# How can a $\sigma$-algebra be “treated” or computed? Example

My question is: I have a random variable $X:\Omega \rightarrow \mathbb{R}$, the $\sigma$-algebra generated by $X$ is: $\sigma(X) := \{X^{-1}(B), B\in \mathcal{B}(\mathbb{R})\}$.

But, imagine now that $X=\exp \{\mu + \sigma Z\}$ with $Z \sim N(0,1)$.

Does this help when trying to compute $\sigma(X)$? Like... can we say something like $\sigma(X) = \sigma(Z)$? or similar? Is this special case easily solvable? I mean, write explicitly $\sigma(X)$.

Thank you very much for your time and help!

-
Your question uses $\sigma$ in two different senses in $\sigma(X)=\sigma(Z)$ and in $X= \exp (\mu + \sigma Z)$. – Henry Sep 13 '15 at 14:21

The Borel $\sigma$-algebra is generated by the bounded closed intervals. This implies that $\sigma(X)$ is generated by the preimages of bounded closed intervals under $X$. Now if $f:\mathbb{R}\to\mathbb{R}$ is continuous, it will map bounded closed intervals to bounded closed intervals. If $f$ is moreover injective (aka 1-to-1), it will map different closed intervals to different closed intervals. So let us show that $\sigma(X)=\sigma(f\circ X)$.

$\sigma(f\circ X)\subseteq\sigma(X)$:

Let $B$ be an arbitrary Borel-subset of $\mathbb{R}$. Then $$(f\circ X)^{-1}(B)=X^{-1}\big(f^{-1}(B)\big).$$ Since $f^{-1}(B)$ is a Borel set, we get $\sigma(f\circ X)\subseteq\sigma(X)$.

$\sigma(f\circ X)\supseteq\sigma(X)$:

Let $C$ be a bounded and closed interval. Then $f(C)$ is a bounded and closed interval and hence a Borel-set. Since $f$ is injective, we have $f^{-1}\big(f(C)\big)=C$. So $$X^{-1}(C)=X^{-1}\Big(f^{-1}\big(f(C)\big)\Big)=(f\circ X)^{-1}\big(f(C)\big).$$ Hence, $\sigma(f\circ X)\supseteq\sigma(X)$.

Note: The distribution of the random variables does not matter at all.

-
I understood everything prefectly! Thank you very much :) – mark Apr 25 '13 at 15:50