1
$\begingroup$

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.

$\endgroup$
7
$\begingroup$

Your discrete stochastic process is defined as:

\begin{equation} x_t = B_1 + B_2t + w_t~~~~~~~, ~~ w_t \sim WN(0,\sigma^2) \end{equation}

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks for the solution. I did manage to solve it, but I still dont understand "covariance depends on tine lag" part. Thank you $\endgroup$ – Hassam Ullah Sheikh Jan 29 '17 at 18:04
  • $\begingroup$ 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. $\endgroup$ – Archimede Jan 31 '17 at 16:49
  • $\begingroup$ 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. $\endgroup$ – Archimede Jan 31 '17 at 16:52
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.