= Waiting time for a new record after a fixed sample
{title2=$P(\tau=k)=\frac n{(n+k-1)(n+k)}$}
For independent identically distributed observations with continuous <cumulative distribution function> $F$, fix the maximum $Y_n$ of the first $n\geq1$ observations and let $\tau$ be the first subsequent observation exceeding it. The <probability integral transform> gives $U=F(Y_n)$ density $nu^{n-1}$ on $(0,1)$. Conditional on $U=u$, $\tau$ has a <geometric distribution> of parameter $1-u$. Therefore
$$
P(\tau=k)=n\int_0^1u^{n+k-2}(1-u)\,du
=\frac n{(n+k-1)(n+k)},\qquad
P(\tau>k)=\frac n{n+k}.
$$
The masses sum to one, so the waiting time is finite almost surely, but the <tail-sum formula for expectation> gives $E\tau=\infty$. These conclusions require only continuity of $F$, not a density or a strictly increasing distribution function.
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