My book outlines a procedure but a preliminary part of it is unclear to me.

Let X be the number of occurences of an event over a unit of time and assume that it has a Poisson distribution with mean $m=\lambda $. Let $T_1, T_2 , T_3, \ldots $ be the interarrival times of the occurences and they are iid with an exponential $\lambda $ distribution. Note that $ X=k $ iff $$ \sum_{j=1}^k T_j \leq 1 \quad \text{and} \quad \sum_{j=1}^{k+1} T_j >1. $$

This is precisely what I do not understand. Why does the total waiting tme until $k$ occurences have to be less than or equal to 1? Any help is greatly appreciated. Thank you.

  • 2
    $\begingroup$ "Let $X$ be the number of occurrences of an event over a unit of time..." $\endgroup$
    – cardinal
    Nov 24, 2013 at 22:25

1 Answer 1


So if $T_1, T_2, \dots$ are the interval times of the occurrences, the first event occurs at $t = T_1$, the second at $t = T_1 + T_2$, etc.

Then $X = {}$ the number of occurrences during the first unit of elapsed time is $$ X = \max_k T_1 + \cdots + T_k \le 1.$$

Hence if $X = k$, $T_1 + \cdots + T_k \le 1$ and $T_1 + \cdots + T_{k+1} > 1$.

PS. I think the simplest/fastest way to simulate a Poisson process with constant rate $\lambda$ on the interval $(0, t)$ is

  1. draw $X \sim \mathcal P (\lambda t)$ the number of events

  2. draw $u_1, \dots, u_X$ uniformly in $(0,t)$ and order them: these are the times of occurrences of the $X$ events.

  • $\begingroup$ Thank you but my book suggests another algorithm, which I have not figured out yet. 1. Set X=0 and T=0 2. Generate U uniform (0,1) and let $Y=-(1/ \lambda) log(1-U)$, 3. SET T=T+Y 4. If T>1, output X, else set X=X+1 and go to step 2. I have not been able to understand it yet. $\endgroup$
    – JohnK
    Nov 25, 2013 at 14:17
  • $\begingroup$ I was assuming that uou can use a function for generation of random Poisson deviates, such as rpois in R. Your book has a more elementary approach. If you let $F(x) = 1 - e^{-\lambda x}$ be the cdf of the exponential distribution $\mathcal E(\lambda)$, then you have $Y = F^{-1}(U)$, from which you should be able to prove that, $U$ being uniform, $Y \sim \mathcal E(\lambda)$. Then the successive values of $T$ along the iterations of steps 2 and 3 are $T_1$, $T_1+ T_2$, etc. You should be able to conclude. $\endgroup$
    – Elvis
    Nov 25, 2013 at 15:00
  • $\begingroup$ Oh that's right. Thanks again man. My book, is an intermediate one in Mathematical Statistics, but I am a novice studying on my own. $\endgroup$
    – JohnK
    Nov 25, 2013 at 15:04

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