Archive for the Books Category

Gaussian hare and Laplacian tortoise

Posted in Books, Kids, pictures, Statistics, University life with tags , , , , , , , , , , , on October 19, 2018 by xi'an

A question on X validated on the comparative merits of L¹ versus L² estimation led me to the paper of Stephen Portnoy and Roger Koenker entitled “The Gaussian Hare and the Laplacian Tortoise: Computability of Squared-Error versus Absolute-Error Estimators”, which I had missed at the time, despite enjoying a subscription to Statistical Science till the late 90’s.. The authors went as far as producing a parody of Granville’s Fables de La Fontaine by sticking Laplace’s and Gauss’ heads on the tortoise and the hare!

I remember rather vividly going through Steve Stigler’s account of the opposition between Laplace’s and Legendre’s approaches, when reading his History of Statistics in 1990 or 1991… Laplace defending the absolute error on the basis of the default double-exponential (or Laplace) distribution, when Legendre and then Gauss argued in favour of the squared error loss on the basis of a defaul Normal (or Gaussian) distribution. (Edgeworth later returned to the support of the L¹ criterion.) Portnoy and Koenker focus mostly on ways of accelerating the derivation of the L¹ regression estimators. (I also learned from the paper that Koenker was one of the originators of quantile regression.)

Le Monde puzzle [#1071]

Posted in Books, Kids with tags , , , , , , , , , , , on October 18, 2018 by xi'an

A “he said she said” Le Monde mathematical puzzle sixth competition problem that reminded me of the “Singapore birthday problem” (nothing to do with the original birthday problem!):

Arwen and Brandwein are privately and respectively given the day and month of Caradoc’s birthday [in the Gregorian calendar] with the information that the month number is at least the day number. Arwen starts by stating she knows Brandwein cannot deduce the birthday, followed by Brandwein who says the same about Arwen. If this “she says he says” goes on for the largest possible number of steps before Arwen says she knows, when is Caradoc’s birthday? Arwen and Brandwein are later given two numbers corresponding to Deirdre’s birthday with no indication of which one is the day and which one is the month. They know both numbers end up with the same digit and that the month number is strictly less than the day number. Arwen states she does not know the date and she knows Brandwein cannot know either. Then Brandwein says he indeed does not the date but he knows whether he got the day or the month. This prompts Arwen to conclude she knows, then Brandwein to do the same. When is Deirdre’s birthday?

Since this was a fairly easy puzzle (and since I had spent too much time debugging the previous R code!), I did not try coding this one but instead drew the possibilities and remove the impossibilities on a blackboard. The first question is quite simple actually since the day numbers stand between 1 and 12 and that each “I cannot know” excludes one of the remaining endpoints, removing first excludes 1 from both lists, then 12, then 2, then …. 8, ending up with 7. And 07/07 as Caradoc’s birthday. The second case sees 13,…,20,23,…,30 eliminated from Arwen’s numbers, then 3,…,10 as well, which eliminates the same numbers from Brandwein’s possibilities. That he knows whether it is a month or a day leaves only 1,2,21,22,31 as possible numbers. Then Arwen’s certainty reduces her numbers to be 2, 21, 22, or 31, and since Brandwein is also sure, the only possible cases are (2,22) and (22,2). Meaning Deirdre’s birthday is on 22/02. I dunno if this symmetry was to be expected! (And I cannot fathom why this puzzle is awarded so many points, when compared with the others.)

back to the Bayesian Choice

Posted in Books, Kids, Statistics, University life with tags , , , , , , , , on October 17, 2018 by xi'an

Surprisingly (or not?!), I received two requests about some exercises from The Bayesian Choice, one from a group of students from McGill having difficulties solving the above, wondering about the properness of the posterior (but missing the integration of x), to whom I sent back this correction. And another one from the Czech Republic about a difficulty with the term “evaluation” by which I meant (pardon my French!) estimation.

severe testing or severe sabotage? [not a book review]

Posted in Books, pictures, Statistics, University life with tags , , , , , , , , , on October 16, 2018 by xi'an

Last week, I received this new book of Deborah Mayo, which I was looking forward reading and annotating!, but thrice alas, the book had been sabotaged: except for the preface and acknowledgements, the entire book is printed upside down [a minor issue since the entire book is concerned] and with some part of the text cut on each side [a few letters each time but enough to make reading a chore!]. I am thus waiting for a tested copy of the book to start reading it in earnest!

 

ABC intro for Astrophysics

Posted in Books, Kids, Mountains, R, Running, Statistics, University life with tags , , , , , , , , , , , on October 15, 2018 by xi'an

Today I received in the mail a copy of the short book published by edp sciences after the courses we gave last year at the astrophysics summer school, in Autrans. Which contains a quick introduction to ABC extracted from my notes (which I still hope to turn into a book!). As well as a longer coverage of Bayesian foundations and computations by David Stenning and David van Dyk.

accelerating HMC by learning the leapfrog scale

Posted in Books, Statistics with tags , , , , , , , , on October 12, 2018 by xi'an

In this new arXiv submission that was part of Changye Wu’s thesis [defended last week],  we try to reduce the high sensitivity of the HMC algorithm to its hand-tuned parameters, namely the step size ε  of the discretisation scheme, the number of steps L of the integrator, and the covariance matrix of the auxiliary variables. By calibrating the number of steps of the Leapfrog integrator towards avoiding both slow mixing chains and wasteful computation costs. We do so by learning from the No-U-Turn Sampler (NUTS) of Hoffman and Gelman (2014) which already automatically tunes both the step size and the number of leapfrogs.

