# If arrivals are not occuring at random, why do arrival times still have a probability distribution?

If the arrival times are, say, uniformly distributed U(0,10), I don't quite understand why the arrivals in this case are said to be not occurring at random (this claim was implied in my textbook chapter on queues), since they aren't deterministic either. Sure, the next arrival is certain to occur within 10 mins, and the arrival time is not memoryless, but it isn't deterministic either...

Does all this mean that a variate can be neither random nor deterministic? I'm not sure whether my comprehension issue is with the logic or with the terminology.

• Why do you say "This much is clear: if arrivals occur at random, then the process is Poisson"? For that to be the case, you are using a particular meaning of random which I would guess includes memorylessness – Henry Oct 12 '13 at 10:24
• @Henry Oh yes thank you for pointing out my misstatement; it has helped me realise that unthinkingly quoting this sentence from my textbook was the root of my confusion leading to this posted question. I've edited the sentence out. – Ryan G Oct 12 '13 at 11:12

It's just a terminology issue. The word "random" is vague and its use should in general be avoided. For example, for a lot of people "a random situation" means "all events are equally likely" -these people would have a hard time understanding why "random arrivals" follow the Poisson and not the Uniform. Sometimes the word "random" is used as a "verbal substitute" for the word "independence".

In most cases, the Poisson distribution models essentially the allocation of probabilities over the possible values of a sum of independent indicator functions, (and this is why it is so closely related to the binomial distribution). To obtain the Poisson, we need to make two assumptions related to stochastic independence:

a) That each indicator function (will I call the call center or not in the current minute?) is independent of all the others (whether you call does not affect the probability that I will) (So for example, when the service is down, you can see that calls to call center will stop being independent from each other, since calls will have a common source)

and

b) Each indicator function, viewed as a stochastic process, is independent of its own past: whether or not I called the call center one minute ago, does not affect the probability that I will call in the current minute. (for this to be realistic one should decide on the length of the interval juidiciously -which is part of the applied art of working with Poisson: it is stupid to say that if I called the last minute does not affect the probability that I will call in the current minute, but perhaps, if you set your time period as "day" or "week" the Poisson becomes acceptable).

It is these two assumptions related to independence, that people express by the use of word "random" in this case, which is very confusing indeed.

Now, what would it mean to say that "arrivals follow the uniform"? It would mean that it is equally probable that we will observe, say, 1 arrival or 1.000 arrivals. Or that it is equally probable that out of the $N$ indicator functions, only one will take the value unity, or that 1.000 of them will take the value unity. Or any other value from 1 to $N$, for that matter.
So what kind of random variables are these, that their sum follows the uniform?

• Oh thank you so so much. " Sometimes the word "random" is used as a "verbal substitute" for the word "independence"." is exactly the answer that I needed, and such answers/discussions can hardly be found in published literature (or on the internet). Your first point bears highlighting too; I've always thought that every first lesson in a probability course should emphasise and explain that the word "random" in mathematics has a different meaning from "radom" in layman usage, so that students don't confuse themselves down the line. – Ryan G Oct 12 '13 at 11:14
• You're welcome Ryan. Glad I helped to clear the fog a bit. – Alecos Papadopoulos Oct 12 '13 at 11:24

With a different distribution of arrival times (that is: the difference between the $n$th arrival and the $(n+1)$st arrival), the arrivals themselves are still random. However, such a distribution could occur only if the arrivals themselves are not independant. For example, if the seventh arrival is at 12:00, the next arrival cannot be after 12:10. This is a valid diestribution for a valid random process, but it does not seem like what we want to model with it.

We may consider this as a limit as follows: To model an average of one arrival per minute, first consider $10$ arrivals uniformly distributed on a $10$ minute interval and investigate how the arrival times (again: Those are the differences between arrivals in sorted order) are distributed as a*consequence*. Next consider $100$ arrivals in a $100$ minute interval. Note that the distribution changes because it is possible that not even a single arrival occurs in the first 10 minuts - or all $100$ arrivals could occur there. Finally investigate the limit for $n$ arrivals in $n$ minutes as $n$ tends to infinity (unfortunately we cannot consider a uniform distribution on an infinite interval directly). This leads to a Poisson distribution for the arrival times.

• You say: "With a different distribution of arrival times, the arrivals themselves are still random." But doesn't this contradict my claim (actually, the claim I'm citing from my textbook) "the arrivals in this case are said to be not occurring at random"? In other words, is it accurate/correct to say that if the inter-arrival time has a uniform distribution, then arrivals aren't occurring at random? Shouldn't the author have instead concluded that the arrivals aren't occurring independently (or at a constant rate) instead of concluding that the arrivals are not occurring randomly? – Ryan G Oct 12 '13 at 10:22