[R] Latin Hypercube Sampling with a condition

[Ravi Varadhan]

I am not sure, but this thread from a couple of months ago might be relevant (and useful):

https://stat.ethz.ch/pipermail/r-help/2011-March/273423.html

Ravi.

________________________________________
From: r-help-bounces at r-project.org [r-help-bounces at r-project.org] on behalf of Rob Carnell [carnellr at battelle.org]
Sent: Thursday, June 02, 2011 8:30 AM
To: r-help at stat.math.ethz.ch
Subject: Re: [R] Latin Hypercube Sampling with a condition

Duarte Viana <viana.sptd <at> gmail.com> writes:

>
> Hello all,
>
> I am trying to do a Latin Hypercube Sampling (LHS) to a 5-parameter
> design matrix. I start as follows:
>
> library(lhs)
>
> p1<-randomLHS(1000, 5)
>
> If I check the distribution of each parameter (column), they are
> perfectly uniformly distributed (as expected).For example,
>
> hist(p1[,1])
>
> Now the hard (maybe strange) question. I want the combination of the
> first three parameters to sum up to 1 (which obviously do not)
>
> s<-p1[,1]+p1[,2]+p1[,3]
>
> s==1
>
> It occurred to me to divide each of these parameters with the sum
> (vector "s" above). However the uniform distribution is lost (example
> for parameter 1 - first column):
>
> par1.transf<-p1[,1]/s
>
> hist(par1.transf)
>
> So, is there a way to maintain the random LHS (with uniformly
> distributed parameters) so that the refered condition is fulfilled?
>
> Any suggestions would be much welcome.
>
> Thanks,
>
> Duarte
>
>

Duarte,

In my experience with Latin hypercube samples, most people draw the sample on
a uniform hypercube and then transform the uniform cube to have new
distributions on the margins.  The transformed distributions are not
necessarily uniform.  It is possible to draw a Latin hypercube with correlated
margins and I hope to add that to my package in the future.  I have also done
transforms such that the transformed marginal distributions are correlated (as
you have in your example).  I have not seen a correlated set of uniform
marginal distributions such that the margins sum to one, however.  I'll make a
quick example argument that explains the difficulty...

In two dimensions, you could draw this which is uniform and correlated.
x <- seq(0.05, 0.95, length=10)
y <- 1-x
all.equal(x+y, rep(1, length(x)))
hist(x)
hist(y)

But in three dimensions, it is hard to maintain uniformity because large
samples on the first uniform margin overweight the small samples on the other
margins.
x <- seq(0.05, 0.95, length=10)
y <- runif(length(x), 0, 1-x)
z <- 1-x-y
hist(x)
hist(y)
hist(z)

If you could explain why you want to maintain the uniformity on the margins, I
might be able to suggest something different.

Rob

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