My background is signal processing and I am fairly new to (financial) time series analysis. I was reading the article about autoregressive conditional heteroskedasticity (ARCH) models on Wikipedia.


I am confused about what AR (I know it means autoregressive :-) ) refers to in an ARCH model. There are two possibilities in my view:

1.) The process of interest (and not its volatility), e.g., log-returns, is assumed to follow an AR-model: $$ y_t = a_0 + \sum_{i=1}^p y_{t-i} a_i $$

2.) We assume that the volatility of the process we want to model (e.g., the above mentioned log-returns) follows an AR process.

The introduction section in Wikipedia seems to support my first hypothesis: "The ARCH model is appropriate when the error variance in a time series follows an autoregressive (AR) model..."

What confuses me though is this formula:

$$\sigma_t^2 = \alpha_0 + \sum_{i=1}^q \alpha_i \epsilon_{t-i}^2$$

I don't see the "autoregression" here... Wouldn't and AR be something like:

$$\sigma_t^2 = \alpha_0 + \sum_{i=1}^q \alpha_i \sigma_{t-i}^2$$.

This would then be a special case what is referred to as GARCH model.

  • 1
    $\begingroup$ The line just above your second to last quote from the Wiki article says $\epsilon_t=\sigma_t z_t$ $\endgroup$ Commented May 6, 2018 at 11:20
  • $\begingroup$ Yes. I'm not sure though what you want to say by that. Do you mean that this implies that the second moments are the same? $$E[\epsilon_t^2] = E[\sigma_t^2 z_t^2] = E[\sigma_t^2].$$ Is the ARCH model not a generative model? $\endgroup$
    – Effesian
    Commented May 6, 2018 at 16:24

1 Answer 1


In the following, what you call sigma is called h. So because you can write e(t) as a function of h(t) and a random variable z(t), when you write h(t) as a function of the past of e(t) you are equivalently writing out an autoregressive process for the error.

enter image description here

  • 1
    $\begingroup$ Wouldn't you need to regress on $h_{t-1}, h_{t-2}, ...$ to have the variance be an AR process? $\endgroup$
    – Effesian
    Commented May 7, 2018 at 21:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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