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


1 Answer 1


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 $\tau\geq 1$.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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