0
$\begingroup$

I'm interested in model a GARCH for a series. The original series is $y_t$ (price index of a stock market), which has a unit root. So I created the returns: $x_t = \ln(y_t) - \ln(y_{t-1})$.

Now, I'm confused about the fact of using $\lvert x_t\rvert$ for my GARCH. Why can I use absolute value, I'm thinking, that because I want to model volatility I'm just interested in how the series deviates from its mean in a period of time?

$\endgroup$
  • $\begingroup$ series in singular is still series. $\endgroup$ – user31264 May 19 '16 at 7:49
  • $\begingroup$ I was going through my old answers and noticed this one was not accepted. Do you perhaps need further clarification? $\endgroup$ – Richard Hardy Feb 15 '17 at 10:58
1
$\begingroup$

If $x_t$ is a series of logarithmic returns, the standard GARCH(1,1) model for it takes the form

$$ \begin{aligned} x_t &= \sigma_t \varepsilon_t, \\ \sigma_t^2 &= \omega + \alpha_1 x_{t-1}^2 + \beta_1 \sigma_{t-1}^2, \end{aligned} $$

where $\varepsilon_t$ is $i.i.d.(0,1)$ random variable. (It is straightforward to extend the model order to obtain GARCH($p$,$q$).) If you care about the volatility (measured by standard deviation or variance) of the returns, no absolute values get involved.

A GARCH model that involves absolute values is absolute value GARCH (AVGARCH or TS-GARCH) due to Taylor and Schwert (Bollerslev, 2009, p. 30) and is formulated as follows:

$$ \begin{aligned} x_t &= \sigma_t \varepsilon_t, \\ \sigma_t &= \omega + \alpha_1 |x_{t-1}| + \beta_1 \sigma_{t-1}. \end{aligned} $$

The model is less sensitive to large errors as compared to the regular GARCH model.

References

  • Bollerslev, Tim. "Glossary to ARCH (GARCH)." Volatility and Time Series Econometrics: Essays in Honour of Robert F. Engle. 2009.
$\endgroup$

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.