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For $0<j_1<\cdots <j_n<t$, the probability of one jump in each infinitesimal interval around $j_1,\dots ,j_n$ and no other jump by time $t$ gives the joint densityfor $(J_1,ldots,J_n)$ together with the event $N_t=n$. Sincethe conditional density ison that ordered simplex.

Now $n$ independent uniform points on $[0,t]$ have joint density $t^{-n}$ on the cube. Each ordered vector has $n!$ permutations, so their order statistics have density $n!/t^n$ on the same simplex. The densities agree, proving Poisson process conditional arrival times.

Use the convention $J_0=0$, so $A(t)=t$ when $N_t=0$. Conditional on $N_t=n\ge 1$, the last jump $J_n$ is the largest of $n$ independent uniform points on $[0,t]$. The stated order statistic formula givesand hence, for every $n\ge 0$,Putting $\mu =\lambda t$ and averaging over $N_t$,This agrees with age of a Poisson process.

For two independent observations uniform on $[\theta -a,\theta +a]$, the displayed interval of possible parameter values contains the true parameter almost surely. Intersecting any confidence interval with it preserves that interval's coverage, while excluding values inconsistent with the observations. The order statistics interval $[U_{(1)},U_{(2)}]$ has coverage $1/2$, since the observations straddle $\theta$ with probability $1/2$. If their separation exceeds $a$, the entire interval of possible parameters lies between the observations; this explains why the nominal one-half procedure can be unnecessarily wide on that event.