Covariance formula to find $\sigma(1)$

Question

Suppose that $x_{t}$ follows and $MA(2)$ process: $$x_{t}=m_{1} \varepsilon _{t-1} + m_{2} \varepsilon _{t-2} + \varepsilon _{t},$$ where $\varepsilon _{t}$ is a white noise with mean zero and variance $\sigma^{2}_{\epsilon}$.

I have to find the autocovariance function $\sigma(k)=cov(x_{t},x_{t+1})$.

So the easiest way to do this would be by trial-and-error, such as finding $\sigma(0)$, $\sigma(1)$, $\sigma(2)$, etc to see a pattern for $\sigma(k)$.

So, $\sigma(0)= Var(x_{t})=Var(\varepsilon _{t-1} + m_{2} \varepsilon _{t-2} + \varepsilon _{t})=m_{1}^{2}\sigma^{2}_{\epsilon}+m_{2}^{2}\sigma^{2}_{\epsilon}+\sigma^{2}_{\epsilon}$

I am stuck here $\sigma(1)=cov(x_{t},x_{t+1})=cov(\varepsilon _{t-1} + m_{2} \varepsilon _{t-2} + \varepsilon_{t}, \varepsilon _{t} + m_{2} \varepsilon _{t-1} + \varepsilon_{t+1})$.

How do I use the covariance formula to find $\sigma(1)$?

Answer 1

Your $\sigma(1)=cov(x_{t},x_{t+1})$ should be $cov(m_1\varepsilon _{t-1} + m_{2} \varepsilon _{t-2} + \varepsilon_{t}, m_1\varepsilon _{t} + m_{2} \varepsilon _{t-1} + \varepsilon_{t+1})$

which can be expanded to $$cov(m_1\varepsilon _{t-1},m_1\varepsilon _{t}) + cov(m_1\varepsilon _{t-1},m_2\varepsilon _{t-1}) + cov(m_1\varepsilon _{t-1},m_2\varepsilon _{t+1}) \\ + cov(m_{2} \varepsilon _{t-2},m_1\varepsilon _{t}) + cov(m_{2} \varepsilon _{t-2},m_2\varepsilon _{t-1}) + cov(m_{2} \varepsilon _{t-2},m_2\varepsilon _{t+1}) \\ + cov(\varepsilon _{t},m_1\varepsilon _{t}) + cov(\varepsilon _{t},m_2\varepsilon _{t-1}) + cov(\varepsilon _{t},m_2\varepsilon _{t+1}) \\ = 0+m_1m_2\sigma^2_\varepsilon+0+0+0+0+m_1\sigma^2_\varepsilon+0+0 \\= m_1(1+m_2)\sigma^2_\varepsilon.$$

Similarly $\sigma(2)=cov(x_{t},x_{t+2}) =cov(m_1\varepsilon _{t-1} + m_{2} \varepsilon _{t-2} + \varepsilon_{t}, m_1\varepsilon _{t+1} + m_{2} \varepsilon _{t} + \varepsilon_{t+2}) = m_2 \sigma^2_\varepsilon$

and $\sigma(3)=cov(x_{t},x_{t+3}) =cov(m_1\varepsilon _{t-1} + m_{2} \varepsilon _{t-2} + \varepsilon_{t}, m_1\varepsilon _{t+2} + m_{2} \varepsilon _{t+1} + \varepsilon_{t+3}) = 0$ etc.

Answer 2

Note that the $\{\varepsilon_t\}$ are independent so $$\operatorname{Cov}(\varepsilon_s,\varepsilon_t) = \begin{cases} \sigma^2_\varepsilon,& s=t\\ 0,& s\ne t.\end{cases}$$ Hence \begin{align} \sigma(0) &= \operatorname{Var}(x_t)\\ &= \operatorname{Var}(\varepsilon_t + m_1\varepsilon_{t-1}+m_2\varepsilon_{t-2})\\ &= \sigma^2_\varepsilon(1 + m_1^2 + m_2^2),\\\\ \end{align}

\begin{align} \sigma(1) &= \operatorname{Cov}(x_t, x_{t+1})\\ &= \operatorname{Cov}(\varepsilon_t + m_1\varepsilon_{t-1}+m_2\varepsilon_{t-2}, \varepsilon_{t+1}, m_1\varepsilon_t, m_2\varepsilon_{t-1})\\ &= \sigma^2_\varepsilon(m_1(1+m_2)),\\\\ \end{align}

\begin{align} \sigma(2) &= \operatorname{Cov}(x_t, x_{t+2})\\ &= \operatorname{Cov}(\varepsilon_t + m_1\varepsilon_{t-1}+m_2\varepsilon_{t-2}, \varepsilon_{t+2}, m_1\varepsilon_{t+1}, m_2\varepsilon_t)\\ &= \sigma^2_\varepsilon m_2. \end{align}

Therefore

$$ \sigma(j) = \begin{cases} \sigma^2_\varepsilon(1 + m_1^2 + m_2^2),& j=0\\ \sigma^2_\varepsilon(m_1(1+m_2)),& |j|=1\\ \sigma^2_\varepsilon m_2,& |j|=2\\ 0,& |j|>2. \end{cases} $$