Poisson process arrival distribution Consider a Poisson process with rate $\lambda$ in a given time interval $[0,T]$. The inter-arrival time between successive arrivals is negative exponential distributed with mean $\frac{1}{\lambda}$ such that $X_1 >0$, and $\sum_{i=1}^\text{Last} X_i < T$, where $X$ represents inter-arrival time.
What about the distribution of time between Last arrival and ending time $T$? Is it also negative exponential distributed and has a mean value of $\frac{1}{\lambda}$? Can we study time segment $[0,T]$ of Poisson process in the backward direction too? In the forward direction, time between $t=0$ and first arrival is negative exponential distributed. In the backward direction, Last arrival is the first arrival and is the time between $t=T$ and Last is also negative exponential distributed. Is there any way to justify this? or some reference?
 A: What appears below answers the question as originally posted, where it was stated that the number of arrivals during $[0,T]$ is equal to a given number $k.$ 
Suppose the number of arrivals during $[0,T]$ is exactly $k\ge 1.$ You've called the inter-arrival times $X_i,$ so that $X_1$ is the time of the first arrival, $X_2+X_2$ is the time of the second arrival, and so on. For $t\ge 0,$ let $N_t$ be the number of arrivals before time $t,$ so you're saying $N_T=k.$
What, then, is the conditional distribution of $X_1$ given that $N_T=k$?
\begin{align}
& \Pr(X_1\le t\mid N_T=k) = \Pr(N_t\ge 1 \mid N_T=k) = 1-\Pr(N_t=0\mid N_T=k) \\[10pt]
= {} & 1 - \frac{\Pr(N_t=0\ \&\ N_T=k)}{\Pr(N_T=k)} = 1-\frac{\Pr(N_t=0 \ \&\ N_T -N_t =k)}{(\lambda T)^k e^{-\lambda T} /k!} \\[10pt]
= {} & 1-\frac{\Pr(N_t=0)\Pr(N_T-N_t = k) }{(\lambda T)^k e^{-\lambda T} /k!} = 1 - \frac{e^{-\lambda t} \cdot (\lambda(T-t))^k e^{-\lambda(T-t)}/k!}{(\lambda T)^k e^{-\lambda T} /k!} \\[10pt]
= {} & 1 - \frac{(T-t)^k}{T^k}.
\end{align}
(This does not depend on $\lambda.$)
In a similar way you can show that $T - \sum_{i=1}^k X_i$ has the same conditional distribution given that $N_t=k.$
A: Here's an answer to the question as last modified.
First suppose that on the entire real line, not just on $[0,\infty),$ we assign to each interval $(a,b)$ a random variable $N_{(a,b)}$ for which


*

*$\Pr(N_{(a,b)} = n) = \dfrac{(\lambda(b-a))^n e^{-\lambda(b-a)}}{n!}$ for $n=0,1,2,3,\ldots,$ and

*For intervals $A,B,C,\ldots$ no two of which intersect each other, $N_A,N_B,N_C,\ldots$ are independent.


Then let $X_1$ be the time of the first arrival after time $0,$ let $X_2$ be the time from then until the next arrival after that, and so on. Then we have
$$
\Pr(X_n > x) = e^{-\lambda x} \text{ for } x\ge0.
$$
Now consider the time $Y$ from the last arrival before time $T>0$ until time $T.$ For $x\ge0$ we have
$$
\Pr(Y>x) = \Pr(\text{no arrivals during } [T-x,T]) = \frac{(\lambda x)^0 e^{-\lambda x}}{0!} = e^{-\lambda x}.
$$
I.e. it has the same distribution that any one of the inter-arrival times has.
And then if you want to truncate it at $x=T,$ as suggested in comments under the question, you can modify that accordingly.
