# Independence of sigma-algebras

1. Jun 25, 2010

### Richard27182

1. The problem statement, all variables and given/known data

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.

2. Relevant 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 $$J\subseteq I$$:
$$\mathbb{P}(\bigcap_{i\in J} A_i )=\prod_{i\in J} \mathbb{P}(A_i )$$

3. 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 $$n$$
$$\mathbb{P}(\bigcap_{i=1}^{n}A_i^c)=\mathbb{P}((\bigcup_{i=1}^{n}A_i )^c)\text{(De Morgan's law)}$$
$$=1-\mathbb{P}(\bigcup_{i=1}^{n} A_i)$$
$$=1-\sum_{1}^{n}\mathbb{P}(A_i)+\sum_{i,j:1\le i<J\le n}\mathbb{P}(A_i \cap A_j)+...-(-1)^{n-1}\mathbb{P}(\bigcap_{i=1}^{n}A_i) \text{(by Inclusion-Exclusion)}$$
$$=1-\sum_{i=1}^{n}\mathbb{P}(A_i)+\sum_{i,j:1\le i<J\le n}\mathbb{P}(A_i)\mathbb{P}( A_j)+...-(-1)^{n-1}\prod_{i-1}^{n}\mathbb{P}(A_i) \text{(by independence of} A_i )$$
$$=\prod_{i=1}^{n}(1-\mathbb{P}(A_i))$$
$$=\prod_{i=1}^{n}\mathbb{P}(A_i^c)$$
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
$$\mathbb{P}(\bigcap_{i=1}^{n}B_i}=\prod_{i=1}^{n}\mathbb{P}(B_i)$$
$$\mathbb{P}(\bigcap_{i=1}^{n}B_{i})=\mathbb{P}((\bigcap_{i=1}^{r}B_i )\cap (\bigcap_{i=r+1}^{n}B_i ))$$

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

2. Jun 28, 2010

### 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 $$\mathbb{P}(\bigcap_{i=1}^{r}A_i )>0$$ (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

3. Jun 29, 2010

### Richard27182

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

4. Jul 5, 2010

### 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 $$\mathbb{P}(\bigcap_{i=1}^{r}A_i )>0$$ (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