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Homework Help: Independence of sigma-algebras

  1. Jun 25, 2010 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant equations

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

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

    3. The attempt at a solution

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

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

    Now if we have that for some [tex]i \in J,B_i =\emptyset[/tex], this is immediate (both sides are zero)
    If we have that for some [tex]i\in J,B_i =\Omega[/tex], it reduces to the case where [tex]\nexists i \in J[/tex] s.t. [tex]B_i=\Omega[/tex]
    So we are left with three cases to check:
    1. [tex]B_i =A_i \forall i \in J[/tex]
    2. [tex]B_i =A_i^c \forall i \in J[/tex]
    3. [tex]B_i =A_i[/tex] for some [tex]i \in J[/tex] and [tex]B_i =A_i^c[/tex] for the rest of [tex]i\in J[/tex]
    Case 1 follows immediately from the independence of the [tex]A_i[/tex]
    Case 2 Labelling the sets in [tex]J[/tex] from [tex]1[/tex] to [tex]n[/tex]
    [tex]
    \mathbb{P}(\bigcap_{i=1}^{n}A_i^c)=\mathbb{P}((\bigcup_{i=1}^{n}A_i )^c)\text{(De Morgan's law)}[/tex]
    [tex]=1-\mathbb{P}(\bigcup_{i=1}^{n} A_i)[/tex]
    [tex]=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)}[/tex]
    [tex]=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 )[/tex]
    [tex]=\prod_{i=1}^{n}(1-\mathbb{P}(A_i))[/tex]
    [tex]=\prod_{i=1}^{n}\mathbb{P}(A_i^c)
    [/tex]
    as required.
    So all that remains is to check case 3. That is, we need to show that if (possibly relabelling the [tex]A_i[/tex]) [tex]B_i =A_i \text{ } i=1,...,r[/tex]
    [tex]B_i=A_i^c\text{ } i=r+1,...,n[/tex] then
    [tex]\mathbb{P}(\bigcap_{i=1}^{n}B_i}=\prod_{i=1}^{n}\mathbb{P}(B_i)[/tex]
    [tex]\mathbb{P}(\bigcap_{i=1}^{n}B_{i})=\mathbb{P}((\bigcap_{i=1}^{r}B_i )\cap (\bigcap_{i=r+1}^{n}B_i ))
    [/tex]

    Unfortunately, I've got stuck here and am unable to make any further progress.
     
  2. jcsd
  3. Jun 28, 2010 #2
    OK, so I have an answer:

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

    as required
     
  4. Jun 29, 2010 #3
    above answer does not work, I've applied De morgan's laws incorrectly
     
  5. Jul 5, 2010 #4
    OK, so I have another answer:

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

    as required
     
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