Volatility models: ARCH, GARCH and realised measures

TS · Chapter 413 min readAsked at Two Sigma, Citadel, Optiver, AQR

Assumes The ARMA family: identification, estimation and forecasting.

After this lesson you should be able to

  • Write down GARCH(1,1) and interpret its parameters.
  • Compute the persistence and the long-run variance.
  • Say why realised volatility measures beat daily-return models when you have intraday data.

Returns are nearly unpredictable in the mean and highly predictable in the variance. GARCH is the standard model of that second fact, and it is worth knowing well because volatility forecasting is one of the few places in finance where a model genuinely works.

Equation 4.1

GARCH(1,1)

Today’s variance is a constant, plus a reaction to yesterday’s squared shock, plus a memory of yesterday’s variance.

σt2=ω+αεt−12+βσt−12\sigma_t^2 = \omega + \alpha\varepsilon_{t-1}^2 + \beta\sigma_{t-1}^2
α\alpha
How sharply volatility reacts to news. Typically around 0.050.05–0.100.10 on daily equity data.
β\beta
How long it remembers. Typically 0.850.85–0.920.92.
α+β\alpha + \beta
Persistence. Below one for stationarity, and usually just below.

Derivation 4.2

The long-run variance

Set the conditional variance equal to its own expectation and solve.

  1. σˉ2=ω+ασˉ2+βσˉ2\bar\sigma^2 = \omega + \alpha\bar\sigma^2 + \beta\bar\sigma^2

    In the stationary state, E[ε2]=σˉ2\mathbb{E}[\varepsilon^2] = \bar\sigma^2.

  2. σˉ2(1−α−β)=ω\bar\sigma^2(1 - \alpha - \beta) = \omega
σˉ2=ω1−α−β,t1/2=ln⁡0.5ln⁡(α+β)\bar\sigma^2 = \frac{\omega}{1 - \alpha - \beta}, \qquad t_{1/2} = \frac{\ln 0.5}{\ln(\alpha + \beta)}

Example 4.3

A GARCH(1,1) fit gives ω=0.000002\omega = 0.000002, α=0.08\alpha = 0.08, β=0.90\beta = 0.90. What is the long-run volatility and the half-life of a shock?

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Worked solution

  1. Formula
    σˉ2=ω1−α−β,σann=σˉ252\bar\sigma^2 = \frac{\omega}{1 - \alpha - \beta}, \qquad \sigma_{\text{ann}} = \bar\sigma\sqrt{252}
  2. Substitute
    1−0.98=0.021 - 0.98 = 0.02
  3. Solve
    σˉ2=2×10−60.02=10−4\bar\sigma^2 = \frac{2 \times 10^{-6}}{0.02} = 10^{-4}
  4. σˉ=0.01 daily⇒0.01×15.87=15.9%\bar\sigma = 0.01 \text{ daily} \Rightarrow 0.01 \times 15.87 = 15.9\%
  5. t1/2=ln⁡0.5ln⁡0.98≈34 dayst_{1/2} = \frac{\ln 0.5}{\ln 0.98} \approx 34 \text{ days}
  6. Answer
    about 16% annualised, with a 34-day half-life\text{about } 16\% \text{ annualised, with a } 34\text{-day half-life}

Sanity check. A persistence of 0.980.98 is entirely typical and means shocks take weeks to fade — which is exactly the volatility clustering everyone observes. Push it to 0.9950.995 and the half-life is about 138 trading days, more than six months.

060120000From a quiet marketFrom a shockDays aheadForecast variance
Figure 4.4 · How a volatility forecast returns to normal. With α+β=0.98\alpha + \beta = 0.98 the pull back to the long-run variance has a half-life of about 3434 days. Both forecasts converge on the same level, which is what makes ω/(1−α−β)\omega/(1-\alpha-\beta) the number to quote when someone asks where volatility settles.

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