It's not too hard to see that any correlation matrix must have certain properties, such as all entries in the range -1 to 1, symmetric, positive semi-definite (excluding pathological cases like singular matrices for the moment).

But I'm wondering about the other direction. If I write down some matrix that is positive semi-definite, is symmetric, has 1's along the main diagonal, and all entries are between -1 and 1, does this guarantee that there exists a set of random variables giving rise to that correlation matrix?

If the answer is easy to see, a helpful hint about how to define the set of random variables that would give rise to a given matrix would be appreciated.


1 Answer 1


Yes, although the restriction that all entries are between $-1$ and $1$ is implied by the other properties (and so is not needed).

Let $\Sigma$ be a $n \times n$, symmetric, positive semidefinite matrix with $1$'s along the main diagonal.

First, $\Sigma$ is a covariance matrix. Since $\Sigma$ is symmetric and positive semidefinite, $\Sigma$ has a nonnegative symmetric square root $\Sigma^{1/2}$. Let $X$ be a $n$-vector of independent random variables, each with variance $1$. (For example, $X$ could be an $n$-vector of independent $N(0,1)$ random variables.) Construct the $n$-vector $\Sigma^{1/2} X$. Then, by properties of covariance matrices, $$\text{cov} (\Sigma^{1/2} X) = \Sigma^{1/2} \text{cov}(X) \Sigma^{1/2} = \Sigma^{1/2} I \Sigma^{1/2} = \Sigma.$$ Thus $\Sigma$ is the covariance matrix for the random vector $\Sigma^{1/2} X$. (This derivation is on the Wikipedia article for covariance matrices.)

Second, since $\Sigma$ has $1$'s on its diagonal, the standard deviation of each random variable in $\Sigma^{1/2} X$ is $1$. Thus the correlation matrix $R$ for the random vector $\Sigma^{1/2} X$ is $$R = I^{-1} \Sigma I^{-1} = \Sigma.$$ Thus $\Sigma$ is a correlation matrix as well.


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