13.4 GARCH(1,1): forecasting volatility

13.4.1Model and stationarity#

Let $\varepsilon_t = r_t - \mathbb E_{t-1}r_t$ (in practice $r_t$ minus a constant, or $r_t$ itself for daily data). GARCH(1,1) (Bollerslev 1986, generalising Engle 1982) is

with $\omega>0$, $\alpha,\beta\ge 0$. $\sigma_t^2$ is $\mathcal F_{t-1}$-measurable. The persistence is $p = \alpha+\beta$.

Proof

$\mathbb E[\varepsilon_t^2] = \mathbb E[\sigma_t^2\,\mathbb E_{t-1}z_t^2] = \mathbb E\sigma_t^2$. Taking expectations of (13.13), $m_t \equiv \mathbb E\sigma_t^2$ satisfies $m_t = \omega + (\alpha+\beta)m_{t-1}$. This linear recursion converges to a finite limit from any start iff $|\alpha+\beta|<1$, and its fixed point is $\omega/(1-\alpha-\beta)$. If $\alpha+\beta\ge 1$, $m_t \ge m_0 + t\omega \to\infty$ (for $=1$) or grows geometrically, so no finite stationary second moment exists. $\varepsilon_t$ is an MDS, so it is uncorrelated at all non-zero lags.∎

Strict stationarity needs only $\mathbb E\ln(\alpha z_t^2+\beta) < 0$ (Nelson 1990), which holds even for $\alpha+\beta = 1$ (IGARCH). An IGARCH process is strictly stationary with infinite variance: its variance forecasts never revert.

Proof

$\mathbb E_{t-1}\eta_t = \sigma_t^2(\mathbb E z_t^2 - 1) = 0$. Substitute $\sigma_t^2 = \varepsilon_t^2 - \eta_t$ and $\sigma_{t-1}^2 = \varepsilon_{t-1}^2 - \eta_{t-1}$ in (13.13): $\varepsilon_t^2 - \eta_t = \omega + \alpha\varepsilon_{t-1}^2 + \beta\varepsilon_{t-1}^2 - \beta\eta_{t-1}$.∎

So squared returns follow an ARMA(1,1) with AR coefficient $\alpha+\beta$. This is why the autocorrelations of $r_t^2$ in M02 decay slowly: with $\alpha+\beta$ near 0.98 the decay rate is 2% a day.

13.4.2Filtering and forecasting#

Given data $r_1,\dots,r_n$, the variance path is computed recursively. The pre-sample values $\varepsilon_0^2$ and $\sigma_0^2$ are unobserved; a standard choice (the arch package's "backcast") sets both equal to a number $b$, an exponentially weighted average of early squared returns. Then

The effect of $b$ decays like $\beta^t$ and is negligible after a few hundred observations.

The half-life of a variance shock is

Proof

$\sigma_{T+1}^2$ is known at $T$, so the claim holds for $k=1$. For $k\ge2$, by the tower property $\mathbb E_T[\varepsilon_{T+k-1}^2] = \mathbb E_T[\sigma_{T+k-1}^2\,\mathbb E_{T+k-2}z^2] = \mathbb E_T[\sigma_{T+k-1}^2]$. Taking $\mathbb E_T$ of (13.13) at $T+k$: $\mathbb E_T\sigma_{T+k}^2 = \omega + (\alpha+\beta)\mathbb E_T\sigma_{T+k-1}^2$. Subtract $\bar\sigma^2 = \omega + (\alpha+\beta)\bar\sigma^2$: the deviation $d_k = \mathbb E_T\sigma_{T+k}^2 - \bar\sigma^2$ satisfies $d_k = (\alpha+\beta)d_{k-1}$, so $d_k = (\alpha+\beta)^{k-1}d_1$. The half-life solves $(\alpha+\beta)^h = 1/2$.∎

The variance of the $H$-day return $\sum_{k=1}^H \varepsilon_{T+k}$ is the sum of the forecasts (the $\varepsilon$ are uncorrelated), so the average variance per day over the next $H$ days is

This is the GARCH term structure of volatility, the object to compare with option-implied volatility of the same maturity (M09).

13.4.3Likelihood and maximum-likelihood estimation#

Proof

By the prediction-error decomposition, $f(\varepsilon_1,\dots,\varepsilon_n) = \prod_t f(\varepsilon_t\mid\mathcal F_{t-1})$, and given $\mathcal F_{t-1}$, $\varepsilon_t \sim N(0,\sigma_t^2)$ with density $(2\pi\sigma_t^2)^{-1/2}\exp(-\varepsilon_t^2/(2\sigma_t^2))$. Take logs and sum. For the gradient, $\partial_\theta[\ln\sigma_t^2 + \varepsilon_t^2/\sigma_t^2] = (1/\sigma_t^2 - \varepsilon_t^2/\sigma_t^4)\,\partial_\theta\sigma_t^2$; multiply by $-\frac12$. Differentiating (13.13) with the chain rule gives the recursion for $\partial_\theta\sigma_t^2$ (the $\alpha\varepsilon_{t-1}^2$ term does not depend on $\theta$ through $\varepsilon$, since $\varepsilon$ is data).∎

At the true $\theta$, each score term has conditional mean $\frac12\mathbb E_{t-1}(z_t^2-1)\cdot(\ldots) = 0$ whatever the distribution of $z_t$, provided $\mathbb E z_t^2 = 1$. So maximising (13.20) is consistent even when returns are fat-tailed: this is quasi-maximum likelihood (QMLE). The standard errors must then use the sandwich form $A^{-1}BA^{-1}$ (Hessian $A$, outer product of scores $B$), as in Bollerslev and Wooldridge (1992). Fat-tailed $z_t$ make $A^{-1}$ alone too small.

