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Motivation a special case

A special case of this question is an inequality between the Spearman's rank correlation and the Pearson correlation (Why is the sum of individual Spearman's rho squared less than 1 as opposed to Pearson's r in a synthetic example? ). The expectation of the squared sample Spearman's rank correlation will approach $\rho_{f(X),g(X+Y)}^2$ with $f$ and $g$ the cumulative distribution function of the variables $X$ and $X+Y$.

By trying out several different distributions for $X$ and $Y$ it seems that in all cases the Spearman's rank correlation coefficient is smaller than the Pearson's correlation coefficient.

Thinking about that question it seemed to me intuitive that the correlation between $X$ and $X+Y$ must be a maximum. When we apply a formula, then information gets lost and the correlation should decrease. I imagine that a correlation between $f(X)$ and $g(X+Y)$ can't be larger than a correlation between the original $X$ and $X+Y$.

General question

So I am wondering whether, beyond that example with the Spearman's rank correlation it is true in general for other functions as well:

If $X$ and $Y$ are independent, then the correlation should decrease or at least stay the same when we apply functions to $X$ and $X+Y$. $\rho_{f(X),g(X+Y)}^2 \leq \rho_{X,X+Y}^2$

Is this true?

Is there a simple proof for this?

Edit: Apparently there can be made many examples relatively easy. Could it also work when $Var(Y) > Var(X)$ and $f$ and $g$ are monotonic functions?

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    $\begingroup$ $f(X) = X+Y, G(X+Y)= X+Y?$ $\endgroup$
    – Dave
    Commented Jul 6, 2022 at 21:31
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    $\begingroup$ @Dave is $f(X) = X+Y$ a function of $X$? $\endgroup$
    – Sextus Empiricus
    Commented Jul 6, 2022 at 21:32
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    $\begingroup$ Choosing $X \sim \mathrm{Expo}(1)$ and $Y \sim \mathcal{N}(0,1)$ as well as $f(X)=X^2$ and $g(X+Y)=(X+Y)^2$ should yield a counterexample. $\endgroup$
    – statmerkur
    Commented Jul 6, 2022 at 22:10
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    $\begingroup$ Consider the empirical distribution of the points $(0,0), (1,1), \ldots, (n-1,n-1),(e^n,-e^n)$ for large integral $n.$ I hope these assertions are obvious: (1) the Spearman coefficient grows arbitrarily close to $1$ and $(2)$ the Pearson coefficient grows arbitrarily close to $-1$ as $n$ grows large. This applies to your situation where $Y=0$ and $g$ is the identity on rational numbers and otherwise multiplies irrationals by $-1.$ $\endgroup$
    – whuber ♦
    Commented Jul 6, 2022 at 23:27
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    $\begingroup$ @whuber that is an interesting example but it makes the Pearson correlation slightly smaller when I apply that function. $\endgroup$
    – Sextus Empiricus
    Commented Jul 7, 2022 at 5:51

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An example:

A distribution $X$ with values $-1$ and $1$ and a distribution $Y$ with values $-0.5$ and $0.5$ (e.g. scaled Rademacher distributed variables).

Then let $g$ and $f$ be a sign function

$$g(x) = f(x) = \begin{cases} -1 & \quad \text{if $x<0$}\\ 0 & \quad \text{if $x=0$}\\ 1 & \quad \text{if $x>0$}\end{cases}$$

This makes $f(X) = g(X+Y)$ and makes the correlation $1$.

Edit: Apparently there can be made many examples relatively easy. Could it also work when $Var(Y) > Var(X)$ and $f$ and $g$ are monotonic functions?

We can adapt the example from statmerkur and use $X \sim Exp(1)$ and $Y \sim N(0,4)$ and the functions $f(x) = g(x) = x^3$.

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