Clarification on the Poisson distribution Given that an event occurs at some rate $\lambda$ per unit time, I know that the probability that n such events occur in unit time is given by the Poisson distribution.
$P(n) = e^{-\lambda}\frac{\lambda^n}{n!}$
However, let's say I measure the time intervals at which the next event occurs. I obtain a histogram since the events happen randomly and this is a PDF because each event occurs at some time between $t=0$ and $t=\infty$. What distribution does this follow? I thought earlier, that it was 
$P(t) = \lambda t e^{-\lambda t}$ 
but this is incorrect. Integrating this from $t=0$ to $t=\infty$ yields $\frac{1}{\lambda}$. I would guess then that
$P(t) = \lambda^2 t e^{-\lambda t}$ but the expectation value of this is $\frac{2}{\lambda}$, which doesn't make sense. If events occur at a rate $\lambda$, the expected time of each event should be $\frac{1}{\lambda}$. 
Any pointers?
 A: The time $T$ until the next event has continuous distribution, with density function $f_T(t)=\lambda e^{-\lambda t}$ for $t\gt 0$, and $f_T(t)=0$ elsewhere.
The distribution of $T$ is called the exponential distribution with parameter $\lambda$. 
The expectation $E(T)$ (mean) of $T$ is given by $E(T)=\int_0^\infty t\lambda e^{-\lambda t}\,dt$. Integration shows this is $\dfrac{1}{\lambda}$, exactly as you expected.
Added: The following is an informal justification of the result. If the number of events in unit time has Poisson distribution with parameter $\lambda$, then the number of events in time $\lambda t$ has Poisson distribution with parameter $\lambda t$. So the probability the number of events is $0$ is $e^{-\lambda t}\frac{(\lambda t)^0}{0!}$, that is, $e^{-\lambda t}$.
Thus the probability that the waiting time $T$ is $\gt t$ is $e^{-\lambda t}$. Thus the cdf $F_T(t)$, that is, the probability that $T\le t$, is given by $1-e^{-\lambda t}$, for positive $t$.
Differentiation now gives us the density function $f_T(t)$ described above. 
