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.

  • $\begingroup$ Please add the relevant self-study tag to your post. $\endgroup$ Commented Dec 13, 2022 at 16:03

1 Answer 1


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 $\operatorname{Cov}(X_0, w_0) = \sigma^2$.


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