# Covariance function of MA(1) process

probably answered before but I would I want to see if my reasoning is correct, as my textbook skips the calculations but the answer coincide.

Q: Let $$Z_t \sim \text{WN}(0, \sigma^2)$$ (white noise), and define the MA(1) process $$X_t = Z_t + \theta Z_{t-1}, t\in \mathbb{Z}, \theta \in \mathbb{R}.$$ Find the covariance function $$\gamma_X(t, t+h)$$.

Sol: $$E(X_t) = 0, E(X_t^2) = \sigma^2(1+\theta^2)$$. Then

$$$$\begin{split} \gamma_X(t,t+h) &= Cov(X_t, X_{t+h}) = E[(X_t - E(X_t))(X_{t+h}-E(X_{t+h})] \\ &= E(X_t X_{t+h}) = E[(Z_t + \theta Z_{t-1})(Z_{t+h} + \theta Z_{t+h-1} )] \\ &= E(Z_tZ_{t+h}) + \theta E(Z_t Z_{t+h-1}) + \theta E(Z_{t-1} Z_{t+h}) + \theta^2 E(Z_{t-1}Z_{t+h-1}) \end{split}$$$$

Since $$Z_t$$ is white noise, it holds that $$\gamma_Z(s,t) = E(Z_s E_t) = \sigma^2 \delta(s-t)$$. Hence

$$$$\begin{split} \gamma_X(t,t+h) &= \sigma^2\delta(h) + \theta \sigma^2\delta(h+1) + \theta \sigma^2\delta(h-1) + \theta^2 \sigma^2 \delta(h) \\ &= \begin{cases} \begin{split} &\sigma^2(1+\theta^2), \quad &h = 0 \\ &\sigma^2 \theta, \quad &h = \pm 1 \\ &0, \quad &\text{else} \end{split} \end{cases} \end{split}$$$$

• Your title should be edited -- it is MA(1) process, not AR(1) process. Commented Apr 5, 2023 at 14:13
• Thank you, fixed it. Commented Apr 5, 2023 at 14:14
• Your solution looks great. Commented Apr 5, 2023 at 15:31