Conditional Normality and Unconditional Variance in a GARCH Model

Summary

The document asks how to interpret the conditioning in a GARCH process where returns are written as a time-varying volatility multiplied by an independent standard normal shock. Since the variance recursion uses past returns and volatility, the conditional variance at the current time is known given the past information set. The accepted response confirms that the current observation is conditionally normal with mean zero and variance equal to that measurable conditional variance.

The response also distinguishes conditional variance from unconditional variance. By taking expectations of the conditional second moment and imposing a stationary constant unconditional variance, it gives a formula in terms of the model parameters. A second response summarizes the result through the scaling property of a normal random variable when the scale is conditionally fixed. These conclusions rely on the stated Gaussian shock assumption and suitable parameter restrictions for a finite stationary variance. The document is a conceptual explanation, not a discussion of estimation, diagnostics, or alternative innovation distributions.

Key ideas

• In a GARCH model, the current volatility is determined by information available through the preceding period.
• Conditional on that past information, scaling a standard normal shock gives a normal observation with mean zero.
• The conditional variance changes over time according to the GARCH recursion.
• The unconditional variance is obtained by taking expectations of the conditional second moment under stationarity.
• The Gaussian conditional distribution depends on the assumed normal innovations, while finite unconditional variance requires appropriate parameter conditions.

Tags

# Understanding the conditioning in a GARCH process


# Understanding the conditioning in a GARCH process












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 $N(0,\omega(1-\alpha-\beta)+\alpha y_{t-1}^2+\beta \sigma_{t-1}^2)$.

## Answer by Kumar (score 4, accepted)

https://quant.stackexchange.com/a/12873

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

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

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

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

## Answer by emcor (score 1)

https://quant.stackexchange.com/a/14031

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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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.