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Assume X orthogonal to Z and Y orthogonal to Z (but X and Y potentially correlated), can you conclude T=XY orthogonal to Z ?

noemie
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    Could you be more specific about what you mean by "orthogonal"? A result like the one you inquire about is plausible when it means *independent* but otherwise is not. – whuber Sep 07 '20 at 19:30

4 Answers4

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No, one cannot conclude that.

Let $ X = \{-1, 0, 1 \}$ with equal probability;

$Y = 2X$;

$Z = X^2$

$Cov(X,Z) = 0 $

$Cov(Y,Z) = 0 $

$Cov(XY,Z) \neq 0 $

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whisperer
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No, assume $Y=X$ be Rademacher distributed, and $Z$ is independent of others having non-zero mean. Then , $E[XYZ]\neq 0$.

gunes
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  • This is not a counterexample - $\mathbb{E}[XYZ] - \mathbb{E}[XY] \mathbb{E}[Z] = 0$ in this case. (Equivalently, $XY$ is a constant, which is of course uncorrelated to anything. Or - $Z$ is independent of $X$ and $Y$, and so of $XY$, and so has to be uncorrelated to $XY$). – stochasticboy321 Sep 08 '20 at 00:48
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    orthogonality does not mean uncorrelated – gunes Sep 08 '20 at 06:20
  • I guess you're right. Also, on re-reading, I find my comment above quite obnoxious. My apologies. – stochasticboy321 Sep 08 '20 at 07:58
  • not at all, it's a discussion – gunes Sep 08 '20 at 08:01
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The answer is No. Consider this counter-example:

Let $Z$ take on 4 possible values $(2, 1, -1, 2)$ with equal probability of 0.25.

Let $X = 1, 0, 1, \frac12$ when $Z = 2, 1, -1, 2$ respectively. Let $Y = X$.

Easy to show that $Cov(X, Z) = E(XZ) = 0$ and likewise $Cov(Y, Z) = E(YZ) = 0$, but $Cov(XY, Z) = E(X^2Z) \ne 0$.

Shang Zhang
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I could be completely misunderstanding this, but consider a book. The spine is orthogonal to the base of the pages, but you can open the book to (almost any angle).

sjb-2812
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