rxode2random (version 2.1.1)

rxnbinom: Simulate Binomial variable from threefry generator

Description

Care should be taken with this method not to encounter the birthday problem, described https://www.johndcook.com/blog/2016/01/29/random-number-generator-seed-mistakes/. Since the sitmo threefry, this currently generates one random deviate from the uniform distribution to seed the engine threefry and then run the code.

Usage

rxnbinom(size, prob, n = 1L, ncores = 1L)
rxnbinomMu(size, mu, n = 1L, ncores = 1L)

Value

negative binomial random deviates. Note that rxbinom2

uses the mu parameterization an the rxbinom uses the prob

parameterization (mu=size/(prob+size))

Arguments

size

target for number of successful trials, or dispersion parameter (the shape parameter of the gamma mixing distribution). Must be strictly positive, need not be integer.

prob

probability of success in each trial. 0 < prob <= 1.

n

number of observations. If length(n) > 1, the length is taken to be the number required.

ncores

Number of cores for the simulation

rxnorm simulates using the threefry sitmo generator;

mu

alternative parametrization via mean: see ‘Details’.

Details

Therefore, a simple call to the random number generated followed by a second call to random number generated may have identical seeds. As the number of random number generator calls are increased the probability that the birthday problem will increase.

The key to avoid this problem is to either run all simulations in the rxode2 environment once (therefore one seed or series of seeds for the whole simulation), pre-generate all random variables used for the simulation, or seed the rxode2 engine with rxSetSeed()

Internally each ID is seeded with a unique number so that the results do not depend on the number of cores used.

Examples

## Use threefry engine

rxnbinom(10, 0.9, n = 10) # with rxbinom you have to explicitly state n
rxnbinom(3, 0.5, n = 10, ncores = 2) # You can parallelize the simulation using openMP

rxnbinom(4, 0.7)

# use mu parameter
rxnbinomMu(40, 40, n=10)