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I am new here so if I violate any rule, please inform.

Consider the stochastic process given by $\{ N(t) : t \geq 0 \}$ which is time homogeneous poisson process with arrival rate $\lambda$. Let $W_n$ be the waiting time, i.e, $W_n = \inf \{t\geq0:N(t)=n \}$.

We want to show that $P(W_k < \infty) = 1 \ \ \forall k=1,2,\ldots $.

Any help would be appreciated.

Thanks.

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    The result is quite general: each interarrival time is almost surely finite hence the sum $W_k$ of the $k$ first interarrival times is almost surely finite.2012-11-26

2 Answers 2

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Clearly we may assume that $N(t)$ is right-continuous with monotone increasing sample path.

Then $t \geq W_n$ if and only if $N(t) \geq n$ and we have

$\Bbb{P}(W_n \leq t) = \Bbb{P}(N(t) \geq n) = \sum_{k=n}^{\infty} \frac{(\lambda t)^{k}}{k!} e^{-\lambda t} = 1 - \sum_{k=0}^{n-1} \frac{(\lambda t)^{k}}{k!} e^{-\lambda t}. $

Now taking $t \to \infty$, the monotone convergence theorem yields the desired result $\Bbb{P}(W_n < \infty) = 1$.

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Inter event times for a poisson process are exponentially distributed with pdf

$X \sim \lambda t e^{-\lambda t}$

(from Wikipedia)

The probability that the interevent waiting time is less than or equal $t \in [0, \infty)$ is given by:-

$\mathbb{P}(t=T) = \int_0^\infty \lambda s e^{-\lambda s} dt\\ \to 1 ~\text{as}~ t \to \infty.$

So the probability of (at least) one event, becomes 1 as $t$ tends to infinity. Hence, the probability of no events (= 1 - $\mathbb{P}(\text{at least}~ 1 ~\text{event})$) tends to $0$ as $t \to \infty$

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    I have missed the part where we generalise to the case where k >1, but did's comment adresses this. I hope this answer is more helpful now anyway.2012-11-26