optimising accept-reject

November 20, 2012
By

(This article was first published on Xi'an's Og » R, and kindly contributed to R-bloggers)

I spotted on R-bloggers a post discussing optimising the efficiency of programming accept-reject algorithms. While it is about SAS programming, and apparently supported by the SAS company, there are two interesting features with this discussion. The first one is about avoiding the dreaded loop in accept-reject algorithms. For instance, taking the case of the truncated-at-one Poisson distribution, the code

rtpois=function(n,lambda){
  sampl=c()
  while (length(sampl)<n){
    x=rpois(1,lambda)
    if (x!=0) sampl=c(sampl,x)}
  return(sampl)
  }

is favoured by my R course students but highly inefficient:

> system.time(rtpois(10^5,.5))
   user  system elapsed
61.600  27.781  98.727

both for the stepwise increase in the size of the vector and for the loop. For instance, defining the vector sampl first requires a tenth of the above time (note the switch from 10⁵ to 10⁶):

> system.time(rtpois(10^6,.5))
   user  system elapsed
 54.155   0.200  62.635

As discussed by the author of the post, a more efficient programming should aim at avoiding the loop by predicting the number of proposals necessary to accept a given number of values. Since the bound M used in accept-reject algorithms is also the expected number of attempts for one acceptance, one should start with something around Mn proposed values. (Assuming of course all densities are properly normalised.) For instance, in the case of the truncated-at-one Poisson distribution based on proposals from the regular Poisson, the bound is 1/1-e. A first implementation of this principle is to build the sample via a few loops:

rtpois=function(n,lambda){
propal=rpois(ceiling(n/(1-exp(-lambda))),lambda)
propal=propal[propal>0]
n0=length(propal)
if (n0>=n)
return(propal[1:n])
else return(c(propal,rtpois(n-n0,lambda)))
}

with a higher efficiency:

> system.time(rtpois(10^6,.5))
   user  system elapsed
  0.816   0.092   0.928

Replacing the expectation with an upper bound using the variance of the negative binomial distribution does not make a significant dent in the computing time

rtpois=function(n,lambda){
  M=1/(1-exp(-lambda))
  propal=rpois(ceiling(M*(n+2*sqrt(n/M)/(M-1))),lambda)
  propal=propal[propal>0]
  n0=length(propal)
  if (n0>=n)
   return(propal[1:n])
  else return(c(propal,rtpois(n-n0,lambda)))}

since we get

> system.time(rtpois(10^6,.5))
   user  system elapsed
  0.824   0.048   0.877

The second point about this Poisson example is that simulating a distribution with a restricted support using another distribution with a larger support is quite inefficient. Especially when λ goes to zero By comparison, using a Poisson proposal with parameter μ and translating it by 1 may bring a considerable improvement: without getting into the gory details, it can be shown that the optimal value of μ (in terms of maximal acceptance probability) is λ and that the corresponding probability of acceptance is

\dfrac{1-\exp\{-\lambda\}}{\lambda}

which is larger than the probability of the original approach when λ is less than one. As shown by the graph below, this allows for a lower bound in the probability of acceptance that remains tolerable.


Filed under: R, Statistics, University life Tagged: accept-reject algorithm, loops, Poisson distribution, R, R course, SAS, system.time

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