Independence of sigma-algebras

[Richard27182]

Homework Statement

Let (A_n : n\in \mathbb{N}) be a sequence of events in a probability space. Show that the events A_n are independent if and only if the \sigma-algebras \sigma(A_n)=\{\emptyset, A_n, A_n^c, \Omega\} are independent.

Homework Equations

For \sigma-algebras \mathcal{A}_i \subseteq \mathcal{F} to be independent, we need that if A_i \in \mathcal{A}_i\forall i, then the A_i must be independent.

For events A_i:i\in I to be independent, we need that , for any finite subset J\subseteq I:
\mathbb{P}(\bigcap_{i\in J} A_i )=\prod_{i\in J} \mathbb{P}(A_i )

The Attempt at a Solution

So first assume that the \sigma-algebras are independent. Now A_i \in \sigma(A_i) \forall i, so the A_i must be independent.

Now assume that the events A_i are independent. Now take events B_i such that B_i \in A_i \forall i Now we need to check that for any finite subset J \subseteq \mathbb{N}, that we have
\mathbb{P}(\bigcap_{i\in J}B_i )=\prod_{i \in J} \mathbb{P}(B_i )

Now if we have that for some i \in J,B_i =\emptyset, this is immediate (both sides are zero)
If we have that for some i\in J,B_i =\Omega, it reduces to the case where \nexists i \in J s.t. B_i=\Omega
So we are left with three cases to check:
1. B_i =A_i \forall i \in J
2. B_i =A_i^c \forall i \in J
3. B_i =A_i for some i \in J and B_i =A_i^c for the rest of i\in J
Case 1 follows immediately from the independence of the A_i
Case 2 Labelling the sets in J from 1 to n
<br /> \mathbb{P}(\bigcap_{i=1}^{n}A_i^c)=\mathbb{P}((\bigcup_{i=1}^{n}A_i )^c)\text{(De Morgan&#039;s law)}
=1-\mathbb{P}(\bigcup_{i=1}^{n} A_i)
=1-\sum_{1}^{n}\mathbb{P}(A_i)+\sum_{i,j:1\le i&lt;J\le n}\mathbb{P}(A_i \cap A_j)+...-(-1)^{n-1}\mathbb{P}(\bigcap_{i=1}^{n}A_i) \text{(by Inclusion-Exclusion)}
=1-\sum_{i=1}^{n}\mathbb{P}(A_i)+\sum_{i,j:1\le i&lt;J\le n}\mathbb{P}(A_i)\mathbb{P}( A_j)+...-(-1)^{n-1}\prod_{i-1}^{n}\mathbb{P}(A_i) \text{(by independence of} A_i )
=\prod_{i=1}^{n}(1-\mathbb{P}(A_i))
=\prod_{i=1}^{n}\mathbb{P}(A_i^c)<br />
as required.
So all that remains is to check case 3. That is, we need to show that if (possibly relabelling the A_i) B_i =A_i \text{ } i=1,...,r
B_i=A_i^c\text{ } i=r+1,...,n then
\mathbb{P}(\bigcap_{i=1}^{n}B_i}=\prod_{i=1}^{n}\mathbb{P}(B_i)
\mathbb{P}(\bigcap_{i=1}^{n}B_{i})=\mathbb{P}((\bigcap_{i=1}^{r}B_i )\cap (\bigcap_{i=r+1}^{n}B_i ))<br />

Unfortunately, I've got stuck here and am unable to make any further progress.

 

[Richard27182]

OK, so I have an answer:

\mathbb{P}((\bigcap_{i=1}^{r}B_i )\cap (\bigcap_{i=r+1}^{n}B_i ))
=\mathbb{P}((\bigcap_{i=1}^{r}A_i )\cap (\bigcap_{i=r+1}^{n}A_i^c ))
=\mathbb{P}((\bigcap_{i=1}^{r}A_i )\cap ((\bigcap_{i=r+1}^{n}A_i )^c))
Now assuming that \mathbb{P}(\bigcap_{i=1}^{r}A_i )&gt;0 (if not, the result follows immediately)
=\mathbb{P}(\bigcap_{i=1}^{r}A_i )\mathbb{P}((\bigcap_{i=r+1}^{n}A_i )^c|(\bigcap_{i=i}^{r}A_i))
=\mathbb{P}(\bigcap_{i=1}^{r}A_i )(1-\mathbb{P}((\bigcap_{i=r+1}^{n}A_i )|(\bigcap_{i=i}^{r}A_i)))
=\mathbb{P}(\bigcap_{i=1}^{r}A_i )\mathbb{P}(\bigcap_{i=r+1}^{n}A_i^c )
=\prod_{i=1}^{r}\mathbb{P}(A_i )\prod_{i=r+1}^{n}\mathbb{P}(A_i^c)
=\prod_{i=1}^{n}\mathbb{P}(B_i)

as required

 

[Richard27182]

above answer does not work, I've applied De morgan's laws incorrectly

 

[Richard27182]

OK, so I have another answer:

\mathbb{P}((\bigcap_{i=1}^{r}B_i )\cap (\bigcap_{i=r+1}^{n}B_i ))
=\mathbb{P}((\bigcap_{i=1}^{r}A_i )\cap (\bigcap_{i=r+1}^{n}A_i^c ))
=\mathbb{P}((\bigcap_{i=1}^{r}A_i )\cap ((\bigcup_{i=r+1}^{n}A_i )^c))
Now assuming that \mathbb{P}(\bigcap_{i=1}^{r}A_i )&gt;0 (if not, the result follows immediately)
=\mathbb{P}(\bigcap_{i=1}^{r}A_i )\mathbb{P}((\bigcup_{i=r+1}^{n}A_i )^c|(\bigcap_{i=i}^{r}A_i))
=\mathbb{P}(\bigcap_{i=1}^{r}A_i )(1-\mathbb{P}((\bigcup_{i=r+1}^{n}A_i )|(\bigcap_{i=i}^{r}A_i)))
=\mathbb{P}(\bigcap_{i=1}^{r}A_i )(1-\mathbb{P}((\bigcup_{i=r+1}^{n}A_i )))
=\mathbb{P}(\bigcap_{i=1}^{r}A_i )\mathbb{P}((\bigcup_{i=r+1}^{n}A_i)^c)
=\mathbb{P}(\bigcap_{i=1}^{r}A_i )\mathbb{P}(\bigcap_{i=r+1}^{n}A_i^c )
=\prod_{i=1}^{r}\mathbb{P}(A_i )\prod_{i=r+1}^{n}\mathbb{P}(A_i^c)
=\prod_{i=1}^{n}\mathbb{P}(B_i)

as required