Maybe it's a simple question but I don't really understand why it is theoretically required. Let's take the standard GARCH(1,1) $$\sigma^2_{t+1}=\omega+\alpha\epsilon^2_{t}+\beta\sigma^2_{t}$$ In most textbooks the conditions are: $\omega>0$, $\alpha,\beta\geq0$ and $\alpha+\beta<1$.

I don't know in other contexts, but for financial series, $\alpha$ and $\beta$ are in most cases strongly significant, positive and rather persistent (sum slightly less than 1). By construction (since they are squared) $\epsilon^2_t$ and $\sigma^2_t$ are always positive. The positivity of both coefficients $\alpha$ and $\beta$ and lagged variables $\epsilon^2_{t}$ and $\sigma^2_{t}$ ensures the positivity of $\sigma^2_{t+1}$. Considering this, why then should $\omega$ be strictly positive and why is the condition $\geq0$ not enough?

I have a few series for which I want to estimate the conditional volatility equation. Most of them have a positive intercept, but some of these have $\omega=0$ with positive coefficients $\alpha$ and $\beta$. What should I do since this is not theoretically contemplated in textbooks? How should I interpret these results?


1 Answer 1


Consider the GARCH(1,1) process \begin{align} r_{t+1} &= \sigma_{t+1} z_{t+1} \\ \sigma^2_{t+1} &= \omega+\alpha r^2_t +\beta \sigma^2_{t} \end{align} for the returns $r_t$, with ${z_t} \sim N (0,1)$ IID.

In what follows, let us distinguish the conditional return variance $$ V [ r_{t+1} \vert \mathcal{F}_t ] = \sigma^2_{t+1} $$ from the unconditional return variance $$ V [ r_{t+1} ]$$ where $\mathcal{F}_t$ denotes the information available up to time $t$ (filtration), in our case all past values $\{\sigma_t,...,\sigma_0,r_t,...,r_0\}$.

First, notice that if both $\alpha$ and $\beta $ happen to be zero, then if $\omega$ is allowed to reach zero as well there is a possibility that the conditional variance will become zero.

Second, because $z_t$ are zero-mean, unit variance and i.i.d, we have that the unconditional returns' variance is strictly equal to (*) \begin{align} V[r_{t+1}] &= E[r_{t+1}^2]-E [r_{t+1}]^2 = E[r_{t+1}^2] \\ &= E [ \sigma_{t+1}^2 z_{t+1}^2] = E [ \sigma_{t+1}^2] E [z_{t+1}^2] \\ &= E[\sigma_{t+1}^2] \end{align}

We can use this relationship along with the GARCH definition to write \begin{align} V[r_{t+1}] &= E[\sigma_{t+1}^2] = E [ (\omega+\alpha r^2_t +\beta \sigma^2_{t}) ] \\ &= \omega + \alpha V [r_t] + \beta V [r_t] \\ &= \omega + V [r_t] (\alpha + \beta) \end{align} Now, assuming weak stationarity we should have that the unconditional variance $\sigma^2_\infty$ is such that $$ \sigma^2_\infty = V [r_{t+1}] = V [r_t] $$ which using the above gives $$ \sigma^2_\infty = \frac {\omega}{1-\alpha-\beta} $$ From which we see why the condition $\alpha + \beta < 1$ is relevant but also that allowing $\omega $ to reach zero would also imply a possibility that the unconditional variance becomes zero.

At the end of the day, we therefore have that if $\omega=0$ while the other coefficients are positive, you will have (conditional) heteroskedasticity in the sense that conditional variance will evolve through time, but the unconditional stationary variance will be zero, with unrealistic consequence that returns become deterministic at some point.

(*) This could have been anticipated since \begin{align} V [r_{t+1}] &= V [r_{t+1} \vert \mathcal{F}_0] \\ &= E [r_{t+1}^2 \vert \mathcal{F}_0] \\ &= E [ E [ r_{t+1}^2 \vert \mathcal{F}_t] \vert \mathcal{F}_0] \\ &= E [ V [ r_{t+1} \vert \mathcal{F}_t] \vert \mathcal{F}_0] \\ &= E [ \sigma_{t+1}^2 \vert \mathcal{F}_0] \\ &= E [ \sigma_{t+1}^2 ] \end{align} where we have made use of the tower law (and the fact that $z_t $ are zero-mean and independently distributed so that $E [r_{t+1}]=0$).

  • $\begingroup$ Thank you Quantuple, I haven't thought of the unconditional variance, that makes perfect sense. However, I'm still unsure about how to proceed in my case. I have nine series, six of them have positive but small $\omega$ (from 0.000001 to 0.000003), three of them have $\omega$=0.000000 (I'm working with rugarch package in R). Should I assume that omega is in the order of 10^{-8}, such that when approximated with 6 decimals it is just 0? In that case, should I use an artificial value (for example 4,9*10^{-8})? If that's not the case, why would R imply an unconditional variance equal to 0? $\endgroup$
    – Kondo
    Commented Jun 1, 2016 at 7:12
  • $\begingroup$ Well zero intercept would mean zero unconditional variance, i.e. your returns become deterministic at some point which is plain wrong. I'm not familiar with that packag sorry. But is it possible that you misinterpret the coefficients it returns? Are your series stationary? $\endgroup$
    – Quantuple
    Commented Jun 1, 2016 at 7:37
  • $\begingroup$ Alright! I solved it using your answer to another of my questions (see <quant.stackexchange.com/questions/26076/…). Since I'm using the variance targeting technique, I reversed the equation of the unconditional variance (substituting with sample variance) and found the implied $\omega$ which is smaller than for other series (in the order of 10^{-7}), and therefore approximated to 0.000000. Thanks for both hints! $\endgroup$
    – Kondo
    Commented Jun 1, 2016 at 8:13
  • $\begingroup$ @Kondo, Glad I could help. indeed $\omega << 1$ is not something infrequent especially if the other coefficients are of the same order. It is really the unconditional and conditional variances at which you should look at. $\endgroup$
    – Quantuple
    Commented Jun 1, 2016 at 9:33

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