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Given three discrete random variables $A, B, C$, for which $A \bot C$ and $B \bot C$, is $(A,B) \bot C$? If so, where can I find a reference on this subject? Someone told me it was true and to search for it in Casella and Berger's book on Statistical Inference, but I could not find it.

jcredberry
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    Please specify the meaning of "joint probability $AB\perp C$". Does $\perp$ mean independent? Does it make sense to say distributions are independent (and not events are inependent or random variables are independent)? – GEdgar Sep 22 '22 at 00:34
  • @GEdgar Out of curiosity, is there a natural Hilbert space that has probability distributions as vectors? – Alan Sep 22 '22 at 00:38
  • @GEdgar Good question! I believe you are right, but what happens with AB? Is it also a (joint) random variable? Yes, $\bot$ means independence. – jcredberry Sep 22 '22 at 00:52
  • The statement is false, I'm sure this question must be a duplicate since I've seen someone post the counterexample quite recently. – Suzu Hirose Sep 22 '22 at 00:54
  • @SuzuHirose It is not a duplicate, at least not an intentional one. Can you point me to this counterexample, please? – jcredberry Sep 22 '22 at 00:57
  • When I say it is a duplicate I am not discussing your intention, I am discussing whether the question has been asked before. This question is certainly a duplicate. Here is one example of a question dealing with this from six years ago. https://math.stackexchange.com/questions/1783225/example-of-pairwise-independent-but-not-jointly-independent-random-variables – Suzu Hirose Sep 22 '22 at 00:58
  • @GEdgar I have corrected the question. Thx. – jcredberry Sep 22 '22 at 00:58
  • @SuzuHirose I am unsure if the question I am asking is exactly answered in the one you suggested me. Please see that I am not asking for pairwise independence. – jcredberry Sep 22 '22 at 01:06
  • It’s still not clear to me what you mean by $AB\perp C$. As the question is written now I assume you mean the product of $A$ and $B$. But before you said “joint” which I would assume means to ask if $(A,B)$ is independent of $C$. (The answer is no in either case.) – spaceisdarkgreen Sep 22 '22 at 01:08
  • @spaceisdarkgreen indeed, it is not the product but (A,B). – jcredberry Sep 22 '22 at 01:09
  • Pairwise independence means that $A$ and $B$ are independent and $B$ and $C$ are independent, so in fact you are asking about pairwise independence. – Suzu Hirose Sep 22 '22 at 01:09
  • Then is the same: let $A=XY$, $B=YZ$ and $C=XZ$, where X,Y,Z are independent Rademacher (though this serves equally well as a counterexample to the alternative interpretation). For a maybe slightly easier example, let $A$ come from any symmetric distribution, $C$ be a Rademacher independent of $A$ and $B=AC$. – spaceisdarkgreen Sep 22 '22 at 01:14
  • Define better what is AB. Are you talking about events or random variables. AB can also mean $A\cap B$ if $A$ and $B$ are events. Or AB is the product of 2 r.vs. Or you want to check whether the joint distribution for A, B and C is the product of the marginal distributions for A, B and C. – bluemaster Sep 22 '22 at 01:18
  • @SuzuHirose yes, but what I mean is that (using your comment's notation), $A$ is not necessarily independent of $C$. So no pairwise independence. In pairwise independence $A \bot B$, $A \bot C$, and $B \bot C$. In that case, it is known that they are not mutually independent. What I am asking seems like a different question. – jcredberry Sep 22 '22 at 01:20
  • Yes but the thing is that if there are counterexamples when all three of them are pairwise independent, then there must be counterexamples when only two of them are pairwise independent, since that is a weaker condition. – Suzu Hirose Sep 22 '22 at 01:33
  • @suzuhirose thx for your time. – jcredberry Sep 22 '22 at 01:44
  • @spaceisdarkgreen thanks, you have been very helpful. – jcredberry Sep 22 '22 at 01:44

1 Answers1

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As noted, this has surely been asked here already.

Example. Toss a fair coin twice in succession.
Let $A$ be the result of the first toss, $A=0$ for heads and $A=1$ for tails, each with probability $1/2$.
Similarly, let $B$ be the result of the second toss, $B=0$ for heads and $B=1$ for tails, each with probability $1/2$.
Finally, let $C$ be agreement of the tosses: that is $C=0$ if $A=B$ and $C=1$ if $A \ne B$.
Then: $(A,B)$ has four possible outcomes: $(0,0), (0,1), (1,0), (1,1)$, each with probability $1/4$.

Compute:
$A$ is independent of $C$;
$B$ is independent of $C$;
[not asked] $A$ is independent of $B$;
but $(A,B)$ is not independent of $C$.

GEdgar
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