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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.


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What would the probability be at infinity? – Simon Hayward Nov 25 '12 at 19:49
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. – Did Nov 26 '12 at 10:59

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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Why does $P(T=\infty)$ equal that limit? In my opinion, your answer needs some explanation. – Stefan Hansen Nov 26 '12 at 9:40
Hmmm, have attempted to edit, but I think I'm hitting against mthe limits of my own knowledge. If you have a better idea of how to edit, then please feel free. – Simon Hayward Nov 26 '12 at 9:58
Something is not right: $N(t+\tau)-N(t)\sim \mathrm{po}(\lambda \tau)$ and also $k$ does not appear in your first formula. – Stefan Hansen Nov 26 '12 at 10:15
Yes. You are correct. Unfortunately I am at work right now and I'm not able to keep editing! I will try to fix at lunchtime. – Simon Hayward Nov 26 '12 at 10:18
I've had a go at simplifying (a lot) and re-writing. I think it is a simpler and better answer now. – Simon Hayward Nov 26 '12 at 13:45

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