# Hull's GARCH vs. Definition in Time Series Literature

I have been reading up on volatility estimation and I encountered GARCH in Hull's "Options, Futures and Other Derivatives" (8e). He defines $u_n = \log{S_n/S_{n-1}}$ where $S_n$ is the price of some financial security, and then mentions how to calculate sample mean and sample variance, then assumes sample mean is zero so as to make calculations easier. I'm not sure he's still assuming zero mean when he mentions GARCH, but here's his rendition of GARCH(1,1):

$\sigma_n^2 = \omega + \alpha u_{n-1}^2+\beta \sigma_{n-1}^2$

All other sources say it's:

$\sigma_n^2 = \omega + \alpha a_{n-1}^2+\beta \sigma_{n-1}^2$,

where $a_{n-1}=\sigma_{n-1} \epsilon_{n-1}$ with $\epsilon_i$ zero mean iid, i.e.

$\sigma_n^2 = \omega + \alpha (\sigma_{n-1} \epsilon_{n-1})^2+\beta \sigma_{n-1}^2$

So my question is, how are these two formulations equal? By now I've read Tsay's "An Introduction to Analysis of Financial Data with R" and Ruppert's "Statistics and Data Analysis for Financial Engineering" and I still don't see how they could be similar.

Thanks in advance for hints, /not a statistician

As far as I know, the usual form is $$\operatorname{Definition:} u_n = \operatorname{log} \frac{S_n}{S_{n-1}}$$ $$\operatorname{Assumption:} \operatorname{E}(u_n|I_{n-1}) = 0$$ where $I_{n-1}$ is the information available at time $n-1$.

GARCH(1,1) model: $$u_n = \sigma_n \epsilon_n$$ $$\operatorname{Assumption:} \epsilon_n \sim \operatorname{i.i.d.} (0,1)$$ $$\sigma_n^2 = \omega + \alpha u_{n-1}^2+\beta \sigma_{n-1}^2$$

I am a bit surprised that "all other sources" have another definition.

In this definition, there is an implicit definition of $\epsilon_n$: $$\epsilon_n := \frac{u_n}{\sigma_n}.$$

If you like you may substitute $u_{n-1}$ with $\sigma_{n-1} \epsilon_{n-1}$ to get $$\sigma_n^2 = \omega + \alpha (\sigma_{n-1} \epsilon_{n-1})^2+\beta \sigma_{n-1}^2.$$

So the two expressions are compatible, but the first one seems to be more usual (at least in my experience).

• Thanks! I'm not sure I'm much wiser though. Sure, working from Bollerslev's original paper we can just put $\epsilon = u/ \sigma$, but that doesn't really tell me anything. Consider an AR(1) process, $u_n-\mu=\phi(u_{n-1}-\mu)+\epsilon_n$. This makes sense to me, we're simply saying that returns depend "somehow" on the previous (period's) return, and an error term. In the above, what are we saying? "Volatility depends on previous return and previous volatility, and return is the product of volatility and an error term". The latter... well, it needs some motivating, at least to me. Commented Jun 4, 2015 at 15:11
• In a GARCH(1,1) model above, returns are characterized as mean-unpredictable (conditional mean prediction is zero) with conditional heteroskedasticity defined roughly as "volatility depends on previous return and previous volatility" (as you put it). So the cond. mean is unpredictable but the cond. variance is. $\epsilon_n$ plays no role in the intuitive explanation, it is there just as a latent variable (if I may call it thus). Commented Jun 4, 2015 at 18:10
• I think I get what you're saying that $\epsilon$ is latent. Now, somehow the error term of the model comes out as $u_n/ \sigma_n$. Going back to Bollerslev, he defines $\epsilon_t \sim N(0,\sigma_t^2)$, and (for G(1,1)) $\sigma_t^2=\omega+\alpha \epsilon_{t-1}^2 + \beta \sigma_{t-1}^2$. From this formulation, how do we reach $u_n/ \sigma_n$? My guess is this: let $\epsilon_n=u_n$ and $u_n \sim N(0,\sigma_n^2)$. In general, if $X \sim N(0,\sigma^2)$ then $aX \sim N(0,a^2 \sigma^2)$. Thus we ought to have $u_n/ \sigma_n \sim N(0,1)$, and now we put $u_n/ \sigma_n=: \epsilon_n$. Is this "right"? Commented Jun 6, 2015 at 2:33
• I think what you wrote is about using alternative notation which is confusing when you refer to different sources at once. In Bollerslev's notation, $\epsilon$ is the raw residual of the conditional mean model; in the notation in your original question $\epsilon$ is the standardized residual that is latent in the GARCH model. Once this is clear, I do not see any more confusion. Commented Jun 6, 2015 at 7:08