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 [1] 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)