Author: Rob Carnell
Date: 2024-06-30

# Examples of Correlated and Multivariate Latin hypercubes

#### Rob Carnell

#### 2024-06-30

Generally, Latin hypercubes are drawn in a way that makes the marginal distributions for the uniform Latin hypercube and transformed Latin hypercube independent. In some cases researches want to use Latin hypercubes to study problem areas where the marginal distributions are correlated or come from a multivariate distribution.

There are a variety of methods to create such a Latin hypercube. The method used in this package is to draw and transform an uncorrelated hypercube and then use the columnwise-pairwise algorithm to adjust to a set of pre-defined conditions. A set of predefined conditions are not always achievable.

## Example 1: Simple Correlation

Assumptions:

- \(X_1 \sim uniform(2, 4)\)
- \(X_2 \sim Normal(1, 3)\)
- \(X_3 \sim Exponential(3)\)
- \(X_4 \sim LogNormal(1, 1)\)
- \(cor(x_1, x_2) = 0.3\)
- \(cor(x_3, x_4) = 0.5\)

    lhs_A <- correlatedLHS(lhs::randomLHS(30, 4),
                           marginal_transform_function = function(W, ...) {
                             W[,1] <- qunif(W[,1], 2, 4)
                             W[,2] <- qnorm(W[,2], 1, 3)
                             W[,3] <- qexp(W[,3], 3)
                             W[,4] <- qlnorm(W[,4], 1, 1)
                             return(W)
                           },
                           cost_function = function(W, ...) {
                             (cor(W[,1], W[,2]) - 0.3)^2 + (cor(W[,3], W[,4]) - 0.5)^2
                           },
                           debug = FALSE, maxiter = 1000)

Check that the desired correlations were created:

    cor(lhs_A$transformed_lhs[,1:2])[1,2]
    #> [1] 0.2999044

    cor(lhs_A$transformed_lhs[,3:4])[1,2]
    #> [1] 0.499209

## Example 2: Dirichlet distribution

Assume that we want \(X\) to be Dirichlet distributed with \(\alpha = 4,3,2,1\)

Therefore the margins of the Dirichlet are:

- \(X_1 ~ beta(4, 10-4)\)
- \(X_2 ~ beta(3, 10-3)\)
- \(X_3 ~ beta(2, 10-2)\)
- \(X_4 ~ beta(1, 10-1)\)

### Method 1: `correlatedLHS`

    lhs_B <- correlatedLHS(lhs::randomLHS(30, 4),
                           marginal_transform_function = function(W, ...) {
                             W[,1] <- qbeta(W[,1], 4, 6)
                             W[,2] <- qbeta(W[,2], 3, 7)
                             W[,3] <- qbeta(W[,3], 2, 8)
                             W[,4] <- qbeta(W[,4], 1, 9)
                             return(W)
                           },
                           cost_function = function(W, ...) {
                             sum((apply(W, 1, sum) - 1)^2)
                           },
                           debug = FALSE,
                           maxiter = 1000)

Check properties

    range(apply(lhs_B$transformed_lhs, 1, sum)) # close to 1
    #> [1] 0.947719 1.093283

    apply(lhs_B$transformed_lhs, 2, mean) # close to 4/10, 3/10, 2/10, 1/10
    #> [1] 0.40105106 0.29898984 0.20216883 0.09855691

### Method 2: `q_dirichlet`

    lhs_B <- lhs::qdirichlet(lhs::randomLHS(30, 4), c(4,3,2,1))

Check properties

    all(abs(apply(lhs_B, 1, sum) - 1) < 1E-9) # all exactly 1
    #> [1] TRUE

    apply(lhs_B, 2, mean) # close to 4/10, 3/10, 2/10, 1/10
    #> [1] 0.3976103 0.2983999 0.2033601 0.1006297

## Example 3: Rejection Sample

Assumptions:

- \(X_1 \sim uniform(1, 4)\)
- \(X_2 \sim uniform(10^{-6}, 2)\)
- \(X_3 \sim uniform(2, 6)\)
- \(X_4 \sim uniform(10^{-6}, 0.1)\)
- \(lower < \prod_{i=1}^4 X_i < upper\)

First build an empirical sample using rejection sampling

    set.seed(3803)
    N <- 100000

    reject_samp <- data.frame(
      v1 = runif(N, 1, 4),
      v2 = runif(N, 1E-6, 2),
      v3 = runif(N, 2, 6),
      v4 = runif(N, 1E-6, 0.1)
    )

    p <- with(reject_samp, v1*v2*v3*v4)
    ind <- which(p < 1 & p > 0.3)
    reject_samp <- reject_samp[ind,]

Now build the correlated sample using the reject sample as an empirical distribution function and the boundaries as a cost function.

    lhs_C <- correlatedLHS(lhs::randomLHS(30, 4),
                           marginal_transform_function = function(W, empirical_sample, ...) {
                             res <- W
                             for (i in 1:ncol(W)) {
                               res[,i] <- quantile(empirical_sample[,i], probs = W[,i])
                             }
                             return(res)
                           },
                           cost_function = function(W, ...) {
                             p <- W[,1]*W[,2]*W[,3]*W[,4]
                             pp <- length(which(p > 0.3 & p < 1)) / nrow(W)
                             return(1-pp)
                           },
                           debug = FALSE,
                           maxiter = 10000,
                           empirical_sample = reject_samp)