Archive for cross validated

simulation fodder for future exams

Posted in Books, Kids, R, Statistics with tags , , , , on February 20, 2019 by xi'an

Here are two nice exercises for a future simulation exam, seen and solved on X validated.The first one is about simulating a Gibbs sampler associated with the joint target

exp{-|x|-|y|-a|y-x|}

defined over IR² for a≥0 (or possibly a>-1). The conditionals are identical and non-standard, but a simple bound on the conditional density is the corresponding standard double exponential density, which makes for a straightforward accept-reject implementation. However it is also feasible to break the full conditional into three parts, depending on the respective positions of x, y, and 0, and to obtain easily invertible cdfs on the three intervals.The second exercise is about simulating from the cdf

F(x)=1-\exp\{-ax-bx^{p+1}/(p+1)\}

which can be numerically inverted. It is however more fun to call for an accept-reject algorithm by bounding the density with a ½ ½ mixture of an Exponential Exp(a) and of the 1/(p+1)-th power of an Exponential Exp(b/(p+1)). Since no extra constant appears in the solution,  I suspect the (p+1) in b/(p+1) was introduced on purpose. As seen in the above fit for 10⁶ simulations (and a=1,b=2,p=3), there is no deviation from the target! There is nonetheless an even simpler and rather elegant resolution to the exercise: since the tail function (1-F(x)) appears as the product of two tail functions, exp(-ax) and the other one, the cdf is the distribution of the minimum of two random variates, one with the Exp(a) distribution and the other one being the 1/(p+1)-th power of an Exponential Exp(b/(p+1)) distribution. Which of course returns a very similar histogram fit:

leave Bayes factors where they once belonged

Posted in Statistics with tags , , , , , , , , , , on February 19, 2019 by xi'an

In the past weeks I have received and read several papers (and X validated entries)where the Bayes factor is used to compare priors. Which does not look right to me, not on the basis of my general dislike of Bayes factors!, but simply because this seems to clash with the (my?) concept of Bayesian model choice and also because data should not play a role in that situation, from being used to select a prior, hence at least twice to run the inference, to resort to a single parameter value (namely the one behind the data) to decide between two distributions, to having no asymptotic justification, to eventually favouring the prior concentrated on the maximum likelihood estimator. And more. But I fear that this reticence to test for prior adequacy also extends to the prior predictive, or Box’s p-value, namely the probability under this prior predictive to observe something “more extreme” than the current observation, to quote from David Spiegelhalter.

I’m getting the point

Posted in Statistics with tags , , , , , , on February 14, 2019 by xi'an

A long-winded X validated discussion on the [textbook] mean-variance conjugate posterior for the Normal model left me [mildly] depressed at the point and use of answering questions on this forum. Especially as it came at the same time as a catastrophic outcome for my mathematical statistics exam.  Possibly an incentive to quit X validated as one quits smoking, although this is not the first attempt

efficiency and the Fréchet-Darmois-Cramèr-Rao bound

Posted in Books, Kids, Statistics with tags , , , , , , , , , , , on February 4, 2019 by xi'an

 

Following some entries on X validated, and after grading a mathematical statistics exam involving Cramèr-Rao, or Fréchet-Darmois-Cramèr-Rao to include both French contributors pictured above, I wonder as usual at the relevance of a concept of efficiency outside [and even inside] the restricted case of unbiased estimators. The general (frequentist) version is that the variance of an estimator δ of [any transform of] θ with bias b(θ) is

I(θ)⁻¹ (1+b'(θ))²

while a Bayesian version is the van Trees inequality on the integrated squared error loss

(E(I(θ))+I(π))⁻¹

where I(θ) and I(π) are the Fisher information and the prior entropy, respectively. But this opens a whole can of worms, in my opinion since

