# Are the epsilons independent from each other?

1. Oct 22, 2008

### 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?

Last edited: Oct 22, 2008
2. Oct 22, 2008

### winterfors

Re: Independent?

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

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

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

3. Oct 23, 2008

### St41n

Re: Independent?

Ok, this makes sense
Thank you very much