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The waiting times for poisson distribution is an exponential distribution with parameter lambda. But I don't understand it. Poisson models the number of arrivals per unit of time for example. How is this related to exponential distribution? Lets say probability of k arrivals in a unit of time is P(k) (modeled by poisson) and probability of k+1 is P(k+1), how does exponential distribution model the waiting time between them?

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    $\begingroup$ A Poisson distribution doesn't have waiting times. Those are a property of a Poisson process. $\endgroup$
    – Glen_b
    May 21, 2016 at 1:30
  • $\begingroup$ Also see here, a better explanation about the difference between these two distributions. $\endgroup$
    – Belter
    Sep 22, 2017 at 6:04

5 Answers 5

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I will use the following notation to be as consistent as possible with the wiki (in case you want to go back and forth between my answer and the wiki definitions for the poisson and exponential.)

$N_t$: the number of arrivals during time period $t$

$X_t$: the time it takes for one additional arrival to arrive assuming that someone arrived at time $t$

By definition, the following conditions are equivalent:

$ (X_t > x) \equiv (N_t = N_{t+x})$

The event on the left captures the event that no one has arrived in the time interval $[t,t+x]$ which implies that our count of the number of arrivals at time $t+x$ is identical to the count at time $t$ which is the event on the right.

By the complement rule, we also have:

$P(X_t \le x) = 1 - P(X_t > x)$

Using the equivalence of the two events that we described above, we can re-write the above as:

$P(X_t \le x) = 1 - P(N_{t+x} - N_t = 0)$

But,

$P(N_{t+x} - N_t = 0) = P(N_x = 0)$

Using the poisson pmf the above where $\lambda$ is the average number of arrivals per time unit and $x$ a quantity of time units, simplifies to:

$P(N_{t+x} - N_t = 0) = \frac{(\lambda x)^0}{0!}e^{-\lambda x}$

i.e.

$P(N_{t+x} - N_t = 0) = e^{-\lambda x}$

Substituting in our original eqn, we have:

$P(X_t \le x) = 1 - e^{-\lambda x}$

The above is the cdf of a exponential pdf.

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    $\begingroup$ Ok this makes it clear. Exponential pdf can be used to model waiting times between any two successive poisson hits while poisson models the probability of number of hits. Poisson is discrete while exponential is continuous distribution. It would be interesting to see a real life example where the two come into play at the same time. $\endgroup$
    – user862
    Aug 25, 2010 at 18:03
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    $\begingroup$ Huh? is $t$ a moment in time or a period of time? $\endgroup$ Mar 8, 2015 at 20:34
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    $\begingroup$ Note, that a poisson distribution does not automatically imply an exponential pdf for waiting times between events. This only accounts for situations in which you know that a poisson process is at work. But you'd need to prove the existence of the poisson distribution AND the existence of an exponential pdf to show that a poisson process is a suitable model! $\endgroup$ Nov 20, 2015 at 6:47
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    $\begingroup$ @CodyBugstein Both: they are interchangeable in this context. Arrivals are independent of each other, which means that it doesn't matter what the offset of time is. The period from time 0 till time t is equivalent to any time period of length t. $\endgroup$
    – chtenb
    Apr 22, 2016 at 8:19
  • $\begingroup$ @user862: It's exactly analogous to the relationship between frequency and wavelength. Longer wavelength; lower frequency analogous to: longer waiting time; lower expected arrivals. $\endgroup$
    – DWin
    May 21, 2016 at 0:29
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For a Poisson process, hits occur at random independent of the past, but with a known long term average rate $\lambda$ of hits per time unit. The Poisson distribution would let us find the probability of getting some particular number of hits.

Now, instead of looking at the number of hits, we look at the random variable $L$ (for Lifetime), the time you have to wait for the first hit.

