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