Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $0<u<s<t$.


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.

share|cite|improve this question
up vote 2 down vote accepted

Robert Israel has already given a counterexample to your first question, and it also gives a counterexample for your second question (let $G_i = \sigma(X_i)$).

As for your question about Brownian motion, I think it isn't worded quite right: to say $W_t$ is adapted to $\{F_t\}$ usually just means $W_t \in F_t$ for each $t$. You also need to know that $W_t - W_s$ is independent of $F_s$, which doesn't follow from adaptedness alone. This stronger assumption is sometimes stated as "$W_t$ is a Brownian motion with respect to the filtration ${F_t}$".

Under this assumption, the answer to your question is yes, and the key is that $F_s$, $W_{t_2} - W_{t_1}$ and $W_{t_3} - W_{t_2}$ are not merely pairwise independent, they are mutually independent. I recommend trying to prove this. Once it is done, it is not too hard to get a proof for your problem. Give it a try, and post back if you have further questions.

Edit: As to your revised questions (presently posted as another answer), the usual assumption is (A), but (A) implies (B). To see why, suppose $s \le t_1 \le t_2 \le t_3 \le t_4$, $A \in F_s$, $B \in \sigma(W_{t_2} - W_{t_1})$, $C \in \sigma(W_{t_4} - W_{t_3})$. We have to show $$P(A \cap B \cap C) = P(A) P(B) P(C).$$ The key is that $A \cap B \cap C = (A \cap B) \cap C$, and $A \cap B \in F_{t_2}$ (verify this). But $C$ is independent of $F_{t_2}$. Now I expect you can finish the argument, and extend it to any number of increments.

share|cite|improve this answer
Nate, thank you very much for the answer, your second paragraph procisely describe the assumptions my question should have been asked under, and I will try to prove the mutual independence! – Joe84 Oct 26 '11 at 20:59
In light of your comments above, I got it! It feels so good to finally see the solution, it has frustrated me for quite sometime, its a shame I didn't think of the path to solution myself, maybe next time! Thank you very much again! PS: sorry for the wrong posting, I should have read the faq. – Joe84 Oct 28 '11 at 17:33

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 $X_1 = -1$.

share|cite|improve this answer
Thank you for the example, indeed its not the case! I was wondering what about the special case of: A Brownian motion $\{W_t\}$ adapted to filtration $\{F_s\}$, if $0<s<t_1<t_2<t_3<\infty$, then is $a_1(W_{t_2}-W_{t_1})+a_2(W_{t_3}-W_{t_2})$ independent of $F_s$ where $a_1,a_2$ are constants. By definition individual future Brownian increments are independent of $F_s$, so what about linear combination of future increments? Are these also independent of $F_s$. – Joe84 Oct 26 '11 at 20:19

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.