Show that $\{\epsilon_t+\eta_s\}$ is a stationary process

Let $\{\epsilon_t\}$ and $\{\eta_s\}$ two non-correlated stationary process $\forall t,s$. Show that $\{\epsilon_t+\eta_s\}$ is a stationary process with autocovariance function equal the sum of the autocovariances of $\{\epsilon_t\}$ and $\{\eta_s\}$.

Proof:

If $\{\epsilon_t\}$ and $\{\eta_s\}$ are stationary then they have mean and variance constants $\forall t,s$ and the covariance is independent of time. Let

$$\begin{cases}E[\epsilon_t]=\mu_1\\Var(\epsilon_t)=\sigma_1^2\\Cov(\epsilon_t,\epsilon_{t+h})=\rho_1\end{cases}\qquad\text{and} \qquad\begin{cases}E[\eta_s]=\mu_2\\Var(\eta_s)=\sigma_2^2\\Cov(\eta_s,\eta_{s+h})=\rho_2\end{cases}$$

where $\mu_1,\mu_2,\sigma_1,\sigma_2,\rho_1,\rho_2\in \mathbb{R}$

For $\{\epsilon_t+\eta_s\}$

$$E[\epsilon_t+\eta_s]=E[\epsilon_t]+E[\eta_s]=\mu_1+\mu_2$$

$$Var(\epsilon_t+\eta_s)=Var(\epsilon_t)+Var(\eta_s)+2cov(\epsilon_t,eta_s)=\sigma_1^2+\sigma_2^2$$

since that the process are non-correlated then $cov(\epsilon_t,\eta_s)=0$

$$Cov(\epsilon_t+\eta_s,\epsilon_{t+h}+\eta_{s+h})=Cov(\epsilon_t,\epsilon_{t+h})+cov(\epsilon_t,\eta_{s+h})+Cov(\eta_s,\epsilon_{t+h})+cov(\eta_s,\eta_{s+h})$$ $$=cov(\epsilon_t,\epsilon_{t+h})+cov(\eta_s,\eta_{s+h})=\rho_1+\rho_2$$

Then $\{\epsilon_t+\eta_s\}$ is a stationary process.

Since the autovariance function $\gamma_\epsilon(h)=Cov(\epsilon_t,\epsilon_{t+h})$ and $\gamma_\eta(h)=Cov(eta_s,eta_{s+h})$, the autovariace function of $\{\epsilon_t+\eta_s\}$ is $$\gamma_{\epsilon+\eta}(h)=\gamma_\epsilon(h)+\gamma_\eta(h)$$

Is this right?

• Even if it is obvious, I think it is god to say that $\epsilon_t+\eta_t$ has a finite second moment.
• You should mention again the fact that the processes $\epsilon$ and $\eta$ are uncorrelated when you compute $\operatorname{ Cov} (\epsilon_t+\eta_s,\epsilon_{t+h}+\eta_{s+h})$.