14
$\begingroup$

I use a standard GARCH model: \begin{align} r_t&=\sigma_t\epsilon_t\\ \sigma^2_t&=\gamma_0 + \gamma_1 r_{t-1}^2 + \delta_1 \sigma^2_{t-1} \end{align}

I have different estimates of the coefficients and I need to interpret them. Therefore I am wondering about a nice interpretation, so what does $\gamma_0$,$\gamma_1$ and $\delta_1$ represent?

I see that $\gamma_0$ is something like a constant part. So it represents kind of an "ambient volatility". The $\gamma_1$ represents the adjustment to past shocks. Also, the $\delta_1$ is not very intuitively for me: It represents the adjustment to pas volatility. But I would like to have a better and more comprehensive interpretation of these parameters.

So can anyone give me a good explanation of what those parameters represent and how a change in the parameters could be explained (so what does it mean if e.g. the $\gamma_1$ increases?).

Also, I looked it up in several books (e.g. in Tsay), but I could not find good information, so any literature recommendation about the interpretation of these parameters would be appreciated.

Edit: I would be also interested in how to interpret the persistence. So what is exactly persistence?

In some books I read, that the persistence of a GARCH(1,1) is $\gamma_1+\delta_1$, but e.g. in the book by Carol Alexander on page 283 he talks about only the $\beta$ parameter (my $\delta_1$) being the persistence parameter. So is there a difference between persistence in volatility ($\sigma_t$) and persistence in shocks ($r_t$)?

vo

$\endgroup$
  • 1
    $\begingroup$ vol-of-vol would be 'volatility of volatility'; the volatility can jump around more. $\endgroup$ – Glen_b Jun 16 '13 at 2:04
  • $\begingroup$ shouldn't this be moved to quant finance beta? $\endgroup$ – Ivanov Jun 20 '13 at 15:56
  • 2
    $\begingroup$ StatTistician, why define $r_t$ at the start only to call the same quantity $a_t$ on the very next line? You don't need two symbols for the same thing. $\endgroup$ – Glen_b Jun 21 '13 at 3:42
  • 1
    $\begingroup$ I think mean equation should be $r_t$ = $\mu$ + $\sigma_t\epsilon_t$ $\endgroup$ – Metrics Jun 21 '13 at 3:51
  • $\begingroup$ I removed $a_t$ from the text, since it is superfluous and makes the GARCH(1,1) definition in the question be a non-standard one. $\endgroup$ – mpiktas Dec 4 '15 at 7:57
4
$\begingroup$

Campbell et al (1996) have following interpretation on p. 483.

$\gamma_1$ measures the extent to which a volatility shock today feeds through into next period’s volatility and $\gamma_1 + \delta_1$ measures the rate at which this effect dies over time.

According to Chan (2010) persistence of volatility occurs when $\gamma_1 + \delta_1 = 1$ ,and thus $a_t$ is non-stationary process. This is also called as IGARCH (Integrated GARCH). Under this scenario, unconditional variance become infinite (p. 110)

Note: GARCH(1,1) can be written in the form of ARMA (1,1) to show that the persistence is given by the sum of the parameters (proof in p. 110 of Chan (2010) and p. 483 in Campbell et al (1996). Also, $a^2_{t-1} - \sigma^2_{t-1}$ is now the volatility shock.

$\endgroup$
  • $\begingroup$ GARCH(1,1) can be written in the form of ARMA (1,1): more precisely, a GARCH(1,1) for $r_t$ can be written as ARMA(1,1) for $r_t^2$ (not for $r_t$). $\endgroup$ – Richard Hardy Jan 27 '17 at 11:48
0
$\begingroup$

the large values of the third coefficient ($\delta_{1}$) means that large changes in the volatility will affect future volatilizes for a long period of time since the decay is slower.

$\endgroup$
  • $\begingroup$ Sandile, I have taken the liberty of making your answer super explicit by including the term your reference. $\endgroup$ – Alexis Jun 10 '14 at 17:41
  • $\begingroup$ What do you think of the preceding answer, then? @Metrics explicitely gave an interpretation for $\gamma_1+\delta_1$, and not $\delta_1$ in isolation. $\endgroup$ – chl Jun 10 '14 at 17:54
0
$\begingroup$

Alpha catches the arch effect Beeta catches the garch effect Sum of both more close to 1, implies volatility remains long

$\endgroup$

protected by kjetil b halvorsen Apr 16 '18 at 19:16

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

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