# The pdf of last interval of a Poisson process in $[0,T]$

Assume a Poisson point process with rate $\lambda$ in time $[0,T]$. Supoose $X$ is the random variable representing the time between the last arrival and $T$. What is the probability density function of $X$ as $T\to \infty$?

The pdf is $\frac{d}{dx}P\left(X\leq x\right)$. We can break up $P\left(X\leq x\right)$ by the number of arrivals in time $[0,T]$:

\begin{align} P\left(X\leq x\right)&=\sum_{k=1}^{\infty}Poisson(k, \lambda T)P\left(X\leq x|k \text{ arrivals}\right)\\ &=\sum_{k=1}^{\infty}Poisson(k, \lambda T)P\left(T-T_k\leq x|k \text{ arrivals}\right)\\ &=\sum_{k=1}^{\infty}Poisson(k, \lambda T)P\left(T-T_k\leq x|T_k \leq T <T_{k+1}\right) \end{align}

where $T_k=\sum_{i=1}^{k}A_i$ is the time $k^{th}$ arrival, and $A_i$ is the $i^{th}$ inter-arrival time. Any idea how to continue? Or, a resource that already has the answer?

• @TheoreticalEconomist Thanks. I edited the problem to give that explanation. $A_i$ is the i^{th} inter-arrival; i.e., time between $(i−1)^{th}$ and $i^{th}$ arrival. Feb 8, 2017 at 18:59
• @blaza thanks. But that is not true. Feb 9, 2017 at 5:35
• yes, sorry, I'm deleting my comment Feb 9, 2017 at 10:16

I'll denote $T_k = A_1 +\dots + A_k$.

If there are exactly $k$ arrivals in time $T$, that means $T_k\leq T < T_{k+1}$.

So you can write \begin{align}P(T_k\geq T-x\mid \text{k arrivals})&=\frac{P(T_k\geq T-x, \ T_k\leq T < T_{k+1})}{P(\text{k arrivals})}\\&=\frac{P(T-x \leq T_k\leq T < T_{k+1})}{P(\text{k arrivals})}\\&=\frac{P(T-x \leq T_k\leq T < T_k+A_{k+1})}{P(\text{k arrivals})}\end{align}

From here, you can use the independence of $A_{k+1}$ and $T_k$ and try to work out the numerator as the integral $$\iint\limits_{\{T-x \leq u\leq T < u+v\}}f_{T_k}(u)f_{A_{k+1}}(v)\,du\,dv,$$noting that $T_k\sim Gamma(k,\lambda)$ and $A_{k+1}\sim Exp(\lambda)$.

On the other hand, there is an easier way to approach the problem.

Denote $N(t)$ the number arrivals in time $t$. Then $T_{N(T)}$ is the time of the last arrival in the interval $[0,T]$. So, for $x\in (0,T)$, we have:

\begin{align}P(X\leq x)&=P(0\leq T-T_{N(T)}\leq x)\\&=P(T-x\leq T_{N(T)}\leq T)\\&=P(N(T)-N(T-x)\geq 1)\\&=1-P(N(x)=0)\\&=1-e^{-\lambda x},\end{align} where the last two equations follow from $N(T)-N(T-x)$ having the same distribution as $N(x)$, which is $Poisson(\lambda x)$.

Thus we get the cdf of $X$ (note that $X$ is a.s. positive and $X\leq T$ a.s.): $$F(x)=\begin{cases}0,&x\leq0\\ 1-e^{-\lambda x},& x\in(0,T)\\ 1,& x\geq T \end{cases}$$ This looks like the cdf of the exponential distribution, but we have a jump at $x=T$, so it's not really the exponential distribution. But when $T\to \infty$ as the question said, we get the exponential distribution.

• Thanks. Can you please specify what integral you mean? (You said we can calculate the integral). About the second argument, something is wrong. $P(T-x\leq T_{N_{T}}\leq T) = P(N(T)-N(T-x) \geq 0)= P(N(x) \geq 0)$ Feb 9, 2017 at 14:51
• @Susan20200 I added the integral I thought of. It's just calculating the probability of a random vector, where the variables in the vector are independent. I'm not sure what you think is wrong for the second argument? There can't be $0$ arrivals in the interval $[T-x,T]$, as there is the $N(T)$-th arrival there, because of $T-x\leq T_{N(T)}\leq T$, so there is at least one. Feb 9, 2017 at 15:11
• Can you please clarify why $P(T-x\leq T_{N(T)}\leq T)=P(N(T)-N(T-x)\geq 1)$? Feb 9, 2017 at 17:17
• Because the time of the $N(T)$-th arrival is between $T-x$ and $T$, there is at least one arrival in the interval $[T-x,T]$, so the difference between the number of arrivals in time $T$ (i.e. $N(T)$) and the number of arrivals in time $T-x$ (i.e. $N(T-x)$) is at least one. Feb 9, 2017 at 18:03
• Thanks. I understand the arguments. Let me thing. If $x=T$, then $P(X\leq x)=(1-P(\text{no arrival in [0,T]}))\cdot 1 = 1-e^{-\lambda T}$, and your answer for $X=T$ is $1-e^{-\lambda T}$ Feb 9, 2017 at 18:14