# What is the correct inter-arrival time distribution in a Poisson process?

Given a Poisson process (e.g. radioactive decay) with rate $\lambda$, then the expression $\exp(-\lambda t)$ is the probability of observing no counts in time interval $t$. This can be interpreted (after normalization) as the probability distribution (exponential distribution) of time intervals during which no counts are observed.

It is often said that this can also be interpreted as the distribution of times between counts. Although I'm sure this is right, I disagree: the distribution of time intervals in which no counts are observed is different than the distribution of time intervals between counts.

I would appreciate an explanation. Here's my reasoning.

I agree the exponential distribution is the time distribution for the first arrival, since the time origin (your stopwatch) was started at a non-specific time. However, the second arrival time must be measured with respect to the first one, which is a known time.

Take for example a low count rate (say 5 per minute) and a short time interval (0.1 sec). It is unlikely that I will get a count in that interval (i.e. probability of observing no counts is high -> exponential distribution). But that does not mean that 0.1 sec is a highly probable time interval between successive counts (quite the contrary actually). In fact, the exponential distribution says that the most likely interval is 0.

If you start your stopwatch at the time you record the first count, then the time interval before you record a second count would follow the distribution $P(1)= (\lambda t) \exp(-\lambda t)$, which seems to me to be the right inter-arrival time distribution.

Thanks!

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I think what you mean is that the distribution of the length of the longest interval with no counts containing a fixed time $t_0$ (i.e. the interval from the last count before $t_0$ to the first count after $t_0$) is not exponential. In fact, it has the Gamma distribution with shape parameter $2$ and rate $\lambda$, i.e. it is the sum of two independent exponential random variables with rate $\lambda$, the time from the last count to $t_0$ and the time from $t_0$ to the next count. – Robert Israel Aug 17 '12 at 21:07
See the "Hitchhiker's paradox". – Robert Israel Aug 17 '12 at 21:13
I am under the impression that the asker find memorylessness of exponential random variables counterintuitive. (Just my guess.) – Tunococ Aug 17 '12 at 22:18
@Robert, could you elaborate on why that interval is a Gamma? I have opened a new question here – medley56 Dec 9 '15 at 23:34

Let $X(t)$ be a Poisson process with rate $\lambda$. For any time $t>0$, $X(t)$ follows a Poisson distribution $\mathcal{Po}(\lambda t)$. The probability that the count remains at zero at time $t$, given that is started with zero is then $$\mathbb{P}\left(X(t)=0\right) = \mathrm{e}^{-\lambda t}$$ Let $T_1$ be the time of the first jump (a random variable). Then the event $\{X(t) = 0\}$ is the same as the event $\{T_1 > t\}$, meaning that the first jump will occur after epoch $t$, i.e. $$\mathbb{P}\left(T_1 > t\right) = \mathbb{P}\left(X(t)=0\right) = \mathrm{e}^{-\lambda t}$$ That means that $T_1$ is an exponential random variable, whose density is $\lambda \mathrm{e}^{-\lambda t}$.

In order to establish that Poisson inter-arrival times are also exponential random variables we need to consider the values of the process at times $s$ and $t$, such that $s<t$. Since Poisson process has independent increments: $$\mathbb{P}\left(X(s) = n, X(t) = m\right) = \mathbb{P}\left(X(s) = n, X(t) - X(s) = m -n\right) \stackrel{\text{indep. incr.}}{=} \\ \mathbb{P}\left(X(s)=n\right) \cdot \mathbb{P}\left(X(t-s)=m-n\right) = \frac{(\lambda s)^n}{n!} \mathrm{e}^{-\lambda s} \frac{(\lambda (t-s))^{m-n}}{(m-n)!} \mathrm{e}^{-\lambda (t-s)} \mathbf{1}_{ m \geqslant n \geqslant 0} = \\\frac{(\lambda t)^m}{m!} \mathrm{e}^{-\lambda t} \cdot \binom{m}{n} \left(\frac{s}{t}\right)^n \left(1-\frac{s}{t}\right)^{m-n} \mathbf{1}_{ m \geqslant n \geqslant 0}$$ This implies, for $s>0$: $$\mathbb{P}\left(X(t+s) = n| X(t) = n\right) = \frac{\mathbb{P}\left(X(t+s) = n, X(t) = n\right)}{\mathbb{P}\left(X(t) = n\right)} = \mathrm{e}^{-\lambda s}$$ That is, given that the state of the process at times $t$, the probability that no jump occurs within $s$ seconds is $\mathrm{e}^{-\lambda s}$, i.e. the time to the next jump is an exponential random variable with rate $\lambda$.

Of course, this is all a consequence of the independence of increments and the property $X(t+s) - X(t) \stackrel{d}{=} X(s)$.

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+1. This is exactly the right way to make the transition from discrete to continuous in a probability course. The continuous distribution of the waiting time follows from the discrete distribution of the counts. (However, I'm not sure if it fully answers the poster's questions, though.) – Michael Hardy Aug 17 '12 at 23:06
Thanks to everyone for their help: yes, counterintuitive, yes, Gamma distribution; Sasha, thanks for the demonstration. – Richard Aug 20 '12 at 20:04

Here is an easy explanation to what you are asking:

Consider the interarrival times of a Poisson process $(A_1, A_2,\dots)$, where $A_i$ is the elapsed time between arrival $i$ and arrival $i+1$.

The 1st arrival occurs after time $t$ iff there are no arrivals in the interval $[0,t]$, hence:

$P(A_1 > t) = P(N(t) = 0) = \exp(-\lambda t)$,

which implies,

$P(A_1 \leq t) = 1 – \exp(-\lambda t)$,

which is the cdf of the exponential distribution.

Thus, the density is

$f(t) = \lambda \exp(-\lambda t)$,

which is what you were asking.

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What is the expected time then that i have to for the second event to finish? – curious Oct 14 '14 at 22:09
Why is $P(N(t) = 0) = exp(-\lambda t)$? – Omar Haque Apr 8 at 10:21