Between covid-19 news and announcements of imminent Russia-Ukraine wars I needed a bit of a distraction. Sooo, here it is how to create an n x n autocorrelation matrix based on a correlation rho, with a simple 5 x 5 example in R:
n <- 5
rho <- 0.9
mat <- diag(n)
mat <- rho^abs(row(mat) - col(mat))This produces the following output:
[,1] [,2] [,3] [,4] [,5] [1,] 1.0000 0.900 0.81 0.729 0.6561 [2,] 0.9000 1.000 0.90 0.810 0.7290 [3,] 0.8100 0.900 1.00 0.900 0.8100 [4,] 0.7290 0.810 0.90 1.000 0.9000 [5,] 0.6561 0.729 0.81 0.900 1.0000
How does this work? Starting from defining n as 5, and the basic rho correlation as 0.9, then mat <- diag(n) creates a diagonal matrix—a square matrix with ones in the diagonal and zeros elsewhere—with 5 rows and 5 columns.
[,1] [,2] [,3] [,4] [,5] [1,] 1 0 0 0 0 [2,] 0 1 0 0 0 [3,] 0 0 1 0 0 [4,] 0 0 0 1 0 [5,] 0 0 0 0 1
Looking at the final line of code, we have that row(mat) and col(mat) create matrices with the row or col coordinates for each location:
> row(mat)
[,1] [,2] [,3] [,4] [,5]
[1,] 1 1 1 1 1
[2,] 2 2 2 2 2
[3,] 3 3 3 3 3
[4,] 4 4 4 4 4
[5,] 5 5 5 5 5
> col(mat)
[,1] [,2] [,3] [,4] [,5]
[1,] 1 2 3 4 5
[2,] 1 2 3 4 5
[3,] 1 2 3 4 5
[4,] 1 2 3 4 5
[5,] 1 2 3 4 5
Then abs(row(mat) - col(mat)) is telling us how far is each cell in the matrix from the diagonal. We use that distance as an exponent for rho (0.9 in our example).
> abs(row(mat) - col(mat))
[,1] [,2] [,3] [,4] [,5]
[1,] 0 1 2 3 4
[2,] 1 0 1 2 3
[3,] 2 1 0 1 2
[4,] 3 2 1 0 1
[5,] 4 3 2 1 0
Putting it all together, rho^abs(row(mat) - col(mat)) produces:
> rho^abs(row(mat) - col(mat))
[,1] [,2] [,3] [,4] [,5]
[1,] 1.0000 0.900 0.81 0.729 0.6561
[2,] 0.9000 1.000 0.90 0.810 0.7290
[3,] 0.8100 0.900 1.00 0.900 0.8100
[4,] 0.7290 0.810 0.90 1.000 0.9000
[5,] 0.6561 0.729 0.81 0.900 1.0000