Leave-one-out identity for empirical distributions (${\hat{p}}_{n+1}=(n+1{)}^{-1}{\sum }_{i=1}^{n+1}{\hat{q}}_{n,i}$)

Given $n+1$ observations, let ${\hat{q}}_{n,i}$ be the empirical distribution of the $n$ observations remaining after deleting observation $i$. Their average is the empirical distribution ${\hat{p}}_{n+1}$ of the full sample. For each symbol, its indicator function for a particular observation appears in exactly $n$ deletion samples. Summing the empirical counts and dividing by $n(n+1)$ therefore gives its full-sample frequency. This identity is deterministic and needs neither independence nor identical distributions.