1
$\begingroup$

Let the model $$y_t=\beta_0+\beta_1t+z_t\qquad t=1,2,\dots$$ $$z_t=\epsilon_t+\theta\epsilon_{t-1}$$ where $\epsilon_t$ is White noise with zero mean and variance $\sigma^2$ and $\beta_0,\beta_1,\theta$ constants.

a) Is $y_t$ stationary?

b) Is $(1-B)y_t$ stationary? (where $B$ is a lag operator)

This model can be written as $$y_t=\delta_t+\epsilon_t+\theta\epsilon_{t-1}$$

a) It is not a stationary process, this process have trend and clearly this have no constant mean, because $E[y_t]$ varies according to $t$.

b) $$(1-B)y_t=\beta_0+\beta_1t+\epsilon_t+\theta\epsilon_{t-1}-\beta_0-\beta_1(t-1)+\epsilon_{t-1}-\theta\epsilon_{t-2}$$ $$=\beta_1+\epsilon_t+\epsilon_{t-1}(\theta-1)+\theta\epsilon_{t-2}$$ $$=\beta_1+\epsilon_t[1+B(\theta-1)-\theta B^2]$$

$$B=\frac{(1-\theta)\pm \sqrt{(\theta-1)^2+4\theta}}{2}$$

The roots are $1$ and $-\theta$, so this process is not stationary?

I have other doubts too:

1) The process $y_t$ is a $MA(1)$ with non-zero mean?

2) When I look the roots of polynomial, both roots need to be $>1$ in modulus or just one is enough?

$\endgroup$
  • $\begingroup$ What is the definition of $B$ in "b) Is $x_t=(1−B)y_t$ stationary?" $\endgroup$ – Dilip Sarwate Dec 28 '16 at 2:11
  • $\begingroup$ B is the backshift operator. B X_t =X_(t-1). First of all y_t is nonstationary. It has a linear trend so the mean changes over time. Why did you ask and answer that question in your Post? $\endgroup$ – Michael Chernick Dec 28 '16 at 2:31
  • $\begingroup$ You are confusing people by changing x_t to y_t in your notation. $\endgroup$ – Michael Chernick Dec 28 '16 at 2:39
  • $\begingroup$ I am confused all you did was reexpress y_t. It is difficult see what point you are trying to make. y_t being nonstationary is settled. $\endgroup$ – Michael Chernick Dec 28 '16 at 2:47
  • 1
    $\begingroup$ @Roland I don't think so. y_t has a linear trend component which is why it is nonstationary.. Added to that trend is a moving average process. That is why I say it has a moving average component. $\endgroup$ – Michael Chernick Dec 28 '16 at 17:15
1
$\begingroup$

1) No, the series $\boldsymbol{Y} \equiv \{ Y_t | t \in \mathbb{Z} \}$ is not stationary. The trend term in $t$ means that a shift in the time index will yield a different distribution.

2) Let $Z_t \equiv (1-B) Y_t$ be the first-difference of the original series and let $\boldsymbol{Z} \equiv \{ Z_t | t \in \mathbb{Z} \}$ be the series of first-differences. As you have already derived (your derivation has errors but the result is correct), this time-series is an MA($2$) process with coefficients $\theta_1 = \theta - 1$ and $\theta_2 = -\theta$ and fixed mean $\mathbb{E}(Z_t) = \beta_1$ . It has stationary distribution:

$$Z_t \sim \text{N} \Big( \beta_1, 2 (\theta^2 - \theta + 1) \sigma^2 \Big) .$$

$\endgroup$

Your Answer

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