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A number (n) Cruise ship passengers disembark at a port of call. n people will arrive on 5400 second interval prior to the ship's departure. then the average inter-arrival time is 5400/n = 54 seconds. Assume that all passengers are anti-social singles and that they do not return in groups. At most 1 passenger is arriving per second.

Questions: Why would the inter-arrival times of passengers be modeled with a Poisson distribution? What are the limitations, issues or assumptions of such a model?

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  • $\begingroup$ At stats.stackexchange.com/a/215253/919 I carefully discuss all these points. $\endgroup$ – whuber Sep 1 '19 at 17:31
  • $\begingroup$ @whuber Since we are told that all $n$ passengers returned to the ship in the given 5400-second period, then, assuming that the arrivals can be modeled as a Poisson process, don't the actual arrival times correspond to the order statistics of a sample of size $n$ from a $U(0,5400)$ distribution? $\endgroup$ – Dilip Sarwate Sep 2 '19 at 14:19
  • $\begingroup$ @Dilip Yes, if you add the implied assumptions. The most important (and likely contrafactual) assumption is that the arrival rate is constant throughout the interval. One explicit assumption bothers me a little, too: one might expect the arrivals of "anti-social singles" to be spaced more uniformly than random (Poisson) arrivals. Indeed, with truly Poisson arrivals there's a $1-\exp(-1/54)\approx 2\%$ chance that the smallest inter-arrival time is less than one second. $\endgroup$ – whuber Sep 2 '19 at 15:03
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We begin with a discussion of connections between Poisson and exponential distributions.

Suppose that $N_t$ random events occur in the time interval $(0,t],$ where $t>0,$ and that $N_t \sim \mathsf{Pois}(\lambda t).$ The probability of seeing none of these Poisson events in $(0,t]$ is $P(N_t = 0) = e^{-\lambda t}.$

Then let $X$ be the waiting time, starting at $t = 0,$ until we see the first event.

$$P(X > t) = 1 - F_X(t) = P(N_t = 0) = e^{-\lambda t},$$

for $t > 0.$ The density function $f_X$ of the CDF $F_X$ is found by differentiation, so that, for $t > 0,$

$$f_X(t) = F_X^\prime(t) = \lambda e^{-\lambda t}.$$

Thus $X$ has an exponential distribution with rate $\lambda.$ That is, $X \sim \mathsf{Exp}(\text{rate} = \lambda).$ One can show, using integration by parts, that $E(X) = SD(X) = 1/\lambda.$

One could run this argument backwards to show that exponential times between passenger arrivals imply a Poisson distribution for the number of passengers arriving in a given time interval. (See @whuber's link, just now posted, for additional detail.)

The exponential distribution is commonly used to model waiting times---perhaps sometimes even when its use does not really match reality. Temptations to use the exponential model are (a) its mathematical simplicity and (b) its no-memory property:

$$P(X > s+t|X > s) = \frac{P(X > s+t, X > s)}{P(X > s)} = \frac{P(X > s+t)}{P(X > s)}\\ = \frac{\exp[-\lambda(s+t)]}{\exp[-\lambda s]} = e^{-\lambda t} = P(X > t).$$

Thus, in order to use an exponential model, we need not take into account how long a passenger has already been waiting for 'arrival' in order to find that passenger's additional waiting time to arrival. Also, as you say, there is a Poisson assumption that customers arrive independently of ona another.

In any one application of Poisson-exponential modeling, it is unlikely that all of the independence and no-memory assumptions are exactly true. However, the exponential distribution has been widely and often successfully used to model the behavior of real-life waiting time and queueing ('waiting line') situations.

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  • $\begingroup$ All this is fine but doesn't take into account the fact that all $n$ passengers do in fact return within the fixed time interval $(0,5400]$. So, all the analysis must be conditioned on the event that $N(5400)$, the number of arrivals in the interval $(0,5400]$, equals the fixed number $n$. Your model would allow for more passengers to return than the number disembarking. $\endgroup$ – Dilip Sarwate Sep 2 '19 at 14:24

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