# The Inspection Paradox and related topics

In the final class for Applied Probability, we discussed the so-called Inspection Paradox for an arrivals process. We assume that buses, sat, arrive as a Poisson process with rate 1, and consider the size of the interval (between buses) containing some fixed time T. The ‘paradox’ is that the size of this interval is larger in expectation than the average time between buses, which of course is given by an exponential random variable.

As with many paradoxes, it isn’t really that surprising after all. Perhaps what is more surprising is that the difference between the expected observed interval and the expected actual interval time is quite small here. There are several points of interest:

1) The Markov property of the Poisson process is key. In particular, this says that the expectation (and indeed the distribution) of the waiting time for a given customer arriving at T is not dependent on T, even if T is a random variable (or rather, a class of random variables, the stopping times). So certainly the inspection paradox property will hold whenever the process has the Markov property, because the inspected interval contains the inspected waiting time, which is equal in distribution to any fixed interval.

2) Everything gets slightly complicated by the fact that the Poisson process is defined to begin at 0. In particular, it is not reversible. Under the infinitesimal (or even the independent Poisson increments) definition, we can view the Poisson process not as a random non-decreasing function of time, but rather as a random collection of points on the positive reals. With this setup, it is clearly no problem to define instead a random collection of points on all the reals. [If we consider this instead as a random collection of point masses, then this is one of the simplest examples of a random measure, but that’s not hugely relevant here.]

We don’t need to worry too much about what value the Poisson process takes at any given time if we are only looking at increments, but if it makes you more comfortable, you can still safely assume that it is 0 at time 0. Crucially, the construction IS now reversible. The number of points in the interval [s,t] has distribution parameterised by t-s, so we it doesn’t matter which direction we are moving in down the real line. In this case, A_t, the time since the previous arrival, and E_t, the waiting time until the next arrival, are both Exp(1) RVs, as the memorylessness property applies in each time direction.

For the original Poisson process, we actually have A_t stochastically dominated by an Exp(1) distribution, because it is conditioned to be less than or equal to t. So in this case, the expected interval time is some complicated function of t, lying strictly between 1 and 2. In our process extended to the whole real line, the expected interval time is exactly 2.

This idea of extending onto the whole real line explains why we mainly consider delayed renewal processes rather than just normal renewal processes. The condition that we start a holding time at 0 is often not general enough, particularly when the holding times are not exponential and so the process is not memoryless.

3) There is a general size-biasing principle in action here. Roughly speaking, we are more likely to arrive in a large interval than in a small interval. The scaling required is proportional to the length of the interval. Given a density function f(x) of X, we define the size-biased density function to be xf(x). We need to normalise to give a probability distribution, and dividing by the expectation EX is precisely what is needed. Failure to account for when an observation should have the underlying distribution or the size-biased distribution is a common cause of supposed paradoxes. A common example is ‘on average my friends have more friends than I do’. Obviously, various assumption on me and my friends, and how we are drawn from the set of people (and the distribution of number of friends) is required that might not necessarily be entirely accurate in all situations.

In the Poisson process example above, the holding times have density function $e^{-x}$, so the size-biased density function if $xe^{-x}$. This corresponds to a $\Gamma(2,1)$ distribution, which may be written as the sum of two independent Exp(1) RVs as suggested above.

4) A further paradox mentioned on the sheet is the waiting time paradox. This says that the expected waiting time is longer if you arrive at a random time than if you just miss a bus. This is not too surprising: consider at least the stereotypical complaint about buses in this country arriving in threes, at least roughly. Then if you just miss a bus, there is a 2/3 chance that another will be turning up essentially immediately. On the sheet, we showed that the $\Gamma(\alpha,1)$ distribution has this property also, provided $\alpha<1$.

We can probably do slightly better than this. The memoryless property of the exponential distribution says that:

$\mathbb{P}(Z>t+s|Z>t)=\mathbb{P}(Z>s).$

In general, for the sort of holding times we might see at a bus stop, we might expect it to be the case that if we have waited a long time already, then we are less likely relatively to have to wait a long time more, that is:

$\mathbb{P}(Z>t+s|Z>t)\leq\mathbb{P}(Z>s),$

and furthermore this will be strict if neither s nor t is 0. I see no reason not to make up a definition, and call this property supermemorylessness. However, for the subclass of Gamma distributions described above, we have the opposite property:

$\mathbb{P}(Z>t+s|Z>t)\geq\mathbb{P}(Z>s).$

Accordingly, let’s call this submemorylessness. If this is strict, then it says that we are more likely to have to wait a long time if we have already been waiting a long time. This seems contrary to most real-life distributions, but it certainly explains the paradox. If we arrive at a random time, then the appropriate holding time has been ‘waiting’ for a while, so is more likely to require a longer observed wait than if I had arrived as the previous bus departed.

In conclusion, before you think of something as a paradox, think about whether the random variables being compared are being generated in the same way, in particular whether one is size-biased, and whether there are effects due to non-memorylessness.

