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:


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!

  • $\begingroup$ One problem here maybe that unconditional variance should not depend on $t$ otherwise it will violate assumption of stationarity. So it must be clarified here that $x_t$ is stationary such that unconditional mean of $x_t$ is also time invariant, thereby making unconditional variance of the process (say, $y_t$) also time invariant. $\endgroup$
    – Dayne
    Dec 7, 2020 at 11:02
  • $\begingroup$ In my case x_t is stationary. So this would imply the expectation above in my expression should just be substituted with the unconditional mean of x_t? $\endgroup$
    – Emil Bille
    Dec 7, 2020 at 11:18
  • $\begingroup$ I think so yes. $\endgroup$
    – Dayne
    Dec 7, 2020 at 13:05
  • $\begingroup$ Note that $\mathbb{E}(x_{t-1}\mid I_{t-1})=x_{t-1}$. $\endgroup$ Dec 19, 2020 at 10:59

1 Answer 1


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

$${\rm Var}(u) = \omega+\alpha\cdot {\rm Var}(u)+\beta\cdot {\rm Var}(u)+\phi\cdot E(x) $$

$$\implies {\rm Var}(u) =\frac{\omega+\phi\cdot E(x)}{1-\alpha-\beta}$$

  • $\begingroup$ That makes alot of sence. Thank you! :) $\endgroup$
    – Emil Bille
    Dec 7, 2020 at 22:10
  • $\begingroup$ The answer is intuitive, but is a proof available anywhere? $\endgroup$ Dec 19, 2020 at 11:00
  • $\begingroup$ @RichardHardy The proof is simple, and I included it in the post. $\endgroup$ Dec 19, 2020 at 14:40
  • $\begingroup$ @AlecosPapadopoulos, appreciated. $\endgroup$ Dec 19, 2020 at 15:04

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