The probability that the waiting time is more than a given time value is $P(L \gt t) = P(\text{no hits in time t})=\frac{\Lambda^0e^{-\Lambda}}{0!}=e^{-\lambda t}$ (by the Poisson distribution, where $\Lambda = \lambda t$).

$P(L \le t) = 1 - e^{-\lambda t}$ (the cumulative distribution function). We can get the density function by taking the derivative of this:

$$f(t) = \begin{cases} \lambda e^{-\lambda t} & \mbox{for } t \ge 0 \\ 0 & \mbox{for } t \lt 0 \end{cases}$$

Any random variable that has a density function like this is said to be exponentially distributed.

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    $\begingroup$ I enjoyed the $P(L>t)=P$ (no hits in time t) explanation. This made sense for me. $\endgroup$ Feb 12, 2014 at 8:02
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    $\begingroup$ Another point, 1 unit time has $\lambda$ hits, so $t$ units time have $\lambda t$ hits. $\endgroup$
    – Belter
    Sep 22, 2017 at 5:32
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The other answers do a good job of explaining the math. I think it helps to consider a physical example. When I think about a Poisson process, I always come back to the idea of cars passing on a road. Lambda is the average number of cars that pass per unit of time, let's say 60/hour (lambda = 60). We know, however, that the actual number will vary - some days more, some days less. The Poisson Distribution allows us to model this variability.

Now, an average of 60 cars per hour equates to an average of 1 car passing by each minute. Again though, we know there's going to be variability in the amount of time between arrivals: Sometimes more than 1 minute; other times less. The Exponential Distribution allows us to model this variability.

All that being said, cars passing by on a road won't always follow a Poisson Process. If there's a traffic signal just around the corner, for example, arrivals are going to be bunched up instead of steady. On an open highway, a slow tractor-trailer may hold up a long line of cars, again causing bunching. In these cases, the Poisson Distribution may still work okay for longer time periods, but the exponential will fail badly in modeling arrival times.

Note also that there is huge variability based on time of day: busier during commuting times; much slower at 3am. Make sure that your lambda is reflective of the specific time period you are considering.

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The Poisson Distribution is normally derived from the Binomial Distribution (both discrete). This you'll find on Wiki.

However, the Poisson distribution (discrete) can also be derived from the Exponential Distribution (continuous).

I've added the proof to Wiki (link below):

https://en.wikipedia.org/wiki/Talk:Poisson_distribution/Archive_1#Derivation_of_the_Poisson_Distribution_from_the_Exponential_Distribution

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  • $\begingroup$ the connection between discrete and continuous was not obvious, thanks for this! $\endgroup$
    – jspacek
    Dec 18, 2018 at 1:23
  • $\begingroup$ I'm not convinced by that wikipedia solution. In particular, higher order calculations include limits on integrals containing 1-x-y terms, which I do not (at least at the present time) understand. Further, the authors p(0;lambda) term does not seem to give the same answer if the integral used here is substituted for 1-int where int is another integral with limits between [0,1] rather than [1,+inf]. I've been working on this for about a week and not made much progress. $\endgroup$ Apr 30, 2020 at 20:57
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While the other answers here go into more explanatory detail, I am going to give you a simple summary of the equation relating a set of IID exponential random variables and a generated Poisson random variable. A Poisson random variable with parameter $\lambda > 0$ can be generated by counting the number of sequential events occurring in time $\lambda/\eta$ where the times between the events are independent exponential random variables with rate $\eta$. (Setting $\eta=1$ gives you a simple way to generate a Poisson random variable from a series of IID unit exponential random variables.)

This means that if $E_1,E_2,E_3,... \sim \text{Exp}(\eta)$ with rate parameter $\eta>0$, and $K \sim \text{Pois}(\lambda)$ with rate parameter $\lambda>0$ then you have:

$$\mathbb{P}(K \geqslant k) = \mathbb{P} \Big( E_1+\cdots+E_k \leqslant \frac{\lambda}{\eta} \Big).$$

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