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Assuming the hint is true, I attempt to prove the latter prop:

Assume on the contrary that $\mathscr{L} = \mathscr{R}$.

If $\sigma(Y_0) \subseteq \mathscr{L}$, then $\sigma(Y_0) \subseteq \mathscr{R}$.

Since $\sigma(Y_0)$ and $\mathscr{R}$ are independent, $\sigma(Y_0)$ is independent of itself.

This means $\forall F \in \sigma(Y_0), P(F \cap F) = P(F)P(F) \ \to \ P(F) \in \{0,1\}$.

Choose $F = (Y_0 = 1)$ or $F = (Y_0 = -1)$. We have $P(F) = 1/2 \notin \{0,1\}$

Is that right? I based it off the K 0-1 Law proof here.

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    $\begingroup$ Is $\sigma(\cal Y, T_n)$ the same thing as $\sigma(\cal Y \cup T_n)$? I don't understand what the comma means. $\endgroup$ – Shalop Jul 16 '15 at 1:52
  • $\begingroup$ @Shalop Um, like this? $\endgroup$ – BCLC Jul 16 '15 at 19:35
  • $\begingroup$ @Shalop Anyway, I think understanding $\mathscr{L}$ and $\mathscr{R}$ is relevant for proving the hint rather than assuming the hint and then proving the second statement $\endgroup$ – BCLC Jul 16 '15 at 19:36
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    $\begingroup$ Do you want a proof of the hint? I thought that you just wanted someone to check what you wrote above, i.e, that the event $\{Y_0=1\}$ is in $\cal L \backslash R$, which is true. BTW, I didn't downvote your question. $\endgroup$ – Shalop Jul 16 '15 at 20:32
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    $\begingroup$ Yeah, it looks fine to me. $\endgroup$ – Shalop Jul 16 '15 at 23:11

Yeah, it looks fine to me. – Shalop Jul 16 at 23:11


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