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Are the epsilons independent from each other?

  1. Oct 22, 2008 #1
    [tex]\varepsilon _t = z_t \sigma _t [/tex]

    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. jcsd
  3. Oct 22, 2008 #2
    Re: Independent?

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

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

  4. Oct 23, 2008 #3
    Re: Independent?

    Ok, this makes sense
    Thank you very much
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