8
$\begingroup$

Given a time series $u_i$ of returns (where $i=1,\dotsc,t$), $\sigma_i$ is calculated from GARCH(1,1) as $$ \sigma_i^2=\omega+\alpha u_{i-1}^2 +\beta \sigma_{i-1}^2. $$ What is the mathematical basis to say that $u_i^2/\sigma_i^2$ will exhibit little autocorrelation in the series?

Hull's book "Options, Futures and Other Derivatives" is an excellent reference. In 6th ed. p. 470, "How Good is the Model?" he states that

If a GARCH model is working well, it should remove the autocorrelation. We can test whether it has done so by considering the autocorrelation structure for the variables $u_i^2/\sigma_i^2$. If these show very little autocorrelation our model for $\sigma_i$ has succeeded in explaining autocorrelation in the $u_i^2$.

Maximum likelihood estimation for variance ends with maximizing $$ -m \space \ln(v) -\sum_{i=1}^{t} u_i^2/v_i $$ 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 (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).

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Marco Breitig May 27 '14 at 16:28
3
$\begingroup$

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.

However

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

Then

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

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.

$\endgroup$
1
$\begingroup$

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.”

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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