Implementation details that matter (all in exercises/modules/M13-forecasting/solutions/m13_garch.py):

• Enforce $\omega>0$, $\alpha,\beta\ge0$, $\alpha+\beta<1$ by reparametrising: $\omega = s\,e^{\vartheta_0}$ ($s$ the sample variance of the returns, a scale), $\alpha+\beta = \operatorname{logistic}(\vartheta_1)$, $\alpha/(\alpha+\beta) = \operatorname{logistic}(\vartheta_2)$, and optimise freely over $\vartheta$.
• Scale returns to percent. The likelihood surface in $\omega$ is badly conditioned when $\omega \sim 10^{-6}$.
• Start from several points; the likelihood is flat along the ridge $\alpha+\beta \approx$ const.

The test suite simulates 10,000 days from $(\omega,\alpha,\beta) = (10^{-6}, 0.08, 0.90)$, recovers $\alpha,\beta$ within 0.03, matches the arch package's log-likelihood to $10^{-8}$ relative at fixed parameters, and checks (13.17) against a 400,000-path Monte Carlo.

13.4.4Kurtosis implied by GARCH#

Proof

Write $\sigma_t^2 = \omega + \sigma_{t-1}^2(\alpha z_{t-1}^2 + \beta)$. Square and take expectations, using independence of $z_{t-1}$ and $\sigma_{t-1}$ and stationarity ($\mathbb E\sigma^4 \equiv m_4$, $\mathbb E\sigma^2 = \bar\sigma^2$): $m_4 = \omega^2 + 2\omega\bar\sigma^2(\alpha+\beta) + m_4\,\mathbb E(\alpha z^2+\beta)^2$. Now $\mathbb E(\alpha z^2+\beta)^2 = 3\alpha^2 + 2\alpha\beta + \beta^2 = (\alpha+\beta)^2 + 2\alpha^2$. With $p = \alpha+\beta$ and $\omega = \bar\sigma^2(1-p)$, the numerator is $\bar\sigma^4[(1-p)^2 + 2p(1-p)] = \bar\sigma^4(1-p^2)$. Hence $m_4 = \bar\sigma^4(1-p^2)/(1-p^2-2\alpha^2)$, and $\mathbb E\varepsilon^4 = 3m_4$.∎

With $(\alpha,\beta) = (0.09, 0.89)$, $\kappa = 3\times0.0396/0.0234 = 5.08$; with $(0.08, 0.90)$, $\kappa = 4.43$. Conditionally Gaussian shocks with GARCH variance already produce fat unconditional tails. Index returns usually show more (M02), which is why Student-$t$ innovations and jump terms are common.

Worked example: a Nifty-like GARCH forecast#

Take illustrative parameters $\alpha = 0.09$, $\beta = 0.89$ (persistence 0.98, inside the 0.95–0.99 range typically reported for equity indices) and a long-run volatility of 15% a year. Daily long-run variance is $\bar\sigma^2 = 0.15^2/252 = 8.93\times10^{-5}$, so $\omega = \bar\sigma^2(1-0.98) = 1.786\times10^{-6}$. The half-life (13.18) is $\ln0.5/\ln0.98 = 34.3$ trading days.

One-step update. Today's conditional volatility was 18% annualised ($\sigma_T^2 = 0.18^2/252 = 1.286\times10^{-4}$) and the index fell 3% ($\varepsilon_T = -0.03$). Then $\sigma_{T+1}^2 = 1.786\times10^{-6} + 0.09\times0.0009 + 0.89\times1.286\times10^{-4} = 1.972\times10^{-4}$, or $\sqrt{1.972\times10^{-4}\times252} = 22.3\%$ annualised.

Term structure. Suppose instead $\sigma_{T+1}$ is 25% annualised. By (13.17) the forecast for day 5 is 24.4% and for day 21 is 22.2% (annualised). By (13.19) the average over the next 21 trading days is $\sqrt{252\,[8.93\times10^{-5} + (2.480\times10^{-4} - 8.93\times10^{-5})(1-0.98^{21})/(21\times0.02)]} = 23.5\%$. A one-month implied volatility well above 23.5% says the options market prices more risk than this model; whether that gap is a variance risk premium or information the model lacks (an election, a budget) is the M09 question.

Evaluating volatility forecasts#

True variance is unobserved. If you evaluate a variance forecast $h_t$ against a noisy unbiased proxy $\hat\sigma_t^2$ (squared return, or intraday realised variance), only some loss functions rank forecasts the same way as they would against the true variance. Patton (2011) shows that MSE, $(\hat\sigma_t^2 - h_t)^2$, and QLIKE, $\hat\sigma_t^2/h_t - \ln(\hat\sigma_t^2/h_t) - 1$, are robust in this sense; MSE on volatilities (square roots) is not. QLIKE is the negative Gaussian log-likelihood up to constants, so (13.20) is a QLIKE fit.

Indian evidence (as reported in abstracts; read the papers before relying on details) Pandey (2003, IIM Ahmedabad working paper) compared GARCH and EGARCH with unconditional estimators against realised volatility on Nifty and reported that conditional models forecast five-day and monthly realised volatility poorly. Rajvanshi, Santra and Basu (IIM Calcutta WPS 797) report that GARCH models augmented with India VIX outperform both plain GARCH and implied volatility alone for Nifty 50, using five-minute realised variance as the target. The two are consistent with a common pattern: implied volatility contains information GARCH lacks (scheduled events), and GARCH's edge is at short horizons.