Inter-arrival time distribution It is known that for a Poisson process the inter-arrival time is exponentially distributed. My question, which may be nonsense, is this. Suppose you want to experimentally evaluate the distribution of inter-arrival time (not necessarily of a Poisson process). You measure the differences in the arrival time of consecutive customers, for example. But the problem is that the inter-arrival time depends on the (absolute) time the proceeding customer arrived. If it arrived earlier, the current inter-arrival would be different. So, does it make sense to measure inter-arrival times and build the distribution with those samples?
 A: If you are looking at an analysis based on inter-arrival times then that is a sensible thing to look at.  
So for a frequent transport service, looking at inter-arrival times can tell you whether the timetable dominates the effect of congestion ("why buses travel around in threes") or not.  But in other cases it makes less sense, for example if there is a higher level of service at peak times and looking at actual times may be more informative.
On your question about the inter-arrival time depending on the time the proceeding customer arrived, that is part of the point.  If (looking from now) the previous customer arrived some time ago, then there was a chance that anorther customer could have arrived in the intervening time, but did not.  Whether this has an effect on distribution of the arrival time of the next customer, making it more likely or less likely the next customer is later, is an interesting object of analysis.
A: 
So, does it make sense to measure inter-arrival times and build the distribution with those samples?

In a sense, yes. A Poisson process can be thought of as something that gives you the number of events within a given "window" of time. So you may think of that window as starting the moment that the previous event occurred.
The way I prefer to look at Poisson processes is by using the more general Markov model:
$$ p(0,t+dt) = p(0,t)\left[1-f(0,t)dt\right] $$
$$ p(r,t+dt) = p(r,t)\left[1-f(r,t)dt\right]+p(r-1,t)f(r-1,t)dt,\ r = 1,2,3,\ldots $$
where $r$ represents the number of events, $p$ represents the probability, and $t$ represents time. (This representation is pulled from Statistical Analysis of Random Dispersion by A. Rogers).
In the above representation, $f(r,t)$ represents the transition rate function. If we take the limit of the above as $dt\to 0$, then we can obtain the probability generating function of the process. And if we set $f(r,t) = \lambda a$, where $\lambda a$ is constant, then we have obtained a standard, completely random Poisson process. Changing $f(r,t)=c+br$ allows us to generate binomial or negative binomial distributions (depending on the sign of $b$) quite easily.
All that is dandy, but what is relevant is that the probability $p$, for a stationary Poisson process (so $f(r,t) = f(r)$, i.e. the transition rate is not dependent on time), is only dependent on the time interval $dt$, and not on the absolute time $t$ at all. And in the case of a true Poisson distribution, the probability is only dependent on the constant intensity term $\lambda a$, and not on the number of prior events $r$ at all.
