Let's say, for the sake of argument, that you want a latin hypercube sample with 1,000 samples.

For a later computational step, it would be more convenient to have a group of ten sets of 100 samples each, rather than a single massive set.

Are those two things equivalent?

My intuition is no, because you don't have 1000 equally probable spaces with one point in them, you have 100 equally probable spaces with ten points in each, but I want to double check that.


3 Answers 3


Your intuition is correct. Here is a more precise mathematical description. I find that this book chapter is a useful reference for material on Latin Hypercube sampling.

Assume that the LHS samples are taken on the unit interval $[0,1)$ as described in Section 10.3 of the book chapter. Let $X$ denote the vector of samples from the 1000 sample LHS. Let $Y$ denote the vector of samples from the ten combined 100 sample LHS. How are $Y$ and $X$ different?

Even in one dimension, the different 'levels' of stratification imply that (the distributions of) these samples differ. Choose a particular dimension - then consider the number of samples falling into the subinterval $[0,0.001)$ within the chosen dimension. The basic properties of LHS (10.6 and 10.7 in the book chapter) imply that, for $X$, the number of samples in this subinterval is fixed at one. However, for $Y$, the number of samples in this subinterval will be $\text{Binomial}(10,0.1)$, which is clearly random, not fixed.

In multiple dimensions, other differences between $X$ and $Y$ will be apparent. Choose two dimensions and consider only the samples residing in a small square such as $[0,0.01)\times[0,0.01)$ within the chosen two dimensions. For $Y$ the location of a sample inside this square places no constraints on the locations of other samples inside this square. However, for $X$ the location of a sample inside this square restricts the locations of other samples inside the square ('unthreatened rooks').

See also this question which approaches the same situation from a slightly different angle (asking whether the estimates obtained from this kind of modified LHS are still unbiased).


Those two methods of drawing a Latin hypercube sample are not equivalent, and the repeated method does not form a LHS when taken as a whole.

In code with R:



# draw 10 LHS's of size 100
X <- vector("list", 10)
for (i in seq_along(X)) X[[i]] <- randomLHS(100, 4)
Z <- do.call(rbind, X)

# draw one LHS of size 1000
Y <- randomLHS(1000, 4)

# margins look uniform
for (i in 1:4) hist(Z[,i], breaks = 50)

# margins look uniform
for (i in 1:4) hist(Y[,i], breaks = 50)

isLatinHypercube <- function(a)
  temp <- floor(a*nrow(a)) + 1
  all(apply(temp, 2, function(z) length(z) == length(unique(z)))) &
    all(apply(temp, 2, sum) == nrow(a)*(nrow(a) + 1) / 2)

isLatinHypercube(Z) # FALSE
isLatinHypercube(Y) # TRUE

Intuitively I would think they're not equivalent either. Although I don't have the math to back it up, I think we can work through this with words.

As the Wikipedia page on LHS puts it, we want our sampled points to be "unthreatened rooks" on a chess board. With the 1000 sample LHS, you know for sure that each sample is an unthreatened rook. However with 10 100 sample LHSs, it is possible you are sampling two or more observations that would be threatening each other on the 1000 sample LHS.

I might be re-stating your intuition with different words, but I think you're intuition is right.


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