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Suppose I know over a window $[0, T)$ that I have observed $n$ samples from a poisson process $N_t \sim p(n|\lambda t) = \frac{1}{n!}(\lambda t)^{n}\exp(-\lambda t)$.

What is the conditional distribution of the arrival times $t_1 < t_2 < ... < t_n$ on $[0, T)$? In particular, what is the marginal distribution of $t_1$?

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An interesting property of Poisson processes is that each event can be considered as "placed" independently and uniformly at a given time $t$ in $[0,T]$ (just like rain drops falling uniformly over the length of a board of length $T$). In other words, if there are $n$ events $\{\tau_i\}_{i=1}^n$, we have $\tau_i \sim \mathrm{Unif}(0,T)$ for all $i$. What you are looking for is the distribution of $t_1=\min \{\tau_i: i=1,\dots,n\}$. Please see this answer for distribution of the minimum of uniform IID variables. In your case: $$ p(t_1|n) = \frac{n}{T^n}(T-t_1)^{n-1} \quad\text{for $t_1\in[0,T]$}. $$

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Since the inter-arrival times are independent exponentially distributed, the joint pdf of the $n$ first arrival times is \begin{align} f(t_1,\dots,t_n) &=f(t_1)f(t_2|t_1)\dots f(t_n|t_{n-1}) \\&=\lambda e^{-\lambda t_1}\lambda e^{-\lambda(t_2-t_1)}\cdots \lambda e^{-\lambda(t_n-t_{n-1})} \\&=\lambda^n e^{-\lambda t_n} \end{align} for $0<t_1<\dots<t_n$ and 0 elsewhere. Conditional on $n$ arrivals in $(0,T)$, the conditional pdf is \begin{align} f(t_1,\dots,t_n|N(T)=n) &=\frac{f(t_1,\dots,t_n,N(T)=n)}{P(N(T)=n)} \\&=\frac{f(t_1,\dots,t_n)P(N(T)=n|t_1,\dots,t_n)}{P(N(T)=n)} \\&=\frac{\lambda^n e^{-\lambda t_n}e^{-\lambda(T-t_n)}}{e^{-\lambda T}(\lambda T)^n/n!} \\&=\frac{n!}{T^n}, \end{align} for $0<t_1<\dots<t_n<T$ and 0 elsewhere. This is the joint pdf of the order statistics of $n$ iid uniform random variables on $(0,T)$.

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