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Independence of sigma algebras

  1. Oct 26, 2011 #1
    I am trying to establish whether the following is true (my intuition tells me it is), more importantly if it is true, I need to establish a proof.

    If $X_1, X_2$ and $X_3$ are pairwise independent random variables, then if $Y=X_2+X_3$, is $X_1$ independent to $Y$? (One can think of an example where the $X_i$ s are Bernoulli random variables, then the answer is yes, in the general case I have no idea how to prove it.)

    A related problem is:

    If $G_1,G_2$ and $G_3$ are pairwise independent sigma algebras, then is $G_1$ independent to the sigma algebra generated by $G_2$ and $G_3$ (which contains all the subsets of both, but has additional sets such as intersection of a set from $G_2$ and a set from $G_3$).

    This came about as I tried to solve the following:
    Suppose a Brownian motion $\{W_t\}$ is adapted to filtration $\{F_s\}$, if $0<s<t_1<t_2<t_3<\infty$, then show $a_1(W_{t_2}-W_{t_1})+a_2(W_{t_3}-W_{t_2})$ is independent of $F_s$ where $a_1,a_2$ are constants.

    By definition individual future increments are independent of $F_s$, for the life of me I don't know how to prove linear combination of future increments are independent of $F_s$, intuitive of course it make sense...

    Any help is greatly appreciated.
     
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  3. Oct 26, 2011 #2

    micromass

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    What you say is indeed true. A ful discussion can be found in Billingsley's "Probability and Measure" on page 50.

    The proof relies on a lemma, which states that

    Lemma: If [itex]\mathcal{A}_1,...,\mathcal{A}_n[/itex] are independent [itex]\pi[/itex]-systems (=stable under finite intersections), then [itex]\sigma(\mathcal{A}_1),...,\sigma(\mathcal{A}_n)[/itex] are independent.

    The proof is as follows:

    Let [itex]\mathcal{B}_i=\mathcal{A}_i\cup \{\Omega\}[/itex] then we still have independent [itex]\pi[/itex]-systems.
    Take [itex]B_2,...,B_n[/itex] be fixed in [itex]\mathcal{B}_2,...,\mathcal{B}_n[/itex] respectively. Denote

    [tex]\mathcal{L}=\{L\in \mathcal{F}~\vert~P(L\cap \bigcap{B_i})=P(L)\prod P(B_i)\}[/tex]

    This is a [itex]\lambda[/itex]-system that contains [itex]\mathcal{A}_1[/itex]. This implies that [itex]\sigma(\mathcal{A_1})[/itex] is independent from [itex]\mathcal{A_2},...,\mathcal{A}_n[/itex]. Now we can proceed by induction.

    Now we can prove the main theorem:

    If [itex]\mathcal{A}_i,i\in I[/itex] are independent [itex]\pi[/itex]-systems. If [itex]I=\bigcup I_j[/itex] is a disjoint union, then [itex]\sigma(\bigcup_{i\in I_j} \mathcal{A}_i),j\in J[/itex] are independent.

    Proof: we put

    [tex]\mathcal{C}_j=\{C~\vert~\exists K\subseteq I_j~\text{finite}, B_k\in \mathcal{A}_k, k\in K: C=\bigcap_{k\in K}{B_k}\}[/tex]

    Then [itex]\mathcal{C}_j,j\in J[/itex] are independent [itex]\pi[/itex]-systems, and [itex]\sigma(\mathcal{C}_j)=\mathcal{B}_j[/itex]. Now apply the lemma.
     
  4. Oct 26, 2011 #3
    Thank you for the reply!
    I was wondering, the condition of the lemma that $A_1, A_2,...,A_n$ are independent pi-systems is too strong, stronger than pairwise independent, not sure how to apply it in case of pairwise independent systems. In any case I will read the sections in Billingsley's book carefully.

    Btw is there a quick explanation for:

    A Brownian motion $\{W_t\}$ adapted to filtration $\{F_s\}$, if $0<s<t_1<t_2<t_3<\infty$, then $a_1(W_{t_2}-W_{t_1})+a_2(W_{t_3}-W_{t_2})$ is independent of $F_s$ where $a_1,a_2$ are constants.

    Many thanks.
     
  5. Oct 26, 2011 #4
    Pretty sure it isn't true. Suppose instead of Y = X_2 + X_3, we have Y = (X_2, X_3). If X_1 is independent of Y, then
    P(X_1=x_1,X_2=x_2,X_3=x_3)) = P(X_1=x_1, Y = (x_2,x_3)) = P(X_1=x_1) P(X_2=x_2, X_3=x_3)
    = P(X_1=x_1) P(X_2=x_2) P(X_3=x_3)
    thus implying that X_1, X_2, and X_3 are all independent (not just pairwise). This isn't necessarily true.

    So to disprove it, find 3 RVs that are pairwise independent but not all independent, and such that all of the pairwise sums of possible values of X_2 and X_3 sum to different values (so that X_2 + X_3 corresponds to (X_2, X_3)).
     
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