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After years of mathematics, I am struggling with this simple question.

If we have 3 r.v. $X,Y,Z$ and we have $X$ independent to $Y$ and to $Z$, then do we have that $X$ is also independent to $YZ$ ?

At first sight, I thought that if $X$ is independent to $Y$ and $Z$, it is also independent to the sigma-algebra generated by $Y$ and to $Z$ and hence $YZ$ but the example below made me confused : https://en.wikipedia.org/wiki/Pairwise_independence

If someone can make this clear... Thank you very much in advance !

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2 Answers 2

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Short answer: No, $X \perp Y , X \perp Z$ doesn't imply $X \perp YZ$

Let's say you do an experiment where you choose a number randomly from:

{1,2,3,4,6,7,8,9}

Let X = 1 if:

(Your chosen number is even AND less than five) OR (your chosen number is odd AND greater than 5)

and 0 otherwise

Let Y = 1 if your chosen number is even, 0 otherwise Let Z = 1 if your chosen number is greater than 5, 0 otherwise

Now we know X = 1 with probability - 0.5 (2,4,7,9) and 0 with probability 0.5 (1,3,6,8)

If we know Y = 1, X is still 1 with p = 0.5 (2,4) and 0 with p = 0.5 (6,8) If we know Y = 0, X is still 1 with p = 0.5 (7,9) and 0 with p = 0.5 (1,3)

If we know Z = 1, X is still 1 with p = 0.5 (7,9) and 0 with p = 0.5 (6,8) If we know Z = 0, X is still 1 with p = 0.5 (2,4) and 0 with p = 0.5 (1,3)

So X is independent of Y and X is independent of Z.

But knowing if a number is even AND knowing if it's greater than 5 (Y & Z), makes us know X with certainty. e.g. Y = 1, Z = 1, then YZ = 1, X has to be 0 with probability 1 (as X is 0 if the number is an even number > 5)

$P(X = 0 | Y ) = P(X = 0) = 0.5$ (same for X = 1)

$P(X = 0 | Z ) = P(X = 0) = 0.5$ (same for X = 1)

But

$P(X = 0 | YZ = 1) = 1 \neq P(X=0) $ (similar for X = 1)

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  • $\begingroup$ This answer might be improved by showing the explicit calculation: $P(X=0|YZ)=P(X=0)$, etc. Definitely a cool simple example! $\endgroup$
    – jdods
    Commented Apr 12, 2016 at 15:13
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    $\begingroup$ @jdods agreed. I've edited the last part of my answer. Will add more calculation as I get time $\endgroup$ Commented Apr 12, 2016 at 15:23
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    $\begingroup$ Thanks @sanketalekar This is a great example :) $\endgroup$
    – Wiles01
    Commented Apr 12, 2016 at 16:15
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    $\begingroup$ @jdods I don't understand. How is (1,1) possible ? If YZ = 1, X has to be 0. $\endgroup$ Commented Apr 12, 2016 at 16:27
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    $\begingroup$ My mistake: I had the variable values mixed up. The corrected numbers should be $f_{X,YZ}=1/4,1/4,1/2,0$. $\endgroup$
    – jdods
    Commented Apr 13, 2016 at 19:18
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You can see it as having information on $Y$ is not enough to determine $X$ an identically with information on $Z$ however if you have combined informations it could determined $X$.

An example I came up with is (may be not a good one but it helps me understand whats going on).

You have a set of peoples on with you can know the colours of the hair, given by the variable $X$, there first name given by $Y$, and there last name given by $Z$.

The independence of $X$ and $Y$ tells you that even if you know there first name that does not help you in knowing the colours of there hairs. Intuitively it says that the color of the hair does not depend on your name.

The same with the last name.

However if you know both the first name and last name you know which person it is exactly hence you know the color of its hairs.

I hope this example is not confusing you :S

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  • $\begingroup$ Thank you for your answer ! I get your point but this means that X could be dependent to YZ. However, the answer of @Andres Mejia seems to indicate we are independent to the "YZ-plane"... So still confused :( $\endgroup$
    – Wiles01
    Commented Apr 12, 2016 at 12:52
  • $\begingroup$ @Wiles01 Sorry I do not see Andres Mejia answers and I dont know what you mean by: :are independent to the YZ-plane" ... $\endgroup$
    – wece
    Commented Apr 12, 2016 at 12:55
  • $\begingroup$ I was unclear by was meant by $YZ$. $\endgroup$ Commented Apr 12, 2016 at 12:56
  • $\begingroup$ Both of you agree that we may find examples where X is independent to Y and Z but X is some way dependent to YZ ? $\endgroup$
    – Wiles01
    Commented Apr 12, 2016 at 12:58
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    $\begingroup$ @wece your example doesn't necessarily work, because knowing Y you do know something about X. Agreed knowing Y or Z individually will not help you know X with certainity, but it does affect the probability distribution of X if you know Y (or Z). Hair color distributions are not necessarily uncorrelated to first names. $\endgroup$ Commented Apr 12, 2016 at 15:52

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