Are the epsilons independent from each other?

[St41n]

$$\varepsilon _t = z_t \sigma _t$$

$$z_t \sim IID\,N\left( {0,1} \right)$$

$$\sigma _t^2 = f\left( {\sigma _{t - 1} ,z_{t - 1} } \right)$$Are the epsilons independent from each other? Why?

 

[winterfors]

$$\varepsilon _t = z_t \sigma _t$$

$$z_t \sim IID\,N\left( {0,1} \right)$$

$$\sigma _t^2 = f\left( {\sigma _{t - 1} ,z_{t - 1} } \right)$$


Are the epsilons independent from each other? Why?

I can't see that they could be independent in the general case of $f (\sigma ,z)$ being any real-valued function, since $p(\varepsilon_{t+1} | \sigma_t ,z_t)$ is dependent on $\sigma_t$ and $z_t$, just like $p(\varepsilon_{t} | \sigma_t ,z_t) = \delta(\varepsilon_{t} - \sigma_t z_t)$ ($p$ representing probability densities and $\delta$ the Dirac delta distribution). This implies that

$$p(\varepsilon _t,\varepsilon _{t+1}) = \int\int{<br /> p(\varepsilon_{t} | \sigma_t ,z_t) <br /> p(\varepsilon_{t+1} | \sigma_t ,z_t)<br /> p(\sigma_t)p(z_t)<br /> d\sigma_t dz_t}$$

cannot, in general, be written as a product $p(\varepsilon _t,\varepsilon _{t+1}) = p(\varepsilon_{t})p(\varepsilon_{t+1})$ as is the case for independent variables.

-Emanuel

 

[St41n]

Ok, this makes sense
Thank you very much