# Independence between innovations and variance in an ARCH model

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

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

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 $E(Z_n^2)=1$?

• No problem. Do you have answer regarding this one?
– Anna
Commented Feb 22, 2018 at 14:25
• @Anna Please check if the answer below is satisfying for you.
– Emil
Commented Feb 23, 2018 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.
• 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.