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The likelihood function for the ordered arrival times of a Poisson process, including the absence of further arrivals before $T$, isThe arrival-time constraint does not depend on $\lambda$. Equivalently, the Poisson distribution of the total count gives $L(\lambda )=e^{-\lambda T}(\lambda T{)}^n/n!$, which differs by a parameter-independent factor. Conditional on the count, the Poisson process conditional arrival times have density $n!/T^n$, independent of $\lambda$. Thus the total count contains all the information about the rate when the exposure is known; it is a sufficient statistic.

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.

Condition on $N_t=n$. By Poisson process conditional arrival times, the jump times are the ordered version of independent $U_1,\dots ,U_n$ uniformly distributed on $[0,t]$. The sum is symmetric in the marked pairs, and the marks are independent, soWrite the bracket as $a$. Averaging over the Poisson distribution of $N_t$ givesThis is the exponential formula for a marked Poisson sum.