Consider a vector $r$ of $n$ random variables. Let $\mu = E[r]$ and $\Sigma$ denote the covariance matrix of $r$; that is,

$$\Sigma_{ij} = \sigma_{ij} = E\Bigl[\bigl(r_i - E[r_i]\bigr)\bigl(r_j - E[r_j]\bigr) \Bigr].$$

We know that $\Sigma$ is a positive semidefinite matrix, because we can show that

$$x^T \Sigma x = E\Bigl[ x^T r - E\bigl[x^T r\bigr] \Bigr]^2 \geq 0.$$

(The proof is by writing out the product in summation notation and doing some algebraic manipulation.)

I am currently proofreading a set of practice problems in which the student is asked to consider $r$ having a given $\mu$ and $\Sigma$. Specifically, $n=2$, $\sigma_{1}^2 = 0.1$, $\sigma_{2}^2 = 0.05$, and $\sigma_{12} = \sigma_{21} = -0.4$. Thus, the covariance matrix is

$$\Sigma = \begin{bmatrix} 0.1 & -0.4 \\ -0.4 & 0.05 \end{bmatrix}$$

which is not PSD: If $x = (1, 1)$, then $x^T A x = -0.65 < 0$ as you can verify.

Am I correct to conclude that it is impossible for two random variables to have the given variances and covariances?

(Or are there additional assumptions that went into the proof that $\Sigma$ is PSD that, when relaxed, enable one to construct random variables with the given variances?)

  • $\begingroup$ Feels like a mistake when specifying the covariance matrix. Just a wild guess. $\endgroup$ Commented Jul 26, 2022 at 2:33
  • $\begingroup$ Because, for instance, the variance $$\pmatrix{1&1}\Sigma\pmatrix{1\\1}=-0.65\lt 0,$$ $\Sigma$ cannot be a covariance matrix. $\endgroup$
    – whuber
    Commented Jul 26, 2022 at 4:54

1 Answer 1


Covariances cannot have arbitrary values in comparison to variances; $|\operatorname{Cov}(X,Y)| \leq \sqrt{\operatorname{Var}(X)\operatorname{Var}(Y)}$. So, Yes, it is not possible to find random variables that have the alleged covariance matrix, which is, as you have discovered, not a positive semidefinite matrix. For the given variances, $|\operatorname{Cov}(X,Y)|$ has maximum value $\frac{\sqrt{2}}{20} \approx 0.0707\cdots$, and the given value $-0.4$ is way out of range,

  • $\begingroup$ That inequality is just the determinant condition for a $2\times 2$ symmetric matrix to be PSD. Is that where that equality comes from, or does it follow from a prior principle? $\endgroup$
    – Max
    Commented Jul 26, 2022 at 23:05
  • 1
    $\begingroup$ @Max The root of the inequality you are asking about is the Cauchy=-Schwarz Inequality which was known long before covariance matrices (or even matrices) were thought of. $\endgroup$ Commented Jul 27, 2022 at 1:39
  • $\begingroup$ Specifically, Cauchy–Schwartz as applied to the respective definitions of the variance and covariance. $\endgroup$
    – Max
    Commented Jul 27, 2022 at 2:20
  • $\begingroup$ That a covariance matrix must be positive semi-definite can be proved simply by using that a variance cannot be negative. $\endgroup$ Commented Oct 22, 2023 at 18:11

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