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  • $\begingroup$ When you estimated equation 4 with ML, I suppose you mean you assumed constant variance for the innovation. Yes? $\endgroup$ Commented Mar 25, 2015 at 11:28
  • $\begingroup$ Yes since I have no knowledge (or so I assume) whether or not there are any ARCH effect in the residuals when estimating (4). I maximize the following likelihood function: $l_{T}\left(\theta\right)=-\frac{T}{2}\log\left(2\pi\right)-\frac{1}{2}\sum_{t=1}^{T}\left(\log\left(\sigma^{2}\right)+\frac{\varepsilon_{t}^{2}}{\sigma^{2}}\right)$ $\endgroup$ Commented Mar 25, 2015 at 12:09
  • $\begingroup$ On another note. How do I get LaTex symbols in the comment box? $\endgroup$ Commented Mar 25, 2015 at 12:17
  • $\begingroup$ The likelihood function from the previous comment should read: $l_{T}\left(\theta\right)=-\frac{T}{2}\log\left(2\pi\right)-\frac{1}{2}\sum_{t=1}^{T}\left(\log\left(\sigma^{2}\right)+\frac{\varepsilon_{t}^{2}}{\sigma^{2}}\right)$ A bit further down I have written: "Another way to estimate the model is to estimate in one go. Doing that I get a model: ... ". There I estimate the full model in one go, i.e. Eq. (4) and (6) at the same time. As you can see the difference between estimates is not that big (I do have 2000 obs.!). I wanted to show that there are different ways to estimate the model. $\endgroup$ Commented Mar 25, 2015 at 16:00
  • $\begingroup$ ctd. and furthermore it is quite important to have a well specified model before testing for ARCH as explained in my answer. $\endgroup$ Commented Mar 25, 2015 at 16:01