Quantitative Finance Stack Exchange is a question and answer site for finance professionals and academics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given a time series of $u_i$ returns where i=1 to t. $\sigma_i$ is calculated from GARCH(1,1) as $\sigma_i^2=w+\alpha u_{i-1}^2 +\beta \sigma_{i-1}^2$ . What is the mathematical basis to say that $u_i^2/\sigma_i^2$ will exhibit little auto-correlation in the series? Hull's book Options, Futures and Other Derivative is an excellent reference. In Hull 6th Ed p470, "How Good is the Model?" he states that "If a GARCH model is working well, it should remove the auto-correlation. We can test whether it has done so by considering the auto-correlation structure for the variables $u_i^2/\sigma_i^2$. If these show very little auto-correlation out model for $\sigma_i$ has succeeded in explaining auto-correlation in the $u_i^2$". Max Log-Likelihood Estimation for variance ends with maximize $$ -m \space ln(v) -\sum_{i=1}^{t} u_i^2/v_i $$ where $v_i$ is variance= $\sigma_i^2$. This function does not really mean $u_i^2/v_i$ being minimized, because $-ln(v_i)$ gets larger and so does $u_i^2/v_i$ as $v_i$ gets smaller. However, it makes intuitive sense that dividing $u_t$ return by its volatility(instant or regime) volatility, explains away volatility related component of the time series. I am looking for a mathematical or logical explanation of this.

I think Hull is not very accurate here as the time series may have trends etc, also, there are better approaches to finding i.i.d from the times series than using $u_i^2/\sigma_i^2$ alone. I particularly like Filtering Historical Simulation- Backtest Analysis by Barone-Adesi(2000) FHS

share|improve this question
In a nutshell, you model the variance process of a time series $u_{i}$ with GARCH(1,1). Return time series have small absolute value, so $u_{i}^{2}$ is a good proxy for $\left(u_{i}-\overline{u}\right)^{2}$ as a variance estimator. Therefore $\frac{u_{i}^{2}}{\sigma_{i}^{2}}$ is a good proxy for a white noise time series if the GARCH(1,1) model was right and you explained all auto-correlation in the initial time series by it. – Marco Breitig May 27 '14 at 16:28

What is the mathematical basis to say that $u^{2}_{t}/\sigma_{t}^{2}$ will exhibit little auto-correlation in the series?

Let's $r_{t}$ be a series of returns and let's assume (Assumption I) it follows a covariance stationary process defined as :

$r_{t}=\sigma_{t} z_{t}$

where $z_{t}$ is i.i.d with $E_{t}(z_{t})=0$ and $Var_{t}(z_{t})=1$ ;

Then $ Var_{t}(r_{t}) =\sigma_{t}^{2}$

Next if we assume (Assumption II) that the conditional variance process of $r_{t}$ follows a GARCH(1,1), it means (Bollerslev (1986)) :

$\sigma_{t}^{2} = w + \alpha r_{t-1}^{2} + \beta \sigma_{t-1}^2 $

It can be shown that the precedent equation can be rewritten using an ARMA(1,1) representation:

$r_{t}^{2} = w + (\alpha +\beta) r_{t-1}^{2} + v_{t} - \beta v_{t-1} $

Where $ v_{t} = r_{t}^{2} - \sigma_{t}^{2}$

without going into the math you see that $r_{t}^{2} $ has some autocorrelations since it has an autoregressive structure.


$r_{t}^{2}=\sigma_{t}^{2} z_{t}^{2}$


$z_{t}^{2}=r_{t}^{2} /\sigma_{t}^{2}$

But we know that $z_{t}$ is IID(0,1),then its squares will also be IID.

So if assumptions I and II are respected the series $z_{t}^{2}=r_{t}^{2} /\sigma_{t}^{2}$ (the standardized residuals) will be IID and will exhibit no autocorrelations at all. If the DGP is not perfectly in line with our assumptions (but still close enough), the series will exhibit little autocorrelation since $z_{t}$ will be almost IID.

share|improve this answer

enter image description here In the paragraph before the one from which you gave the quotation is written such a thing: “…when $u_i^2$ is high, there is a tendency for $u_{i+1}^2$, $u_{i+2}^2$, … to be high; when $u_i^2$ is low, there is a tendency for $u_{i+1}^2$, $u_{i+2}^2$, … to be low.” This means that they $u_{i+1}^2,u_{i+1}^2, u_{i+2}^2,…$ are correlated. Which by itself means that $u_i^2$ exhibits autocorrelation. If we predicted well the $σ_i^2-s$ (and $u_i^2$ is an approximation of it), after dividing $σ_i^2$ on $u_i^2$ we should no more have the pattern described in quotation. Which means we are “removing autocorrelation.”

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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