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I am looking at the ARCH model where we have $\hat{\varepsilon}_t^2=\alpha_0 + \alpha_1\hat{\varepsilon}_{t-1}^2 + \alpha_2\hat{\varepsilon}_{t-2}^2 + \cdots + \alpha_q\hat{\varepsilon}_{t-q}^2 +v_t$

Any significant alpha 1 to q would mean that there is evidence of ARCH effects. So far so good.

What I am struggling with is the white noise process $v_t$

1) Why do we need it?

2) Will the mean always be zero? Isn't the white noise estimated along with the other parameters. How can we be sure that the mean is zero?

3) In my textbook this is a unit variance. What would be the implications of a variance 5 for example or a very low at 0.05? My suggestion> A higher variance would simply lead to the alpha coefficients would being lower.

The conclusions of the ARCH model really builds on the white noise process, that the mean is zero and the unconditional variance is one. So I guess that my main question really is. How can we add this to the process? Is it because that this has the properties that will make the ARCH framework fit with the stylized facts of returns?

Hope you can help me out.

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  • $\begingroup$ Note that the maximization of the arch likelihood is not trivial so you should look for a package in R or python that already does it for you. And, even then, you may run into problems. $\endgroup$ – mark leeds May 9 at 18:05
  • $\begingroup$ @Richard Hardy: I have no problem being incorrect ( happens quite often ) but could you let me know what's incorrect. thanks. $\endgroup$ – mark leeds May 15 at 2:21
  • $\begingroup$ The claim the $v_t$ term is assumed to be normal with mean $0$ and variance $\sigma^2$ is incorrect. An easy way to see it is noting that the left-hand side of the equation is nonnegative while $v_t$ can take negative values up to negative infinity. $\endgroup$ – Richard Hardy May 15 at 6:03
  • $\begingroup$ I don't see why the left hand side being non-negative precludes an error term that is normally distributed. On the other hand, I looked around and I can't find that particular ARCH formulation anywhere ( there is no error term in the ones I looked at ) so I think that I will take my comment out. Thanks. ams.sunysb.edu/~zhu/ams586/ARCH_GARCH.pdf $\endgroup$ – mark leeds May 15 at 13:33
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  1. Whether you use gross returns, $R_t := \frac{P_{t}}{P_{t-1}}$, or continuously compounded returns, $r_t := ln P_t - ln P_{t-1}$, for stock market prices, you will find that their squared and absolute values are extremely persistent at all data frequencies. Moreover, if for example, you computed the average of $r_t$ each day using high frequency data (such as, returns computed every 5 minute), you would get what we call "realized volatility." In a very general continuous time context, this is a valid estimate of the conditional entropy of your returns under the physical measure (see Martin's 2017 QJE paper for details). You may think of it as an estimator of conditional variance polluted by higher moments like (conditional) skewness and kurtosis. Setting this technical detail aside, if you took a look at those realized volatility series, you'd find out that they also exhibt a lot of temporal dependance. Finally, if you estimate GARCH models (a generalization of ARCH models), you would learn that the maximum likelihood estimator wants a very high degree of persistence in the conditional variance equation. Using daily frequency for the S&P500 since 1990, I estimated the Heston and Nandi (2000) GARCH model under the physical measure and got a persistence of about 0.98 (that's the first order autocorrelation of the filtered conditional variance process)... The bottom line: there is a huge amount of evidence comming from all angles that says that while returns themselves appear absurdly hard to predict over short time spans, the magnitude of their changes has an extremely persistent dynamic.

  2. Your ARCH model generally has this form: \begin{align} r_{t+1} &= \mu_{t+1} + \sqrt{h_{t+1}} z_{t+1}, \; z_t \sim N(0,1) \\ h_{t+1} &= \alpha_0 + \sum_{i=1}^q \alpha_i h_{t-i+1} z_{t-i+1}^2 \end{align} where $h_t$ is the conditional variance of the return process between time $t-1$ and $t$, $z_t$ is a white noise process, $(\alpha_i)_{i=0}^q$ are parameters and $\mu_t$ is some mean process. When you write the maximum likelihood, you can take advantage of the fact that $h_t$ is known at time $t-1$ to filter out the processes $(z_t, h_t)$ recursively from some starting values. For example, for the case of $q=1$, you can pick $h_1 := V(r_t)$ which can be approximated using the sample estimator of the variance of $r_t$. Given parameter values and $r_1$, this uniquely pins down the value of $z_1$. Then, using $h_1$ and $z_1$, you get $h_2$ and do the same thing to get $z_2$ and so on. Once you have obtained series for $(h_t,z_t)$, you can compute the likelihood associated with the parameter values you choose, exploiting the conditional normality of returns.

  3. If you change the variance of $z_t$ when you write the likelihood, you are effectively just scaling $h_t$. This should be obvious given the model I wrote above.

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  • $\begingroup$ The formula for $h_{t+1}$ is incorrect. The sum should contain squares of raw innovations, not levels of standardized innovations. $\endgroup$ – Richard Hardy May 14 at 17:21
  • $\begingroup$ @RichardHardy Thanks. $\endgroup$ – Stéphane May 15 at 0:48
  • $\begingroup$ Still incorrect, because you now have standardized innovations, but need raw ones. $\endgroup$ – Richard Hardy May 15 at 6:05
  • $\begingroup$ Better now. The more common way of representing the model is to have the raw errors $\varepsilon_t:=\sqrt{h_t}z_t$ explicitly in the first equation, then add the equation $\varepsilon_t:=\sqrt{h_t}z_t$ and then use $\varepsilon_t^2$ in the last equation. This makes it more transparent and easier on the eye. $\endgroup$ – Richard Hardy May 16 at 10:15

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