Can we construct a pair of random variables having any given covariance? 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?)

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,
