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Let $X,Y,Z$ be random variables such that

  • $X,Y,Z$ have mean $0$ and variance $1$.
  • $X$ and $Y$ are pairwise independent.
  • $Z$ can be arbitrarily dependent on $X$ and $Y$.

My question is: What can we say about the dependence of $Z$ on $X$ and $Y$? For example, if $\operatorname{Cov}(X,Z)=1$ (so they are perfectly correlated), then $Z$ and $Y$ must be independent. In general, can we say anything about how much $Z$ can depend on (or be correlated with) $X$ and $Y$? Maybe something like $|\operatorname{Cov}(X,Z)+\operatorname{Cov}(Y,Z)|\leq 1$ holds?

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  • $\begingroup$ Are you sure that $Y$ and $Z$ are independent if $Cov(X,Z) = 1$? $\endgroup$
    – Dieter Kadelka
    Commented Feb 10 at 18:24
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    $\begingroup$ @DieterKadelka: $\textrm{Cov}(X,Z)=EXZ=1=(EX^2EZ^2)^{1/2}$. Equality in Cauchy-Schwarz implies that $X=aZ$ (and then in fact $X=Z$ from the other conditions). $\endgroup$
    – Christian Remling
    Commented Feb 10 at 18:37
  • $\begingroup$ If you type 3\text{Cov}(X,Y), the result will be different from that of 3\operatorname{Cov}(X,Y), thus $3\text{Cov}(X,Y)$ versus $3\operatorname{Cov}(X,Y).$ $\endgroup$
    – Michael Hardy
    Commented Feb 10 at 21:03
  • $\begingroup$ @MichaelHardy Thanks, I've changed \text to \operatorname. $\endgroup$
    – user108
    Commented Feb 10 at 23:38
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    $\begingroup$ Bessel's inequality gives $\mathrm{Cov}(X,Z)^2 + \mathrm{Cov}(Y,Z)^2 \leq \mathrm{Var}(Z) = 1$. $\endgroup$
    – Terry Tao
    Commented Feb 11 at 18:59

2 Answers 2

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No, this is is hopeless, as becomes clear when you write $$ \operatorname{Cov}(X,Z)+\operatorname{Cov}(Y,Z)=E(X+Y)Z . \tag{1}\label{463874_1} $$ Your assumptions don't restrict $U=X+Y$ much, you only know that $EU=0$, $\operatorname{Var}(U)=2$ and $U$ is 2-divisible.

More to the point perhaps, you can just take $Z=(X+Y)/\sqrt{2}$ to obtain a counterexample, then \eqref{463874_1} equals $\sqrt{2}$. This is also the largest possible value, by Cauchy–Schwarz.

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  • $\begingroup$ Thank you! I agree this is as much as you can say when it comes to the sum of covariances. $\endgroup$
    – user108
    Commented Feb 10 at 23:34
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Let $X,Y$ by i.i.d. with $\Pr(X=+1) = \Pr(X=-1) = 1/2,$ and let $Z=XY.$ Then $Z$ has the same distribution as $X$ or $Y,$ and $X$ and $Y$ are independent and $X$ and $Z$ are independent and $Y$ and $Z$ are independent, and all three have the same distribution, but $X,Y,Z$ are not independent.

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  • $\begingroup$ Yes, I've seen this example before and the fact that pairwise independence does not imply joint independence. $\endgroup$
    – user108
    Commented Feb 10 at 23:33

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