I want to create a climate model ensemble, testing 5 parameters (real, uniformly distributed between two values), using a latin hypercube approach. The problem is that I'm not sure how many replications I want to do. Is it viable to do one latin hypercube of 20 samples, and then another of 10? How would I make sure the 30 samples are relatively evenly distributed? Or would it be possible to do 10-sized LHCs, and do multiples? ie. do 3 or 4 LHC samples of size 10, making sure that each LHC was independent of the others?


1 Answer 1


With Latin hypercube samples, you have to decide on the number of samples, so that you break your range into either 10 or 20 bins to begin with. Otherwise, you will likely miss some parts of your space. My understanding is that the only quasi-Monte Carlo method that allows to take the next sample (or next $N$ samples) easily without trying to figure out their dependence on the previously collected samples is Halton sequence. See the encompassing treatment in Niederreiter (1992).

In fact, as far as I recall from computational physics literature (most likely, it was in Morokoff and Caflisch (1995)), for the sequence of length up to about $6^d$ where $d$ is the dimension of your space, quasi Monte Carlo sequences do not show appreciable gains over the standard pseudo-random number generators. So you may not have to bother with LHC and agonize over the choice between 10 and 20 samples -- you can just start with any random number generator you have at hand, and keep adding new ones if you are not satisfied with the achieved precision.

  • $\begingroup$ Thanks, great answer, even if it did scare me a bit with it's depth at first glance (I have little experience with Monte-carlo methods). $\endgroup$
    – naught101
    Mar 20, 2012 at 0:44
  • $\begingroup$ The Halton sequence is especially cool, although I find it odd that you use a separate prime for each dimension. I guess that any discrepancy between distributions in different dimensions would clear up pretty quickly though. $\endgroup$
    – naught101
    Mar 20, 2012 at 0:53
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    $\begingroup$ No, the standard Halton sequence gets a bit weird in higher dimensions/bigger primes: the subsequent point lie on parallel lines, so to get the unit square covered for 41 and 43, you need to have 41*43 = 1700 points. The solutions to this problem involve scrambling operations, although I am not sure whether Niederreiter covers it. I just published a little paper explaining them :) $\endgroup$
    – StasK
    Mar 23, 2012 at 4:30
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    $\begingroup$ Yeah, I see that. I think I've decided to use the Sobol' sequence for that reason. I asked a separate question about which sequence is preferable in which situation $\endgroup$
    – naught101
    Mar 23, 2012 at 5:03

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