Sigma algebra respect to random variables.

Question

Let $\Omega=\{1,2,3,4,5,6\}$ and $X(\omega)=\omega \mod 2$, which is a random variable. Then, my first question is, what is $\sigma(X)$? In my understanding, $\sigma(X)=\{\{\omega\mid X(\omega)\in B\}\mid B\subset R\}$, so $\sigma(X)=\{\emptyset,\{1,2,3\},\{4,5,6\},\Omega\}$.

Let $Y(\omega)=1_{\omega\geq 4}$. Then, my second question is, what is $\sigma(X,Y)$? In my thought, $\sigma(X,Y)=\{\{\omega\mid (X(\omega),Y(\omega))\in B_1\times B_2\}\mid B_1,B_2\subseteq R\}$. $\sigma(X,Y)=\left\{\{2\},\{5\},\{1,3\},\{4,6\},\{2,5\},\{1,2,3\},\{2,4,6\},\{1,3,5\},\{4,5,6\},\{1,3,4,6\},\{1,2,3,5\},\{2,4,5,6\},\{1,2,3,4,6\},\{1,3,4,5,6\},\Omega\right\}$ Is this true?

Answer 1

I don't beleive you've identified $\ \sigma\left(X\right)\ $ correctly. Consider your expression, $\ X^{-1}\left(B\right) = \{\omega\mid X(\omega)\in B\}\ $, for the elements of $\ \sigma\left(X\right)\ $. This is correct. Strictly speaking, $\ B\ $ should be restricted to be a measurable subset of $\ R\ $, but in this case it makes no difference if you take all subsets of $\ R\ $ to be measurable.

Since the only possible values of $\ X\left(\omega\right)\ $ are $\ 0\ $ and $\ 1\ $, then unless $\ 0\in B\ $ or $\ 1\in B\ $, we must have $\ X^{-1}\left(B\right) = \emptyset\ $. If $\ 0\in B\ $ but $\ 1\not\in B\ $, then $\ X\left(\omega\right)\in B\ $ if and only if $\ \omega= 2, 4, \mbox{ or }\ 6\ $, so $\ X^{-1}\left(B\right) = \left\{2,4,6\right\}\ $. Similarly, if $\ 1\in B\ $ but $\ 0\not\in B\ $, then $\ X^{-1}\left(B\right) = \left\{1,3,5\right\}\ $, and if both $\ 0\in B\ $ and $\ 1\not\in B\ $, then $\ X^{-1}\left(B\right) = \left\{1,2,3,4,5,6\right\}= \Omega\ $, thus giving $\sigma(X)=\{\emptyset,\{1,3,5\},\{2,4,6\},\Omega\}$.

Your definition of $\ \sigma\left(X,Y\right)\ $ is also not quite correct. It should be \begin{align} \sigma(X&,Y)=\\ &\{\{\omega\mid (X(\omega),Y(\omega))\in B \}\mid B\subseteq R\times R, \mbox{measurable}\}\ . \end{align}That is, the elements of $\ \sigma(X,Y)\ $ are the inverse images not merely of product sets, but all measurable subsets of $\ R\times R\ $. Nevertheless, your list of sets in $\ \sigma(X,Y)\ $ is nearly right. You've inadvertently omitted the empty set from the list

Answer 2

Thanks. The definition of $\sigma(X,Y)$ is $$\{\{(\omega_1,\omega_2)\mid (X(\omega_1),Y(\omega_2))\in B\}\mid B\subseteq R\times R, mesurable\}$$, and for the case $(0,1)\in B$ and others not in $B$, $$\{(2,4),(2,5),(2,6),(4,4),(4,5),(4,6),(6,4),(6,5),(6,6)\}$$. I need to consider $2^4$ cases, like $(0,0)\in B$, $(0,1)\in B$, $(1,0)\notin B$, $(1,1)\in B$..., right?