ಬೆಂಗಳೂರು ಸ್ನ್ಯಾಪ್‌ಶಾಟ್¹⁰ [jatp]

tch

simulating from the joint cdf

An X validated question (what else?!) brought back (to me) the question of handling a bivariate cdf for simulation purposes. In the specific case of a copula when thus marginals were (well-)known…. And led me to an erroneous chain of thought, fortunately rescued by Robin Ryder! When the marginal distributions are set, the simulation setup is indeed equivalent to a joint Uniform simulation from a copula

In specific cases, as for instance the obvious example of Gaussian copulas, there exist customised simulation algorithms. Looking for more generic solutions, I turn to the Bible, where Chapter XI[an], has two entire sections XI.3.2. and XI.3.3 on the topic (even though Luc Devroye does not use the term copula there despite them being introduced in 1959 by A, Sklar, in response to a query of M. Fréchet). In addition to a study of copulas, both sections contain many specific solutions (as for instance in the [unnumbered] Table on page 585) but I found no generic simulation method. My [non-selected] answer to the question was thus to propose standard solutions such as finding one conditional since the marginals are Uniform. Which depends on the tractability of the derivatives of C(·,·).

However, being dissatisfied with this bland answer, I thought further about the problem and came up with a fallacious scheme, namely to first simulate the value p of C(U,V) by drawing a Uniform, and second simulate (U,V) conditional on C(U,V)=p. Going as far as running an R code on a simple copula, as shown above. Fallacious reasoning since (as I knew already!!!), C(U,V) is not uniformly distributed! But has instead a case-dependent distribution… As a (connected) aside, I wonder if the generator attached with Archimedean copulas has any magical feature that help with the generation of the associated copula.

1 / duh?!

An interesting case on X validated of someone puzzled by the simulation (and variance) of the random variable 1/X when being able to simulate X. And being surprised at the variance of the ratio being way larger than the variances of both numerator and denominator.

Monte Carlo in the convent

Last week, at the same time as the workshop on retrospective Monte Carlo in Warwick, there was a Monte Carlo conference in Paris, closing a Monte Carlo cycle run by Institut Louis Bachelier from October 2015 till June 2016. It took place in the convent of Les Cordeliers, downtown Paris [hence the title] and I alas could not attend the talks. As I organised a session on Bayesian (approximate) computations, with Richard Everitt, Jere Koskela, and Chris Sherlock as speakers (and Robin Ryder as chair), here are the slides of the speakers (actually, Jere most kindly agreed to give Chris’ talk as Chris was to sick to travel to Paris):

a Bernoulli factory of sorts?

A nice question was posted on X validated as to figure out a way to simulate a Bernoulli B(q) variate when using only a Bernoulli B(p) generator. With the additional question of handling the special case q=a/b, a rational probability. This is not exactly a Bernoulli factory problem in that q does not write as f(p), but still a neat challenge. My solution would have been similar to the one posted by William Huber, namely to simulate a sequence of B(p) or B(1-p) towards zooming on q until the simulation of the underlying uniforms U allows us to conclude at the position of U wrt q. For instance, if p>q and X~B(p) is equal to zero, the underlying uniform is more than p, hence more than q, leading to returning zero for the B(q) generation. Else, a second B(p) or B(1-p) generation means breaking the interval (0,p) into two parts, one of which allows for stopping the generation, and so on. The solution posted by William Huber contains an R code that could be easily improved by choosing for each interval between p and (1-p) towards the maximal probability of stopping. I still wonder at the ultimate optimal solution that would minimise the (average or median) number of calls to the Bernoulli(p) generator.