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I have some trouble solving the following problem:

Given are the stationairy processes $X_t$ and $Y_t$:

$X_t = Z_t*\sqrt{7+0.5X_{t-1}^2}$

$Y_t = 2+(2/3)*Y_{t-1}+X_t$

Where $Z_t$ is distributed IID $N(0,1)$

Now I simply need to find the ACF's (autocorrelation functions) $\rho_X(h)$ and $\rho_Y(h)$ for both $X_t$ and $Y_t$, but I only know how to do this when the processes are recognizable as simple AR or MA processes. Help is appreciated.

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  • $\begingroup$ do you know that Y is an AR(1) process with ARCH noise ? $\endgroup$ – mike Nov 14 '13 at 14:43
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Note that $E[X_t\mid\mathcal F_{t-1}]=\sqrt{7+\frac12X_{t-1}^2}\cdot E[Z_t]=0$ hence $\rho_X(h)=0$ for every $h\ne0$. Furthermore, $X_t^2=Z_t^2(7+\frac12X_{t-1}^2)$ and $E[Z_t^2]=1$ hence $E[X_t^2]=7+\frac12E[X_{t-1}^2]$ by independence. If the sequence $(X_t)$ is at stationarity, this implies that $\rho_X(0)=E[X_t^2]=14$.

Note that $E[Y_t]=2+\frac23E[Y_{t-1}]$ because $E[X_t]=0$, hence, at stationarity, $E[Y_t]=6$. Let $U_t=Y_t-6$ then $U_t$ is centered and $\rho_Y=\rho_U$. Furthermore, $U_t=\frac23U_{t-1}+X_t$ hence $E[U_t\mid\mathcal F_{t-1}]=\frac23U_{t-1}$ since $E[X_t\mid\mathcal F_{t-1}]=0$. This implies that $\rho_U(h)=\left(\frac23\right)^{|h|}\rho_U(0)$ where $\rho_U(0)=E[U_t^2]$. Note that $U_t^2=\frac49U_{t-1}^2++\frac43U_{t-1}X_t+X_t^2$ hence $\rho_U(0)=\frac95\rho_X(0)$.

Finally, $\rho_Y(h)=14\cdot\frac95\cdot\left(\frac23\right)^{|h|}$ for every $h$.

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