\[ X_{t}=\frac{1}{3}(Z_{t-1}+Z_{t}+Z_{t+1});\quad Z_{t}\stackrel{i.i.d.}{\sim}WN(\mu,\sigma^2) \] We wish to find the ACVF \[ \begin{align} (h=0) & \quad Cov(X_{t}, X_{t})=\text{Var}(X_{t})=\text{Var}\left( \frac{1}{3}(Z_{t-1}+Z_{t}+Z_{t+1}) \right)=\frac{1}{9}(\sigma^2+\sigma^2+\sigma^2)=\frac{\sigma^2}{3} \\ (h=1) & \quad Cov(X_{t}, X_{t+1}) = Cov\left( \frac{1}{3}(Z_{t-1}+Z_{t}+Z_{t+1}),~ \frac{1}{3}(Z_{t}+Z_{t+1}+Z_{t+2}) \right) \\ & \quad = \frac{1}{9}Cov(Z_{t-1}+Z_{t}+Z_{t+1},~ Z_{t}+Z_{t+1}+Z_{t+2}) = \frac{1}{9}(Cov(Z_{t},Z_{t})+Cov(Z_{t+1},Z_{t+1}))\\ & \quad = \frac{1}{9}(\sigma^{2}+\sigma^2)=\frac{2\sigma^2}{9} \\ (h=2)& \quad Cov(X_{t},X_{t+2})=\frac{\sigma^2}{9} \\ (h\geq 3)&\quad Cov(X_{t}, X_{t+h})=0. \end{align} \]
For the ACF can be found with \[ Cor(X_{t}, X_{t+h})= \frac{Cov(X_{t},X_{t+h})}{\sqrt{ \text{Var}(X_{t})\text{Var}(X_{t+h}) }} \] To compute \(\text{Var}(X_{t})\) we write \[ \text{Var}(X_{t})=\text{Var}\left( \frac{1}{3}(Z_{t-1}+Z_{t}+Z_{t+1}) \right)=\frac{1}{9}(\text{Var}(Z_{t-1})+\text{Var}(Z_{t})+\text{Var}(Z_{t+1}))=\frac{\sigma^2}{3}. \] Therefore \[ \sqrt{ \text{Var}(X_{t})\text{Var}(X_{t+h}) }=\sqrt{ \frac{\sigma^2}{3}\cdot \frac{\sigma^2}{3} }=\frac{\sigma^2}{3}. \] The ACF is therefore \[ \begin{align} (h=0) & \quad Cor(X_{t}, X_{t})=\frac{\sigma^3}{3}\cdot \frac{3}{\sigma^2}=1 \\ (h=1) & \quad Cor(X_{t}, X_{t+1})=\frac{2\sigma^2}{9}\cdot \frac{3}{\sigma^2}=\frac{2}{3} \\ (h=2)& \quad Cor(X_{t}, X_{t+1})=\frac{\sigma^2}{9}\cdot \frac{3}{\sigma^2}=\frac{1}{3} \\ (h\geq 3)& \quad Cor(X_{t}, X_{t+h})=0 \end{align} \]
\[ Cov(X_{t}, X_{t+h})=\mathbb{E}[X_{t}X_{t+h}]-\mathbb{E}[X_{t}]\mathbb{E}[X_{t+h}] \]
\[ \text{Var}(X)=\mathbb{E}X^2+(\mathbb{E}X)^2 \] \[ \mathbb{E}\left( \frac{1}{3}X \right)^2= \frac{1}{9} \mathbb{E}X^2 \]