Given sample space $\omega = \left\{(x,x,x), (y,y,y), (z,z,z), (x,y,z), (x,z,y), (y,z,x), (y,x,z), (z,x,y), (z,y,x)\right\}$ with equal probability $\frac{1}{9}$. Show events on $(\omega, \mathscr{F})$ which are pairwise independent but not mutually independent.

My thought I don't quite understand why it's possible to set up such events. The definition of pairwise independent events (for example, $A$ and $B$) is that $P(A\cap B) = P(A)P(B)$. But this is completely false if $A\cap B =\emptyset$, which is quite possible if we take $A=(x,x,x)$ and $B = (y,y,y)$. So how could we create an event that ever avoids this case? And of course, if we could create one satisfying pairwise independent, then the mutually independent would be much easier, because we only need the intersections of at least $\geq 3$ elements among those $9$ elements to be equal to an empty set. But this is quite possible with $A=(x,x,x), B = (y,y,y)$ and $C = (z,z,z)$. I also note that the 3 basis elements of the sample space are $(x,x,x), (y,y,y), (z,z,z)$, but I'm completely stuck on creating such a good events.

Could someone please help me with this problem? Would really appreciate any input.


closed as unclear what you're asking by Did, Claude Leibovici, Namaste, Shailesh, Michael Lugo Mar 13 '17 at 17:50

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  • $\begingroup$ I'm not sure what your concern is with the first part. The two events $\{(x,x,x)\}$ and $\{(y,y,y)\}$ are not independent, that's all. If you want a pair of independent events you'll simply have to come up with something else. Also, carefully distinguish $(x,x,x)$ (an outcome, i.e. an element of the sample space $\Omega$) from $\{(x,x,x)\}$ (an event, i.e. a subset of the sample space $\Omega$ that happens to only contain one element). $\endgroup$ – Nate Eldredge Sep 15 '16 at 2:33
  • $\begingroup$ And I don't understand your comment about "basis". There are no vector spaces here. $\endgroup$ – Nate Eldredge Sep 15 '16 at 2:34
  • $\begingroup$ @NateEldredge: I understand, but I could not find a way to create such event to avoid those empty intersection cases:P Thanks for your reminder as well. Regards to the "basis," I was completely wrong. Could you give this problem a try? $\endgroup$ – user177196 Sep 15 '16 at 2:43
  • $\begingroup$ Maybe you should warm up with a simpler question. Consider the sample space $\{a,b,c,d\}$ where all outcomes have equal probability $1/4$. Can you find two events which are independent? $\endgroup$ – Nate Eldredge Sep 15 '16 at 2:45
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    $\begingroup$ Ok, but you should notate more carefully: it's $\{a,b,c,d\}$ and $\{a\}$ (use \{ \} to get braces). Can you find a different pair, where neither one is either $\{a,b,c,d\}$ nor empty? $\endgroup$ – Nate Eldredge Sep 15 '16 at 3:09

You are confusing events and outcomes.

An event is a set of outcomes.

What you've been given is set of all possible outcomes, which is a sample space.

An event is a subset of the sample space.

That is: Some events may consist of more than one outcome.

EG: The event of "$x$ in first position" is $\{(x,x,x),(x,y,z),(x,z,y)\}$. Call it, $A$ so $\mathsf P(A)=\tfrac 39$

So should I define other events as B:="X in second position" and C:="X is in third position." . Then $\mathsf P(B)=1/3$ and $\mathsf P(C)=1/3$ as well. But now, if I take the intersection of these 2 events, I would get the empty set:p

No, the intersection is not empty. $(A\cap B)=(A\cap C)=(B\cap C)=(A\cap B\cap C)=\{(x,x,x)\}$, and $\mathsf P(\{(x,x,x)\})=1/9$.

Now test for independencies; pairwise and mutual.

PS: There are several other such examples in this space.   Can you find a less obvious one?

  • $\begingroup$ Are you steering towards Bernstein's example? $\endgroup$ – Andris Birkmanis Sep 15 '16 at 4:16
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    $\begingroup$ @Graham Kemp: So should I define other events as $B: = \text{"X in second position"}$ and $C: = \text{"X is in third position."}$. Then $P(B) = \frac{1}{3}$ and $P(C) = \frac{1}{3}$ as well. But now, if I take the intersection of these 2 events, I would get the empty set:p $\endgroup$ – user177196 Sep 15 '16 at 4:19
  • $\begingroup$ @user177196 The intersection of your three events is not empty, it is $\{(x,x,x)\}$, but $\mathsf P(A,B,C) = \tfrac 19$, and $\mathsf P(A)~\mathsf P(B)~\mathsf P(C) = \tfrac 1{27}$, so... $\endgroup$ – Graham Kemp Sep 15 '16 at 4:24
  • $\begingroup$ PS: There are many other collections of three events in this space with these properties. Can you find a less obvious one? $\endgroup$ – Graham Kemp Sep 15 '16 at 4:29
  • $\begingroup$ @GrahamKemp: thank you very much for your further patience. I sincerely appreciated it. I finally got another case as well!! Btw, could you help review my solution to the following problem: math.stackexchange.com/questions/1928313/… $\endgroup$ – user177196 Sep 15 '16 at 19:14

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