# Covariance of a process and its noise in ARMA(1,1)

I need to calculate $$\operatorname{Cov}(X_0, w_0)$$ where $$X_t$$ is an ARMA(1,1) process given by $$X_t = a_1X_{t-1} + w_t + b_1w_{t-1}, \quad w_t\sim\mathcal{N}(0, \sigma^2) \text{ independently}.$$ So far I have calculated $$\operatorname{Cov}(X_0, w_0)=\sigma^2$$ which gives unexpected behavior so I suspect is wrong.

Denote $$\phi(z) = 1 - a_1z$$, $$\theta(z) = 1 + b_1z$$. Under the assumption that $$\phi(z) \neq 0$$ for all $$z \in \mathbb{C}$$ such that $$|z| \leq 1$$ (i.e., $$|a_1| < 1$$. Of course, $$\phi(z)$$ and $$\theta(z)$$ should have no common zeros, i.e., $$a_1 \neq -b_1$$), $$X_t$$ has the representation (see Theorem 3.1.1 and Definition 3.1.3 in Time Series: Theory and Methods (Second Edition) by P. J. Brockwell and R. A. Davis): \begin{align} X_t = \sum_{j = 0}^\infty \psi_j w_{t - j}, \quad t = 0, \pm 1, \ldots, \tag{1} \end{align} i.e., $$\{X_t\}$$ is causal.
It then follows by $$(1)$$ and the assumption of $$\{w_t\}$$ that \begin{align} \operatorname{Cov}(X_0, w_0) = \psi_0\operatorname{Var}(w_0) = \psi_0\sigma^2, \end{align} where $$\psi_0$$ is determined by (see Equation (3.3.5) in the same reference above) \begin{align} (1 - a_1z)(\psi_0 + \psi_1 z + \psi_2 z^2 + \cdots) = 1 + b_1 z, \end{align} i.e., $$\psi_0 = 1$$. Therefore, your answer is correct (provided the time series is causal).
Alternatively, if you are told that $$\{X_t\}$$ is causal, then by \begin{align} X_0 - a_1X_{-1} = w_0 + b_1w_{-1}, \end{align} we have \begin{align} \operatorname{Cov}(X_0 - a_1X_{-1}, w_0) = \operatorname{Cov}(w_0 + b_1w_{-1}, w_0) = \sigma^2. \tag{2} \end{align} Since $$\operatorname{Cov}(X_{-1}, w_0) = 0$$ by causality, $$(2)$$ implies $$\operatorname{Cov}(X_0, w_0) = \sigma^2$$.