4

Find an example where the random variables $X_1,X_2,X_3$ are pairwise independent, but not all together. I can't really understand how I am to do so. How is it done? There cannot be any multiplication. Is it a sum? I searched for it but didn't understand how it is interpreted mathematically. I could really use your help.

Meitar
  • 2,911

2 Answers2

4

Let $X_1,X_2$ be independent, uniform variables in $\{-1,1\}$ and $X_3 = X_1X_2$.

Then any pair is independent, but any two completely determine the third.

Eric Tressler
  • 4,309
  • 1
    Note that this can be naturally generalized to $n$ variables, any $n-1$ of which are independent. – vadim123 May 10 '15 at 19:30
  • How do I show the independence? I didn't understand it fully... – Meitar May 10 '15 at 19:31
  • @Meitar You need to show that $X_1$ and $X_3$ are independent; you should try it for a while first, it will make more sense if you prove it yourself. It follows directly from any definition of independence that you want to use. – Eric Tressler May 10 '15 at 19:32
  • Okay, now I understand a bit more. – Meitar May 10 '15 at 19:41
  • Does $X_1$ and $X_2$ have any probability of equaling $1$ or $-1$ I choose? – Meitar May 10 '15 at 19:43
  • @Meitar No, they are both uniformly distributed, which means they have probability $1/2$ of being $1$ and probability $1/2$ of being $-1$ (in general, it means that they have an equal probability of taking on any value in their set of outcomes, when that set is finite). – Eric Tressler May 10 '15 at 19:44
  • I didn't notice the "uniformly distributed", but I am glad I thought myself of distributing them that way, because it makes me feel like I am on the right track. – Meitar May 10 '15 at 19:45
  • Okay. I got it. Thank you for your help. – Meitar May 10 '15 at 19:48
3

The example can be:

Experiment - two coins are tossed.

$X_1$ - first toss results in head.

$X_2$ - second toss results in tail.

$X_3$ - one of the tosses is head, the other is tail. $$P(X_1)=P(X_2)=P(X_3)=\frac{1}{2}$$ Easy to check that they are pairwise independent: $$P(A\cap B)=P(A) \cdot P(B)=\frac{1}{4}$$ $$P(A\cap C)=P(A) \cdot P(C)=\frac{1}{4}$$ $$P(B\cap C)=P(B) \cdot P(C)=\frac{1}{4}$$

But: $$P(A\cap B \cap C)=P(A \cap B) = \frac{1}{2}$$ And: $$P(A)\cdot P(B) \cdot P(C) = \frac{1}{8}$$ Hence these events are pairwise independent but not mutually independent.

marco11
  • 757