Connection between sum of uniform (0,1) random variables and Poisson arrival times

Question

Is there any underlying reason for the commonality between the following two distributions, where we see that the CDF of the first is the reciprocal of the PDF of the second? I imagine it has something to do with order statistics but am interested if there is an elegant connection

Sum of Uniform (0,1) Variables

$$P(X_1+\dots+X_n\leq t)=\frac{t^n}{n!}$$ where $0<t<1$ and ${X_1,\dots,X_n}$ are n i.i.d. uniform (0,1) random variables. Proven here

Poisson Process: Conditional Arrival Times given n arrivals

If we know n arrivals occurred in the interval $[0,t]$, we find that the PDF for the vector of arrival instants $(T_1, T_2, \dots, T_n)$ is $f_{T_1, \dots, T_n}(t_1, \dots, t_n|N(t)=n)=\frac{n!}{t^n}$

This can be found by applying Baye's Rule and using the memoryless property $$F_{T_1,\dots,T_n}(T_1\leq t_1, \dots T_n\leq t_n|N(t)=n)=\frac{P(T_1\leq t_1) \dots P(t_{n-1}<T_n\leq t_n)}{P(N(t)=n)}$$ then differentiating.

Answer 1

Poisson processes have independent increments. Let $0 \le t_1 < \dots < t_n < t_{n + 1} = t$. The independent increments assumption allows us to write the following probability as follows: $$ P(\cap_{i = 1}^{n}\{T_i \in \space (t_i, t_i + \delta]\}|\{N_t = n\}) $$ $$ =\frac{P(N_{t_1} = 0)\Pi_{i = 1}^{n}P(N_{t_i + \delta} - N_{t_i} = 1)P(N_{t_{i + 1}} - N_{t_{i} + \delta} = 0)}{P(N_t = n)} $$ $$ = \frac{e^{-\lambda t_1} \Pi_{i = 1}^{n}\delta\lambda e^{-\lambda \delta}e^{-\lambda(t_{i + 1} - t_{i} - \delta)}}{(\lambda t)^{n}e^{-\lambda t}/n!} $$

The multiplication over $e^{-\lambda (t_{i + 1} - t_i - \delta)}$ will yield $e^{-\lambda \Sigma_{i = 1}^{n}(t_{i + 1} - t_{i} - \delta)} = e^{-\lambda t + \lambda t_1 + \lambda n \delta}$ when combined with the other expressions, the numerator thus becomes $\delta^{n}\lambda^{n}e^{-\lambda t}$, which when divided by the denominator gives us $\delta^{n}\frac{n!}{t^{n}}$. The open sets generated by $\delta$ has volume $\delta^{n}$, which if we divide the expression by then take the limit of $\delta$ to zero, gives us the density of the expression, which is $f_{T_1, \dots, T_n|N_t}(t_1, \dots, t_n | n) = \frac{n!}{t^{n}}\mathbb{1}_{\{0 \le t_1 < t_2 < \dots < t_n\}}$, which is the expression you got. In the second line of the expression, we are calculating the probability of seeing arrivals within the $\delta$ arrival and not anywhere else, the division and multiplications also follow bayes, then we use poisson distribution with rate $\lambda t $ in the denominator and respective parameters in the numerator, so everything we got is valid.