Abstract

In this vignette, I demonstrate how to calculate the MLE and run a likelihood ratio test in the mean-zero array normal model.

Maximum Likelihood Estimation

First, I simulate some data. X will be generated with identity covariance along all modes. Y will have identity covariance along the first three modes, and an AR-1(0.9) covariance along the fourth mode. Z will have diagonal covariance along the first two modes, and an AR-1(0.9) covariance along the third mode.

library(tensr)
p <- c(10, 10, 10, 10)
X <- array(rnorm(prod(p)),dim = p)

cov_Y <- start_ident(p)
cov_Y[[4]] <- 0.9^abs(outer(1:p[4],1:p[4],"-"))
cov_Y[[4]] <- cov_Y[[4]] / det(cov_Y[[4]])^(1/p[4]) #scale to have unit determinant.
Y <- atrans(array(rnorm(prod(p)),dim = p), lapply(cov_Y, mhalf))


cov_Z <- start_ident(p)
diag(cov_Z[[1]]) <- 1:p[1] / prod(1:p[1])^(1/p[1])
diag(cov_Z[[2]]) <- p[2]:1 / prod(1:p[2])^(1/p[2])
cov_Z[[3]] <- 0.9^abs(outer(1:p[3],1:p[3],"-"))
cov_Z[[3]] <- cov_Z[[3]] / det(cov_Z[[3]])^(1/p[3])
Z <- atrans(array(rnorm(prod(p)),dim = p), lapply(cov_Z, mhalf))

The holq is used to calculate the maximum likelihood estimates. Assuming we know the covariance structure of each mode, we set the the appropriate modes to have identity or diagonal covariance matrices by using the mode_rep or mode_diag options.

holq_X <- holq(X,mode_rep = 1:4, print_diff = FALSE)
holq_Y <- holq(Y,mode_rep = 1:3, print_diff = FALSE)
holq_Z <- holq(Z,mode_rep = 4, mode_diag = c(1,2), print_diff = FALSE)

mle_X <- mle_from_holq(holq_X)
mle_Y <- mle_from_holq(holq_Y)
mle_Z <- mle_from_holq(holq_Z)

Estimates of the scale are pretty close to the true value of 1.

cat("Estimates of scale:\n",
    "From X:", mle_X$sig_mle,"\n",
    "From Y:", mle_Y$sig_mle,"\n",
    "From Z:", mle_Z$sig_mle,"\n"
)

## Estimates of scale:
##  From X: 1.005876 
##  From Y: 0.9890368 
##  From Z: 0.9900916

And the estimates of the covariances are pretty close to their truth. For example, when we look at the true and estimated variances of the first mode covariance in Z, we get

cat("     True:", paste("(", paste(format(diag(cov_Z[[1]]), digits = 2),
                                   collapse = ", "), ")",sep = ""), "\n",
    "Estimate:", paste("(", paste(format(diag(mle_Z[[1]][[1]]), digits = 2),
                                  collapse = ", "), ")", sep = ""), "\n")

##      True: (0.22, 0.44, 0.66, 0.88, 1.10, 1.32, 1.55, 1.77, 1.99, 2.21) 
##  Estimate: (0.21, 0.46, 0.65, 0.89, 1.05, 1.36, 1.65, 1.72, 1.94, 2.27)

Likelihood Ratio Testing

Let’s test for all modes having the identity covariance matrix versus the first three modes having the identity covariance matrix and the fourth is unconstrained. Of course, in this situation we need not resort to tensor methods. Using X as our data should only reject the null 5% of the time. But for Y where the alternative is true, we should have more power.

First, we’ll calculate the null distribution of the likelihood ratio test statistic. Then we’ll calculate these test statistics using both X and Y as our data.

null_distribution <- lrt_null_dist_dim_same(p = p, null_ident = 1:4,
                                            alt_ident = 1:3, itermax = 500)

sig_k <- holq(X,mode_rep = 1:3, print_diff = FALSE)$sig
sig_h <- holq(X,mode_rep = 1:4, print_diff = FALSE)$sig
lrt_stat_val_X <- lrt_stat(sig_null = sig_h,sig_alt = sig_k,p = p)

sig_k <- holq(Y,mode_rep = 1:3, print_diff = FALSE)$sig
sig_h <- holq(Y,mode_rep = 1:4, print_diff = FALSE)$sig
lrt_stat_val_Y <- lrt_stat(sig_null = sig_h,sig_alt = sig_k,p = p)

We can calculate p-values.

p_value_x <- mean(null_distribution > lrt_stat_val_X)
p_value_y <- mean(null_distribution > lrt_stat_val_Y)

cat(" p-value using X:", p_value_x,"\n",
    "p-value using Y:", p_value_y,"\n")

##  p-value using X: 0.206 
##  p-value using Y: 0

We can also plot the null distribution along with the observed statistics. If you can’t see the line for Y, that’s because the likelihood ratio test statistic is so large.

par(mar = c(2.5, 2.5, 2, 0), mgp = c(1.5, .5, 0))
hist(null_distribution, xlab = "LRT Stat", main = "Null Distribution")
abline(v = lrt_stat_val_X, col = 2, lwd = 2, lty = 2)
abline(v = lrt_stat_val_Y, col = 4, lwd = 2, lty = 2)
legend("topright", c("From X","From Y"), col = c(2,4), lty = 2, lwd = 2, cex = 0.6)

As a second example, we run a likelihood ratio test using Z, testing for diagional covariance along the first mode, identity covariance along the second and fourth modes, and unstructured along the third, against the correct model of diagonal along the first two modes, identity along the fourth mode, and unstructured along the third.

null_distribution <- lrt_null_dist_dim_same(p = p,null_ident = c(2,4),
                                            alt_ident = 4, null_diag = 1,
                                            alt_diag = c(1,2), itermax = 500)

sig_k <- holq(Z, mode_rep = 4, mode_diag = c(1,2), print_diff = FALSE)$sig
sig_h <- holq(Z, mode_rep = c(2,4), mode_diag = 1, print_diff = FALSE)$sig
lrt_stat_val_Z <- lrt_stat(sig_null = sig_h, sig_alt = sig_k, p = p)

p_value_z <- mean(null_distribution > lrt_stat_val_Z)

cat("P-value:", p_value_z,"\n")

## P-value: 0

References

Gerard, D., & Hoff, P. (2016). A higher-order LQ decomposition for separable covariance models. Linear Algebra and its Applications, 505, 57-84. [Link to LAA] [Link to arXiv]