Generate numbers with specific correlation [closed]

Question

How to generate a sequence of numbers, which would have a specific correlation (for example 0.56) and would consist of.. say 50 numbers with R program? Ty.

Answer 1

Assuming you mean two normal/Gaussian vectors of values with correlation 0.56

We can use mvrnorm() from package MASS

require(MASS)
out <- mvrnorm(50, mu = c(0,0), Sigma = matrix(c(1,0.56,0.56,1), ncol = 2),
               empirical = TRUE)

which gives

> cor(out)
     [,1] [,2]
[1,] 1.00 0.56
[2,] 0.56 1.00

The empirical = TRUE bit is important otherwise the actual correlation achieved is subject to randomness too and will not be exactly the stated value with larger discrepancies for smaller samples.

Assuming you mean a lag 1 correlation of 0.56 & Gaussian random variables

For this one you can use the arima.sim() function:

> arima.sim(list(ar = 0.56), n = 50)
Time Series:
Start = 1 
End = 50 
Frequency = 1 
 [1]  0.62125233 -0.04742303  0.57468608 -0.07201988 -1.91416757 -1.11827563
 [7]  0.15718249  0.63217365 -1.24635896 -0.22950855 -0.79918784  0.31892842
[13]  0.33335688 -1.24328177 -0.79056890  1.08443057  0.55553819  0.33460674
[19] -0.33037659 -0.65244221  0.70461755  0.61450122  0.53731454  0.19563672
[25]  1.73945110  1.27119241  0.82484460  1.58382861  1.81619212 -0.94462052
[31] -1.36024898 -0.30964390 -0.94963216 -3.75725819 -1.77342095 -1.20963799
[37] -1.76325350 -1.20556172 -0.94684678 -0.85407649  0.14922226 -0.31109945
[43]  0.39456259  0.89610859 -0.70913792 -2.27954408 -1.14722464  0.39140446
[49]  0.66376227  1.63275483

Answer 2

If you don't want to specify those matrices, other options are corgen from ecodist:

library("ecodist")
xy <- corgen(len = 50, r = 0.56, epsilon = 0.01)

Or rolling your own:

simcor <- function (n, xmean, xsd, ymean, ysd, correlation) {
    x <- rnorm(n)
    y <- rnorm(n)
    z <- correlation * scale(x)[,1] + sqrt(1 - correlation^2) * 
             scale(resid(lm(y ~ x)))[,1]
    xresult <- xmean + xsd * scale(x)[,1]
    yresult <- ymean + ysd * z
    data.frame(x=xresult,y=yresult)
}

Test

> r <- simcor(n = 50, xmean = 12, ymean = 30, xsd = 3, ysd = 4, correlation = 0.56)
> cor(r$x,r$y)
[1] 0.56

Answer 3

Use rmvnorm from the mvtnorm package to sample from the multivariate normal distribution. For example for correlation of 0.56:

library("mvtnorm")
foo <- rmvnorm(10000,c(0,0),matrix(c(1,0.56,0.56,1),2,2))

Test:

> cor(foo[,1],foo[,2])
[1] 0.5611207