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Let $\{v_\lambda:~\lambda\in\Lambda\}$ be a family of real-valued random variables on a (complete) probability space $(\Omega, \sigma, \mathbb{P})$. Assume the variables lie in $\bigcap_{p=1}^\infty L^p(\Omega)$ and denote by $\mathcal{A}$ the unital algebra generated by them (along with the constants), and $\alpha$ the $\sigma$-algebra generated by them (i.e. the minimal complete sub-sigma-algebra with respect to which they are measurable).

If the variables are Gaussian then it is a consequence of the Wiener chaos decomposition (see e.g. Janson - Gaussian Hilbert Spaces) that $\mathcal{A}$ is dense in $L^2(\Omega, \alpha, \mathbb{P}|_\alpha)$. In particular, $L^2(\Omega,\alpha,\mathbb{P}|_\alpha)$ is generated by $\mathcal{A}$ as a subspace of $L^2(\Omega, \sigma, \mathbb{P})$.

But what can we say in the general case? Is it possible that the algebra and $\sigma$-algebra generate different subspaces? Relevant references would be appreciated - mostly I'm having trouble searching for literature about actual algebras of random variables because the term '$\sigma$-algebra' appears almost everywhere.

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  • $\begingroup$ I suppose the $\sigma$-algebra you refer to is the smallest sub-$\sigma$-algebra of measurable subsets of $\Omega$ relative to which all of the $\nu_\lambda$ are measurable functions. Is that right? $\endgroup$
    – Ruy
    Commented Nov 21, 2016 at 13:46
  • $\begingroup$ When you say "the algebra they generate" do you intend to throw in the constants? I am not sure if it is strictly necessary, but I think you need it if the Wiener chaos argument is going to work. $\endgroup$ Commented Nov 21, 2016 at 14:01
  • $\begingroup$ Edited to make these things a little clearer: but yes, the unital algebra and the minimal (complete) $\sigma$-algebra they generate. $\endgroup$
    – Ollie
    Commented Nov 21, 2016 at 15:15
  • $\begingroup$ If your random variables were in $L^\infty$ this would follow from the multiplicative system theorem. There might be another version of it that works for unbounded functions, I'm not sure. $\endgroup$ Commented Nov 21, 2016 at 16:43
  • $\begingroup$ Thanks, I'll look that up, might yield some generalisation :) - I am more interested in the unbounded case, as the Gaussian prototype suggests. $\endgroup$
    – Ollie
    Commented Nov 21, 2016 at 17:42

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