Talk:Dynkin system

λ system
The equivalent λ system is defined as follows: Definition. A family L of subsets of C is called a λ-system if
 * (1) Ω belongs to L,
 * (2) L is closed under complementation,
 * (3) L is closed under countable unions of pw disjoint sets.

Given any class C of sets, L(C) denotes the λ-system generated by C. Jackzhp 23:50, 28 October 2006 (UTC)

asdf
I changed a slight mistake. Sorry, no latex improvement. September 13 / 2006 (USF)

A λproof for Dynkin's Lemma?
I suggest adding in a proof for Dynkin's Lemma.--A 20:00, 5 October 2007 (UTC)

I wonder, would be an appropriate source?--A 20:04, 5 October 2007 (UTC)


 * If the site is self-published, as it appears to be, then it would not qualify unless the author is already a recognized (and published) expert on the topic. The author calls himself a derivatives trader. While he cold have a degree in statistics, we really don't know anything else about him or the website. See WP:RS and WP:V for the definitive rules. ·:· Will Beback  ·:· 22:41, 5 October 2007 (UTC)

I came up with proof(not sure if it is right) during preparation for exam. It uses transfinite induction. For every ordinal $$\lambda < \omega_1 $$ we define new set $$ A_{\lambda} $$. $$ A_0=P\cup \emptyset \cup \Omega $$, $$ A_{\lambda} = \left \{ \bigcup_{\alpha \in \omega_0} G_\alpha \setminus \bigcup_{\beta \in \omega_0} H_\beta | G_i,H_j \in \bigcup_{\mu < \lambda} A_\mu \right \} $$ Than one can show that every $$ A_{\lambda} \subset D, A_{\lambda} \subset \sigma (P), A_{\lambda} \subset A_{\mu} $$ iff $$ \lambda < \mu$$, every $$ A_{\lambda} $$ is π-system, $$ \bigcup_{\lambda \in \omega_1} A_\lambda = \sigma(P), \bigcup_{\lambda \in \omega_1} A_\lambda \subset D $$.

Ok the motivation. I want to generate $$\sigma (P)$$ from P. So you can do it with transfinite induction that in ever step you add new sets in form $$ \bigcup A_i $$ and $$ A \setminus B$$. But than it is hard to show that all these new sets are still in Dynkin's system. So you want in every step create pi-system and than it is easy to show that new set generated from pi-system is still in Dynkin's system. So in every step you don't use operation $$\bigcup$$ and $$ \setminus $$ but insted you use $$ \bigcup \setminus \bigcup $$ to generate new sets.

$$A_{\lambda} \subset \sigma (P) $$ This is obvious.

every $$ A_{\lambda} $$ is π-system. $$(\bigcup_j A_j \setminus \bigcup_j B_j ) \cap (\bigcup_j C_j \setminus \bigcup_j D_j) = \bigcup_{i,j} A_i \cap C_j \setminus (  \bigcup_j B_j \cup \bigcup_j D_j ) $$. This is again in form $$ \bigcup \setminus \bigcup $$ because every previous $$ A_\mu, \mu < \lambda $$ are already pi-systems.

$$ A_{\lambda} \subset D $$ Can be show thanks to that every previous $$ A_\mu, \mu < \lambda $$ are pi-systems. You can than convert sum of sets to sum of disjoint sets.

$$ \bigcup_{\lambda \in \omega_1} A_\lambda = \sigma(P) $$ This is quite easy. But you have to use fact that cofinality of $$ \omega_1$$ is $$ \omega_1 $$

So if anyone would have time a will to check it I would be happy to rewrite it properly and post it.

Not the Doob-Dynkin lemma
Maybe the "Dynkin's π-λ Theorem" is sometimes called "Dynkin's lemma", but surely it is not the "Doob–Dynkin lemma" (and not related to it). I correct the text accordingly. Boris Tsirelson (talk) 08:21, 7 September 2012 (UTC)