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My understanding of the Poisson Distribution is that its PMF $P(x=k) = \dfrac {\lambda^k e^{-\lambda}} {k!}$ refers to the probability of finding k events given an expected arrival expectancy $\lambda$. This gives me, rather trivially, that the expected value for the number of arrivals is equal to the average number of arrivals $\lambda$. However, suppose I know $\lambda$ is 3 events per day. How can I calculate the expected number of days before $n$ events happen? Can I just invert my $\lambda$, so that my units are now days/event, and use the same distribution?

A supplemental question: Currently, the units in my exponent appears to be events/time. Shouldn't I have to multiply by some time $t$, so that the distribution looks like $P(x=k) = \dfrac {(\lambda t)^k e^{-\lambda t}} {k!}$? (I'm taking "events" to be unitless...) If so, I would expect my new distribution to be $P(t=k) = \dfrac {(\frac n {\lambda})^{k} e^{ \frac {-n} {\lambda}}} {k!}$, where $n$ is the number of events, $k$ is the amount of time, and $\lambda$ is still in events/time. Thus if, in the above example, I want to know the probability that it would take 1 day for 5 arrivals, I would set $k$ = 1, $n$ = 5, and $\lambda$ = 3. Is there anything wrong with this formulation?

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If the number of events per unit time has Poisson distribution with parameter $\lambda$, then the waiting time for the first event has exponential distribution with parameter $\lambda$, and therefore expectation $\frac{1}{\lambda}$.

The waiting time until the $n$-th event is the sum of $n$ exponentials with parameter $\lambda$. It therefore has mean $\frac{n}{\lambda}$.

Added: The following may deal with your second question. Let the number of events per unit time have Poisson distribution with parameter $\lambda$. Then the number $Y$ of events in time $t$ has Poisson distribution with parameter $\lambda t$. This is a special property of the Poisson.

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  • $\begingroup$ Can you help me understand why it's exponential rather than also Poisson? Also, am I right in my understanding that "Poisson with parameter $\lambda$" is the same as saying that the expected number of events per unit time is $\lambda$? Then, is there a continuous version of the Poisson distribution? It seems to me that an exponential distribution is equivalent to a Poisson distribution with k = 1, but is continuous... $\endgroup$ – Teofrostus Nov 2 '13 at 5:13
  • $\begingroup$ Second question: Poisson with parameter $\lambda$ implies that the expected number of events per unit time is $\lambda$. The rest: The number of events in time $t$ is Poisson with parameter $\lambda t$, so the probability of $0$ events in time $t$ is $e^{-\lambda t}$. That says that the probability that the waiting time is $\ge t$ is $e^{-\lambda t}$. You recognize this as the probability that an exponential is $\ge t$. (More) $\endgroup$ – André Nicolas Nov 2 '13 at 5:23
  • $\begingroup$ (More) Proving that the number of events in time $t$ is Poisson takes a while. It is intuitively reasonable if we think of the Poisson as a binomial with $p$ very small, $n$ large, and $np=\lambda$. But the right way is to derive the Poisson from fundamental principles. Takes a while, done in most good courses. $\endgroup$ – André Nicolas Nov 2 '13 at 5:26

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