# Does exponential waiting time for an event imply that the event is Poisson-process?

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?

• waiting times should be $exp(\lambda)$, not $exp(t\lambda)$ to create a Poisson process. – Xiaomi Oct 29 '18 at 10:32
• @Xiaomi thx for the remark, edited – moreblue Oct 29 '18 at 10:46
• 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. – Rasmus Oct 29 '18 at 13:59
• 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. – whuber Oct 29 '18 at 14:06