1

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.

BCLC
  • 13,459
  • 1
    Is $\sigma(\cal Y, T_n)$ the same thing as $\sigma(\cal Y \cup T_n)$? I don't understand what the comma means. – shalop Jul 16 '15 at 01:52
  • @Shalop Um, like this? – BCLC Jul 16 '15 at 19:35
  • @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 – BCLC Jul 16 '15 at 19:36
  • Explain downvote? – BCLC Jul 16 '15 at 19:52
  • 1
    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. – shalop Jul 16 '15 at 20:32
  • @Shalop I didn't tag you in my 'Explain downvote?' comment :P I don't yet want proof of the hint. I am attempting to prove it myself. I do want a proof check (hence the proof-verification tag). What do you mean such event is in $\mathscr{L}$ \ $\mathscr{R}$ ? An event in the empty set has undefined rather than zero probability? – BCLC Jul 16 '15 at 20:37
  • I mean that the event ${Y_0=1}$ is in $\cal L$ but not in $\cal R$. Actually your proof (together with the hint) shows that. – shalop Jul 16 '15 at 20:40
  • @Shalop Um, so the proof is right? – BCLC Jul 16 '15 at 21:04
  • 1
    Yeah, it looks fine to me. – shalop Jul 16 '15 at 23:11

1 Answers1

0

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

BCLC
  • 13,459