How does explanatory variables in variance affect unconditional variance of GARCH(1,1) I have a question about the unconditional variance of a GARCH process, where exogenous explanatory variables are included in the variance.
The usual GARCH models the variance using: $$\sigma^2_t=\omega+\alpha\cdot\epsilon_{t-1}^2+\beta\cdot\sigma_{t-1}^2$$
The usual GARCH unconditional variance, without additional explanatory variables, is given by:
$$\sigma^2_=\frac{\omega}{1-\alpha-\beta}$$
My question is, if we include an explanatory variable $x_t$ in the variance equation, how does this change the unconditional variance?
If the model is:
$$\sigma^2=\omega+\alpha\cdot\epsilon_{t-1}^2+\beta\cdot\sigma_{t-1}^2+\phi\cdot x_{t-1}$$
My guess would be that on some day $t$, the unconditional variance would change with the previous days value of $x_t$, so something like:
$$\sigma^2_t=\frac{\omega+\phi\cdot E[x_{t-1}|I_{t-1}]}{1-\alpha-\beta}$$
Hope it makes sence!
Your last expression is not correct because, as noted in the comments, you are after the unconditional variance, which is constant in these models. it should be
$$\sigma^2=\frac{\omega+\phi\cdot E(x)}{1-\alpha-\beta}$$
PS: Also, you are missing a $t$-subscript in your penultimate expression.
--
RESPONSE to comment
Using more explicit notation, we assume that the conditional variance is
$${\rm Var}(u_t \mid I_{t-1}) = \omega+\alpha\cdot u_{t-1}^2+\beta\cdot {\rm Var}(u_{t-1} \mid I_{t-2})+\phi\cdot x_{t-1}$$
Taking expectations through out
$$E\big[{\rm Var}(u_t \mid I_{t-1})\big] = \omega+\alpha\cdot E(u_{t-1}^2)+\beta\cdot E\big[{\rm Var}(u_{t-1} \mid I_{t-2})\big]+\phi\cdot E(x_{t-1}) \tag{1}$$
Now, by the Law of Total Variance,
$${\rm Var}(u_t) = E\big[{\rm Var}(u_t \mid I_{t-1})\big] + {\rm{Var}}\big[E(u_t \mid I_{t-1})\big]$$
Under the assumption $E(u_t \mid I_{t-1}) = 0 \implies E(u_t) = 0$, we obtain the relation
$${\rm Var}(u_t) = E\big[{\rm Var}(u_t \mid I_{t-1})\big] \tag{2}$$
and lagging once,
$${\rm Var}(u_{t-1}) = E\big[{\rm Var}(u_{t-1} \mid I_{t-2})\big] \tag{3}$$
Another assumption is that $u$ is homoskedastic unconditionally. This together with $E(u_t)=0$ means
$${\rm Var}(u_t) = {\rm Var}(u_{t-1}) = E(u^2) \tag{4}$$
Finally, another assumption is that the $x$'s are identically distributed over time, so $E(x_{t-1}) = E(x)$.
Using all these into $(1)$ we get
$${\rm Var}(u) = \omega+\alpha\cdot {\rm Var}(u)+\beta\cdot {\rm Var}(u)+\phi\cdot E(x) $$
$$\implies {\rm Var}(u) =\frac{\omega+\phi\cdot E(x)}{1-\alpha-\beta}$$