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I'm trying to generate correlated data (preferably multinormal) with predefined correlations (e.g. 0.35 or 0.9). Any idea how I can do it? I'm using R and I did find a way to generate this (using mvrnorm), but you need to supply a covariance matrix. I have a covariance matrix with correlations around 0.9; however, I don't know how I can modify its entries to change the correlation. If I can do that, I'll be able to generate correlated data with the correlations I need.

Regards,

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  • $\begingroup$ You just need to play with the values in the covariance matrix in mvrnorm and relate them with the definition of correlation matrix. $\endgroup$
    – user10525
    Commented Apr 23, 2012 at 12:00
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    $\begingroup$ If you post the code for you covariance matrix we can tell you how to modify it to get other correlations. $\endgroup$
    – MånsT
    Commented Apr 23, 2012 at 12:11
  • $\begingroup$ Procrastinator, I can't just change the values in the matrix to whatever I want, changing any number has an effect on other entries in the matrix and I must know how the other entries change (inc. or dec.) before changing anything. For example, changing the variance of any variable will change its covaraince with the other variables. $\endgroup$
    – Jawad
    Commented Apr 23, 2012 at 12:30
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    $\begingroup$ The correlation between $X_i$ and $X_j$ is given $$ Cor(X_i,X_j) = \frac{Cov(X_i, X_j)}{sd(X_i)sd(X_j)}. $$ If your correlation matrix is V this is $$ Cor(X_i,X_j) = \frac{V_{ij}}{\sqrt{V_{ii}}\sqrt{V_{jj}}}. $$ Maybe this can help you set up your covariance matrix, especially if you are able to simplify your problem by standardizing each variable. $\endgroup$
    – Erik
    Commented Apr 23, 2012 at 13:25
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    $\begingroup$ This question has been discussed on here before. For example, look here: stats.stackexchange.com/questions/13382/… $\endgroup$
    – Macro
    Commented Apr 23, 2012 at 14:52

2 Answers 2

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Actually this is a trap question: it sounds easy but it is not (+1). The short answer to your question is you can't.

I will give an example. Imagine you have 3 Gaussian variables $X_1, X_2$ and $X_3$. You want the correlation between $X_1$ and $X_2$ to be 0, and all correlations with $X_3$ to be 1. This is obviously impossible because $X_1 = X_3$ and $X_3 = X_1$ says that $X_1 = X_2$ (up to shifting and scaling), which contrasts with the assumption that they are independent!

You would have the same situation if you replace 0 by "close to 0" and 1 by "close to 1" in the previous example. The issue here is that not every matrix is a correlation matrix. The requirement for being a correlation matrix is to be symmetric and positive definite.

You cannot choose arbitrary correlation values, but you can check whether they define a valid correlation matrix. Say that you have a symmetric square matrix mat with required correlation coefficients. You can test that it is positive definite as shown below.

all(eigen(mat)$values >= 0)

For symmetric real matrices, positive definite is equivalent to having all eigenvalues positive.

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  • $\begingroup$ It might be good to make the inequality in the code nonstrict to allow for perfect correlations between linear combinations of variables. $\endgroup$
    – cardinal
    Commented Jun 3, 2012 at 14:41
  • $\begingroup$ @cardinal Done. But that is purely for demonstration purposes. Testing strict equality of real numbers is something R cannot do as (.3-.2) == (.2-.1) shows. $\endgroup$
    – gui11aume
    Commented Jun 3, 2012 at 14:47
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    $\begingroup$ Good point; it was actually the larger conceptual point I was trying to address. That "limitation" has more to do with floating point representation, than R itself, though. Testing against zero is a bit special. Some related routines in R will truncate small values to zero if they fall below a tolerance. $\endgroup$
    – cardinal
    Commented Jun 3, 2012 at 14:54
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    $\begingroup$ @cardinal 'That "limitation" has more to do with floating point representation, than R itself, though' Yes of course. Apologies to the R team :-) $\endgroup$
    – gui11aume
    Commented Jun 3, 2012 at 15:01
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The MASS package has a function called mvrnorm() that can generate a group or random numbers to a specified level of correlation. An example of the setup can be found in the beginning of the example here: http://menugget.blogspot.de/2011/11/propagation-of-error.html

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  • $\begingroup$ Sorry, didn't see that Jawad had already pointed you to the same function. In any case, the example posted might help you understand how to set it up. $\endgroup$
    – Marc in the box
    Commented Apr 23, 2012 at 12:47
  • $\begingroup$ Thanks Marc, from the page I understand that all I have to do is set the diagonal elements of my covariance matrix to rho and the off-diagonal elements to 1 and I should get the data I need correlated by rho? $\endgroup$
    – Jawad
    Commented Apr 23, 2012 at 13:16
  • $\begingroup$ Not exactly - the covariance matrix will depend on your defined standard deviations. If sd=1 for all series, then you are correct. Otherwise, you will need to define your std. devs for each series. $\endgroup$
    – Marc in the box
    Commented Apr 23, 2012 at 13:26
  • $\begingroup$ No. The variances of the variables should be along the diagonal and the off-diagonal elements should be rho (if $\sigma^2=1$). $\endgroup$
    – MånsT
    Commented Apr 23, 2012 at 13:38

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