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In a GARCH model like the following

$$y_t=\sigma_tz_t,\\ \sigma_t^2=\omega(1-\alpha-\beta)+\alpha y_{t-1}^2+\beta \sigma_{t-1}^2$$ 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?

  • So 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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2 Answers 2

up vote 3 down vote accepted

$$ E\left[ {{y_t}|{{\cal F}_{t - 1}}} \right] = E\left[ {{\sigma _t}{z_t}|{{\cal F}_{t - 1}}} \right] = {\sigma _t}E\left[ {{z_t}} \right] = 0 $$

$$ {\mathop{\rm var}} \left[ {{y_t}|{{\cal F}_{t - 1}}} \right] = {\mathop{\rm var}} \left[ {{\sigma _t}{z_t}|{{\cal F}_{t - 1}}} \right] = \sigma _t^2{\mathop{\rm var}} \left[ {{z_t}} \right] = \sigma _t^2 $$

$$ {y_t}|{{\cal F}_{t - 1}} \sim {\cal N}\left( {0,\sigma _t^2} \right) $$

So you are right in your conclusion. Unconditional variance is a constant $$ E\left[ {y_t^2} \right] = E\left[ {E\left[ {y_t^2|{{\cal F}_{t - 1}}} \right]} \right] = E\left[ {\sigma _t^2} \right] = {\sigma ^2} \\ {\sigma ^2} = \omega \left( {1 - \alpha - \beta } \right) + \alpha {y ^2} + \beta {\sigma ^2}\\ {\sigma ^2} = \frac{{\omega \left( {1 - \alpha - \beta } \right)}}{{1 - \left( {\alpha + \beta } \right)}} $$

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What is the unconditional variance of the above GARCH model? –  Sunv Jul 14 at 12:17

You can use the known result, that when $X\sim N(0,1)$, then $aX\sim N(0,a^2)$ where $a=\sigma_t$ is conditionally constant.

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