## spacings on a torus

Posted in Books, Kids, R, Statistics, University life with tags , , , , , , , , on March 22, 2018 by xi'an

While in Brussels last week I noticed an interesting question on X validated that I considered in the train back home and then more over the weekend. This is a question about spacings, namely how long on average does it take to cover an interval of length L when drawing unit intervals at random (with a torus handling of the endpoints)? Which immediately reminded me of Wilfrid Kendall (Warwick) famous gif animation of coupling from the past via leaves covering a square region, from the top (forward) and from the bottom (backward)…

The problem is rather easily expressed in terms of uniform spacings, more specifically on the maximum spacing being less than 1 (or 1/L depending on the parameterisation). Except for the additional constraint at the boundary, which is not independent of the other spacings. Replacing this extra event with an independent spacing, there exists a direct formula for the expected stopping time, which can be checked rather easily by simulation. But the exact case appears to be add a few more steps to the draws, 3/2 apparently. The following graph displays the regression of the Monte Carlo number of steps over 10⁴ replicas against the exact values:

## no country for old biases

Posted in Books, Kids, Statistics with tags , , , , , on March 20, 2018 by xi'an

Following a X validated question, I read a 1994 paper by Phil Dawid on the selection paradoxes in Bayesian statistics, which first sounded like another version of the stopping rule paradox. And upon reading, less so. As described above, the issue stands with drawing inference on the index and value, (i⁰,μ⁰), of the largest mean of a sample of Normal rvs. What I find surprising in Phil’s presentation is that the Bayesian analysis does not sound that Bayesian. If given the whole sample, a Bayesian approach should produce a posterior distribution on (i⁰,μ⁰), rather than follow estimation steps, from estimating i⁰ to estimating the associated mean. And if needed, estimators should come from the definition of a particular loss function. When, instead, given the largest point in the sample, and only that point, its distribution changes, so I am fairly bemused by the statement that no adjustment is needed.

The prior modelling is also rather surprising in that the priors on the means should be joint rather than a product of independent Normals, since these means are compared and hence comparable. For instance a hierarchical prior seems more appropriate, with location and scale to be estimated from the whole data. Creating a connection between the means… A relevant objection to the use of independent improper priors is that the maximum mean μ⁰ then does not have a well-defined measure. However, I do not think a criticism of some priors versus other is a relevant attack on this “paradox”.

## uniform on the sphere [or not]

Posted in pictures, R, Statistics with tags , , , , , , , , , , , , on March 8, 2018 by xi'an

While looking at X validated questions, I came upon this comment that simulating a uniform distribution on a d-dimensional unit sphere does not proceed from generating angles at random on (0,2π) and computing spherical coordinates… Which I must confess would have been my initial suggestion! This is obvious, nonetheless, when computing the Jacobian of the spherical coordinate transform, which involves powers of the sines of the angles, in a decreasing sequence from d-1 to zero. This means that the angles should be simulated according to their respective sine-power densities. However, except for the d=3 case, where simulating from the density sin(φ) is straightforward by inverse cdf, i.e. φ=acos(1-2u), the cdfs for the higher powers are combinations of sines and cosines, and as such are not easily inverted. Take for instance the eighth power:

F⁸(φ)=(840 φ – 672 sin(2 φ) + 168 sin(4 φ) – 32 sin(6 φ) + 3 sin(8 φ))/3072

While the densities are bounded by sin(φ), up to a constant, and hence an accept-reject can be easily derived, the efficiency decreases with the dimension according to the respective ratio of the Wallis’ integrals, unsurprisingly. A quick check for d=4 shows that the Normal simulation+projection-by-division-by-its-norm is faster.

Puzzling a bit further about this while running, I wondered at the simultaneous simulations from sin(φ), sin(φ)², sin(φ)³, &tc., but cannot see a faster way to recycle simulations from sin(φ). Points (φ,u) located in-between two adjacent power curves are acceptable simulations from the corresponding upper curve but they need be augmented by points (φ,u) under the lower curve to constitute a representative sample. In the end, this amounts to multiplying simulations from the highest power density as many times as there are powers. No gain in sight… Sigh!

However, a few days later, while enjoying the sunset over Mont Blanc(!), I figured out that there exists a direct and efficient way to simulate from these powers of the sine function. Indeed, when looking at the density of cos(φ), it happens to be the signed root of a Beta(½,(d-1)/2), which avoids the accept-reject step. Presumably this is well-known, but I have not seen this proposal associated with the uniform distribution on the sphere.

## an interesting identity

Posted in Books, pictures, Statistics, University life with tags , , , , , , , , on March 1, 2018 by xi'an

Another interesting X validated question, another remembrance of past discussions on that issue. Discussions that took place in the Institut d’Astrophysique de Paris, nearby this painting of Laplace, when working on our cosmostats project. Namely the potential appeal of recycling multidimensional simulations by permuting the individual components in nearly independent settings. As shown by the variance decomposition in my answer, when opposing N iid pairs (X,Y) to the N combinations of √N simulations of X and √N simulations of Y, the comparison

$\text{var} \hat{\mathfrak{h}}^2_N=\text{var} (\hat{\mathfrak{h}}^1_N)+\frac{mn(n-1)}{N^2}\,\text{var}^Y\left\{ \mathbb{E}^{X}\left\{\mathfrak{h}(X,Y)\right\}\right\}$

$+\frac{m(m-1)n}{N^2}\,\text{var}^X\left[\mathbb{E}^Y\left\{\mathfrak{h}(X,Y)\right\}\right]$

unsurprisingly gives the upper hand to the iid sequence. A sort of converse to Rao-Blackwellisation…. Unless the production of N simulations gets much more costly when compared with the N function evaluations. No wonder we never see this proposal in Monte Carlo textbooks!

