Consider the time series
$$
x_t = \beta_1 + \beta_2t + w_t,
$$
where $\beta_1, \beta_2$ are known constants and $w_t$ is a white noise process with variance $\sigma^2_w$.
I want to show that the process $y_t = x_t – x_{t-1}$ is stationary. I have the following definition that I can use
Def: A stationary time series $x_t$ is a finite variance process such that
- i) the mean value function, $\mu_t$, is constant and does not depend on time $t$, and
- ii) the autocovariance function $\gamma(s,t)$ depends on $s$ and $t$ only through their difference $\left|s -t\right|$.
What I've tried: We have $y_t = x_t – x_{t-1} = \beta_2 + w_t – w_{t-1}$. Therefore $\mu_t =\operatorname{E}[y_t] = \beta_2$, which is constant and does not depend on time $t$, so that the first condition is satisfied. If I look at the autocovariance function I get
\begin{align}
\gamma(s, t) = \operatorname{cov}(y_s, y_t) &= \operatorname{E}[(y_s – \mu_s)(y_t – \mu_t)] = \operatorname{E}[w_sw_t – w_sw_{t-1} – w_{s-1}w_t + w_{s-1}w_{t-1}]\\
&= \operatorname{E}[w_s w_t] – \operatorname{E}[w_s w_{t-1}] – \operatorname{E}[w_{s-1}w_t] + \operatorname{E}[w_{s-1}w_{t-1}]
\end{align}
If I look at this result then it makes a lot of sense to conclude that the autocovariance only depends on $s$ and $t$ through their difference. Since $w_t$ is i.i.d. the autocovariance function should be the same for different $s,t$ as long as their difference was the same. Unfortunately I'm not sure I should explain this in more detail and make a clear conclusion.
Question: How do I show that the autocovariance function depends on $s$ and $t$ only through their difference $\left|s-t\right|$?
Best Answer
We use the fact that $(w_s,w_t)$ has the same distribution as $(w_0,w_{t-s})$ (which is the same as that of a couple of independent centered random variables with variance $\sigma_w^2$). Therefore, $$ \mathbb E\left[w_sw_t\right]=\mathbb E\left[w_0w_{t-s}\right]. $$ Do a similar reasoning for the other three terms in the expression of $\gamma(s,t)$.