“Geyer (1994) established the asymptotic properties of the MC-MLE estimates under general conditions; in particular that the x’s are realisations of an ergodic process. This is remarkable, given that most of the theory on M-estimation (i.e.estimation obtained by maximising functions) is restricted to iid data.”

Michael Guttman and Aapo Hyvärinen also use additional simulated data in another likelihood of a logistic classifier, called noise contrastive estimation. Both methods replace the unknown ratio of normalising constants with an unbiased estimate based on the additional simulated data. The major and impressive result in this paper [now published in the Electronic Journal of Statistics] is that the noise contrastive estimation approach always enjoys a smaller variance than Geyer’s solution, at an equivalent computational cost when the actual data observations are iid. And the artificial data simulations ergodic. The difference between both estimators is however negligible against the Monte Carlo error (Theorem 2).

This may be a rather naïve question, but I wonder at the choice of the alternative distribution *ψ*. With a vague notion that it could be optimised in a GANs perspective. A side result of interest in the paper is to provide a minimal (re)parameterisation of the truncated multivariate Gaussian distribution, if only as an exercise for future exams. Truncated multivariate Gaussian for which the normalising constant is of course unknown.

“(…) I respect your personal reasons not to have an abortion — no one is forcing you to have one. I respect your choice. I’m pro-choice — often called pro-abortion by the anti-abortion crusaders, although *no one* is pro-abortion. What’s unequal about the argument is the choice; the difference between pro-life and pro-choice is the choice. Pro-life proponents have no qualms about forcing women to go through childbirth — they give women no choice (…)

I must remind the Roman Catholic Church of the First Amendment to the United States Constitution: “Congress shall make no law respecting an establishment of religion, or prohibiting the free exercise thereof.” In other words, we are free to practice the religion of our choice, *and* we are protected from having someone else’s religion practiced on us. Freedom of religion in the United States also means freedom *from *religion (…)

The prevailing impetus to oppose abortion is to punish the woman who doesn’t want the child. The sacralizing of the fetus is a ploy. How can “life” be sacred (and begin at six weeks, or at conception), if a child’s life *isn’t* sacred after it’s born? Clearly, a woman’s life is never sacred; as clearly, a woman has no reproductive rights (…)

Of an unmarried woman or girl who got pregnant, people of my grandparents’ generation used to say: “She is paying the piper.” Meaning, she deserves what she gets — namely, to give birth to a child. That cruelty is the abiding impetus behind the dishonestly named right-to-life movement. *Pro-life* always was (and remains) a marketing term. Whatever the anti-abortion crusaders call themselves, they don’t care what happens to an unwanted child — not after the child is born — and they’ve never cared about the mother.”

Given N ocre balls, N aquamarine balls, and two urns, what is the optimal way to allocate the balls to the urns towards drawing an ocre ball with no urn being empty?

Both my reasoning and a two line exploration code led to having one urn with only one ocre ball (and no acquamarine ball) and all the other balls in the second urn.

odz<-function(n,m,t) 2*m/n+(t-2*m)/(t-n) probz=matrix(0,trunc(N/2)-1,N-1) for (n in 1:(N-1)) for (m in 1:(trunc(N/2)-1)) probz[m,n]=odz(n,m,N)]]>

In the last issue of Bayesian Analysis, Lukasz Rajkowski studies the most likely (MAP) cluster associated with the Dirichlet process mixture model. Reminding me that most Bayesian estimates of the number of clusters are not consistent (when the sample size grows to infinity). I am always puzzled by this problem, as estimating the number of clusters sounds like an ill-posed problem, since it is growing with the number of observations, by definition of the Dirichlet process. For instance, the current paper establishes that the number of clusters intersecting a given compact set remains bounded. (The setup is one of a Normal Dirichlet process mixture with constant and known covariance matrix.)

Since the posterior probability of a given partition of {1,2,…,n} can be (formally) computed, the MAP estimate can be (formally) derived. I inserted *formally* in the previous sentence as the derivation of the exact MAP is an NP hard problem in the number n of observations. As an aside, I have trouble with the author’s argument that the convex hulls of the clusters should be disjoin: I do not see why they should when the mixture components are overlapping. (More generally, I fail to relate to notions like “bad clusters” or “overestimation of the number of clusters” or a “sensible choice” of the covariance matrix.) More globally, I am somewhat perplexed by the purpose of the paper and the relevance of the MAP estimate, even putting aside my generic criticisms of the MAP approach. No uncertainty is attached to the estimator, which thus appears as a form of penalised likelihood strategy rather than a genuinely Bayesian (Analysis) solution.

The first example in the paper is using data from a Uniform over (-1,1), concluding at a “misleading” partition by the MAP since it produces more than one cluster. I find this statement flabbergasting as the generative model is not the estimated model. To wit, the case of an exponential Exp(1) sample that cannot reach a maximum of the target function with a finite number of sample. Which brings me back full-circle to my general unease about clustering in that much more seems to be assumed about this notion than what the statistical model delivers.

]]>As for the validity of the latest release of R (currently R-3.6.1 which came out on 2019-07-05, named Action of the Toes!), I figure the bazillion R programs currently running should be able to detect any defect pretty fast, although awareness of the incredible failure of sample() reported in an earlier post took a while to appear.

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“Increasing export capacity from the Freeport LNG project is critical to spreading freedom gas throughout the world by giving America’s allies a diverse and affordable source of clean energy” M. Menezes, US Secretary of Energy

“NASA should NOT be talking about going to the Moon – We did that 50 years ago. They should be focused on the much bigger things we are doing, including Mars (of which the Moon is a part)” DT,, 7 June

“I just met with the Queen of England (U.K.) [sic], the Prince of Whales [re-sic]” DT, 13 June

“[Sarah Sanders] is going to be leaving the service of her country and she’s going to be going (…) She’s a very special person, a very, very fine woman, she has been so great, she has such heart, she’s strong but with great, great heart, and I want to thank you for an outstanding job.” DT, 13 June

“…when I asked, ‘How many will die?’ ‘150 people, sir’, was the answer from a General. 10 minutes before the strike I stopped it, not … proportionate to shooting down an unmanned drone.” DT, 21 June

“The reason we have tragedies like that on the border is because that father didn’t wait to go through the asylum process in the legal fashion and decided to cross the river and not only died but his daughter died tragically as well,” K. Cuccinelli, head of US Immigration and Citizenship Services, 28 June

“If Japan is attacked, we will fight World War III. But if we’re attacked, Japan doesn’t have to help us at all. They can watch it on a Sony television.” DT, 24 June

**A** light birthday problem as Le Monde mathematical puzzle:

Each member of a group of 35 persons writes down the number of those who share the same birth-month and the number of those who share the same birth-date [with them]. It happens that these 70 numbers include all integers from 0 to 10. Show that at least two people share a birth-day. What is the maximal number of people for this property to hold?

Which needs no R code since the result follows from the remark that the number of individuals sharing a birth-month with just one other, n¹, is a multiple of 2, the number of individuals sharing a birth-month with just two others, n², a multiple of 3, and so on. Hence, if no people share a birth-day, n¹,n²,…,n¹⁰>0 and

n¹+n²+…+n¹⁰ ≥ 2+3+…+11 = 6·11-1=65

which means that it is impossible that the 10 digits n¹,…,n¹⁰ are all positive. All the way up to 65 people. As an aside, no correction of the wrong solution to puzzle #1105 was published in the subsequent editions.

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