The core idea behind NUTS is to pick the step size via primal-dual averaging in a burn-in (warmup, Andrew would say) phase and to build at each iteration a proposal based on following a locally longest path on a level set of the Hamiltonian. This is achieved by a recursive algorithm that, at each call to the leapfrog integrator, requires to evaluate both the gradient of the target distribution and the Hamiltonianitself. Roughly speaking an iteration of NUTS costs twice as much as regular HMC with the same number of calls to the integrator. Our approach is to learn from NUTS the scale of the leapfrog length and use the resulting empirical distribution of the longest leapfrog path to randomly pick the value of  L at each iteration of an HMC scheme. This obviously preserves the validity of the HMC algorithm.

While a theoretical comparison of the convergence performances of NUTS and this eHMC proposal seem beyond our reach, we ran a series of experiments to evaluate these performances, using as a criterion an ESS value that is calibrated by the evaluation cost of the logarithm of target density function and of its gradient, as this is usually the most costly part of the algorithms. As well as a similarly calibrated expected square jumping distance. Above is one such illustration for a stochastic volatility model, the first axis representing the targeted acceptance probability in the Metropolis step. Some of the gains in either ESS or ESJD are by a factor of ten, which relates to our argument that NUTS somewhat wastes computation effort using a uniformly distributed proposal over the candidate set, instead of being close to its end-points, which automatically reduces the distance between the current position and the proposal.

Le Monde puzzle [#1070]

Posted in Books, Kids, R, University life with tags , , , , , , , on October 11, 2018 by xi'an

Rewording Le Monde mathematical puzzle  fifth competition problem

For the 3×3 tables below, what are the minimal number of steps to move from left to rights when the yellow tokens can only be move to an empty location surrounded by two other tokens?

In the 4×4 table below, there are 6 green tokens. How many steps from left to right?

Painful and moderately mathematical, once more… For the first question, a brute force simulation of random valid moves of length less than 100 returns solutions in 4 steps:

     [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]
     1    1    1    0    0    1    0    0    1
     1    0    1    0    1    1    0    0    1
     0    0    1    1    1    1    0    0    1
     0    0    1    1    0    1    0    1    1
     0    0    1    0    0    1    1    1    1

But this is not an acceptable move because of the “other” constraint. Imposing this constraint leads to a solution in 9 steps, but is this the lowest bound?! It actually took me most of the weekend (apart from a long drive to and from a short half-marathon!) to figure out a better strategy than brute force random exploration: the trick I eventually figured out is to start from the finishing (rightmost) value F of the grid and look at values with solutions available in 1,2,… steps. This is not exactly dynamic programming, but it keeps the running time under control if there is a solution associated with the starting (leftmost) value S. (Robin proceeded reverse-wise, which in retrospect is presumably equivalent, if faster!) The 3×3 grid has 9 choose 5, ie 126, possible configurations with 5 coins, which means the number of cases remains under control. And even so for the 4×4 grid with 6 coins, with 8008 configurations. This led to a 9 step solution for n=3 and the proposed starting grid in yellow:

[1] 1 1 1 0 0 1 0 0 1
[1] 1 1 0 0 1 1 0 0 1
[1] 1 1 0 1 1 0 0 0 1
[1] 0 1 0 1 1 1 0 0 1
[1] 0 1 1 1 0 1 0 0 1
[1] 1 1 1 1 0 0 0 0 1
[1] 0 1 1 1 1 0 0 0 1
[1] 0 0 1 1 1 0 0 1 1
[1] 0 0 1 1 0 0 1 1 1
[1] 0 0 1 0 0 1 1 1 1

and a 19 step solution for n=4:

[1] 1 0 0 0 0 1 0 0 0 1 1 1 0 0 0 1
[1] 1 0 0 0 0 1 0 0 0 0 1 1 0 0 1 1
[1] 1 0 0 0 0 1 1 0 0 0 1 1 0 0 1 0
[1] 1 1 0 0 0 1 1 0 0 0 0 1 0 0 1 0
[1] 1 1 0 0 0 0 1 1 0 0 0 1 0 0 1 0
[1] 1 1 1 0 0 0 1 1 0 0 0 0 0 0 1 0
[1] 1 0 1 1 0 0 1 1 0 0 0 0 0 0 1 0
[1] 1 1 1 1 0 0 1 0 0 0 0 0 0 0 1 0
[1] 1 1 0 1 0 1 1 0 0 0 0 0 0 0 1 0
[1] 1 0 0 1 1 1 1 0 0 0 0 0 0 0 1 0
[1] 0 0 0 1 1 1 1 0 0 0 1 0 0 0 1 0
[1] 0 0 0 1 1 1 0 0 0 1 1 0 0 0 1 0
[1] 0 0 0 1 1 1 0 0 1 1 0 0 0 0 1 0
[1] 0 0 0 1 0 1 0 0 1 1 0 0 0 1 1 0
[1] 0 0 0 1 0 1 0 0 1 0 0 0 1 1 1 0
[1] 0 0 0 1 1 1 0 0 1 0 0 0 1 0 1 0
[1] 0 0 0 1 0 1 0 0 1 1 0 0 1 0 1 0
[1] 0 0 0 1 0 1 0 0 0 1 1 0 1 0 1 0
[1] 0 0 0 1 0 1 1 0 0 1 1 0 1 0 0 0

The first resolution takes less than a minute and the second one just a few minutes (or less than my short morning run!). Surprisingly, using this approach does not require more work, which makes me wonder at the solution Le Monde journalists will propose. Given the (misguided) effort put into my resolution, seeing a larger number of points for this puzzle is no wonder.