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Suppose $X:\Omega\to \mathbb{R}$ is a Random Variable. I am interested in the relationship of $\sigma(X)$ and $\sigma(A:A=\{\omega:X(\omega)\leq x\},x\in\mathbb{R})$. Intuitively, it seems to me that $\sigma(A:A=\{\omega:X(\omega)\leq x\},x\in\mathbb{R})\subset \sigma(X)$ as both sigma algebras contain information about X, but the latter requires more precise information than the former, being then finer.

I am, however, a little bit unsure of my proof. My attempt is the following:

If $C=\{\omega:X(\omega)\leq c\}$ for some $c\in \mathbb{R}$, $C\in \sigma(X)$. Therefore, $\sigma(X)$ contains the collection $\{A:A=\{\omega:X(\omega)\leq x\},x\in\mathbb{R}\}$. As $\sigma(A:A=\{\omega:X(\omega)\leq x\},x\in\mathbb{R})$ is the smallest sigma algebra containing such a collection and as $\sigma(X)$ is also a sigma algebra, $\sigma(A:A=\{\omega:X(\omega)\leq x\},x\in\mathbb{R})\subset \sigma(X)$.

Is this argument valid?

Heatconomics
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1 Answers1

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Your proof is okay, but it should be mentioned that we also have: $$\sigma\left(\left\{ \left\{ X\leq x\right\} \mid x\in\mathbb{R}\right\} \right)\supseteq\sigma(X)$$ so that the two $\sigma$-algebras are just the same.

In short:$$\sigma\left(\left\{ \left\{ X\leq x\right\} \mid x\in\mathbb{R}\right\} \right)=\sigma\left(X^{-1}\left(\left\{ \left(-\infty,x\right]\mid x\in\mathbb{R}\right\} \right)\right)=X^{-1}\left(\sigma\left(\left\{ \left(-\infty,x\right]\mid x\in\mathbb{R}\right\} \right)\right)$$$$=X^{-1}\left(\mathcal{B}\right)=\sigma\left(X\right)$$where the third equality is an application of the rule that: $$f^{-1}\left(\sigma\left(\mathcal{V}\right)\right)=\sigma\left(f^{-1}\left(\mathcal{V}\right)\right)\tag1$$For a proof of that rule see here.

Let me add that $(1)$ is very handsome to know.

drhab
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  • Thanks! Indeed it is a very useful property. I guess that the tortuous way of proving the reverse inclusion would be by using the pi-lambda lemma or some monotone-class type of result. I will definitely keep that shortcut in my mind! – Heatconomics Oct 29 '19 at 14:16
  • Proof of inverse inclusion is not really difficult and does not depend on pi-lambda lemma (or other results like that). Just have a look at the link in my answer. – drhab Oct 29 '19 at 15:29