# Relationship between poisson and exponential distribution

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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A Poisson distrbution doesn't have waiting times. Those are a property of a Poisson process. – Glen_b May 21 at 1:30

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 simplifies to:

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

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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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. – user862 Aug 25 '10 at 18:03
Huh? is $t$ a moment in time or a period of time? – CodyBugstein Mar 8 '15 at 20:34
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! – Jan Nov 20 '15 at 6:47
@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. – Chiel ten Brinke Apr 22 at 8:19
@user862: It's exactly analogous to the relationship between frequency and wavelength. Longer wavelength; lower frequency analogous to: longer waiting time; lower expected arrivals. – DWin May 21 at 0:29

For a Poisson process, hits occur at random independent of the past, but with a known long term average rate $\lambda$ of hits per unit time. 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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I enjoyed the $P(L>t)=P$ (no hits in time t) explanation. This made sense for me. – user1603548 Feb 12 '14 at 8:02
The f(x) should be f(t). – bsbk Sep 30 '15 at 22:19

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).