Arch models: dependence and squared residuals I would like a mathematical and intuitive answer to those questions:


*

*Why a dynamics like arch in the volatility of a time series implies that the time series is dependent (although not autocorrelated) ?

*Why do we look at squared residual to infer if arch effects are present?
 A: *

*Well, it does not imply it is dependent, it justs models possible dependency and serial auto-correlation.

*Not sure what context are you talking about. If you just fit a regular model that doesn't take into account potential serial (time-series) auto-correlation, you can just look at the residuals to check if these are serially auto-correlated. If you don't find significant auto-correlation then you might argue that you don't need to consider the auto-correlation in the model.
A: As you are probably referring to the difference between lack of correlation and independence, we can generally say the following.
Correlation measures linear association between two given variables. Independence must be understood in broader sense, also non-linear (i.e. the definition of independence requires no relationship between any function of the two variables).. In the case of time series, if for example the Arma structure modeling the conditional mean is not significant, then there is not a linear relationship between a variable and its past values (no serial correlation), but there may exist a relationship between the squared variable and its squared past values (as we are supposing the Arma structure is 0, otherwise we would have said the “a relationship between the squared difference between the variable and its Arma conditional mean and its past values”) . If this happens, the Arch structure is significant. Which is why Arch refers to the modeling of the second moment of the conditional distribution. Which answers also your second question.
A: 1.
Intuitive answer:
an ARCH model (as its name suggests) implies variance-dependency, not mean-dependency, i.e. if variance at time 1 is high  chances are that variance at time 2 is high too. This does not say anything about the value of the process itself (in the mean), but only about how strong it will be 'jumping around' that mean. Autocorrelation means linear dependency in the mean, which is not assumed for ARCH.
Mathematical answer: (wlog) let $\varepsilon_t$ follow an ARCH(1) process, i.e. $\varepsilon_t = \sigma_t z_t \land \sigma_t^2=\alpha_0 + \alpha_1 \varepsilon_{t-1}^2 \land z_t\sim \mathcal{N}(0,1)$. It can be shown that $Cov(\varepsilon_t,\varepsilon_{t-j})=0\ \forall \ j \ne 0 $ by substituting the terms. Thus no autocorrelation. At the same time, $\varepsilon_t$ clearly depends on $\varepsilon_{t-1}$ through $\sigma_t$. Thus depedence.
2.
Intuitive answer: an ARCH model (as its name suggests) implies variance-dependency, not mean-dependency. The residuals $\varepsilon_t$ are assumed to have zero mean, hence if we look at these, we would check if the model exhibits autocorrelation - which it doesn't by assumption. Hence, $\varepsilon_t^2$ is a point-estimator for the error variance $\sigma_t^2$. If we now check for a relation between $\varepsilon_t^2$ we implicitly check for a relation between variances - which is what ARCH implies.
Mathematical answer: let $\varepsilon_t$ be the residuals following an ARCH(1) model. Then, since $\sigma_t$ and $z_t$ are independent, $E[\hat \varepsilon_t]=0$ and $Var[z_t]=1$, it holds that $Var[\varepsilon_t]=E[\varepsilon_t^2]=\sigma_t^2$. Hence $\varepsilon_t^2$ is an unbiased estimator for $\sigma_t^2$, i.e. $\varepsilon_t^2\approx\sigma_t^2$. If there is no relation between $\varepsilon_t^2$ then there is no relation between $\sigma_t^2$, hence, (in an ARCH(1) constellation) if $\sigma_t^2=\sigma^2$ (constant), then $\alpha_1=0, \alpha_0=\sigma^2$, and the ARCH(1) model simplifies to just homoskedasticity with $\varepsilon_t \sim \mathcal{N}(0,\sigma^2)$.
