Can anyone suggest a method for generating random correlation matrix with $90\%$ of the off-diagonal entries between $[-0.3, 0.3]$. The other $10\%$ should be larger than $0.3$ or smaller than $-0.3$.

  • $\begingroup$ You have to be aware that you can't get arbitrary negative correlations between variables, for one thing. $\endgroup$
    – cardinal
    Commented May 7, 2011 at 1:23
  • $\begingroup$ Can you define what you mean by "random"? This seems related to your previous question. $\endgroup$
    – cardinal
    Commented May 7, 2011 at 1:29
  • 1
    $\begingroup$ Is the 90/10 requirement "hard"? In lower dimensions you might be able to get close by drawing from a Wishart centered at $I$, computing the correlation matrix, and rejecting samples that aren't within some tolerance. Though I suspect this won't scale well at all... $\endgroup$
    – JMS
    Commented May 7, 2011 at 1:52
  • $\begingroup$ @JMS We're starting to get some clarification in new comments to the preceding question linked to by @Cardinal: you might want to check there. $\endgroup$
    – whuber
    Commented May 7, 2011 at 2:19
  • $\begingroup$ @whuber Good, thanks. Might we close this then? It seems "duplicate in intention" as it were and doesn't contain much beyond my foolishness. $\endgroup$
    – JMS
    Commented May 7, 2011 at 3:25

3 Answers 3


Here's a heuristic that I coded up quickly that seems to do quite well:

  1. Initialize a matrix with 1 on the diagonals.
  2. Fill out the upper triangular sub-matrix according to your distribution (90% are uniform on (-.3,.3) and 10% outside that).
  3. Make the matrix symmetric.
  4. Now iterate between
    • Project the matrix onto the PSD cone.
    • Project the matrix onto the set of matrices with diagonal 1.
  5. Alternating projections converges, so we just hope that the matrix we get out has values according to your distribution (see simulation for the check).
   pickone <- function(x){
  } else {

generateMat <- function(x){
  X <- matrix(0,nrow=10,ncol=10)
  diag(X) <- rep(1,10)
  X[upper.tri(X)] <- sapply(1:45,pickone)
  X <- X + t(X)-diag(rep(1,10))
  Xnew <- X

  for(i in 1:50){
    eig <- eigen(Xnew)
    ##project onto the PSD cone
    Xnew <- eig$vectors%*%diag(sapply(eig$values,max,0))%*%t(eig$vectors)
    ##project onto the set of matrices with diagonal 1
    diag(Xnew) <- rep(1,10)

  vals <- Xnew[upper.tri(Xnew)]
  return(mean(vals < .3 & vals > -.3))


   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
 0.7556  0.8667  0.8889  0.8960  0.9333  0.9778

It seems like most of the values after simulating 100 times are close to 90% within (-.3,.3).

  • 2
    $\begingroup$ I don't see the point of doing these alternating projects. It seems like a waste. Let $\newcommand{\Xm}{\mathbf{X}}\newcommand{\Dm}{\mathbf{D}}\Xm$ be your initial matrix and $\widetilde{\Xm}$ the first projection onto the PSD cone. Let $\Dm = \mathrm{diag}(\widetilde{\Xm})$, i.e., the diagonal matrix of the diagonal entries of $\widetilde{\Xm}$. Then $\hat{\Xm} = \Dm^{-1/2} \widetilde{\Xm} \Dm^{-1/2}$ is a positive semidefinite correlation matrix. No iteration needed. $\endgroup$
    – cardinal
    Commented May 12, 2011 at 23:14
  • 1
    $\begingroup$ Also, a note on R. You can use pmax(eig$values,0) instead of the more cumbersome sapply call. $\endgroup$
    – cardinal
    Commented May 12, 2011 at 23:16
  • $\begingroup$ @cardinal Right on both counts. Thanks for the suggestions. $\endgroup$
    – ncray
    Commented May 13, 2011 at 0:17
  • $\begingroup$ (+1) For the basic idea. But, I don't think the distribution is going to be at all uniform on $[-3,3]$, especially for larger dimensions. $\endgroup$
    – cardinal
    Commented May 13, 2011 at 1:14

If all you care about is the proportion of entries between $\pm 0.3$ then sure - generate a random correlation matrix, compute the proportion of entries which are greater than $0.3$ in absolute value, and if there are too many pick some at random and reassign them to random values between $\pm 0.3$. Similarly if there are too few.

Edit: Never you mind, this won't work; see the comments...

  • $\begingroup$ Thanks for your answer. I have a question. First how to generate a random correlation matrix. Secondly, after reassign some numbers, the new matrix might not be a correlation matrix any more. How do I make sure it is still a correlation matrix? $\endgroup$
    – Richard
    Commented May 7, 2011 at 0:50
  • $\begingroup$ It is still positive (semi)definite if you do the reassignment in the way I describe, unless I'm terribly mistaken. There are all sorts of ways that you might generate a correlation matrix; see some of the links in the sidebar. A simple way is to draw a random matrix $A$ which is $n\times p$ with $n<p$, compute $A^TA$ and then normalize it to have ones on the diagonal. $\endgroup$
    – JMS
    Commented May 7, 2011 at 1:02
  • 1
    $\begingroup$ I am very interested in seeing this. Say I generate a random corr matrix that happens to have 20% of them between -0.3 and 0.3. Then I need to resample 70% of the entries unfiormly between -0.3 to 0.3. How do we see the new matrix is still psd? $\endgroup$
    – Richard
    Commented May 7, 2011 at 1:20
  • 2
    $\begingroup$ @JMS, Why would the matrix still be positive semidefinite necessarily? I don't believe your statement is true. $\endgroup$
    – cardinal
    Commented May 7, 2011 at 1:21
  • $\begingroup$ @cardinal, thanks for your comment. We had some discussion about the normally distributed random corr matrix. I don't think i can get the number large enough. So I want to give up normal and see if there is a general way to generate such correlation matrix . $\endgroup$
    – Richard
    Commented May 7, 2011 at 1:29

Here's an older answer to a similar question on SO. It has some code that you could try/modify:

Similar Question

Some other links:

Forecasting Covariance Matrices

Various Matrix Techniques

Matrix Shrinkage Technique


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