time between transitions in continous time discrete state Markov process Problem Statement:
I want to compute the time between transitions in a birth-death model. 
As a simple example, consider that individuals are born with rate $\lambda$ and they die at rate $\sigma n$. The master equation for the number $n$ of individuals is 
$$ \frac{d}{dt}P_n(t) = \lambda P_{n-1}(t) + \sigma (n+1) P_{n+1}(t) - (\lambda + \sigma n)P_n(t). $$
In steady state, the solution is a Poisson distribution: 
$$ P_n = \frac{(\lambda/\sigma)^n}{n!}e^{-\lambda/\sigma}.$$
My target is the time between successive births, deaths, or transitions of any kind. How can I compute this? 
Attempt:
First I'll try to get the distribution of waiting times between successive births. 
Consider that a birth has just occurred at $t=0$. 
Then defining $F_n(t)$ as the probability that there are $n$ individuals and no birth has occurred since $t=0$, one can write the master equation 
$$F_n(t+\delta t) =\sigma (n+1)\delta t F_{n+1}(t) + [1-(\lambda + \sigma n)\delta t]F_n(t)$$
The first term represents death occurring in $\delta t$. The second term represents nothing occurring in $\delta t$. 
As $\delta t \rightarrow 0$, 
$$\frac{d}{dt}F_n(t) = \sigma(n+1)F_{n+1}(t)-(\lambda + \sigma n)F_n(t)$$
This should be the master equation for the probability distribution I'm looking for. 
Introducing a generating function $H(z,t) = \sum_{n=0}^\infty z^n F_n(t)$ leads to the differential equation 
$$ \frac{\partial H}{\partial t} + \sigma (z-1)\frac{\partial H}{\partial z} = -\lambda H.$$
Using the method of characteristics implies the solution 
$$ H(z,t) = [1-e^{\sigma t}(1-z)]e^{-\frac{\lambda}{\sigma}e^{\sigma t}(1-z)-\lambda t}.$$
Crucially, the initial condition is $H(z,0)=\sum_{n=0}^{\infty} z^n F_n(0) = \kappa \lambda \sum z^n P_{n-1} = \kappa \lambda z e^{\lambda(z-1)\sigma}. $ $\kappa$ is a normalization factor. This point I don't fully understand. 
The initial condition is intended to indicate a birth transition just occurred at $t=0$. 
Incorporating the normalization $H(1,0)=1$ sets $\kappa = 1/\lambda, $ so $H(z,0) = z e^{-\lambda(z-1)/\sigma}. $
The probability that no birth occurs from any state in time $t$ is then the sum over all possible states $n$: $\sum_{n=0}^\infty F_n$. The pdf is $f_T(t) = -\frac{d}{dt} \sum_{n=0}^\infty F_n$. Summing every difference equation for $F_n$ gives this quantity as 
$$f_T(t) = \lambda \sum_{n=0}^\infty F_n(t).$$
Using the generating function $H(z,t)$, this can be written as 
$$ f_T(t) = \lambda \sum_{n} z^n F_n(t) |_{z=1} = \lambda H(1,t) = \lambda e^{-\lambda t}.  $$
So there's the distribution of time between births. It's the same as the distribution of time for one birth, given there were initially no individuals at $t=0$. 
 A: You are describing an M/M/$\infty$ queue with arrival rate $\lambda$ and server rate $\sigma$.
The arrival process is Poisson with rate $\lambda$, so inter-arrival times are iid exponential with rate $\lambda$.
The system is reversible, so if we start out in the steady state distribution $P[N(0)=k] =\frac{(\lambda/\sigma)^k}{k!}e^{-\lambda/\sigma}$ for all nonnegative integers $k$ then the departure process is Poisson with rate $\lambda$, so inter-departure times are iid exponential with rate $\lambda$.

Edit: For any stable queue we must have the long term arrival rate equals the long term departure rate:
$$ \lim_{t\rightarrow\infty}\frac{arrive[0,t]}{t}=\lim_{t\rightarrow\infty}\frac{depart[0,t]}{t}$$
where $arrive[0,t]$ is the total number of arrivals during the interval $[0,t]$, and $depart[0,t]$ is the total number of departures during that time. Since the long term arrival rate is $\lambda$, the long term departure rate is also $\lambda$. Indeed, if not, then the queue would either be creating new jobs, or eating existing jobs.
The fact that inter-departure times are (in steady state) iid exponential is not obvious and comes from the detail equation reversible properties (which hold for every birth-death chain that is stable). It is not true if we do not start in steady state.  Assuming initial steady state, your calculations could verify the marginal inter-departure time is exponential. However (assuming initial steady state) it further holds that each inter-departure time is independent of all prior ones.
The inter-event time is more complicated. In fact, even if we start in steady state, the first event is more likely to be an arrival, rather than a departure:
$$  \sum_{k=0}^{\infty} \left(\frac{\lambda}{\lambda + k\sigma}\right) \frac{(\lambda/\sigma)^k}{k!}e^{-\lambda/\sigma} > 1/2$$
The inter-event times are not identically distributed. An arrival sees the system in steady state (and then adds 1 to that) while a departure leaves the system in steady state.  So the expected time to the next event, given we just had an arrival, is less than the expected time to the next event, given we just had a departure. 
The fact that Poisson Arrivals See Time Averages is often called PASTA.
