failures and uses of Jaynes’ principle of transformation groups
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.
Xi'an's Og 
April 15, 2015 at 3:54 am
Well I don’t doubt different physical experiments can result in different answers, and the idea that symmetry arguments can be coerced into modelling them doesn’t strike me as implausible either, but I didn’t read beyond p5 because I was put off by the discussion of those Bertrand solutions. Aren’t they all simply miscounting the chords by using “random selections” from clearly inappropriate coordinate systems on the disc? (R)B3 at least gives us a 1-1 mapping between points and chords and equal weighting of them on the punctured disc, so that solution looks good. Using pairs of points on the perimeter to name and then count the chords I get 4/9 for the solution on the ordinary disc.
April 19, 2015 at 5:26 pm
Or, when I’m not miscounting myself, 1/3.
April 14, 2015 at 3:10 am
I’m going to read the paper now, but I have to say that it strikes me as implausible that it was only a happy mathematical coincidence that Jaynes the physicist was able to use pure reason to predict the frequency distribution of straws in his experiment — which was the proof of the pudding…
April 14, 2015 at 4:09 am
Drory seems to have missed that fact that Jaynes states up front (page 3, before the math) that he’s going to make the problem well-posed by analyzing a specific physical experiment:
Drory thinks he’s arguing against Jaynes, but the position he takes in the paper is the same as Jaynes’s. I don’t think there’s anything mathematical in Drory’s paper that Jaynes would object to — but he would certainly have objected to Drory’s claim that Jaynes does the analysis backwards, and to Drory’s attribution to Jaynes of various opinions that Jaynes never held.
April 14, 2015 at 4:10 am
(In the blockquote, emphasis is as in the original text.)