# Show that a time series process is stationary

Consider the time series $$x_t = B_1 + B_{2}t + w_t,$$ where $B_1$ and $B_2$ are known constants and $w_t$ is a white noise process with variance $\sigma^2$.

Show that the process $y_t = x_t - x_{t-1}$ is stationary.

For finding Covariance I cant seem to understand what to do.

Your discrete stochastic process is defined as:

$$x_t = B_1 + B_2t + w_t~~~~~~~, ~~ w_t \sim WN(0,\sigma^2)$$

Clearly it is not stationary since:

$$E[x_t] = B_1+B_2t$$

Now we consider the differentiated process of $x_t$, using the lag operator ($LY_t=Y_{t-1}$):

$$\Delta Y_t = (1-L)Y_t = Y_t - Y_{t-1}$$

$$= B_1 + B_2t + w_t - (B_1 + B_2(t-1) + w_{t-1})$$

$$= B_1 + B_2t + w_t - B_1 - B_2t + B_2 - w_{t-1}$$

$$\Delta Y_t = B_2 + w_t - w_{t-1}$$

Now it is clearly stationary since we have:

$$E[\Delta Y_t] = B_2~~,~~VAR[\Delta Y_t]=2\sigma^2$$

and the covariance depends on time lag only.

• Thanks for the solution. I did manage to solve it, but I still dont understand "covariance depends on tine lag" part. Thank you – Hassam Ullah Sheikh Jan 29 '17 at 18:04
• The fact that covariance depends only on time lag means that its value changes only if the selected lag k changes. This feature is required in order to define a stochastic process stationary. For example, the covariance function of an AR(1), $k >0$, is: $COV(X_t, X_{t-k}) = \gamma^k \cdot \sigma^2$ clearly depends only on time lak k. It is not affected by the time point in which the time series is. On the other hand, non-stationary process have autocovariance functions that do depend on the time point. – Archimede Jan 31 '17 at 16:49
• As an example take the well known random walk, its autocovariance function, if $s < t$ is defined as: $s \cdot \sigma^2$. As you can see, the greater is s the greater is the autocovariance. – Archimede Jan 31 '17 at 16:52

Since you only need help with the covariance of the process $y_t=x_t-x_{t-1}=\Delta x_t$.

Note that $y_{t+k} = \Delta x_{t+k}=B_2+w_{t+k}-w_{t+k-1} \quad \forall k \in \mathbb R$

Now to find the covariance start by its definition and use its properties i.e. \begin{align*} cov(y_t,y_{t+k})&=cov(B_2+w_{t}-w_{t-1},B_2+w_{t+k}-w_{t+k-1})\\ &= cov(w_t-w_{t-1},w_{t+k}-w_{t+k-1}) \end{align*}

From here its straightforward to find what you are looking for.