# What is the difference between GARCH and ARMA

I am confused. I don't understand the difference a ARMA and a GARCH process.. to me there are the same no ?

Here is the (G)ARCH process

$\sigma_t^2 = \underbrace{ \underbrace{ \alpha_0 + \sum_{i=1}^q \alpha_ir_{t-i}^2} _{ARCH} + \sum_{i=1}^p\beta_i\sigma_{t-i}^2} _{GARCH}$

And here the ARMA

$X_t = c + \varepsilon_t + \sum_{i=1}^p \varphi_i X_{t-i} + \sum_{i=1}^q \theta_i \varepsilon_{t-i}.\,$

Is the ARMA simply an extension of the GARCH; GARCH being used only for returns and with the assumption $\sigma = r\epsilon$ where $\epsilon$ follows a strong white process ...

Thanks

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In addition to fg nu's answer, the variance process in GARCH is time-varying. However, there is a trick here is that given a time-series of log-return of SP500, then to obtain the volatility process what should we do? Some people say that we need using the ARMA model to withdraw the residual series, then plug this residual series into the GARCH model to obtain the conditional variance process? Or directly plug the log-return plug the log-return process of SP500 into the GARCH model to obtain the conditional variance? –  joey May 25 at 16:09

You are conflating the features of a process with its representation. Consider the (return) process $(Y_t)_{t=0}^\infty$.

• An ARMA(p,q) model specifies the conditional mean of the process as

\begin{align} \mathbb{E}(Y_t \mid \mathcal{I}_t) &= \alpha_0 + \sum_{j=1}^p \alpha_j Y_{t-j}+ \sum_{k=1}^q \beta_k\epsilon_{t-k}\\ \end{align} Here, $\mathcal{I}_t$ is the information set at time $t$, which is the $\sigma$-algebra generated by the lagged values of the outcome process $(Y_t)$.

• The GARCH(r,s) model specifies the conditional variance of the process \begin{alignat}{2} & \mathbb{V}(Y_t \mid \mathcal{I}_t) &{}={}& \mathbb{V}(\epsilon_t \mid \mathcal{I}_t) \\ \equiv \,& \sigma^2_t&{}={}& \delta_0 + \sum_{l=1}^r \delta_j \sigma^2_{t-l} + \sum_{m=1}^s \gamma_k Y^2_{t-m} \end{alignat}

Note in particular the first equivalence $\mathbb{V}(Y_t \mid \mathcal{I}_t)= \mathbb{V}(\epsilon_t \mid \mathcal{I}_t)$.

Aside: Based on this representation, you can write $$\epsilon_t \equiv \sigma_t Z_t$$ where $Z_t$ is a strong white noise process, but this follows from the way the process is defined.

• The two models (for the conditional mean and the variance) are perfectly compatible with each other, in that the mean of the process can be modeled as ARMA, and the variances as GARCH. This leads to the complete specification of an ARMA(p,q)-GARCH(r,s) model for the process as in the following representation \begin{align} Y_t &= \alpha_0 + \sum_{j=1}^p \alpha_j Y_{t-j} + \sum_{k=1}^q \beta_k\epsilon_{t-k} +\epsilon_t\\ \mathbb{E}(\epsilon_t\mid \mathcal{I}_t) &=0,\, \forall t \\ \mathbb{V}(\epsilon_t \mid \mathcal{I}_t) &= \delta_0 + \sum_{l=1}^r \delta_l \sigma^2_{t-l} + \sum_{m=1}^s \gamma_m Y^2_{t-m}\, \forall t \end{align}
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Thanks very much fg nu. It is exactly the kind of answer I needed. Now things are getting much clearer :) –  Johan Oct 30 '12 at 16:30
Shouldn't you be conditioning on the information at time $t-1$ if all of the regressors are lagged? –  Jase Dec 10 '12 at 6:32
@Jase Note the definition, "Here, $\mathcal{I}_t$ is the information set at time $t$, which is the $\sigma$-algebra generated by the lagged values of the outcome process $(Y_t)$." That is, $\mathcal{I}_t = \sigma(Y_{t-1}, Y_{t-2}\ldots,)$. Some authors write this as $\mathcal{I}_{t-1}$ but that is counter to the notion of an information set at time $t$. –  fg nu Dec 10 '12 at 7:04
Nice! Do you know why we use the sigma-algebra and not a filtration? –  Jase Dec 10 '12 at 7:17
@Jase, the sequence of information sets $(\mathcal{I}_t)_{t=0}^\infty$ constitutes a filtration. –  fg nu Dec 10 '12 at 7:20