Just would like to clarify my understanding of random variable composition.

Here is my problem statement:

Given a measurable space $(\Omega,\mathcal{B})$ let $X$ be a random variable such that: $X:(\Omega,\mathcal{B}) \rightarrow (\mathbb{R},\mathcal{B}(\mathbb{R})) $

Now, let $\sigma(X)$ be the $\sigma$-algebra generated by the random variable $X$.

FIRST QUESTION: I understand that $\sigma(X)$ contains all the sets that can be created from doing the inverse mapping $X^{-1}(\{ B \in \mathcal{B}(\mathbb{R}) \})$. In other words, all the preimages in $\Omega$ which can be mapped back from $\mathbb{R}$.

Then, can we say that $ \sigma(X) \subset \mathcal{B} $ ?


Considering answer to FIRST QUESTION, then If we let a measurable function $g$, do a subsequent mapping, (as I understand, $g$ can be any measurable function of $X$: $g=f(X)$).


$ ( g \text{ o } X )$ is $\sigma(X)|\mathcal{B}(\mathbb{R})$ measurable.

Can somebody please, explain to me why we use $\sigma(X)|\mathcal{B}(\mathbb{R})$ to denote that the composition is measurable:

$ ( g \text{ o } X ) \in \sigma(X)|\mathcal{B}(\mathbb{R})$ ?

I think it would be better to use: $\mathcal{B}|\mathcal{B}(\mathbb{R})$ to denote measurability composition.

Thank you all.

  • $\begingroup$ The first question is a classical definition of the field. In the second question, you are switching from $g=f\circ X$ to $g\circ X$. Which one are you actually considering? $\endgroup$ – Did Nov 19 '17 at 21:37
  • $\begingroup$ Thanks for your reply. It seems I don't have it clear. Could you please clarify it to me? I thought that: We do the first mapping (RV): $X:(\Omega,\mathcal{B})→(\mathbb{R},\mathcal{B}(\mathbb{R}) )$ Then, we do another mapping: $g:((\mathbb{R},\mathcal{B}(\mathbb{R}) ) →(\mathbb{R},\mathcal{B}(\mathbb{R}) )$ So, the total composition is $(g \text{ o } X)$. Then, I thought that $(g \text{ o } X) = f(X)$ (where f represents any function, in this case $g$) $\endgroup$ – kentropy Nov 19 '17 at 21:52
  • $\begingroup$ Then $g\ne f\circ X$. $\endgroup$ – Did Nov 19 '17 at 21:55

Of course, $X$ is measurable implies that for any set $B\in\mathcal{B}(\mathbb{R})$, $f^{-1}(B)\in\mathcal{B}$. And if $G$ is a collection of sets, such that $G\subset\mathcal{G}$, where $\mathcal{G}$ is a $\sigma$-algebra, then $\sigma(G)\subset\mathcal{G}$. This should answer your first question.

For the second question note that $\sigma(X)$ is the smallest $\sigma$-algebra for which $X$ is measurable with respect to the $\sigma$-algebra, $\mathcal{B}(\mathbb{R})$.

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  • $\begingroup$ Thanks for your reply. So, for your second statement, does it mean that $X \in \sigma(X)|\mathcal{B}(\mathbb{R}) \neq X \in \mathcal{B}|\mathcal{B}(\mathbb{R}) $ ?? $\endgroup$ – kentropy Nov 19 '17 at 21:49
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    $\begingroup$ No, it says that $X\in\sigma(X)|\mathcal{B}(\mathbb{R})=X\in\mathcal{B}|\mathcal{B}(\mathbb{R})$ for any $\sigma$-algebra $\mathcal{B}$ containing $\sigma(X)$ $\endgroup$ – QED Nov 20 '17 at 3:39

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