Imagine that we
have the owner of a pizzeria, let's call him Bertrand, who wants to break with
tradition. For centuries the pizzeria
that he is the current owner of has made round pizzas (what Americans call
"pizza pies" - thus allowing some sense to be made of Dean Martin's
"That's Amore":

*When the moon hits your eye, like a big pizza pie*- which always sounded to me like*When the moon hits your eye, like a big piece o' pie*… who throws around bits of pie, let alone big bits?)
But Bertrand is now
heartily sick of circles and he wants to make the transition to rectangular
pizzas. However, he has a minor
problem. His customer base is accustomed
to a pizza base based pricing scheme - and they they don't want their pizzas to
shrink (or grow) as a consequence of this shape change. Bertrand already has a range of boxes which
fit his circular pizzas perfectly, so he knows how

**his rectangular pizzas will be … all he needs to do is work out how***long***they have to be to keep his loyal customers happy.***wide*
Here's a graphic to
illustrate his conundrum:

This is reasonably
easy to work out. The area of the
rectangle is 2R times the width (w) while the area of the circle is πR

^{2}, so we make those areas equal:
w.2R = πR

^{2}
w = πR/2

Thus we could say
that the "average width" of a circle of length 2R (which is true of
all circles of radius R) is πR/2. All Bertrand needs to do is plug his values of
R (10cm (bambino), 15cm (piccolo), 20cm (medio), 40cm (grandi), 60cm
(ridicolo)) and

*Roberto's his uncle*.
But let's say that
Bertrand did not have a mathematician handy and he was casting around for
another way to work out the area of his pizza in rectangular form. How could he do it?

One way would be to
use a form of integration. Being a very
precise person, and skilful in the ways of pizza, Bertrand could slice his
pizza up into 1mm wide slivers, use those slivers to reassemble the pizza in
rectangular form and then measure the resultant rectangle. This is equivalent to how we find the area
under a line (

*using Reimann sums*). By arranging the slivers in a 2R.w rectangle, Bertrand is effectively "adding them up".
Essentially, if not
practically, Bertrand could do this with infinitesimally narrow slivers and
doing so would only make his result more accurate (

*-*see*Wikipedia's article on Reimann sums*which has animations that show something similar to my arbitrarily large value of N approaching infinity). The infinitesimally narrow slivers would be equivalent to chords and as a consequence, the "average" length (or even*width*) of these chords would be πR/2 - where, by "average" I mean "mean", and this "average" would be the same as the "average" (mean) width of a circle of radius R … but not the "*mean width*" which means something else.
---

I have to go into
an aside here. Or rather two asides. Or maybe three (and three asides do not atriangle
make … oops that'd be four asides now).

The existence of
the mathematical term "

*mean width*" provides us with an example of how English could, at least occasionally, benefit from more concatenation. Such concatenation would allow us to clearly distinguish between mean width in a general sense and what would become meanwidth … in much the same way as we can distinguish between Donald Trump's*wetback*(I think it's the second from the left) and Donald Trump's wet back (fortunately, no image was available).
Even so, I've
effectively used the concept of "

*mean width*" without referring to it. The length of Bertrand's new rectangular pizzas is the*mean width*of a circle of radius R. Because I am not a professional mathematician, just a person who uses the more useful aspects of mathematics on a daily basis, I tend to think of three dimensional objects having three features: length, width (or breadth) and height. Height relates to the object's orientation with respect to gravity, width can be used for both the other dimensions under many circumstances (think of a tower with a rectangular base, we know how high it is, but it seems wrong to think of its longer base as defining its "length"). However, for things which are not particularly high (like rectangular pizzas), it certainly feels like there is a convention such that the longer side gives the length and the shorter side gives the width.
I'm not saying that
people who think differently are necessarily wrong, but I would surely be
forgiven for thinking of them as being

*as thick as two short planks*.
Another little
aside, I tend to over complicate this allusion by thinking that someone who is talking
about short planks doesn't know how to use planks properly, say we have a 10
foot 1x6

*plank*. To me that is a plank that is 10 foot long, 6 inches wide and 1 inch thick. We could make these "short" planks by considering the 1 inch to be its length, but such a plank is at least 6 inches thick. That's thick for a plank, right, and so is the guy who thinks you can*legitimately*think of a 10 foot 1x6 plank as being 1 inch long …
I would be tempted
to challenge such a person with a tale about a chicken who crossed a road, on
one side of which was London while Dover was on the other side - so the obvious
answer to the ancient riddle would thus be "to go on holidays". It just happens to be a fact that, in length,
the A2 is ridiculously short (perhaps even less than 10m in places), but it
makes up for this by virtue of its incredible width (in the order of 115,000
metres).

Then I'd snort
contemptuously and go back to arranging my pencils.

Anyway … I did use
the concept of

*mean width*, but I used it as if it were "mean length". Oops.
---

Bertrand the shape-shifting
pizzeria owner has now been able to find out the necessary width of his new
rectangular pizzas, so let's leave him behind now and look again the other
Bertrand and his problem with chords.

