What's the lower bound of the correlation coefficient?

Question

Suppose a random variable $X \in \mathbb{R}$ follows a discrete distribution $p$ and takes $n$ values. We assume $E[X]=0$ and $|X|\le M$, where $M$ is a constant. Given a smooth and monotonic increasing function $f$, e.g. $f(x)=e^x$, what's the lower bound of the correlation coefficient $$ \rho :=\frac{\operatorname{Cov}( f( X),X)}{\sqrt{\operatorname{Var} (f(X)) \operatorname{Var}(X)}}. $$ Since $f$ is a monotonic increasing, the trivial lower bound is $\rho \ge 0$. However, different choices of $f$ result different $\rho$. If $f$ is a linear function, $\rho=1$. So I wonder can we give a non-trivial lower bound of $\rho$ using $n$, $M$ and the information of $f$?

Answer 1

$\newcommand\P{\operatorname P}\newcommand\E{\operatorname E}\newcommand\Var{\operatorname{Var}}\newcommand\Cov{\operatorname{Cov}}$As you noted, necessarily $\rho\ge0$, so that $0$ is a lower bound on $\rho$.

Let us show that, given just the conditions on $X$ and $f$ stated in your post, $0$ is actually the best (greatest) lower bound on $\rho$.

Indeed, by rescaling, without loss of generality (wlog) $M=3$. Moreover, since any nondecreasing function can be uniformly approximated by smooth (strictly) increasing functions, wlog $f$ can be any nondecreasing function, as long as $\Var f(X)\ne0$. Of course, for the correlation $\rho$ to exist, we also need to require that $\Var X\ne0$.

Let now $X$ be a random variable such that $$\P(X=1)=\P(X=-1)=\frac12-2p$$ and $$\P(X=2)=\P(X=-2)=\P(X=2+p)=\P(X=-2-p)=p,$$ where $p\downarrow0$. For real $x$, let $$f(x):=\max(0,x-2+p),$$ so that $f$ is a nondecreasing function. Then $$\begin{gathered} \E X=0, \quad \Var X\to1, \\ \Cov(f(X),X)=\E f(X)X=2p^2(3+p)\sim6p^2,\\ \E f(X)=3p^2, \quad \E f(X)^2=5p^3,\quad \Var f(X)\sim5p^3,\\ \end{gathered} $$ and hence $$\rho\sim\frac{6p^2}{\sqrt{5p^3}}\to0.\quad\Box$$