# 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. $$\gamma_j = Cov(Y_t, Y_{t-j}$$)).

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.