Are the epsilons independent from each other?

[St41n]

\varepsilon _t = z_t \sigma _t

<br /> z_t \sim IID\,N\left( {0,1} \right)<br />

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

 

[winterfors]

\varepsilon _t = z_t \sigma _t

<br /> z_t \sim IID\,N\left( {0,1} \right)<br />

<br /> \sigma _t^2 = f\left( {\sigma _{t - 1} ,z_{t - 1} } \right)<br />


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

<br /> 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}<br />

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