1
$\begingroup$

enter image description here


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.

$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ Yeah, it looks fine to me. $\endgroup$ – Shalop Jul 16 '15 at 23:11
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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