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Say I have a process, $\{N_t : t \ge 0\}$, which denotes the number of the event that occurred until the time $t$.

And let me define $W = \min \{t : N_t = 1\}$ which is denotes the time until the first event happens.

My question is, does the fact that $W \sim Exp(\lambda)$ (where $\lambda$ : rate), implies $\{N_t : t \ge 0\}$ is a Poisson process with rate $\lambda t$?

My try

I have found that the following equation holds.

$$ \int_0^t \lambda e^{-\lambda y}dy = \sum_{k=1}^\infty \frac{e^{-\lambda t} (\lambda t)^k}{k!} $$

But how should I proceed?

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  • $\begingroup$ waiting times should be $exp(\lambda)$, not $exp(t\lambda)$ to create a Poisson process. $\endgroup$
    – Xiaomi
    Commented Oct 29, 2018 at 10:32
  • $\begingroup$ @Xiaomi thx for the remark, edited $\endgroup$
    – moreblue
    Commented Oct 29, 2018 at 10:46
  • $\begingroup$ I can't comment, however i can refer you to the book "Non-Life Insurance Mathematics" by Thomas Mikosch. Here a proof and the relevant assumptions are given in details. Edit: I am aware that this answer might not meet the requirements, so please do not hesitate to remove or delete. I do however want to provide OP with alternative sources of where to find answers. $\endgroup$
    – Rasmus
    Commented Oct 29, 2018 at 13:59
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    $\begingroup$ Certainly not without more assumptions! After all, one could define another point process simply by stopping after the first event. That would not change the waiting time, but the result certainly wouldn't be a Poisson process. $\endgroup$
    – whuber ♦
    Commented Oct 29, 2018 at 14:06
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    $\begingroup$ Your equation $$\int_0^t \lambda e^{-\lambda y}dy = \sum_{k=1}^\infty \frac{e^{-\lambda t} (\lambda t)^k}{k!}$$ seems like an attempt to compute the cumulative distribution function of the exponential distribution. It should have gone like $$\int_0^t \lambda e^{-\lambda y}dy = -e^{-\lambda y} \Big|_0^t = 1-e^{-\lambda t} $$ In relation to my answer below you may note that this expression $1-e^{-\lambda t}$ is the probability $k>0$ for a Poisson distribution (the probability that the first hit is before time $t$ is the probability to have at least $k>0$ for a Poisson distribution). $\endgroup$
    – Sextus Empiricus
    Commented Sep 9, 2019 at 10:41

1 Answer 1

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Not necessarily a Poisson process

The answer is already sort of given by WHuber in the comments. You need more (restricting) assumptions before the exponential waiting time is to be considered a Poisson process.

In your question you explicitly ask whether the process is Poisson when the waiting time for the first event follows an exponential distribution. So whatever happens after that is not specified. (You might wanted to imply that the exponential distribution is true for the waiting time between all neighboring events, although you should specify this explicitly)

This is not just a pedantic point, because one may think of other types of processes (non-Poisson) that have this exponential distributed first waiting time. And therefore it needs to be explicitly specified. This is also not just solved by saying that the other waiting times are also exponential distributed.

If you extend your statement, say that every waiting time $W_k$ to get from the event $N = k$ and $N = k-1$ is exponential distributed, then you still do not, necessarily, end up with a Poisson process. It will be necessary that the distribution of waiting times are identical and independent distributed.

  • Independent: A Poisson process has independent increments.
  • Identical: In the case of exponential waiting time it must be homogeneous.

Memorylessness

However, interesting to note is that, one thing can still be said about the exponential distribution, without referring to a Poisson process, and this is that it relates to a process that has memorylessness. The answer here, relating to the train paradox and showing a difference between exponential distributed and constant distributed waiting times, may explain this further.

Deriving Poisson distribution from the Exponential distribution

Note: for a derivation in the other direction (deriving the exponential distribution from the Poisson distribution) see here: Relationship between poisson and exponential distribution

I am not sure what you were trying to do with the last equation, but possibly you were trying to derive the expression for the Poisson distribution by using the Exponential distribution. Anyway, it might be interesting to show that derivation in order to explicitly show that exponential distributed waiting times (identical and independent) correspond to a Poisson distribution or Poisson process.

Let the i.i.d variables $W_k \sim Exp(\lambda)$ be the waiting times between the $k$-th and $k-1$-th event. Then the sum of those waiting times $T_n = \sum_{k=1}^n W_k \sim Erlang(n,\lambda)$ is the waiting time for the $n$-th event to occur, and this follows an Erlang distribution (see for derivation here). Let $K$ be the number of events observed within time $T$. The key is that:

The 'probability that you observe less than $k$ events within time $t$', is equal to the 'probability that observing $k$ events takes more than $t$ time'

$$\mathbb{P}(K< k \vert T=t) = \mathbb{P}(T>t \vert K = k)$$

So we can relate the CDF of the erlang distribution:

$$1- \mathbb{P}(T>t \vert K = k) = F(T \leq t \vert K=k, \lambda) = 1 - e^{-\lambda t} \sum_{n=0}^{k=1} \frac{(\lambda t)^n}{n!} $$

with the CDF of the Poisson distribution:

$$\mathbb{P}(K< k \vert T=t) = F(K \leq k \vert \lambda_{p} = \lambda t) = e^{- \lambda_p} \sum_{n=0}^{\lfloor k \rfloor} \frac{\lambda_p^n}{n!}$$

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  • $\begingroup$ Beautifully detailed and careful answer to a not-so-well developed question. I hardly remember the mental trace of myself questioning this post, but this helps a lot. Thank you! $\endgroup$
    – moreblue
    Commented Sep 10, 2019 at 3:57

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