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 $\alpha_0, \alpha_1, ...,\alpha_p$ 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, is there any result which shows that $Z_n$ is independent of $\epsilon_n$ or $\sigma_n$?


How can that be? $Z_n$ is not independent of $\epsilon_n$ since $Z_n=\frac{\epsilon_n}{\sqrt{\sigma_n}}$ and $Z_n$ is not independent of $\sigma_n$ since $Z_n=\frac{\epsilon_n}{\sqrt{\sigma_n}}$. $Z_n$ is a function of both $\epsilon_n$ and $\sigma_n$, so it cannot be independent of them. Change $\epsilon_n$ and keep $\sigma_n$ fixed, and $Z_n$ will change. Change $\sigma_n$ and keep $\epsilon_n$ fixed, and $Z_n$ will change.

Perhaps you are thinking in terms of what comes first and what generates what, but that is irrelevant when we talk about independence in the statistical sense of the word.

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  • $\begingroup$ So if we have $E(\sigma_n^2 \eta_n^2)$ this is not equivalent to $E(\sigma_n^2)$ $\endgroup$ – Anna Feb 21 '18 at 21:14
  • $\begingroup$ @Anna, What is $\eta_n^2$ $\endgroup$ – Richard Hardy Feb 22 '18 at 5:41
  • $\begingroup$ Sorry I meant $Z_n^2$, where $E(Z_n^2)$=1 $\endgroup$ – Anna Feb 22 '18 at 9:16
  • $\begingroup$ @Anna, By $\sigma^2$ do you mean the fourth moment? Because you used $\sigma$ for the second moment in your question. Also, this is actually a new question, so consider posting it as such. The answer to it is not implied by the original question or the original answer. I.e. independence or lack thereof is not informative about the equality vs. inequality there. $\endgroup$ – Richard Hardy Feb 22 '18 at 10:28
  • $\begingroup$ I edited the question, I think the same answer applies right? $\endgroup$ – Anna Feb 22 '18 at 10:48

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