Non-zero covariance between degenerate random variable (RV) and non-degenerate RV?

Question

Can a degenerate random variable (RV) have a non-zero covariance with a non-degenerate RV (or even degenerate, too)?

My intuition says "no" because that would imply (would it?) that values sampled from the degenerate variable and the non-degenerate have some mutual information, i.e. one can draw conclusions about samples from non-degenerate RV from samples from degenerate RV.

Now, I have to admit that there is a chance that a question is actually about me making mistakes in my algebra, however, I came across a case where it appears so:

I examine two linear models, namely: $$(1)\ y = X \beta + \epsilon_1+ g \ \ and \ \ (2) \ y = \epsilon_2 + g\\ \epsilon_1 \sim \mathbb{N}(0, \sigma_1^2 I_n), \epsilon_2 \sim \mathbb{N}(0, \sigma_2^2 I_n),\\ g \sim \mathbb{N}(0, \Omega), $$ where $\Omega$ is some known covariance matrix. Furthermore, $X$ is a square $n \times n$ matrix, i.e. $(1)$ is just a system of linear equations and therefore $\sigma_1 = 0$, so $\epsilon_1$ is degenerate. Now, I want to study covariance between $\epsilon_1$ and $\epsilon_2$: $$cov(\epsilon_1, \epsilon_2) = E[(\epsilon_1 - E[\epsilon_1])(\epsilon_2 - E[\epsilon_2])^T] = E[\epsilon_1 \epsilon_2^T] = E[(y- X \beta - g)(y- g)^T] = \\= E[y y^T - X \beta y^T +g g^T] =y y^T - X \beta y^T + \Omega$$ I could argue that as $X \beta$ is numerically $y$, then $y y^T -X \beta y^T = 0$, but there is still $\Omega$.

Answer 1

1. In relation to your $y$ I suggest you not use the exact same symbol for distinct quantities; a subscript would do

2. However, we can readily answer the title question; if a variable doesn't vary, it can't co-vary (vary with something else). The covariance is definitely 0, the correlation is 0/0 (undefined). The covariance of $\epsilon_1$ and $\epsilon_2$ doesn't involve $\Omega$; fixing your notation problem in 1 will probably help you identify that issue.