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Here is the question:

Let $(\Omega, \mathcal{F}, \mathrm{P})$ be a probability space. Let $T$ be an arbitrary (possibly uncountable) index set, and $\forall t\in T$, let $X_t:(\Omega,\mathcal{F})\rightarrow (E_t, \mathcal{E}_t)$ be a random variable with state space $(E_t, \mathcal{E}_t)$. Then a function $F:\Omega\rightarrow \mathrm{R}$ is measurable with respect to $\sigma({\{X_t\mid t \in T\}})$ if and only if there is a sequence $(t_n)_{n\in N}\subset T$ and a function $g:\times_{n\in \mathrm{N}} E_{t_n} \rightarrow \mathrm{R} $ that is $(\bigotimes_{n\in\mathrm{N}}\mathcal{E}_{t_n})$- measurable (i.e., whith respect to the product $\sigma$-algebra of $(E_{t_n}\mid n\in \mathrm{N})$) such that $F=g(X_{t_1}, X_{t_2},\dots)$.

Any thoughts on how to attack this (one direction is straightforward, but the other???)?. Any help would be much appreciated! Thanks in advance...

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    $\begingroup$ By the way, welcome to MSE! $\endgroup$ Oct 21, 2012 at 14:54

1 Answer 1

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It's just a sketch of proof: write $F=\lim_{n\to \infty}f_n$ (a pointwise limit), where $\{f_n\}$ are $\sigma(\{X_t,t\in T\})$ measurable. If we have done it for each $f_n$ (that is, written $f_n=g_n(X_{t_{j,n}},j\geq 1))=g_n(t_j,j\in D)$, where $D$ is countable and $g_n$ is $\bigotimes_{j\in D}\mathcal E_{t_j}$ measurable), then take $g$ given by the simple limit of $g_n$. So we are reduced to prove it when $F$ is the indicator of a set in $\sigma(\{X_t,t\in T\})$.

The collection of subsets of $\Omega$ whose characteristic function can be expressed as in the second part of the assertion is a $\sigma$-algebra containing $\sigma(X_t)$ for all $t\in T$.

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  • $\begingroup$ many thanks for the help, I really appreciate it! I think I can take if from here. $\endgroup$
    – s_2
    Oct 22, 2012 at 0:03

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