= 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 $U=F(Y_n)$ <probability density function>[density] $nu^{n-1}$ on $(0,1)$. Conditional on $U=u$, $\tau$ has a <geometric distribution> of <statistical parameter>[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 <probability density function>[density] or a strictly increasing <cumulative distribution function>[distribution function].
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