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I am embarrassed to be stuck on this seemingly simple question.

Suppose that $X,Y$ are mean-zero real-valued random variables and $\tilde X,\tilde Y$ are their "independent copies": $\tilde X,\tilde Y$ are mutually independent and independent of $(X,Y)$, and $\tilde X$ (resp., $\tilde Y$) is distributed identically to $X$ (resp., $Y$).

Here is the inequality I am trying to prove/disprove: for some universal constant $c>0$, $$ \mathbb{E}|\tilde X-\tilde Y| \le c\left( \mathbb{E}|X-Y| + \sqrt{|\mathbb{E}XY|} \right). $$

Update. Note that the related inequality, $$ \mathbb{E}|\tilde X-\tilde Y|^2 \le \mathbb{E}|X-Y|^2 + 2|\mathbb{E}XY| , $$ is trivially true.

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2 Answers 2

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[EDITED to ensure that the random variables have expectation 0] I think the answer is no.

Let $Z$ be a random variable taking the values $\pm 1$ with probability $p$ each and 0 with probability $1-2p$; and let $N$ be a standard normal independent of $Z$.

Set $X=ZM+(1-|Z|)N$ and $Y=-ZM+(1-|Z|)N$, where $M=\sqrt{(1-2p)/(2p)}$. Now $\mathbb E|X-Y|=2pM=O(\sqrt p)$ and $\mathbb EXY=-2pM^2+(1-2p)=0$. On the other hand, $\mathbb E|\tilde X-\tilde Y|=\Omega(1)$.

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  • $\begingroup$ Excellent, thank you! What if we add the additional condition that all variables are in the range $[-1,1]$? $\endgroup$
    – Aryeh Kontorovich
    Commented Apr 21, 2022 at 13:16
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    $\begingroup$ I think you can scale everything without changing anything here (and truncate the normals) $\endgroup$
    – Anthony Quas
    Commented Apr 21, 2022 at 13:17
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    $\begingroup$ Here $EX=pM\ne0$, whereas the OP requested zero means for $X$ and $Y$. Can this example be modified accordingly? $\endgroup$
    – Iosif Pinelis
    Commented Apr 21, 2022 at 13:34
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    $\begingroup$ @AryehKontorovich : I think $EXY$ was OK. $\endgroup$
    – Iosif Pinelis
    Commented Apr 21, 2022 at 13:40
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    $\begingroup$ Sorry about that. The idea remains the same: $\mathbb EXY$ is an $L^2$ quantity that is much more sensitive to large changes on a small measure set than the $L^1$ quantity $\mathbb E|X-Y|$. So you can make $\mathbb E|X-Y|$ zero by making them equal. But $\mathbb E|\tilde X-\tilde Y|$ is $\Omega(1)$. Of course this doesn't give a counterexample because $\mathbb EXY$ is also $\Omega(1)$. But then you can make a perturbation that is small in $L^1$, but large in $L^2$ such that $\mathbb EXY=0$. $\endgroup$
    – Anthony Quas
    Commented Apr 21, 2022 at 15:13
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Let $\varepsilon \to 0$ and $M= \sqrt{(1-\varepsilon)/\varepsilon}$. Define $X$ and $Y$ as follows:

  • w.p. $\frac{1-\varepsilon}2$: $X=Y=1$
  • w.p. $\frac{1-\varepsilon}2$: $X=Y=-1$
  • w.p. $\frac{\varepsilon}2$: $X=M$ and $Y=-M$
  • w.p. $\frac{\varepsilon}2$: $X=-M$ and $Y=M$

Then $\mathbf{E}[XY] = (1-\varepsilon) - \varepsilon M^2 = 0$. Also, $\mathbf{E}|X-Y| = \varepsilon \cdot (2M) = 2 \sqrt{\varepsilon(1-\varepsilon)}$. Finally, $\mathbf{E}[X] = \mathbf{E}[Y] = 0$.

However, $\mathbf{E}|\tilde X- \tilde Y| \geq \mathbf{Pr}(\tilde X = - \tilde Y) \cdot 2\geq \frac{(1-\varepsilon)^2}{2} \cdot 2 = (1-\varepsilon)^2$.

On a separate note, observe that $\mathbf{E}|\tilde X - \tilde Y| = \Theta(\mathbf{E}|\tilde X| + \mathbf{E}|\tilde Y|)$.

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    $\begingroup$ Very well done, great! I already accepted Anthony Quas's answer, but this is also very elegant (and bad for my plans)... $\endgroup$
    – Aryeh Kontorovich
    Commented Apr 21, 2022 at 15:15

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