Assume a Poisson process with rate $\lambda$.

Let $T_{1}$,$T_{2}$,$T_{3}$,.... be the time until the first, second, third,......(so on) arrivals following exponential distribution.

If I consider the fixed time interval $[0-T]$, WHAT is the expectation value of the arrival time $T_{1}$,$T_{2}$,$T_{3}$???,....... i.e.

  1. $E[T_{1}|T_{1}\le T]$ ?
  2. $E[T_{2}|T_{1}<T_{2}\le T]$ ?
  3. $E[T_{3}|T_{2}<T_{3}\le T]$ ?
  • $\begingroup$ what is your question? $\endgroup$
    – Deep North
    Dec 15, 2017 at 11:29
  • $\begingroup$ what are the expressions for the expectation values numbered as 1,2, and 3? $\endgroup$ Dec 15, 2017 at 11:30

1 Answer 1


For Poisson process with rate $\lambda$, each time interval correponds to a random variable $X_i$ with an exponeitial distribution.

Therefore, \begin{align*} &T_1=X_1 \\& T_2=X_1+X_2\\&T_3=X_1+X_2+X3\\&.... \end{align*}

$T_i$ has Gamma distribuiotn $\Gamma(i,\lambda)$

I show you the first soltuion. Here $T$ is a fix value not a random variable. \begin{align*} E[T_{1}|T_{1}\le T]&=\int_{0}^\infty t_1f_{T_1|T_1 \le T}(t_1)dt_1\\&=\int_{0}^\infty t_1 \frac{f_{T_1}(t_1)\bf{1}_{(t_1 \le T)}}{P(T_1\le T)}dt_1\\&=\frac{1}{P(T_1 \le T)}\int_{0}^T t_1 f_{T_1}(t_1)dt_1\\&=\frac{1}{1-e^{-\lambda T}}\int_{0}^T t_1 \lambda e^{-\lambda t_1}dt_1\\&=\frac{-1}{1-e^{-\lambda T}}\int_0^T t_1de^{-\lambda t_1}\\&\text{(Integration by parts)}\\&=\frac{-(Te^{-\lambda T}+\frac{1}{\lambda}e^{-\lambda T}-\frac{1}{\lambda})}{1-e^{-\lambda T}} \end{align*}

  • $\begingroup$ Many Many thanks @Deep North! I fully agree this. I also developed a simulator to check it. But for T1 it fully matches. According to the above your solution, $E[T_{2}|T_{1}<T_{2}\le T]$=$E[T_{1}|T_{1}\le T]+E[T_{1}|T_{1}\le T]$? When I used this approach, this is actually not working for T2 and the results for lower lambda mismatches. PLEASE see my added picture in my above post. $\endgroup$ Dec 15, 2017 at 13:08
  • $\begingroup$ Actually for $T_{1}$, I also derived the above equation. and many thanks for you as well, as now it also confirms to me its right. The problem for me is mainly to derive for second and above, But for calculating $T_{2}$, I assumed it to be $T_{1}+T_{1}$. But, simulator says no for lower arrival rates and when constant time $T$ is small as I used 10$ms$ for the above graph. $\endgroup$ Dec 15, 2017 at 13:26
  • $\begingroup$ Dear @Deep North, is it right to do the following, \begin{align*} E[T_{2}|T_{1} \le T_{2} \le T]&=\frac{1}{P(T_{1} \le T_{2} \le T)}\int_{T_{1}}^T t_2 f_{T_2}(t_2)dt_2 \\&=\frac{1} {1-e^{- \lambda (T-T_{1})}} \int_{T_{1}}^T t_2 f_{T_2}(t_2)dt_2 \end{align*} $\endgroup$ Dec 17, 2017 at 17:57
  • $\begingroup$ Hi Halian, I am not sure on that part, my idea is that since it is arrivals, I would say, $T_1$ is always less than $T_2$, probably you only need to focus on $T_2<T$, but note, now $T_2$ has a $\Gamma(2,\lambda)$ distribtion. I am on holiday, sorry to response you late. $\endgroup$
    – Deep North
    Dec 20, 2017 at 11:43
  • $\begingroup$ Dear @Deep North. ok thank you I will see it again now. Many thanks and have good time $\endgroup$ Dec 20, 2017 at 11:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.