# Can the Poisson Distribution be used to find the expected value of time of arrival given an expected arrivals per unit time?

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?

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.
• 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... Nov 2, 2013 at 5:13
• 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) Nov 2, 2013 at 5:23
• (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. Nov 2, 2013 at 5:26