Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 2 / 27J (Applied Probability)

After a record of value $v$, the first subsequent observation exceeding it has the original distribution conditioned on $X>v$. Indeed, summing over the number of failed observations gives the densityIndependence makes this conditional transition depend only on the latest record. Multiplying these transitions, starting with the density $f(v_1)$, provesThe ordering indicator is part of the density; it is zero elsewhere. Continuity of the observation distribution avoids ties.

Take $R_t=0$ when no record has occurred below $t$, the needed convention for the maximum of an empty set. Assume $F(t)<1$ and write $S(t)=1-F(t)$. Given the latest record $v_k<t$, the chance that the next one exceeds $t$ is $S(t)/S(v_k)$. Multiplying this by the joint density from part (a) leaveson the ordered simplex $0<v_1<\cdots <v_k<t$. Its integral is $S(t)\Lambda (t{)}^k/k!$, where $\Lambda (t)={\int }_0^{t}f(s)/S(s)ds=-\log S(t)$. The $k=0$ probability is directly $S(t)=e^{-\Lambda (t)}$. HenceThis is Poisson record counts in cumulative hazard coordinates; the instantaneous rate is the original distribution's hazard function. If the distribution has a finite upper endpoint and $t$ reaches it, $F(t)=1$ and infinitely many record values lie below $t$ almost surely; there is then no finite-parameter Poisson law. The finite-hazard assumption is thus necessary for the literal assertion.