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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$\begingroup$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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