For any $A\subseteq \mathbb{N}$ we let the (lower) density of $A$ be defined by $$d(A) = \liminf\_{n\to\infty}\frac{|A\cap\{0,\ldots,n\}|}{n+1}.$$
If $\pi:\mathbb{N}\to\mathbb{N}$ is a permutation (bijection), let $\text{ex}(\pi) = \{n\in\mathbb{N}: \pi(n) > n\}$ be the set of [*exceedances*](https://mathoverflow.net/questions/359684/why-excedances-of-permutations), and let $\text{negex}(\pi)=\{n\in\mathbb{N}: \pi(n) < n\}$ be the set of "negative exceedances".


To me it seems inconceivable that there is a permutation $\pi:\mathbb{N}\to\mathbb{N}$ with $d\big(\text{ex}(\pi)\big) \neq d\big(\text{negex}(\pi)\big)$ -- but my intuition has let me down many times.


Is my intuition correct this time? ${}$



Is there a lop-sided permutation $\pi:\mathbb{N}\to\mathbb{N}$?
Let $\pi(n) = n+1$ for all $n$ except that $\pi(2^{k}-1) = 2^{k-1}$. In other words, permute cyclically $(2,3)$, $(4,5,6,7)$, $(8,9,10,\ldots,15)$, and so on. Then $d(\operatorname{pos}(\pi))=1$ whereas $d(\operatorname{neg}(\pi))=0$.


