# Can there be functions $g$ and $f$ such that $\rho_{f(X),g(X+Y)}^2 > \rho_{X,X+Y}^2$

### 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?

• $f(X) = X+Y, G(X+Y)= X+Y?$
– Dave
Commented Jul 6, 2022 at 21:31
• @Dave is $f(X) = X+Y$ a function of $X$? Commented Jul 6, 2022 at 21:32
• 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. Commented Jul 6, 2022 at 22:10
• 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 $-1.$
– whuber
Commented Jul 6, 2022 at 23:27
• @whuber that is an interesting example but it makes the Pearson correlation slightly smaller when I apply that function. Commented Jul 7, 2022 at 5:51

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 $$f(x) = g(x) = x^3$$.