Is $X_1$ Independent of $Y = X_2 + X_3$ Given Pairwise Independence?

[Arthur84]

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.

 

[micromass]

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 \mathcal{A}_1,...,\mathcal{A}_n are independent \pi-systems (=stable under finite intersections), then \sigma(\mathcal{A}_1),...,\sigma(\mathcal{A}_n) are independent.

The proof is as follows:

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

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

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

Now we can prove the main theorem:

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

Proof: we put

\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}\}

Then \mathcal{C}_j,j\in J are independent \pi-systems, and \sigma(\mathcal{C}_j)=\mathcal{B}_j. Now apply the lemma.

 

[Arthur84]

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.

 

[mXSCNT]

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)).