GARCH(1,1) Stationarity, Unconditional Variance, and Shock Propagation

Worked Solution

How to Think About It: GARCH(1,1) is the workhorse volatility model on every trading desk. The intuition is simple: tomorrow's variance is a weighted average of a long-run baseline ($\omega$), yesterday's squared return (the "news" or shock, weighted by $\alpha$), and yesterday's variance forecast (the "memory," weighted by $\beta$). The key number is $\alpha + \beta$ -- that is the persistence. If it equals or exceeds 1, shocks never die out and there is no well-defined long-run variance. If it is less than 1, shocks decay geometrically and you get mean-reversion in volatility.

Quick Sanity Checks: For equity indices, typical fitted values are $\alpha \approx 0.05$-$0.10$, $\beta \approx 0.85$-$0.90$, giving $\alpha + \beta \approx 0.95$-$0.99$. This means volatility is highly persistent but stationary. The unconditional variance should be a finite positive number, and shocks should have a half-life on the order of weeks to months.

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(a) Conditions for Covariance Stationarity

The conditional variance must be positive almost surely and finite in expectation. The conditions are:

• $\omega > 0$
• $\alpha \geq 0$
• $\beta \geq 0$
• $\alpha + \beta < 1$

The first three ensure $\sigma_t^2 > 0$ for all $t$. The critical condition is the fourth: $\alpha + \beta < 1$ guarantees that the variance process is mean-reverting and the unconditional moments exist.

To see why, note that GARCH(1,1) can be written as an ARMA(1,1) in $r_t^2$. Define $v_t = r_t^2 - \sigma_t^2$ (the variance "surprise"). Then:

$r_t^2 = \omega + (\alpha + \beta) \, r_{t-1}^2 + v_t - \beta \, v_{t-1}$

This is an ARMA(1,1) process for $r_t^2$ with AR coefficient $(\alpha + \beta)$. Covariance stationarity of an ARMA process requires the AR root to lie inside the unit circle, i.e., $\alpha + \beta < 1$.

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(b) Unconditional Variance

Under stationarity, the unconditional expectation satisfies $E[\sigma_t^2] = E[\sigma_{t-1}^2]$. Take expectations of the GARCH equation:

$E[\sigma_t^2] = \omega + \alpha \, E[r_{t-1}^2] + \beta \, E[\sigma_{t-1}^2]$

Since $E[r_t^2] = E[E[r_t^2 \mid \mathcal{F}_{t-1}]] = E[\sigma_t^2]$, write $\bar{\sigma}^2 = E[\sigma_t^2]$:

$\bar{\sigma}^2 = \omega + \alpha \, \bar{\sigma}^2 + \beta \, \bar{\sigma}^2 = \omega + (\alpha + \beta) \bar{\sigma}^2$

Solving:

$\boxed{\bar{\sigma}^2 = \frac{\omega}{1 - \alpha - \beta}}$

This is well-defined and positive precisely when $\alpha + \beta < 1$ and $\omega > 0$.

Numerical check: With $\omega = 0.00001$, $\alpha = 0.08$, $\beta = 0.90$ (persistence $= 0.98$), the unconditional variance is $0.00001 / 0.02 = 0.0005$, corresponding to a daily vol of $\sqrt{0.0005} \approx 2.2\%$, or roughly $35\%$ annualized. That is in the right ballpark for a volatile equity index.

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(c) Shock Propagation and Half-Life

Suppose at time $t-1$ there is a volatility shock -- a large squared return $r_{t-1}^2$ that pushes $\sigma_t^2$ above its long-run mean. How does this excess variance decay?

Iterate the GARCH equation forward. The deviation from the mean $\bar{\sigma}^2$ at horizon $k$ satisfies:

$E_t[\sigma_{t+k}^2] - \bar{\sigma}^2 = (\alpha + \beta)^k \, (\sigma_t^2 - \bar{\sigma}^2)$

So the impact of any shock decays geometrically at rate $(\alpha + \beta)$ per period. This is the persistence parameter.

Propagation mechanism: The shock enters through the $\alpha \, r_{t-1}^2$ term (direct impact). On subsequent days, it is carried forward through the $\beta \, \sigma_{t-1}^2$ term (the "memory" channel). Together, each day the remaining impact is multiplied by $(\alpha + \beta)$.

Half-life: The half-life $h$ is the number of periods until the shock's impact drops to half its initial size:

$(\alpha + \beta)^h = \frac{1}{2}$

$\boxed{h = \frac{\ln 2}{-\ln(\alpha + \beta)} = \frac{\ln 2}{\ln\!\left(\frac{1}{\alpha + \beta}\right)}}$

Example: If $\alpha + \beta = 0.98$, then $h = \ln 2 / \ln(1/0.98) = 0.693 / 0.0202 \approx 34$ trading days, or about 7 weeks. At persistence $0.95$, the half-life drops to about 14 days. At $0.99$ it climbs to 69 days. This is why equity vol feels "sticky" -- typical persistence is in the $0.95$-$0.99$ range.

Answer: (a) Require $\omega > 0$, $\alpha \geq 0$, $\beta \geq 0$, and $\alpha + \beta < 1$. (b) The unconditional variance is $\bar{\sigma}^2 = \omega / (1 - \alpha - \beta)$. (c) Shocks decay geometrically at rate $(\alpha + \beta)^k$, with half-life $h = \ln 2 \,/\, \ln(1/(\alpha + \beta))$.