<p>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</p>
<p><span class="math-container">$$\sigma^2=\frac{\omega+\phi\cdot E(x)}{1-\alpha-\beta}$$</span></p>
<p>PS: Also, you are missing a <span class="math-container">$t$</span>-subscript in your penultimate expression.</p>
<p>--<br />
<strong>RESPONSE</strong> to comment<br />
Using more explicit notation, we <em>assume</em> that the conditional variance is</p>
<p><span class="math-container">$${\rm Var}(u_t \mid I_{t-1}) = \omega+\alpha\cdot u_{t-1}^2+\beta\cdot {\rm Var}(u_{t-1} \mid I_{t-2})+\phi\cdot x_{t-1}$$</span></p>
<p>Taking expectations through out</p>
<p><span class="math-container">$$E\big[{\rm Var}(u_t \mid I_{t-1})\big] = \omega+\alpha\cdot E(u_{t-1}^2)+\beta\cdot E\big[{\rm Var}(u_{t-1} \mid I_{t-2})\big]+\phi\cdot E(x_{t-1}) \tag{1}$$</span></p>
<p>Now, by the Law of Total Variance,</p>
<p><span class="math-container">$${\rm Var}(u_t) = E\big[{\rm Var}(u_t \mid I_{t-1})\big] + {\rm{Var}}\big[E(u_t \mid I_{t-1})\big]$$</span></p>
<p>Under the assumption <span class="math-container">$E(u_t \mid I_{t-1}) = 0 \implies E(u_t) = 0$</span>, we obtain the relation</p>
<p><span class="math-container">$${\rm Var}(u_t) = E\big[{\rm Var}(u_t \mid I_{t-1})\big] \tag{2}$$</span></p>
<p>and lagging once,</p>
<p><span class="math-container">$${\rm Var}(u_{t-1}) = E\big[{\rm Var}(u_{t-1} \mid I_{t-2})\big] \tag{3}$$</span></p>
<p>Another assumption is that <span class="math-container">$u$</span> is homoskedastic unconditionally. This together with <span class="math-container">$E(u_t)=0$</span> means</p>
<p><span class="math-container">$${\rm Var}(u_t) = {\rm Var}(u_{t-1}) = E(u^2) \tag{4}$$</span></p>
<p>Finally, another assumption is that the <span class="math-container">$x$</span>'s are identically distributed over time, so <span class="math-container">$E(x_{t-1}) = E(x)$</span>.</p>
<p>Using all these into <span class="math-container">$(1)$</span> we get</p>
<p><span class="math-container">$${\rm Var}(u) = \omega+\alpha\cdot {\rm Var}(u)+\beta\cdot {\rm Var}(u)+\phi\cdot E(x) $$</span></p>
<p><span class="math-container">$$\implies {\rm Var}(u) =\frac{\omega+\phi\cdot E(x)}{1-\alpha-\beta}$$</span></p>

CC BY-SA 4.0
2020-12-07T22:10:54.620Z
That makes alot of sence. Thank you! :)
Emil Bille
CC BY-SA 4.0
2020-12-19T11:00:55.810Z
The answer is intuitive, but is a proof available anywhere?
Richard Hardy
CC BY-SA 4.0
2020-12-19T14:40:18.930Z
@RichardHardy The proof is simple, and I included it in the post.
Alecos Papadopoulos
CC BY-SA 4.0
2020-12-19T15:04:57.417Z
@AlecosPapadopoulos, appreciated.
Richard Hardy
CC BY-SA 4.0
2020-12-07T21:30:24.123Z
2020-12-19T14:39:58.853Z
2020-12-19T14:39:58.853Z
Alecos Papadopoulos
<p>I have a question about the <strong>unconditional</strong> variance of a GARCH process, where exogenous explanatory variables are included in the variance.</p>
<p>The usual GARCH models the variance using: <span class="math-container">$$\sigma^2_t=\omega+\alpha\cdot\epsilon_{t-1}^2+\beta\cdot\sigma_{t-1}^2$$</span></p>
<p>The usual GARCH <strong>unconditional</strong> variance, without additional explanatory variables, is given by:</p>
<p><span class="math-container">$$\sigma^2_=\frac{\omega}{1-\alpha-\beta}$$</span></p>
<p>My question is, if we include an explanatory variable <span class="math-container">$x_t$</span> in the variance equation, how does this change the <strong>unconditional</strong> variance?</p>
<p>If the model is:</p>
<p><span class="math-container">$$\sigma^2=\omega+\alpha\cdot\epsilon_{t-1}^2+\beta\cdot\sigma_{t-1}^2+\phi\cdot x_{t-1}$$</span></p>
<p>My guess would be that on some day <span class="math-container">$t$</span>, the unconditional variance would change with the previous days value of <span class="math-container">$x_t$</span>, so something like:</p>
<p><span class="math-container">$$\sigma^2_t=\frac{\omega+\phi\cdot E[x_{t-1}|I_{t-1}]}{1-\alpha-\beta}$$</span></p>
<p>Hope it makes sence!</p>

CC BY-SA 4.0
2020-12-07T11:02:04.580Z
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.
Dayne
CC BY-SA 4.0
2020-12-07T11:18:11.083Z
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?
Emil Bille
CC BY-SA 4.0
2020-12-07T13:05:18.300Z
I think so yes.
Dayne
CC BY-SA 4.0
2020-12-19T10:59:45.450Z
Note that $\mathbb{E}(x_{t-1}\mid I_{t-1})=x_{t-1}$.
Richard Hardy
<econometrics><mathematical-economics><statistics><financial-economics>
How does explanatory variables in variance affect unconditional variance of GARCH(1,1)
Emil Bille
# How does explanatory variables in variance affect unconditional variance of GARCH(1,1)

(Asked by: username_0 on 2020-12-07)

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!

**Tags:** econometrics, mathematical-economics, statistics, financial-economics
> **Comments:**
> **username_1**: 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.  
> **username_0**: 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?  
> **username_1**: I think so yes.  
> **username_2**: Note that $\mathbb{E}(x_{t-1}\mid I_{t-1})=x_{t-1}$.  

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## 1 answers

### Answer by username_3 on 2020-12-07. Score: 2 | Accepted answer

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

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

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

> **Comments:**
> **username_0**: That makes alot of sence. Thank you! :)  
> **username_2**: The answer is intuitive, but is a proof available anywhere?  
> **username_3**: @RichardHardy The proof is simple, and I included it in the post.  
> **username_2**: @AlecosPapadopoulos, appreciated.  

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