= Leave-one-out identity for empirical distributions
{title2=$\widehat p_{n+1}=(n+1)^{-1}\sum_{i=1}^{n+1}\widehat q_{n,i}$}
Given $n+1$ observations, let $\widehat q_{n,i}$ be the <empirical distribution> of the $n$ observations remaining after deleting observation $i$. Their average is the <empirical distribution> $\widehat p_{n+1}$ 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.
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