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?
Best Answer
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
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.