## O’Bayes in action

**M**y next-door colleague [at Dauphine] François Simenhaus shared a paradox [to be developed in an incoming test!] with Julien Stoehr and I last week, namely that, when selecting the largest number between a [observed] and b [unobserved], drawing a random boundary on a [meaning that a is chosen iff a is larger than this boundary] increases the probability to pick the largest number above ½2…

When thinking about it in the wretched RER train [train that got immobilised for at least two hours just a few minutes after I went through!, good luck to the passengers travelling to the airport…] to De Gaulle airport, I lost the argument: if a<b, the probability [for this random bound] to be larger than a and hence for selecting b is 1-Φ(a), while, if a>b, the probability [of winning] is Φ(a). Hence the only case when the probability is ½ is when a is the median of this random variable. But, when discussing the issue further with Julien, I exposed an interesting non-informative prior characterisation. Namely, if I assume a,b to be iid U(0,M) and set an improper prior 1/M on M, the conditional probability that b>a given a is ½. Furthermore, the posterior probability to pick the right [largest] number with François’s randomised rule is also ½, no matter what the distribution of the random boundary is. Now, the most surprising feature of this coffee room derivation is that these properties only hold for the prior 1/M. Any other power of M will induce an asymmetry between a and b. (The same properties hold when a,b are iid Exp(M).) Of course, this is not absolutely unexpected since 1/M is the invariant prior and since the “intuitive” symmetry only holds under this prior. Power to O’Bayes!

When discussing again the matter with François yesterday, I realised I had changed his wording of the puzzle. The original setting is one with two cards hiding the unknown numbers a and b and of a player picking one of the cards. If the player picks a card at random, there is indeed a probability of ½ of picking the largest number. If the decision to switch or not depends on an independent random draw being larger or smaller than the number on the observed card, the probability to get max(a,b) in the end hits 1 when this random draw falls into (a,b) and remains ½ outside (a,b). Randomisation pays.

November 10, 2017 at 10:34 pm

Is this not just the Blackwell “paradox,” discussed for example here: http://www.paseman.com/public/AmericanScientist/Knowing%20When%20to%20Stop_Hill.pdf ?

November 10, 2017 at 10:54 pm

Thanks, I did not know about this paradox. You are right, this is exactly the thing!