Unconditional variance of AR(2) + GARCH(1, 1) I am being asked to derive the unconditional variance for stochastic process $\{Y_t\}$, where:
$$Y_t = \phi_1 Y_{t-1} + \phi_2 Y_{t-2} + \varepsilon_t$$
$$\varepsilon_t = V_t \sigma_t$$
$$\sigma_t^2 = \alpha_0 + \alpha_1 \varepsilon_{t-1}^2 + \beta \sigma_{t-1}^2$$
$$V_t \sim N(0, 1)$$
I don't mind expressing this in terms of autocorrelations (i.e. $\gamma_j = Cov(Y_t, Y_{t-j}$)).
 A: The GARCH-part
The following holds for every GARCH(1,1) regardless of the assumed distribution of $V_t$, as long as $E(V_t)=0$, $E(V_t^2)=1$ and $E(V_t^4)<\infty$. Let's start to derive the first two unconditional moments of $\epsilon_t$ because we need them to calculate the unconditional variance. A useful trick is to first calculate the conditional moments and then use the law of iterated expectation to derive the unconditional moments.
Let $\mathcal F_{t-1}$ be the past history of the process, then the conditional expectation of $\epsilon_t$ is given by:
$$
E(\epsilon_t \vert \mathcal F_{t-1})=E(\sigma_tV_t \vert \mathcal F_{t-1})=\sigma_tE(V_t \vert \mathcal F_{t-1})=0
$$
Thus:
$$
E(\epsilon_t)=E(E(\epsilon_t \vert \mathcal F_{t-1}))=E(0)=0
$$
To derive the second moment, note that you can express $\epsilon_t^2$ as an ARMA(1,1)-process
$$
\epsilon_t^2=\alpha_0+(\alpha_1+\beta_1)\epsilon_{t-1}^2+w_t-\beta_1w_{t-1}
$$
with $w_t=\epsilon_t^2-\sigma_t^2$ being a WN process.
It holds that:
$$
E(\epsilon_t^2\vert \mathcal F_{t-1})=E(\sigma_t^2V_t^2\vert \mathcal F_{t-1})=\sigma_t^2E(V_t^2\vert \mathcal F_{t-1})=\sigma_t^2\cdot 1=\sigma_t^2
$$
Thus:
$$
E(w_t \vert \mathcal F_{t-1})=E(\epsilon_t^2 \vert \mathcal F_{t-1})-\sigma_t^2=\sigma_t^2-\sigma_t^2=0
$$
Therefore:
$$
E(w_t)=E(E(w_t \vert \mathcal F_{t-1}))=E(0)=0
$$
This allows us to write:
\begin{align*}
E(\epsilon_t^2)&=\alpha_0+(\alpha_1+\beta_1)E(\epsilon_{t-1}^2)+E(w_t)-\beta_1E(w_{t-1}) \\
&=\alpha_0+(\alpha_1+\beta_1)E(\epsilon_{t-1}^2)
\end{align*}
Assuming stationarity, we conclude that $E(\epsilon_{t-1}^2)=E(\epsilon_t^2)$ and hence:
$$
Var(\epsilon_t)=E(\epsilon_t^2)-(E(\epsilon_t))^2=E(\epsilon_t^2)=\frac{\alpha_0}{1-(\alpha_1+\beta_1)}
$$
If we impose the restriction $\alpha_1+\beta_1<1$, $Var(\epsilon_t)$ exists and is finite.
Now, observe that:
\begin{align*}
Cov(\epsilon_t,\epsilon_{t-\tau})&=E(\epsilon_t\epsilon_{t-\tau})-E(\epsilon_t)E(\epsilon_{t-\tau})\\
&=E(\epsilon_t\epsilon_{t-\tau})\\
&=E(E(\epsilon_t\epsilon_{t-\tau})\vert \mathcal F_{t-1})=E(\epsilon_{t-\tau} E(\epsilon_t\vert \mathcal F_{t-1}))\\
&=0
\end{align*}
The valuable insight is that $\epsilon_t$ - despite following a GARCH(1,1) process - is a weak white noise because the mean is constant and zero, the variance is constant and finite, and the $\epsilon_t$ are uncorrelated for $\tau\geq 1$.
AR-part
Now, focus on the AR(2) part:
$$
Y_t=\phi_1Y_{t-1}+\phi_2Y_{t-2}+\epsilon_t \quad ,\epsilon_t \sim WN\left(0, \frac{\alpha_0}{1-(\alpha_1+\beta_1)}\right)
$$
Which can be written as:
$$
(1-\phi_1B-\phi_2B^2)Y_t=\epsilon_t
$$
If the conditions $\vert \phi_2\vert<1$, $\phi_2+\phi_1<1$ and $\phi_2-\phi_1<1 $ are fulfilled, the roots lie within the unit circle and hence the process is stationary.
It is not to hard to show that for a general AR(p)-process:
$$
\gamma(k)=
\begin{cases}
\sum_{i=1}^p\phi_i\gamma(i)+Var(\epsilon_t) &, k=0 \\
\sum_{i=1}^p\phi_i\gamma(k-i) &, k \neq 0
\end{cases}
$$
Thus, for $p=2$ and $k=1$, we get:
$$
\gamma(1)=\phi_1\gamma(0)+\phi_2\gamma(-1)=\phi_1\gamma(0)+\phi_2\gamma(1)
$$
Or:
$$
\gamma(1)=\frac{\phi_1\gamma(0)}{1-\phi_2}
$$
And for $k=2$:
$$
\gamma(2)=\phi_1\gamma(1)+\phi_2\gamma(2)=\phi_1\frac{\phi_1\gamma(0)}{1-\phi_2}+\phi_2\gamma(2)
$$
Or equivalent:
$$
\gamma(2)=\frac{\phi_1^2+\phi_2(1-\phi_2)}{1-\phi_2}\gamma(0)
$$
But $\gamma(0)$ is simply given by:
$$
\gamma(0)=\phi_1\gamma(1)+\phi_2\gamma(2)+Var(\epsilon_t)
$$
Thus, we can solve for $\gamma(0)$, which is a rather tedious calculation. The result is:
$$
Var(Y_t)=\gamma(0)=\frac{(1-\phi_2)Var(\epsilon_t)}{(1+\phi_2)(1-\phi_1-\phi_2)(1+\phi_1-\phi_2)}
$$
Where $Var(\epsilon_t)$ was calculated above. Assuming stationarity for the AR-part and the GARCH-part, this is the solution.
