3
$\begingroup$

The ARCH model is: $$\left\{ \begin{align*}& X_t=\sigma_t Z_t, \ \{Z_t\} \sim IIDN(0,1) \\ & \sigma_t ^2 =\alpha _0 +\alpha _1X_{t-1}^2+\ldots+\alpha _p X_{t-p}^2 \end{align*} \right. $$ After fitting such a model we can forecast $\sigma_t^2$, but (I think) the process $\{ X_t \}$ is of interest, not $\{ \sigma_t ^2 \}$. So, actually why do we fit (G)ARCH model, since $Z_t$ has expectation 0 hence the best forecast for $X_t$ will be always 0?

Maybe we fit (G)ARCH model because we always fit it along with, for example, an ARMA model (so called ARMA-GARCH model)?

$\endgroup$
  • $\begingroup$ You are right that the process $\{X_t\}$ is of interest. Now, $\{\sigma_t^2\}$ is one of its properties, making it a natural candidate to be of interest as well. $\endgroup$ – Richard Hardy Aug 18 at 15:55
4
$\begingroup$

The purpose of GARCH models is not typically to make point forecasts. When it is, even if the point forecast doesn't change over time, the width of the prediction interval will, which is generally of value for decision making.

The variance process can be of direct, independent interest, or even of more interest than $X_t$, depending on the application domain. For example, in the original Bollerslev paper about GARCH, an application to inflation forecasting is discussed; the fact that the difficulty in forecasting inflation changes over time and how this ties in to policy and so on is of apparently timeless interest to Economists. An ARMA-GARCH model is used, but it is the variance process that is mainly of interest there.

One of the main applications of GARCH models is in Finance, for stock returns. There, the full predictive distribution is required to make investment decisions; two assets which have the same point forecast but vastly different risk levels are not equally interesting investments. Typically, returns will have a small, difficult to estimate non-zero mean, little to no ARMA structure, but significant time variation in volatility. GARCH is a relatively simple way of getting a decent estimate of volatility, which is crucial in being able to correctly balance risk and reward.

| cite | improve this answer | |
$\endgroup$
3
$\begingroup$

Another very important application of (G)ARCH models is conditional Value-at-Risk (CVaR) estimation. At some level $\alpha$, it is implicitly defined, at time $T$, via $$ P(Y_{T+1}\leq\text{CVaR}_{\alpha, T+1}\mid Y_T=y_T,Y_{T-1}=y_{T-1},\ldots)=\alpha. $$

Take a $GARCH(1,1)$ model with $\epsilon_t\sim N(0,1)$ and \begin{equation}\label{eq:vol GARCH(1,1)} \sigma_{T+1}^2=\sigma_{T+1}^2(\theta)=\omega+\alpha_1Y_{T}^2+\beta_1\sigma_{T}^2. \end{equation} We have $$ P(Y_{T+1}\leq x\mid Y_T=y_T,Y_{T-1}=y_{T-1},\ldots)=\Phi(x/\sigma_{T+1}). $$

Hence, $\Phi(\text{CVaR}_{\alpha, T+1}/\sigma_{T+1})=\alpha$, from which $$ \text{CVaR}_{\alpha, T+1}=\sigma_{T+1}\cdot\Phi^{-1}(\alpha). $$ In practice, we may substitute GARCH estimates $\sigma_t^2(\widehat{\theta})$ to get $$ \widehat{\text{CVaR}}_{\alpha, T+1}=\sigma_{T+1}(\widehat{\theta})\cdot\Phi^{-1}(\alpha). $$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ One may ask, why is CVaR important? It was little known a decade or two ago, but now it is core to Basel III regulation, so the banks must report it. $\endgroup$ – Richard Hardy Aug 18 at 15:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.