Let $\epsilon_n$ denote a real-valued discrete-time stochastic process of residuals, the ARCH($p$) specification is given by

\begin{equation} \label{1.1} \epsilon_n=Z_n\sqrt{\sigma_n} \end{equation} \begin{equation}\label{1.2} \sigma_n=\alpha_0+\sum \limits_{i=1}^p\alpha_i\epsilon_{n-i}^2\,, \end{equation}

where $\alpha_0, \alpha_1, \dots,\alpha_p$ are scalar parameters to be estimated, $\mu_n$ is the fitted model. $Z_n$, are a sequence of independent, identically distributed random variables with mean zero and variance one.

Now, can we say that $E(Z_n^2\sigma_n)=E(\sigma_n)$, since $E(Z_n^2)=1$?

  • $\begingroup$ No problem. Do you have answer regarding this one? $\endgroup$ – Anna Feb 22 '18 at 14:25
  • $\begingroup$ @Anna Please check if the answer below is satisfying for you. $\endgroup$ – Emil Feb 23 '18 at 9:09

Yes, you can say that, but you perhaps need to be a bit clear as to why that result holds. In other words, $\mathbb{E}[Z_n^2 \sigma_n] = 1\cdot\mathbb{E}[\sigma_n] $ because $\sigma_n$ is actually a function of the previous values of the process, i.e. $\sigma_n = \sigma_n(\epsilon_{n-1},\ldots,\epsilon_{n-p}) $, and $Z_n$ is by definition assumed to be independent of the previous values of the process, i.e. independent of $\epsilon_{n-1},\epsilon_{n-2},$ etc.

  • $\begingroup$ The first equality (the "middle step") does not mean independence, it only means zero covariance. $\endgroup$ – Richard Hardy Feb 22 '18 at 15:23
  • $\begingroup$ You are absolutely right. My transgression was due to the fact that in Econometrics courses (where a question such as this one would usually pop up), there is a common, even if simply practical, motive where no covariance and independence are conflated. However, the set of dependent variables with zero covariance is, of course, non-empty :) $\endgroup$ – Emil Feb 22 '18 at 15:32
  • $\begingroup$ Great. I am a little confused by your formulation, though. I guess you wanted to say that even though the first equation happens to hold it cannot be used in a proof, and so you propose an alternative proof. Do I get you right? $\endgroup$ – Richard Hardy Feb 22 '18 at 16:07
  • $\begingroup$ And you were completely correct in being confused, because my claim was in itself wrong (I have thus completely changed the answer). $Z_n$ is indeed independent from $\sigma_n$, and by design at that, since it's assumed in the model specifications that $Z_n$ is independent to the $\sigma$-algebra $\mathcal{F}_{n-1}$ generated by the process up to time $n-1$. Thank you for noticing my mistake. $\endgroup$ – Emil Feb 22 '18 at 19:28
  • 1
    $\begingroup$ Finally. +1 now :) $\endgroup$ – Richard Hardy Feb 22 '18 at 20:02

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