## Bertrand’s tartine

Posted in Books, Kids, pictures, Statistics with tags , , , , , , , , , , on November 25, 2022 by xi'an

A riddle from The Riddler on cutting a square (toast) into two parts and keeping at least 25% of the surface on each part while avoiding Bertrand’s paradox. By defining the random cut as generated by two uniform draws over the periphery of the square. Meaning that ¼ of the draws are on the same side, ½ on adjacent sides and again ¼ on opposite sides. Meaning one has to compute

P(UV>½)= ½(1-log(2))

and

P(½(U+V)∈(¼,¾))= ¾

Resulting in a probability of 0.2642 (checked by simulation)

## Bertrand-Borel debate

Posted in Books, Statistics with tags , , , , , , , , , , , , , on May 6, 2019 by xi'an

On her blog, Deborah Mayo briefly mentioned the Bertrand-Borel debate on the (in)feasibility of hypothesis testing, as reported [and translated] by Erich Lehmann. A first interesting feature is that both [starting with] B mathematicians discuss the probability of causes in the Bayesian spirit of Laplace. With Bertrand considering that the prior probabilities of the different causes are impossible to set and then moving all the way to dismiss the use of probability theory in this setting, nipping the p-values in the bud..! And Borel being rather vague about the solution probability theory has to provide. As stressed by Lehmann.

“The Pleiades appear closer to each other than one would naturally expect. This statement deserves thinking about; but when one wants to translate the phenomenon into numbers, the necessary ingredients are lacking. In order to make the vague idea of closeness more precise, should we look for the smallest circle that contains the group? the largest of the angular distances? the sum of squares of all the distances? the area of the spherical polygon of which some of the stars are the vertices and which contains the others in its interior? Each of these quantities is smaller for the group of the Pleiades than seems plausible. Which of them should provide the measure of implausibility? If three of the stars form an equilateral triangle, do we have to add this circumstance, which is certainly very unlikely apriori, to those that point to a cause?” Joseph Bertrand (p.166)

“But whatever objection one can raise from a logical point of view cannot prevent the preceding question from arising in many situations: the theory of probability cannot refuse to examine it and to give an answer; the precision of the response will naturally be limited by the lack of precision in the question; but to refuse to answer under the pretext that the answer cannot be absolutely precise, is to place oneself on purely abstract grounds and to misunderstand the essential nature of the application of mathematics.” Emile Borel (Chapter 4)

Another highly interesting objection of Bertrand is somewhat linked with his conditioning paradox, namely that the density of the observed unlikely event depends on the choice of the statistic that is used to calibrate the unlikeliness, which makes complete sense in that the information contained in each of these statistics and the resulting probability or likelihood differ to an arbitrary extend, that there are few cases (monotone likelihood ratio) where the choice can be made, and that Bayes factors share the same drawback if they do not condition upon the entire sample. In which case there is no selection of “circonstances remarquables”. Or of uniformly most powerful tests.

## failures and uses of Jaynes’ principle of transformation groups

Posted in Books, Kids, R, Statistics, University life with tags , , , , on April 14, 2015 by xi'an

This paper by Alon Drory was arXived last week when I was at Columbia. It reassesses Jaynes’ resolution of Bertrand’s paradox, which finds three different probabilities for a given geometric event depending on the underlying σ-algebra (or definition of randomness!). Both Poincaré and Jaynes argued against Bertrand that there was only one acceptable solution under symmetry properties. The author of this paper, Alon Drory, argues this is not the case!

“…contrary to Jaynes’ assertion, each of the classical three solutions of Bertrand’s problem (and additional ones as well!) can be derived by the principle of transformation groups, using the exact same symmetries, namely rotational, scaling and translational invariance.”

Drory rephrases as follows:  “In a circle, select at random a chord that is not a diameter. What is the probability that its length is greater than the side of the equilateral triangle inscribed in the circle?”.  Jaynes’ solution is indifferent to the orientation of one observer wrt the circle, to the radius of the circle, and to the location of the centre. The later is the one most discussed by Drory, as he argued that it does not involve an observer but the random experiment itself and relies on a specific version of straw throws in Jaynes’ argument. Meaning other versions are also available. This reminded me of an earlier post on Buffon’s needle and on the different versions of the needle being thrown over the floor. Therein reflecting on the connection with Bertrand’s paradox. And running some further R experiments. Drory’s alternative to Jaynes’ manner of throwing straws is to impale them on darts and throw the darts first! (Which is the same as one of my needle solutions.)

“…the principle of transformation groups does not make the problem well-posed, and well-posing strategies that rely on such symmetry considerations ought therefore to be rejected.”

In short, the conclusion of the paper is that there is an indeterminacy in Bertrand’s problem that allows several resolutions under the principle of indifference that end up with a large range of probabilities, thus siding with Bertrand rather than Jaynes.