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.
Thank you for the hints, I think I have made progress in my understanding, please confirm these if possible.
In one of the assumptions of the Brownian motion $\{W_t\}$ is adapted to a filtration $F_t$, do we assume:
A) Each increment $(W_t-W_s)$ is independent of $F_u$ for $0<u<s<t$.
OR
B) Any number of disjoint increments (in the future of time $s$) and $F_s$ are mutually independent.
Under B) then the assumption answers my problem. Under A), if the filtration
is the one generated by the Brownation motion, then the required mutual independence can be deduced from the mutual independence of the Brownian
increments. However, under A) if the filtration is not necessarily the one generated by the Brownian motion, is it still possible to prove
the required mutual independence? If so please help, I spent a longggg time trying to work it out.
Many thanks.
Best Answer
No, $X_1$ and $Y$ need not be independent. Consider a standard example of three pairwise independent random variables that are not independent: $X_2$ and $X_3$ independent, each having values $-1$ and $1$ with probabilities $1/2$, $X_1 = X_2 X_3$. Then $Y = X_2 + X_3$ is not independent of $X_1$, in fact $Y = 0$ if and only if $X_1 = -1$.