Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question

1 Answer 1

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.

share|improve this answer
    
I understood everything prefectly! Thank you very much :) –  mark Apr 25 '13 at 15:50

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.