Blog Archives

another riddle with a stopping rule

May 26, 2016
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another riddle with a stopping rule

A puzzle on The Riddler last week that is rather similar to an earlier one. Given the probability (1/2,1/3,1/6) on {1,2,3}, what is the mean of the number N of draws to see all possible outcomes and what is the average number of 1’s in those draws? The second question is straightforward, as the proportions

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occupancy rules

May 22, 2016
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occupancy rules

While the last riddle on The Riddler was rather anticlimactic, namely to find the mean of the number Y of empty bins in a uniform multinomial with n bins and m draws, with solution , this led

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ABC random forests for Bayesian parameter inference

May 19, 2016
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ABC random forests for Bayesian parameter inference

Before leaving Helsinki, we arXived the paper Jean-Michel presented on Monday at ABCruise in Helsinki. This paper summarises the experiments Louis conducted over the past months to assess the great performances of a random forest regression approach to ABC parameter inference. Thus validating in this experimental sense the use of

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Using MCMC output to efficiently estimate Bayes factors

May 18, 2016
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Using MCMC output to efficiently estimate Bayes factors

As I was checking for software to answer a query on X validated about generic Bayes factor derivation, I came across an R software called BayesFactor, which only applies in regression settings and relies on the Savage-Dickey representation of the Bayes factor when the null hypothesis writes as θ=θ⁰ (and possibly additional nuisance parameters with

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reversible chain[saw] massacre

May 15, 2016
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reversible chain[saw] massacre

A paper in Nature this week that uses reversible-jump MCMC, phylogenetic trees, and Bayes factors. And that looks at institutionalised or ritual murders in Austronesian cultures. How better can it get?! “by applying Bayesian phylogenetic methods (…) we find strong support for models in which human sacrifice stabilizes social stratification once stratification has arisen, and

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AISTATS 2016 [#1]

May 10, 2016
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AISTATS 2016 [#1]

Travelling through Seville, I arrived in Càdiz on Sunday night, along with a massive depression . Walking through the city from the station was nonetheless pleasant as this is an town full of small streets and nice houses. If with less churches than Seville! Richard Samworth gave the first plenary talk of AISTATS 2016  with

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a Simpson paradox of sorts

May 5, 2016
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a Simpson paradox of sorts

The riddle from The Riddler this week is about finding an undirected graph with N nodes and no isolated node such that the number of nodes with more connections than the average of their neighbours is maximal. A representation of a connected graph is through a matrix X of zeros and ones, on which one

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gap frequencies [& e]

April 28, 2016
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gap frequencies [& e]

A riddle from The Riddler where brute-force simulation does not pay: For a given integer N, pick at random without replacement integers between 1 and N by prohibiting consecutive integers until all possible entries are exhausted. What is the frequency of selected integers as N grows to infinity? A simple implementation of the random experiment

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Le Monde puzzle [#960]

April 27, 2016
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Le Monde puzzle [#960]

An arithmetic Le Monde mathematical puzzle: Given an integer k>1, consider the sequence defined by F(1)=1+1 mod k, F²(1)=F(1)+2 mod k, F³(1)=F²(1)+3 mod k, &tc. For which value of k is the sequence the entire {0,1,…,k-1} set? This leads to an easy brute force resolution, for

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an integer programming riddle

April 20, 2016
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an integer programming riddle

A puzzle on The Riddler this week that ends up as a standard integer programming problem. Removing the little story around the question, it boils down to optimise 200a+100b+50c+25d under the constraints 400a+400b+150c+50d≤1000, b≤a, a≤1, c≤8, d≤4, and (a,b,c,d) all non-negative integers. My first attempt was a brute force R code since there are only

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