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Let $B$ the lag operator and $\{y_t\}$ the following model $$(1-0.6B^4)y_t=(1+0.2B)\epsilon_t$$ where $\epsilon_t\sim N(0,16)$.

a) Is it a stationary process?

b) Find the autocorrelation function of $\{y_t\}$

c)Let $y_{97}=196$, $y_{98}=195$, $y_{99}=198$ and $y_{100}=200$. Find the forecast three-steps ahead $y_{100}(3)$. What is the variance of the forecast error when $k\rightarrow \infty$?

The process can be written as $$y_t=0.6y_{t-4}+\epsilon_t+0.2\epsilon_{t-1}$$ that looks like a Seasonal ARMA process.

a) To check if it is stationary I looked to the roots of the autoregressive polynomial $$1-0.6B^4=0\Rightarrow B=\Big(\frac{1}{0.6}\Big)^{\frac{1}{4}}>1$$ so the process is stationary.

b) Since that $\rho(h)=\frac{\gamma(h)}{\gamma(0)}$ first I found the variance of $y_t$

$$\gamma(0)=var(y_t)=var(0.6y_{t-4}+\epsilon_t+0.2\epsilon_{t-1})$$ $$=0.6^2var(y_{t-4})+var(\epsilon_t)+0.2^2var(\epsilon_{t-1})=\sigma^2(0.6^2+1+0.2^2)=22.4.$$

$$\gamma(h)=E[y_t|y_{t-h}]=[0.6y_{t-4}y_{t-h}+\epsilon_ty_{t-h}+0.2\epsilon_{t-1}|y_{t-h}],$$ $$\gamma(h)=\gamma(h-4), \qquad h=4,8,12,\dots$$ Then the autocorrelation function is $$\rho(h)=\frac{\gamma(h-4)}{22.4},\qquad h=4,8,12,\dots$$

c) For $t=100$ $$y_{t}(3)=E[y_{t+3}|y_1,\dots,y_t]=E[0.6y_{t-1}+\epsilon_{t+3}+0.2\epsilon_{t+2}|y_1,\dots,y_t]$$ $$=y_{t}(3)=E[0.6y_{t-1}|y_1,\dots,y_t]$$ $$y_{100}(3)=E[0.6y_{99}|y_1,\dots,y_{100}]=0.6\times198=118.8.$$

The forecast error is given by $$e_t(k)=y_{t+k}-\hat{y}_t(k)=0.6y_{t+k-4}+\epsilon_{t+k}+0.2\epsilon_{t+k-1}-0.6y_{t+k-4}$$ $$=\epsilon_{t+k}+0.2\epsilon_{t+k-1}$$ $$Var(e_t(k))=Var(\epsilon_t)+0.04Var(\epsilon_{t+k-1})=16(1+0.04)$$

I don't know, it looks wrong, can anyone help me?

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  • $\begingroup$ How do you obtain the right side of the last formula without the $k$? $\endgroup$ – Nick Dec 24 '16 at 1:06
  • $\begingroup$ @Nick You mean without know $k$? $\endgroup$ – user72621 Dec 24 '16 at 1:10
  • $\begingroup$ As for me it is not clear how the passage to the limit, $k\to\infty$, was carried out. $\endgroup$ – Nick Dec 24 '16 at 2:13
  • $\begingroup$ @Nick That is exactly why I think that it's wrong. I just used the fact that all white noises have variance $\sigma^2$ independent of value of $k$. $\endgroup$ – user72621 Dec 24 '16 at 2:20

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