It has been argued
that the problem arises from the fact that there is no single obviously natural
method to select (identify or get) chords and (either consequently or on the
basis that) there is no single obviously natural probability measure.

I suggest that this
mean chord length might be another

**if not exactly obvious (except perhaps in retrospect) was to arrive at a natural probability measure. That is, if you arrive at a mean chord length which is not equal to πR/2, then you have a problem.***useful*
More specifically,
I am suggesting that a method that arrives at a set of chords the mean length
of which is

**πR/2 then as a consequence, we have discovered that there may be something unnatural or skewed about that set, even if, prior to the discovery, the method appeared to be natural and unbiased.***not*
---

I did do some
modelling and proved (to my own satisfaction) that chords selected "at
random" using the 1/2 method have a length, on average, of πR/2.

Chords selected
using the 1/3 method have a length, on average, of 4R/π while chords selected using the 1/4 method have a length, on
average, of what appears to be πR/3.

Note that these averages were calculated on
the same basis - I generated a large number of random chords using each of the
methods and then obtained the arithmetic mean of the resultant chords.

Dear Neopolitan,

ReplyDeleteI think we are getting to the point where you don't even understand when I agree with you and when I don't. So let me be clear on the exact point in your post where our views diverge.

> I suggest that this mean chord length might be another useful if not exactly obvious (except perhaps in retrospect) was to arrive at a natural probability measure

Yes. I agree. It's a natural way to construct a probability measure on the set of chords. There is absolutely no doubt about it. Your result is correct. Every single statement you said before that point is true.

> That is, if you arrive at a mean chord length which is not equal to πR/2, then you have a problem.

No. I disagree. This is exactly the whole point of Bertrand paradox. There is no problem in finding a different mean chord length for a different probability measure.

> More specifically, I am suggesting that a method that arrives at a set of chords the mean length of which is not πR/2 then as a consequence, we have discovered that there may be something unnatural or skewed about that set, even if, prior to the discovery, the method appeared to be natural and unbiased.

First of all, I have to say it once again : it's the same set (of chords), it's only the probability measure that is different. It might seem like a minor nitpicking again, but I think this is the exact point that you don't understand, and that I'm trying to explain.

Now, let's talk math :

The average of a function on a set can be defined ONLY if you have a measure on that set. And in that case, the computation of that average is the integration with respect to this measure.

In the very usual context of finite sets, when no precision is given, it is implicit that the measure is chosen to be the counting measure. (so that it coincides with the everyday notion of average)

In the usual context of functions defined over R^n, when no precision is given, it is implicit that the measure is chosen to be the Lebesgue measure, which can be seen as a nice generalization of the counting measure.

On more general sets, these two choices are not available. And there is no canonical choice that would generalize the counting measure or the Lebesgue measure. (Note that a parametrization of a given set gives me an identification between the set and R^n and hence, the Lebesgue measure becomes available. However, the induced measure on the set will depend on this parametrization.)

Now, the length defines a function on the set of chords. It means that the term "mean length" does not only depend on the set but also on the measure that I put on this set. The set of chord is not a subset of R^n, hence the Lebesgue measure is not available.

So let me be even more concrete. If I give you an infinite collection of line segments, with no additional information, then there is absolutely no way to compute the "mean length" of these segments. No way. It simply does not make sense. Because you need a measure on this set if you want to define a "mean length" over an infinite set, and no measure is given. So actually, "mean length" does not even make sense if you are not given more informations.

But wait, how could you compute the average width of infinitesimally narrow slivers on a pizza without a measure ? Because without even realizing it, you chose a measure on the set ! It happened exactly when you said that you can compute the average by integration. Remember that integration only makes sense if I already have a measure. Here you implicitely used the Lebesgue measure on some parameter space [-R,R]. While this makes perfect sense in the world of pizza, there is no reason to do that in general.

I characterise our point of departure differently. I think our views diverse as soon as I consider using a probability measure for something in the real world, the world I live in, the world of pizza.

DeleteIf Bertrand the pizzeria owner had intuited that the width of his 2R pizza had to be the average length of the chords (which is true given the process he has in mind) and instructed his novice (a pizza padawan, we could say) to collect a representative sample of pizza slivers, one each from N pizzas, then there would be a "correct" way to collect these slivers and a "wrong" way to collect them, right?

So, my question to you is - is there a use, in the real world, the world of pizza, to which we could put the probability measures associated with the 1/3 and 1/4 methods? Note I understand that mathematics is everywhere, even in the real world, but there is "pure" maths and there is "applied" maths - I am not counting any "pure" maths applications here.

And is it not so that I am possibly guilty of no more than hitting on the something that already exists when you say "In the usual context of functions defined over R^n, when no precision is given, it is implicit that the measure is chosen to be the Lebesgue measure, which can be seen as a nice generalization of the counting measure."

So the complaints against me should be along the lines of "yes, it might be implicit that the Lebesgue measure, which results in 1/2 is the one to use, but it's not the only one available and we could use other measures, such as X which results in 1/3 and Y which results in 1/4 and these have utility when we want to do A and B, so your use of the word 'wrong' may be ill-advised". Right?

