Conditional volatility and GARCH

Definition Fix a probability model with physical measure $\mathbb{P}$, under which every expectation on this page is taken, and let $\mathcal{F}_{t-1}$ be the information at date $t-1$, generated by the initial values and the innovations observed up to $t-1$. For an innovation $\varepsilon_t$ with $\mathbb{E}[\varepsilon_t^2] < \infty$ and conditional mean $\mu_t = \mathbb{E}[\varepsilon_t \mid \mathcal{F}_{t-1}]$, the conditional variance is

a nonnegative random variable whose value is known at date $t-1$. The volatility models of this lesson write the innovation as

where $h_t > 0$ is $\mathcal{F}_{t-1}$-measurable and the standardized shock $z_t$ satisfies

Taking out what is known gives $\mathbb{E}[\varepsilon_t \mid \mathcal{F}_{t-1}] = 0$ and $\operatorname{Var}(\varepsilon_t \mid \mathcal{F}_{t-1}) = h_t$. Both statements concern the conditional distribution of $\varepsilon_t$ given the past, in which $h_t$ is a known number, so they stay true when $\mathbb{E}[\varepsilon_t^2]$ is infinite. When the innovations have finite second moments, the law of total variance gives the unconditional variance $\operatorname{Var}(\varepsilon_t) = \mathbb{E}[h_t]$, because the conditional mean is zero, and the zero conditional mean also gives $\operatorname{Cov}(\varepsilon_t, \varepsilon_{t-k}) = 0$ for every $k \ge 1$.

Definition An ARCH model of order $q$ sets $\varepsilon_t = \sqrt{h_t}\, z_t$, where the standardized shocks $z_t$ are independent and identically distributed with mean $0$ and variance $1$ and are independent of the initial values, so that $z_t$ is independent of $\mathcal{F}_{t-1}$. The conditional variance is

with $\omega > 0$ and $\alpha_j \ge 0$. These signs give $h_t \ge \omega > 0$, and because only squares enter, a shock of $-c$ raises the next variance exactly as much as a shock of $+c$. For the ARCH(1) model $h_t = \omega + \alpha \varepsilon_{t-1}^2$, suppose a strictly stationary solution with $\sigma^2 = \mathbb{E}[\varepsilon_t^2] < \infty$ exists. Then $\mathbb{E}[\varepsilon_t^2] = \mathbb{E}[h_t] = \omega + \alpha\, \mathbb{E}[\varepsilon_{t-1}^2]$, so $\sigma^2 = \omega + \alpha \sigma^2$ and

Since $\sigma^2 (1 - \alpha) = \omega > 0$, this forces $\alpha < 1$. The calculation finds the only possible value and shows that $\alpha < 1$ is necessary. That a stationary solution with finite variance exists when $0 \le \alpha < 1$ is a separate result about the stochastic recursion, stated here without proof. If moreover $\mathbb{E}[\varepsilon_t^4] < \infty$, the differences $\nu_t = \varepsilon_t^2 - h_t$ are martingale differences with finite variance and

so the squared innovations follow an AR(1) model with coefficient $\alpha$ and $\operatorname{Corr}(\varepsilon_t^2, \varepsilon_{t-1}^2) = \alpha$. With Gaussian shocks, the stationary solution has a finite fourth moment exactly when $3\alpha^2 < 1$.

Definition A GARCH(1,1) model keeps $\varepsilon_t = \sqrt{h_t}\, z_t$ with the same shock assumptions as the ARCH model and lets the variance depend on its own past,

with $\omega > 0$, $\alpha \ge 0$, $\beta \ge 0$ and a given starting value $h_0 > 0$. The coefficient $\alpha$ is the response to the latest squared shock, $\beta$ carries forward part of the previous variance, and the signs give $h_t \ge \omega > 0$ for every $t \ge 1$. Substituting $\varepsilon_{t-1}^2 = h_{t-1} z_{t-1}^2$ gives the random-coefficient form

in which the multiplier $\alpha z_{t-1}^2 + \beta$ is independent of $h_{t-1}$ and has mean $\alpha + \beta$. For a strictly stationary solution with $\beta < 1$, repeated substitution writes the variance as an ARCH model of infinite order,

GARCH models with more lags add further past squared innovations and further past variances, all with nonnegative coefficients.