  • establishing that a given estimator is efficient requires computing both the bias and the variance of that estimator, not an easy task when considering a Bayes estimator or even the James-Stein estimator. I actually do not know if any of the estimators dominating the standard Normal mean estimator has been shown to be efficient (although there exist results for closed form expressions of the James-Stein estimator quadratic risk, including one of mine the Canadian Journal of Statistics published verbatim in 1988). Or is there a result that a Bayes estimator associated with the quadratic loss is by default efficient in either the first or second sense?
  • while the initial Fréchet-Darmois-Cramèr-Rao bound is restricted to unbiased estimators (i.e., b(θ)≡0) and unable to produce efficient estimators in all settings but for the natural parameter in the setting of exponential families, moving to the general case means there exists one efficiency notion for every bias function b(θ), which makes the notion quite weak, while not necessarily producing efficient estimators anyway, the major impediment to taking this notion seriously;
  • moving from the variance to the squared error loss is not more “natural” than using any [other] convex combination of variance and squared bias, creating a whole new class of optimalities (a grocery of cans of worms!);
  • I never got into the van Trees inequality so cannot say much, except that the comparison between various priors is delicate since the integrated risks are against different parameter measures.

more concentration, everywhere

Posted in R, Statistics with tags , , , , , , , , , , on January 25, 2019 by xi'an

Although it may sound like an excessive notion of optimality, one can hope at obtaining an estimator δ of a unidimensional parameter θ that is always closer to θ that any other parameter. In distribution if not almost surely, meaning the cdf of (δ-θ) is steeper than for other estimators enjoying the same cdf at zero (for instance ½ to make them all median-unbiased). When I saw this question on X validated, I thought of the Cauchy location example, where there is no uniformly optimal estimator, albeit a large collection of unbiased ones. But a simulation experiment shows that the MLE does better than the competition. At least than three (above) four of them (since I tried the Pitman estimator via Christian Henning’s smoothmest R package). The differences to the MLE empirical cd make it clearer below (with tomato for a score correction, gold for the Pitman estimator, sienna for the 38% trimmed mean, and blue for the median):I wonder at a general theory along these lines. There is a vague similarity with Pitman nearness or closeness but without the paradoxes induced by this criterion. More in the spirit of stochastic dominance, which may be achievable for location invariant and mean unbiased estimators…

unbiased estimators that do not exist

Posted in Statistics with tags , , , , , , , on January 21, 2019 by xi'an

When looking at questions on X validated, I came across this seemingly obvious request for an unbiased estimator of P(X=k), when X~B(n,p). Except that X is not observed but only Y~B(s,p) with s<n. Since P(X=k) is a polynomial in p, I was expecting such an unbiased estimator to exist. But it does not, for the reasons that Y only takes s+1 values and that any function of Y, including the MLE of P(X=k), has an expectation involving monomials in p of power s at most. It is actually straightforward to establish properly that the unbiased estimator does not exist. But this remains an interesting additional example of the rarity of the existence of unbiased estimators, to be saved until a future mathematical statistics exam!

a question from McGill about The Bayesian Choice

Posted in Books, pictures, Running, Statistics, Travel, University life with tags , , , , , , , on December 26, 2018 by xi'an

I received an email from a group of McGill students working on Bayesian statistics and using The Bayesian Choice (although the exercise pictured below is not in the book, the closest being exercise 1.53 inspired from Raiffa and Shlaiffer, 1961, and exercise 5.10 as mentioned in the email):

There was a question that some of us cannot seem to decide what is the correct answer. Here are the issues,

Some people believe that the answer to both is ½, while others believe it is 1. The reasoning for ½ is that since Beta is a continuous distribution, we never could have θ exactly equal to ½. Thus regardless of α, the probability that θ=½ in that case is 0. Hence it is ½. I found a related stack exchange question that seems to indicate this as well.

The other side is that by Markov property and mean of Beta(a,a), as α goes to infinity , we will approach ½ with probability 1. And hence the limit as α goes to infinity for both (a) and (b) is 1. I think this also could make sense in another context, as if you use the Bayes factor representation. This is similar I believe to the questions in the Bayesian Choice, 5.10, and 5.11.

As it happens, the answer is ½ in the first case (a) because π(H⁰) is ½ regardless of α and 1 in the second case (b) because the evidence against H⁰ goes to zero as α goes to zero (watch out!), along with the mass of the prior on any compact of (0,1) since Γ(2α)/Γ(α)². (The limit does not correspond to a proper prior and hence is somewhat meaningless.) However, when α goes to infinity, the evidence against H⁰ goes to infinity and the posterior probability of ½ goes to zero, despite the prior under the alternative being more and more concentrated around ½!