1
$\begingroup$

Consider the processes $X_t = \phi X_{t-1} + v_t$ and $Y_t = \phi Y_{t-1} + X_t + e_t$, in which $|\phi| < 1$ and $v_1$ and $e_t$ are non-correlated random errors with zero mean and variances equal to $\sigma^2$. Based on these informations, find the autocovariance function of the process $Y_t$.


I already know that if $|\phi| < 1$ than $Y_t$ should follow a stationary process and it’s autocovariance is:

$\begin{equation} E[Y_{t+h}Y_t] - \mu^2 \end{equation}$

but I do not know how to go further

$\endgroup$

1 Answer 1

0
$\begingroup$

Note that all the processes have zero mean, i.e. $E[Y_t]=E[X_t]=E[v_t]=E[\epsilon_t]=0$.

Let $h>0$: $$\begin{align}E[Y_{t+h}Y_t]&=E[(\phi Y_{t+h-1}+X_{t+h}+\epsilon_{t+h})Y_t]=\phi E[Y_{t+h-1}Y_t]+E[X_{t+h}Y_t]+E[\epsilon_{t+h}Y_t]\\&=\phi r_y(h-1)+\phi^h r_{xy}(0)+E[\epsilon_{t+h}]E[Y_t]=\phi r_y(h-1)+\phi^h r_{xy}(0)\end{align}$$

Note that: $r_{xy}(h)=E[X_{t+h}Y_t]=E[\phi^hX_tY_t+(\sum_{j=0}^{h-1} v_{t+h-j}\phi^j)Y_t]=\phi^h r_{xy}(0)$

For $h=0$: $$\begin{align}r_y(0)&=E[Y_t^2]=E[(\phi Y_{t-1}+X_t+\epsilon_t)^2]\\&=\phi ^2r_y(0)+E[X_t^2]+E[\epsilon_t^2]+2\phi \underbrace{E[Y_{t-1}X_t]}_{r_{xy}(1)}+2E[X_t\epsilon_t]+2E[Y_{t-1}\epsilon_t]\\&=\phi^2r_y(0)+r_x(0)+r_e(0)+2\phi^2 r_{xy}(0)\rightarrow r_y(0)=\frac{r_x(0)+\sigma^2+2\phi^2 r_{xy}(0)}{1-\phi^2}\end{align}$$

Similarly, we can find that $$r_x(0)=\frac{\sigma^2}{1-\phi^2}, \ \ r_x(h)=\phi r_x(h-1)$$

It just remains to find $r_{xy}(0)$: $$\begin{align}r_{xy}(0)&=E[X_tY_t]=[(\phi X_{t-1}+v_t)(\phi Y_{t-1}+X_t+\epsilon_t)]\\&=\phi^2E[X_{t-1}Y_{t-1}]+\phi r_x(1)+E[X_tv_t]\\&=\phi^2r_{xy}(0)+\phi r_x(1)+\sigma^2\end{align}$$

$\endgroup$
6
  • $\begingroup$ Why did you say that $E[X_{t+h}Y_t] = E[X_{t+h}][Y_t]$ and $E[\epsilon_{t+h} Y_t] = E[\epsilon_{t+h}] E[Y_t]$? $\endgroup$
    – motipai
    Oct 14, 2019 at 20:39
  • $\begingroup$ $\epsilon_{t+h}$ doesn't depend on $\epsilon_t$, but $X_{t+h}$ has a separate recursion, therefore $E[X_{t+h}Y_t]\neq E[X_{t+h}]E[Y_t]$, which was a mistake; now corrected. it was a fast deploy, and I'll review the algebra again. The solution implies joint stationarity by the way. $\endgroup$
    – gunes
    Oct 15, 2019 at 11:01
  • $\begingroup$ why $\phi^h r_{xy}(0) = 0$? $\endgroup$
    – motipai
    Oct 15, 2019 at 11:25
  • $\begingroup$ It's not. Where do you see that? $\endgroup$
    – gunes
    Oct 15, 2019 at 11:30
  • $\begingroup$ For $\phi r_y(h-1)+\phi^h r_{xy}(0)+E[\epsilon_{t+h}]E[Y_t]=\phi r_y(h-1)$ then $\phi^h r_{xy}(0)$ must be zero. $\endgroup$
    – motipai
    Oct 15, 2019 at 11:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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