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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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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.

Thus, it follows that:

$P(X_t < x) = 1 - P(X_t > x)$

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

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

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:

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

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
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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 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:

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

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 at 8:02
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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):

http://en.wikipedia.org/wiki/Talk:Poisson_distribution#Derivation_of_the_Poisson_Distribution_from_the_Exponential_Distribution

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