What exactly is $\cap$-stable here?

Question

From my lecture notes:

Theorem: Let $(\Omega, \mathcal A, P)$ be a probability space, $A \in \mathcal A, \mathcal M := \{ M_1, \ldots, M_n \} \subset \mathcal A$. The following statements are equivalent:

a) for all $E \in \sigma(\mathcal M): A$ and $E$ are independent;

b) for all $J\subset \{ 1, \ldots, n \}$: $A$ and $\cap_{j\in J} M_j$ are independent.

Proof: One direction is trivial. For the other direction note that $$\mathcal D_A:=\{ E \in \mathcal A \ | \ A \text{ and } E \text{ are independent} \}$$ is a Dynkin system. It contains all sets of type $\cap_{j\in J} M_j$. And since it is a $\cap$-stable system, $\delta(\mathcal M)=\sigma(\mathcal M)$. Thus, $\sigma(\mathcal M)\subset D_A$.

I don't understand the proof. What exactly is $\cap$-stable here? If $\mathcal M$ were $\cap$-stable, we could conclude that $\delta(\mathcal M)=\sigma(\mathcal M)$. But how does the assumption imply that $\mathcal M$ is $\cap$-stable?

If the proof is indeed false, is there any fix that uses $\mathcal D_A$?

Answer 1

1. A class $\mathcal{C}$ of sets is $\cap$-stable if it is closed under $\cap$. It means that, if $E,F\in\mathcal{C}$ then $E\cap F\in\mathcal{C}$.

2. In the proof of the theorem it is used that the class $\mathcal{B}$ of all sets of type $\cap_{j\in J} M_j$, where $J\subset \{ 1, \ldots, n \}$, is $\cap$-stable.

It is easy to prove such statement.

3. The final steps of the proof need some fixing/clarifcation. Here is, in detail, how they should be:

$\mathcal{D}_A$ is a Dynkin system and $\mathcal{B}\subseteq \mathcal{D}_A$ so $\delta(\mathcal{B})\subseteq \mathcal{D}_A$, but since $\mathcal{B}$ is $\cap$-stable, we have $\delta(\mathcal{B})=\sigma(\mathcal{B})$, so we have that $\sigma(\mathcal{B})\subseteq \mathcal{D}_A$. Now, note that $\sigma(\mathcal{B})=\sigma(\mathcal{M})$. So we can conclude that $\sigma(\mathcal{M})\subseteq \mathcal{D}_A$.

Answer 2

Here is some detail you might find useful

Consider $$\mathcal D_A:=\{ E \in \mathcal A \ | \ A \text{ and } E \text{ are independent} \}$$

Let $E_1,E_2 \in \mathcal D_A$ therefore $$P(A \cap E_1) = P(A)P(E_1)\\ P(A \cap E_2) = P(A)P(E_2) $$

We would like to see that $$P(A \cap E_1 \cap E_2) = P(A)P(E_1)P(E_2) $$

This follows for any pair of sets $E_1,E_2 \in \cap M_j$?
$$E_1 = A_{j_1}^1\cap A_{j_2}^1\ldots \cap A_{j_k}^1 \\ E_2 = A_{i_1}^2\cap A_{i_2}^2\ldots \cap A_{i_{k'}}^2$$

Consider

$\{r_1, \ldots r_{k''}\} = \{j_1,\ldots, j_k\} \cap \{i_1,\ldots i_{k'} \}$

$\{J_1, \ldots J_{s'}\} = \{j_1,\ldots, j_k\} \setminus \{i_1,\ldots i_{k'} \}$

$\{I_1, \ldots I_{s'}\} = \{i_1,\ldots i_{k'} \} \setminus \{j_1,\ldots, j_k\}$

rewrite to obtain $$E_1 = A_{r_1}^1\cap\ldots \cap A_{r_k''}^1\cap A^1_{J_1}\ldots \cap A_{J_s}^1 \\ E_2 = A_{r_1}^2\cap \ldots A_{r_{k''}}^2\cap A^2_{I_1}\ldots \cap A_{I_{s'}}^2$$

Therefore

$$E_1\cap E_2 = \big(A_{r_1}^1\cap A_{r_1}^2\cap\big)\ldots \cap \big(A_{r_k''}^1\cap A_{r_{k''}}^2\big)\cap A^1_{J_1}\ldots \cap A_{J_s}^1 \cap A^2_{I_1}\ldots \cap A_{I_{s'}}^2 $$

So in order to assure that $E_1\cap E_2$ you must check that each $\big(A_{r_i}^1\cap A_{r_i}^2\cap\big) \in M_{r_i}$ which will follow if $M_{r_i}$ is $\cap$- stable (in case $M_{r_i}$ is a sigma algebra that follows).