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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.

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2 Answers 2

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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.

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  • $\begingroup$ Thanks for the solution. I did manage to solve it, but I still dont understand "covariance depends on tine lag" part. Thank you $\endgroup$ Jan 29, 2017 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, 2017 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, 2017 at 16:52
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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.

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