# The Poisson Process – Distributions of Holding Times

So, I was planning to conclude my lecture course on Markov Chains at the Cambridge-Linyi summer school with an hour devoted to the Poisson Process. Unfortunately, working through the discrete time material and examples took longer than I’d expected, so we never got round to it. As I’d put a fair bit of work into the preparation, I figured I might as well write it up here instead.

We need a plausible mathematical model for a situation where people or events arrive randomly in continuous time, in a manner that is as close as possible to the notion of iid random variables. In discrete time, a natural candidate would be a set of iid Bernoulli random variables. For example, with probability p a single bus will arrive in an time interval of a minute. With probability 1-p, no bus will arrive. We might have some logistical motivation for why it is not possible that two or more arrive in a given interval, or we could instead choose a more complicated distribution.

One way to proceed would be to specify the distribution of the times between arrivals. These should be independent and identically distributed, at least intuitively. However, although we might be able to give a sensible guess right now, it is not immediately clear what this distribution should be. For now, we merely remark that the arrival times are called $X_1,X_2,\ldots$, and the holding times between arrivals are defined by
$S_1=X_1, S_n=X_n-X_{n-1},n\geq 2$.

In fact the motivating discrete example gives us much of the machinery we will actually need. Recall that when we define probability distributions for continuously-valued random variables we need a different plan of attack than for discrete RVs. Whereas for the discrete case, it is enough to specify the probability of each outcome, for a continuous random variable, we have to specify the probabilities of intervals, and take care that they have the obvious additive and nesting properties that we want. Taking the integral (whether Riemannian or Lebesgue) of a so-called density function is a natural way to do this.

Similarly here, we build up from small time intervals. The first remark is this: it is natural that the expected number of arrivals in the first minute is equal to the expected number of arrivals in the second minute. After all, we are considering the most general process possible. If there are an infinite number of potential arriving agents, then behaviour in the first minute should be independent and equal (in distribution) to behaviour in the second minute. We can naturally extend this idea to a linear relation. If $N_s$ is the number of arrivals in the time [0,s], then we should have $\mathbb{E}N_s=\lambda s$, where $\lambda$ is some constant of proportionality, equal to $\mathbb{E}N_1$.

The key remark is that as $s\rightarrow 0$, $\mathbb{P}(N_s=1)$ becomes small, and $\mathbb{P}(N_s\geq 2)$ becomes very small. In fact it suffices that $\mathbb{P}(N_s \geq 2)=o(s)$, as this implies:

$\mathbb{P}(N_s=0)=1-\lambda s+o(s),\quad \mathbb{P}(N_s=1)=\lambda s+o(s).$

Note that we are not currently attempting a formal construction of the process. As always, finding a probability space with enough freedom to equip a continuous process is a fiddly task. We are for now just trying to work out what the distribution of the holding times between arrivals should be. There are obvious advantages to defining the process as a collection of iid random variables, for example that we can construct it on a product space.

To do this, we split the time interval [0,1] into blocks

$[0,1]=[0,\frac{1}{n}]\cup[\frac{1}{n},\frac{2}{n}]\cup\ldots\cup[\frac{n-1}{n},1]$

So the probability that someone arrives in the time $[\frac{k}{n},\frac{k+1}{n}]$ is $\frac{\lambda}{n}$. So

$\mathbb{P}(\text{no-one arrives in time }[0,1])=(1-\frac{\lambda}{n})^n\approx e^{-\lambda}.$

As we are working in an $n\rightarrow\infty$ regime, we can replace 1 by general time t, to obtain:

$\mathbb{P}(\text{no-one arrives in time }[0,t])=(1-\frac{\lambda t}{n})^n\approx e^{-\lambda t}.$

So the distribution function of the first arrival time is $F(t)=1-e^{-\lambda t}$ in the conventional notation.
Thus $X_1\sim \text{Exp}(\lambda)$.

However, to emphasis how useful the infinitissimal definition is for actual problems, consider these examples.

1) If we have two arrivals processes at the same object, for example arriving from the left and the right at the same shop, say with rates $\lambda,\mu$, then we want to show that the first arrival time is still exponential. Because of the linearity of expectation property, it is clearly from the definition that the total arrivals process is Poisson with rate $\lambda+\mu$, and so the result follows. Showing this by examining the joint distribution of two exponential random variables is also possible, but much less elegant.

2) Similarly, if we have two shops, A and B, and each arriving person chooses one at random, then the first arrival time at A is: with probability 1/2 distributed as Exp($\lambda$), with probability 1/4 as $\Gamma(2,\lambda)$, and so on. A fairly non-trivial calculation is required to show that this is the same as $Exp(\frac{1}{2}\lambda)$, whereas this follows almost instantly using the the infinitissimal definition.

Moral: with the infinitissimal case, the difficulties are all in the probability space. However, once we have settled those problems, everything else is nice as the key property is linear. Whereas for a construction by iid jump times, the existence and well-definedness is clear, but even for a distribution as tractable as the exponential random variable, manipulation can be tricky.