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$X$ and $Y$ are random independent variables such that: $$P(X=1)=P(X=-1)=\frac{1}{2}$$ $$P(Y=1)=P(Y=-1)=\frac{1}{2}$$
We have that $Z=X\cdot Y$. Are $X$, $Y$, and $Z$ independent?

MY SOLUTION:

$X, Y, Z$ are independent if $P(X\cap Y\cap Z) = P(X)P(Y)P(Z)$. $$Z = \left \{ (1)(1), (1)(-1), (-1)(1), (-1)(-1) \right \} = \left \{ -1,1 \right \}.$$ So, $$X = \left \{ -1,1 \right \}. P(X) = \frac{1}{2}.$$ $$Y=\left \{ -1,1 \right \}.P(Y) = \frac{1}{2}.$$ $$Z=\left \{ -1,1 \right \}.P(Z) = \frac{1}{2}.$$
$$X\cap Y\cap Z = \left \{ -1,1 \right \}.$$ $$P(X\cap Y\cap Z) = \frac{1}{2}.$$
So, since $P(X\cap Y\cap Z) = \frac{1}{2} \neq \frac{1}{8} = P(X)P(Y)P(Z)$, $X,Y,Z$ are NOT independent.

Is this correct?

Thanks for answering!

beepboopbeepboop
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1 Answers1

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These three random variables are pairwise independent, i.e. any two of them are independent, and hence they are uncorrelated (i.e. the covariance, and hence the correlation, between any two of them is $0$) but they are not independent. The fact that knowing $X$ and $Y$ completely determines $Z$ means they cannot be independent, and the argument given in the question also proves that.

Thus

  • $X,Y$ are independent,
  • $X,Z$ are independent,
  • $Y,Z$ are independent,
  • $X,Y,Z$ are not independent.

These three random variables are also exchangeable, i.e. their probability distribution is invariant under permutations of this set of three random variables.

Michael Hardy
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