> then there would be a "correct" way to collect these slivers and a "wrong" way to collect them, right?

ReplyDeleteYes, as I said, in the world of pizza this makes perfect sense.

> So the complaints against me should be along the lines of : "yes, it might be implicit that the Lebesgue measure, which results in 1/2 is the one to use, but it's not the only one available and we could use other measures, such as X which results in 1/3 and Y which results in 1/4 and these have utility when we want to do A and B, so your use of the word 'wrong' may be ill-advised"

Sorry if I wasn't clear enough :

That is (almost) exactly what my complaints are !!

I'm saying "almost" because there is still a fundamental thing that you seem to be misunderstanding :

> "yes, it might be implicit that the Lebesgue measure, which results in 1/2 is the one to use

Here it is once again in three steps :

1 : There is no Lebesgue measure on the set of chords.

2 : There are several, equally useful, parametrizations of the set of chords.

3 : For each of these parametrizations, there is a Lebesgue measure on the set of parameters.

Your pizza example makes you think that the "(c,theta)-parametrization" is the good parametrization and that the other ones are skewed. And when it comes to pizza, you are right.

But what about Bertrand the bicycle wheel maker, who has to put spokes inside a wheel? Or Bertrand the sound engineer, who wants to see how sound distributes in a circular auditorium ? They probably should not use this parametrization ....

Okay, I do think we are getting somewhere. What I need to push me over the line is a scenario in which the 1/3 and 1/4 methods of identifying chords, directly as chords (not other features that can subsequently be turned into chords, such as the random placement of two spokes on a wheel, between the ends of which a chord could be drawn, or the sound density at random points within a circular auditorium, which could then used as midpoints of chords) - have utility. Hopefully you understand what I mean here, I used the chords directly in my pizza example. In your examples there aren't any (obvious) chords being used (except perhaps in the music being played in the auditorium, ha ha ha - I give myself a C for that pun, but a High-C).

DeleteThe integration being using in the pizza reshaping is natural and obvious (to me) and I can't see the

being obvious in either wheel making or auditorium sound checking.chords> What I need to push me over the line is a scenario in which the 1/3 and 1/4 methods of identifying chords, directly as chords have utility

DeleteOk, here is a natural and explicit example :

Bertrand the doctor lives on the Huvadhoo atoll, and for sake of clarity we will assume that it is a perfectly circular atoll. He gets emergency calls (randomly) during the day and has to take his personal boat to get from one place to the other. He wants to compute the mean length of his boat trips. (if you want to imagine a very real scenario ... he actually wants to know his average response time to the emergencies, because if it's greater than 1h on average, he should buy a faster boat)

Obviously, he will go in straight line segments with his boat, and hence will always follow a chord inside this atoll. So the previous question could be interpreted as the average length of a chord inside this atoll, right ?

If he computes the average length of his boat trips for the past 10 years, he will probably arrive at a result very close to (2/pi)*D where D is the diameter of the atoll. (Unsurprisingly, it coincides with the expected length of a random chord using the 1/3 method)

I hope this is natural enough for you.

You cut off the quotation too early. I wrote:

Delete> What I need to push me over the line is a scenario in which the 1/3

1/4 methods of identifying chords, directly as chords (and, between the ends of which a chord could be drawn, or the sound density at random points within a circular auditorium, which could then used as midpoints of chords) - have utility.not other features that can subsequently be turned into chords, such as the random placement of two spokes on a wheelWhat you have given me is the average shortest distance between two points on the circle (or circular atoll). I think we are back to where we were at this exchange:

> > The question as posed talks about choosing chords "at random", it does not talk about choosing end points or midpoints "at random"

> You got me there. But again, I'm not saying that it's wrong to ask for this property. What I'm saying is that this is a rather arbitrary choice for a property. I mean let's phrase it that way : why should a "uniform" probability measure on the set of chords satisfy this property and not some other very nice property?

And (your words, in the same comment): "a chord (which is the intersection of a line and a disc)".

I totally agree that the average shortest distance between (two randomly selected) points on the rim of a disc is going to be 4.R/π and the associated probability measure results in a likelihood that the shortest distance between any two (randomly selected) points is √3.R of 1/3. I even agree that this (the shortest distance between two points on the rim of a disc) gives rise to a chord, or maybe even

a chord. However, I think you have, once again, taken a circle/disc out of its context (ie sitting on a plane) - and placed it in a different context where we have to make a few more assumptions (such as ignoring the admittedly negligible curvature of the planet and assuming not only random timings of call outs but also random location of call outs).isPerhaps this is the true core of our disagreement. I am loathe to extricate the circle/disc from its context (note that on a chart, I could conceptually carve up the atoll to obtain the average length of chords through it, arriving at π.R/2 - a wrong answer for Bertrand M.D.) and you don't seem to mind doing it, and by permitting this, you can arrive comfortably at the 1/3 answer while hanging onto the idea that you are thinking chords rather than points on the rim.

What about the 1/4 answer?