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Given two random variables (say standard Normals) that are not necessarily independent, are there non-linear functions for which $$ \text{Cov}(g(X), g(Y)) \le c(g) \text{Cov}(X,Y), $$ where $c(g)$ is some constant that can depend on the function $g$. In other words, are there classes of functions $g$ that have covariance not too different from the covariance of the underlying variables?

update: Given the comments, my question is ill-posed, what I would like is an analog of Lipschitz continuity, so it should read that for any two gaussian random variables, the above bound holds, so the function $g$ can alter the covariance of the original variables by at most a fixed constant.

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  • $\begingroup$ Your question asks for non-linear functions, so this is just a comment: covariance is invariant to translations by a constant. $\endgroup$
    – Galen
    Jun 18, 2022 at 23:13
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    $\begingroup$ A function $g$ that preserves covariance would satisfy $\operatorname{Cov}[g(X),g(Y)] = \operatorname{Cov}[X,Y]$, which not same at the inequality you give. $\endgroup$
    – Galen
    Jun 18, 2022 at 23:20
  • $\begingroup$ @Galen - 'preserve' was not a good word to describe what I mean, I've edited the question now $\endgroup$ Jun 19, 2022 at 2:21
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    $\begingroup$ Is there any additional requirement on $c(g)$? It seems that since covariances are finite, you can surely find such an upper bound for any transformation. $\endgroup$ Jun 19, 2022 at 5:05
  • $\begingroup$ @whoknowsnot you're right, I've added an edit to try to hopefully make the problem more well posed. $\endgroup$ Jun 19, 2022 at 16:13

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Consider, as you say, standard Normal variates and $g(x) = x/10$ if $|x| < 100$, $0$ otherwise. It should be intuitively obvious that the covariance between $g(x)$ and $g(y)$ is less than that of $x$ and $y$, so $c(g) = 1$ will do.

Clearly this approach is generalizable to other distributions; rescale the variables to be much smaller over the region where virtually all the probability mass lies, then impose some trivial nonlinearity in the region where there is essentially no probability mass. The covariance of the transformed variables will (with a little care) be less than that of the original variables, so $c(g) = 1$ will do.

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