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I came across this notation of which I am unfamiliar;

$\mathscr{F}=\mathscr{G}_{1}\vee \mathscr{G}_{1}$

where $\mathscr{G}_{1}$ and $\mathscr{G}_{2}$ are both sigma-fields of subsets of $\Omega$. It is claimed $\mathscr{F}$ is larger than both of $\mathscr{G}_{1}$ and $\mathscr{G}_{2}$ suggesting to me that

$\mathscr{F}=\mathscr{G}_{1}\vee \mathscr{G}_{1}=\mathscr{G}_{1}\cup \mathscr{G}_{1}$

but I know this last union is not always a sigma-field so perhaps this is not the meaning here?

Staying on this subject, would the notation $\mathscr{G}_{1}\subset\mathscr{G}_{2}$ mean the same as $\mathscr{G}_{1}\leq\mathscr{G}_{2}$, the latter statement which (I assume anyway) means $\mathscr{G}_{2}$ is finer than $\mathscr{G}_{1}$?

For some reason I cannot seem to find a clear definition of this for sigma-fields - sets, partitions etc yes, but not sigma fields - for example here;

https://math.stackexchange.com/questions/1345598/does-meet-of-two-partitions-of-a-set-always-exist

Any help as ever appreciated.

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    $\begingroup$ Offhand, I would expect this to be defined within the context of a lattice, as explained at Wikipedia. The natural lattice in this context would be that of all sigma-fields defined on $\Omega;$ the (partial) ordering would be that inherited from the power set $\mathcal{P}(\Omega).$ In other words, the meet would be the coarsest sigma-field of which both components are subfields. That's not usually their union--it will be bigger. It would help to know the context in which you came across this notation in case some other lattice is understood. $\endgroup$
    – whuber
    Commented Dec 11, 2018 at 23:22
  • $\begingroup$ Thank you @whuber. As I responded to Jonas I believe his answer is correct given the context - I think you are describing the same thing in the sense that $\sigma(\mathcal{G}_{1}\cup \mathcal{G}_{2})$ is this coarsest sigma-field you mention? The context I found this in is "coarsening at random" - essentially where loss of information occurs with censoring or missing values in repeated measures. In this context $\mathcal{G}_{1}$ represents the sigma-field generated by the indicator variable denoting if a random variable is observed, and $\mathcal{G}_{2}$ is for the random variable itself. $\endgroup$
    – dandar
    Commented Dec 12, 2018 at 8:28
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    $\begingroup$ Yes, he is describing the same thing in a more basic way. $\endgroup$
    – Jonas
    Commented Dec 12, 2018 at 8:30

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It is difficult to answer this question, since it is essentially about notation. So my answer contains a bit of clever guessing.

Indeed, $\mathcal{G_1} \cup \mathcal{G_2}$ is in general no $\sigma$-field. Therefore, one may set $$\mathcal{G_1} \vee \mathcal{G_2} = \sigma( \mathcal{G_1} \cup \mathcal{G_2}),$$ where $\sigma(C)$ denotes the $\sigma$-field that is generated by the set system $C \subseteq 2^\Omega$. This function is defined by $$\sigma(C) = \bigcap \left\lbrace \mathcal{G} : \mathcal{G} \supseteq C, \mathcal{G} \text{ is $\sigma$-field on $\Omega$}\right\rbrace. $$

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  • $\begingroup$ Thank you @Jonas for your answer - this makes sense to me given the context I encountered the notation. So essentially $\mathcal{F}=\mathcal{G}_{1}\vee \mathcal{G}_{2}$ is the smallest sigma-field containing $\mathcal{G}_{1}\cup \mathcal{G}_{2}$ in the sense that if $\mathcal{F}'$ is another sigma-field $\mathcal{G}_{1}\cup \mathcal{G}_{2}\subset\mathcal{F}'$ then $\mathcal{F}'\subset\mathcal{F}$? $\endgroup$
    – dandar
    Commented Dec 12, 2018 at 8:14
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    $\begingroup$ In the last line, I would say $\mathcal{F}' \supseteq \mathcal{F}$, since $ \mathcal{F}$ is indeed the smallest $\sigma$-field that contains $\mathcal{G_1}\cup \mathcal{G_2}$. $\endgroup$
    – Jonas
    Commented Dec 12, 2018 at 8:26
  • $\begingroup$ of course - my mistake! Thank you for the confirmation $\endgroup$
    – dandar
    Commented Dec 12, 2018 at 8:43

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