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In a GARCH model like the following $$ \begin{aligned} y_t &= \sigma_tz_t,\\ \sigma_t^2 &=\omega(1-\alpha-\beta)+\alpha y_{t-1}^2+\beta \sigma_{t-1}^2 \end{aligned} $$ where $z_t$ is assumed to be iidN(0,1), we say that conditional on past information $y_t$ has the Gaussian density $$f(y_t|y_{t-1},\sigma_{t-1}^2)=\frac{1}{\sqrt{2\pi\sigma^2_t}}\exp\left(\frac{1}{2\sigma^2_t}y_t^2\right)$$ Am I correct in making the following conclusion?

When conditioning on the past, we know what the value of $\sigma^2_t$ is.
Consequently, we can state that $y_t$ is conditionally normally distributed as $N(0,\omega(1-\alpha-\beta)+\alpha y_{t-1}^2+\beta \sigma_{t-1}^2)$.

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  • $\begingroup$ to make this more clear I recommend stating what is being conditioned in the "conditionally normally distributed" statement. For example, conditioning on $y_{t-1}, \sigma^2_{t-1}$ makes your conclusion correct. $\endgroup$
    – Lucas Roberts
    Commented Mar 24, 2018 at 14:13
  • $\begingroup$ @Sunv: Richard makes a good point and another thing is that, $\sigma^2$ is not observable, so you never know it. You're always estimating it. If this is obvious to you, then apologies for noise. $\endgroup$
    – mlofton
    Commented Mar 17, 2020 at 13:48

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In the density for $y_t$ you should substitute $\sigma_t$ with its expression above.
Once you have done that, it becomes more transparent that your conclusion holds.

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  • $\begingroup$ Thanks. I moved it. $\endgroup$
    – mlofton
    Commented Mar 17, 2020 at 13:49

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