Say I have two random variables $X$ and $Y$. Their respective $\sigma$-algebras are $$\sigma(X) = \{ X^{-1}(B) \mid B \in \mathscr{B} \}$$ and $$\sigma(Y) = \{ Y^{-1}(B) \mid B \in \mathscr{B} \}.$$

Their product $XY$ has sigma-algebra: $$\sigma(XY) = \{ (XY)^{-1}(B) \mid B \in \mathscr{B} \}.$$

Is there a way to define $\sigma(XY)$ in terms of $\sigma(X)$ and $\sigma(Y)$ (not necessarily using unions, intersections or complements)?

For instance, we can define $$\sigma(A_1,A_2) = \sigma(\sigma(A_1) \cup \sigma(A_2))$$ instead of $\sigma(A_1,A_2)$ the smallest $\sigma$-algebra containing $A_1$ and $A_2$ and generated by the partition $\{B_1, B_2, B_3, B_4\}$ where \begin{align}B_1 &= A_1 \setminus A_2,\\ B_2 &= A_2 \setminus A_1,\\ B_3 &= A_1 \cap A_2,\\ B_4 &= (A_1 \cup A_2)^{C}.\end{align}

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    $\begingroup$ Answer: No (as two-valued random variables show). $\endgroup$ – Did Jul 24 '15 at 16:13
  • $\begingroup$ @Did Thanks. What is it that two-valued random variables show? $\endgroup$ – BCLC Jul 24 '15 at 16:44

Let $\Omega = \{a,b\}$. Define $X,Y:\Omega\to\mathbb R$ by $$X(\omega) = \begin{cases}0,& \omega=a\\ 1,&\omega=b. \end{cases} $$ and $$Y(\omega) = \begin{cases}1,& \omega=a\\ 0,&\omega=b. \end{cases} $$ Then $\sigma(X)=\sigma(Y) = 2^\Omega$, but $XY$ is identically zero so $\sigma(XY)=\{\varnothing,\Omega\}$.

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    $\begingroup$ @Did Clever way of circumventing the lower bound on characters in a comment.. $\endgroup$ – Math1000 Jul 24 '15 at 19:57
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    $\begingroup$ I guess you mean to say that it is easy to construct pathological examples and hence 'no'. Thanks Math1000 ^-^ $\endgroup$ – BCLC Jul 25 '15 at 9:19

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