# Why are the standardised residuals of a GARCH process a white noise process?

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?

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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.