## infinite mixtures are likely to take a while to simulate

Posted in Books, Statistics with tags , , , , , , , , on February 22, 2018 by xi'an

Another question on X validated got me highly interested for a while, as I had considered myself the problem in the past, until I realised while discussing with Murray Pollock in Warwick that there was no general answer: when a density f is represented as an infinite series decomposition into weighted densities, some weights being negative, is there an efficient way to generate from such a density? One natural approach to the question is to look at the mixture with positive weights, f⁺, since it gives an upper bound on the target density. Simulating from this upper bound f⁺ and accepting the outcome x with probability equal to the negative part over the sum of the positive and negative parts f⁻(x)/f(x) is a valid solution. Except that it is not implementable if

1.  the positive and negative parts both involve infinite sums with no exploitable feature that can turn them into finite sums or closed form functions,
2.  the sum of the positive weights is infinite, which is the case when the series of the weights is not absolutely converging.

Even when the method is implementable it may be arbitrarily inefficient in the sense that the probability of acceptance is equal to to the inverse of the sum of the positive weights and that simulating from the bounding mixture in the regular way uses the original weights which may be unrelated in size with the actual importance of the corresponding components in the actual target. Hence, when expressed in this general form, the problem cannot allow for a generic solution.

Obviously, if more is known about the components of the mixture, as for instance the sequence of weights being alternated, there exist specialised methods, as detailed in the section of series representations in Devroye’s (1985) simulation bible. For instance, in the case when positive and negative weight densities can be paired, in the sense that their weighted difference is positive, a latent index variable can be included. But I cannot think of a generic method where the initial positive and negative components are used for simulation, as it may on the opposite be the case that no finite sum difference is everywhere positive.

## Gibbs for kidds

Posted in Books, Kids, Statistics, University life with tags , , , , , , , , , , , , , , , on February 12, 2018 by xi'an

A chance (?) question on X validated brought me to re-read Gibbs for Kids, 25 years after it was written (by my close friends George and Ed). The originator of the question had difficulties with the implementation, apparently missing the cyclic pattern of the sampler, as in equations (2.3) and (2.4), and with the convergence, which is only processed for a finite support in the American Statistician paper. The paper [which did not appear in American Statistician under this title!, but inspired an animal bredeer, Dan Gianola, to write a “Gibbs for pigs” presentation in 1993 at the 44th Annual Meeting of the European Association for Animal Production, Aarhus, Denmark!!!] most appropriately only contains toy examples since those can be processed and compared to know stationary measures. This is for instance the case for the auto-exponential model

$f(x,y) \propto exp(-xy)$

which is only defined as a probability density for a compact support. (The paper does not identify the model as a special case of auto-exponential model, which apparently made the originator of the model, Julian Besag in 1974, unhappy, as George and I found out when visiting Bath, where Julian was spending the final year of his life, many years later.) I use the limiting case all the time in class to point out that a Gibbs sampler can be devised and operate without a stationary probability distribution. However, being picky!, I would like to point out that, contrary, to a comment made in the paper, the Gibbs sampler does not “fail” but on the contrary still “converges” in this case, in the sense that a conditional ergodic theorem applies, i.e., the ratio of the frequencies of visits to two sets A and B with finite measure do converge to the ratio of these measures. For instance, running the Gibbs sampler 10⁶ steps and ckecking for the relative frequencies of x’s in (1,2) and (1,3) gives 0.685, versus log(2)/log(3)=0.63, since 1/x is the stationary measure. One important and influential feature of the paper is to stress that proper conditionals do not imply proper joints. George would work much further on that topic, in particular with his PhD student at the time, my friend Jim Hobert.

With regard to the convergence issue, Gibbs for Kids points out to Schervish and Carlin (1990), which came quite early when considering Gelfand and Smith published their initial paper the very same year, but which also adopts a functional approach to convergence, along the paper’s fixed point perspective, somehow complicating the matter. Later papers by Tierney (1994), Besag (1995), and Mengersen and Tweedie (1996) considerably simplified the answer, which is that irreducibility is a necessary and sufficient condition for convergence. (Incidentally, the reference list includes a technical report of mine’s on latent variable model MCMC implementation that never got published.)

## when perfect correlation just means… perfect!

Posted in Statistics with tags , , on February 6, 2018 by xi'an

When looking at an X validated question on generating two perfectly negatively correlated Bernoulli variates last week, my intuition was that one had to be the opposite of the other, which means their parameters had to sum up to one. Intuition that was plain easy to back up by solving the equation

corr(C¹,C²)=-1

in terms of the joint distribution of (C¹,C²). That perfect correlation implies strong constraints on the parameter of the Bernoulli is not highly surprising given its binary support. Although I had no time to pursue the issue, I idly wondered at the generalisation to, say, a Binomial case, i.e., whether or not this case still is the only possible one for the above to hold. But again a perfect correlation can only occur with perfect prediction, i.e., when the Binomial variates have the same number of trials and complementary probability parameters. (Of no particular relevance is the fact that the originator of the question preferred an answer that showed how to simulate two Bernoulli such that C¹+C²=1!)