trick or treat?!

Two weeks ago, we went to a local restaurant, connected to my running grounds, for dinner. While the setting in a 16th building that was part of the original Sceaux castle was quite nice, the fare was mediocre and the bill more suited for a one star Michelin than dishes I could have cooked myself. The height (or rather bottom) of the meal was a dish of sardines consisting in an half-open pilchard can… Just dumped on a plate with a slice of bread. It could have been a genius stroke from the chef had the sardines been cooked and presented in the can, alas it sounded more like the act of an evil genie! Or more plainly a swindle. As those tasty sardines came straight from the shop!

morning run

41ième cross de Sceaux [5km/19:11/15th/2nd V2]

the raven, the cormorant, and the heron

This morning, on my first lap of the Grand Bassin in Parc de Sceaux, I spotted “the” heron standing at its usual place, on the artificial wetland island created at one end of the canal. When coming back to this spot during the second lap, I could hear the heron calling loudly and saw a raven repeatedly diving near it and a nearby cormorant, who also seemed unhappy with this attitude, judging from the flapping of its wings… After a few dozens of those dives, the raven landed at the other end of the island and this was the end of the canal drama! Unless there was a dead fish landed there, I wonder why the raven was having a go at those two larger birds.

sakura [#3]

sakura [#2]

simulating correlated Binomials [another Bernoulli factory]

This early morning, just before going out for my daily run around The Parc, I checked X validated for new questions and came upon that one. Namely, how to simulate X a Bin(8,2/3) variate and Y a Bin(18,2/3) such that corr(X,Y)=0.5. (No reason or motivation provided for this constraint.) And I thought the following (presumably well-known) resolution, namely to break the two binomials as sums of 8 and 18 Bernoulli variates, respectively, and to use some of those Bernoulli variates as being common to both sums. For this specific set of values (8,18,0.5), since 8×18=12², the solution is 0.5×12=6 common variates. (The probability of success does not matter.) While running, I first thought this was a very artificial problem because of this occurrence of 8×18 being a perfect square, 12², and cor(X,Y)x12 an integer. A wee bit later I realised that all positive values of cor(X,Y) could be achieved by randomisation, i.e., by deciding the identity of a Bernoulli variate in X with a Bernoulli variate in Y with a certain probability ϖ. For negative correlations, one can use the (U,1-U) trick, namely to write both Bernoulli variates as

in order to minimise the probability they coincide.

I also checked this result with an R simulation

> z=rbinom(10^8,6,.66)
> y=z+rbinom(10^8,12,.66)
> x=z+rbinom(10^8,2,.66)
cor(x,y)
> cor(x,y)
[1] 0.5000539

Searching on Google gave me immediately a link to Stack Overflow with an earlier solution with the same idea. And a smarter R code.