Lemma: Let $T_n$ be the time of the nth arrival in a Poisson process and $U_k$, $k=1,2....n$ be independent uniform on $(0,1)$. Then the order statistics of $U_1, U_2,....,U_n$ have the same distribution of $(T_1/T_{N+1},T_2/T_{N+1},...,T_N/T_{N+1})$.

I could prove the lemma by simply find out the joint density of the above vector which is $n!$ for both case. However, I would like to know is there any intuition that I could understand the lemma without directly calculate the answer.

I have read this post, but I think that it may be unrelated to the problem, though not for sure.

  • $\begingroup$ Your linked question is effectively the same with $Y_j=T_j-T_{j-1}$ $\endgroup$ – Henry Jul 14 '16 at 12:21

My intuition:

  • Consider a radioactive source of particles treated as producing particles in a Poisson process
  • Conditioning on the $n+1$ particle being emitted at time $t_{n+1}$, the previous $n$ earlier particles can each have been emitted at any time in the interval $[0,t_{n+1}]$
  • Ignoring order, the emission time of each of the $n$ earlier particles has an independent uniform distribution time in that interval, as the cause of each emission is independent of the others
  • Hence the two equivalent distributions in the question

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