Definition In the GARCH(1,1) model, write

for the variance forecast made at date $t-1$ for the horizon $k \ge 0$, where $h_{t \mid t-1} = h_t$ because $h_t$ is already known. It is also the forecast of the squared innovation, since the tower property and $\mathbb{E}[z_{t+k}^2 \mid \mathcal{F}_{t+k-1}] = 1$ give $\mathbb{E}[\varepsilon_{t+k}^2 \mid \mathcal{F}_{t-1}] = h_{t+k \mid t-1}$, which is the conditional variance of $\varepsilon_{t+k}$ given $\mathcal{F}_{t-1}$. Taking conditional expectations of the recursion replaces the unknown $\varepsilon_{t+k}^2$ by $h_{t+k \mid t-1}$, so

The number $\phi$ is the variance persistence. If $\phi \ne 1$ and $\bar h = \omega/(1 - \phi)$, solving the recursion gives

and if $\phi = 1$ it gives $h_{t+k \mid t-1} = h_t + k\omega$. The recursion uses only conditional expectations, so it holds whether or not the process is stationary. When $\phi < 1$ the forecast reverts geometrically to $\bar h$, and for $0 < \phi < 1$ the distance to $\bar h$ halves every $\ln(1/2)/\ln \phi$ steps. When $\phi \ge 1$ the forecast grows without bound. The same number governs the squared innovations. With $\nu_t = \varepsilon_t^2 - h_t$,

and for a stationary solution with a finite fourth moment this is an ARMA(1,1) model with autoregressive coefficient $\phi$ and white noise $\nu_t$.

Definition For a strictly stationary solution with $\mathbb{E}[\varepsilon_t^2] < \infty$, the unconditional variance is the same at every date and equals the mean conditional variance,

In the GARCH(1,1) model, taking expectations of the recursion gives $\mathbb{E}[h_t] = \omega + (\alpha + \beta)\, \mathbb{E}[h_t]$, so

and because $\omega > 0$, a finite variance forces $\alpha + \beta < 1$. The converse is a standard result, stated here without proof. If $\alpha + \beta < 1$, the recursion has a strictly stationary solution with this finite variance, and that solution is weakly stationary with zero autocovariances at every nonzero lag. The boundary case $\alpha + \beta = 1$ is called integrated GARCH, and whenever $\alpha + \beta \ge 1$ no stationary solution has a finite variance. Higher moments need more. If $\kappa = \mathbb{E}[z_t^4] < \infty$, the stationary solution has a finite fourth moment exactly when

which says that the squared multiplier $(\alpha z_t^2 + \beta)^2$ has mean below one. A unit variance of $z_t$ therefore does not make the fourth moment of $\varepsilon_t$ finite.

Definition A process $(X_t)$ is strictly stationary if for every $n$, all dates $t_1, \dots, t_n$ and every shift $s$, the vector $(X_{t_1 + s}, \dots, X_{t_n + s})$ has the same joint distribution as $(X_{t_1}, \dots, X_{t_n})$. The definition needs no moments. A strictly stationary process with finite second moments is weakly stationary, while a strictly stationary process can have an infinite variance. For the GARCH(1,1) model with $\omega > 0$, $\alpha \ge 0$, $\beta \ge 0$ and independent, identically distributed shocks with $\mathbb{E}[z_t^2] = 1$, write $a_t = \alpha z_t^2 + \beta$ and

which is well defined, possibly $-\infty$, because $\mathbb{E}[z_t^2] < \infty$ makes the positive part of $\ln a_t$ integrable. The following sufficient condition is a standard result, stated here without proof. If $\gamma < 0$, the series

converges almost surely and defines a strictly stationary solution of the recursion, and then $\varepsilon_t = \sqrt{h_t}\, z_t$ is strictly stationary too. The logarithm is the right scale because the strong law of large numbers gives $\frac{1}{k} \sum_{i=1}^{k} \ln a_{t-i} \to \gamma$ almost surely, so the products go to zero exponentially fast, while their expectations $(\alpha + \beta)^k$ govern the moments instead. Jensen’s inequality gives $\gamma \le \ln(\alpha + \beta)$, so $\alpha + \beta < 1$ implies $\gamma < 0$, but $\gamma < 0$ also allows $\alpha + \beta \ge 1$.