y$ a mapping. Then the following statements are equivalent.

  1. $g$ is $\mathcal{A}/\sigma(f_i,i \in I)$-measurable
  2. $\forall i \in I: f_i \circ g$ is $\mathcal{A}/\mathcal{A}_i$-measurable

(Here: $Y:=S^T$, $I:=T$, $Y_i := S$, $f_i := \Pi_i$ for $i \in T$.)

$\endgroup$

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.