Suppose I have a GARCH process:

$$X_t = \mu_t+\varepsilon_t$$

$$\varepsilon_t=\sigma_tz_t$$ where $z_t$ is some iid zero mean, unit variance random variable, and: $$\sigma_t^2=α_1 \varepsilon_{t−1}^2 + βσ_{t-1}^2$$

Why is it the case that the standardised residuals, $\frac{X_t-\mathbb{E}(X_t)}{\sqrt{\text{Var}(X_t)}}$ form a white noise process, and how could I analytically show this was the case?

  • 2
    $\begingroup$ What do you think about my answer? If it is helpful and clear, you may accept it by clicking on the tick mark to the left. Otherwise, you may ask for further clarification. This is how Cross Validated works. $\endgroup$ Apr 17, 2021 at 19:13

1 Answer 1


First, $$X_t-\mathbb{E}_{t-1}(X_t)=X_t-\mu_t=\varepsilon_t$$ where the first equality uses the definition of $\mu_t$, namely, $\mu_t=\mathbb{E}_{t-1}(X_t)$*.

Second, $$\text{Var}_{t-1}(X_t)=\text{Var}_{t-1}(\mu_t+\varepsilon_t)=\text{Var}_{t-1}(\varepsilon_t)$$ where the second equality uses the fact that $\mu_t$ is a constant given the information set $I_{t-1}$, and additive constants do not affect variance.

Thus $$\frac{X_t-\mathbb{E}_{t-1}(X_t)}{\sqrt{\text{Var}_{t-1}(X_t)}}=\frac{\varepsilon_t}{\sqrt{\text{Var}_{t-1}(\varepsilon_t)}}=\frac{\varepsilon_t}{\sigma_t}=z_t$$ where the second equality uses the definition of $\sigma_t^2$, namely, $\sigma_t^2=\text{Var}_{t-1}(\varepsilon_t)$, and the third equality uses the definition of $z_t$, namely, $z_t=\varepsilon_t/\sigma_t$.

Finally, $z_t$ is an i.i.d. sequence by assumption of the GARCH model, and an i.i.d. sequence is a white noise process.

  • 1
    $\begingroup$ Regarding the downvote: I would appreciate some constructive feedback so that I could improve the answer. $\endgroup$ Dec 7, 2022 at 10:14
  • $\begingroup$ I am not seeing anything that warrants a downvote. $\endgroup$ Dec 7, 2022 at 13:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.