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An ARMA(p,q) model is given by

$ \qquad \qquad Y_t = c + \sum\limits_{i=1}^{p}\varphi_iY_{t-i}+\sum\limits_{i=1}^{q}\theta_i\varepsilon_{t-i} + \varepsilon$

with $\varepsilon_t \sim N(0,\sigma^2)$.

Let's say our model is simply an ARMA(1,1) model. The expected value for tomorrow's forecast then is

$\qquad \qquad E[Y_{t+1}] = E[c+\varphi_1Y_t+\theta_1\varepsilon_t + \varepsilon_{t+1}]$

$\qquad \qquad \qquad \qquad = c+\varphi_1Y_t+\theta_1\varepsilon_t + E[\varepsilon_{t+1}]$

$\qquad \qquad \qquad \qquad = c+\varphi_1Y_t+\theta_1\varepsilon_t$

with $Y_t$ being our time series. Correct so far?

Now, a GARCH(p,q) model is given by

$\qquad \qquad X_t = \sigma_t\varepsilon_t$

$\qquad \qquad \sigma_t^2=\omega + \sum\limits_{i=1}^{q}\alpha_iX_{t-i}^2+\sum\limits_{i=1}^{p}\beta_i\sigma_{t-i}^2$

with $\varepsilon_t \sim N(0,\sigma^2)$ again and with $X_t$ being our time series here right?

Now my questions arise when looking at an ARMA(n,m)-GHARCH(p,q) model:

$\qquad \qquad Y_t = c + \sum\limits_{i=1}^{n}\varphi_iY_{t-i}+\sum\limits_{j=1}^{m}\theta_jX_{t-j}+X_t$.

I understand that $X_t$ ist the underlying time series of the garch model, but isn't that simply the time series we are looking at and trying to just fit to a GARCH-Model?

Next, when looking at tomorrow's expected value from the ARMA(1,1)-GARCH(1,1) model, it is

$\qquad \qquad E[Y_{t+1}] = ... = c+\varphi_1Y_t+\theta_1X_t + E[X_{t+1}]$

$\qquad \qquad \qquad \qquad = c+\varphi_1Y_t+\theta_1X_t + E[\sigma_{t+1}\varepsilon_{t+1}]$

$\qquad \qquad \qquad \qquad = c+\varphi_1Y_t+\theta_1X_t + E[\sigma_{t+1}]E[\varepsilon_{t+1}]$

$\qquad \qquad \qquad \qquad = c+\varphi_1Y_t+\theta_1X_t + E[\sigma_{t+1}]*0$

$\qquad \qquad \qquad \qquad = c+\varphi_1Y_t+\theta_1X_t $

Aren't $X_t$ and $Y_t$ basically the same time series were we tried to apply these models to? And where is $\sigma^2$ now, I thought our goal was to use tomorrow's forecasted variance in our combined model.

Thanks in advance for reading and taking your time.

Billy Klein
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  • See ["What is the difference between GARCH and ARMA?"](https://stats.stackexchange.com/questions/41509). – Richard Hardy Jan 18 '21 at 12:19
  • The link has helped me a lot, thanks! So basically for my data above the X_t is the "time series" of the variance and the Y_t the time series of the returns of financial data? – Billy Klein Jan 18 '21 at 14:09
  • Looking at your equations above, $X_t$ is the random part of $Y_t$, and $X_t$ is modelled by GARCH as you have specified above. – Richard Hardy Jan 18 '21 at 18:41
  • Thanks a lot, I think I understood the difference between ARMA and GARCH now, I was a little irritated because of the switched parameters $Y_t$ and $\varepsilon_t$. I understand both on their own now I guess, but I still don't get what the combination of them does, because that $\varepsilon$ from the ARMA-GARCH Model is for tomorrows forecast then $E[\varepsilon_{t+1}]=0$ because that follows from $\varepsilon = \sigma_tZ_t$ and $E[Z_t]=0$. I don't get how that $\varepsilon$ helps me with tomorrows forecast. – Billy Klein Jan 19 '21 at 00:46
  • Pay attention to what information we are allowed to use at which point. $\mathbb{E}_t(\varepsilon_{t+1})=0$ but $\mathbb{E}_t(Z_t)=Z_t$ (a concrete realization of $Z_t$). Here I used subscript $_t$ for the expectation operator to denote it is conditional on information available at time $t$ (and similarly with $t+1$). – Richard Hardy Jan 19 '21 at 08:49
  • I understand that at timestamp $t$ we have real realisations for every coefficient. But for $t+1$ I thought it is $E[\varepsilon_{t+1}] = E[\sigma_{t+1}*Z_{t+1}]$ and it is $Z_t \sim N(0,\sigma^2)$, so our $E[Z_{t+1}]$ becomes $0$ here. And here I don't understand how this helps me for the ARMA-GARCH model. – Billy Klein Jan 19 '21 at 10:28
  • I am kind of irritated, because for example for an MA(q) set we can calculate (for a given data set with given observations) the error terms $\varepsilon_t$ recursively like [here](https://stats.stackexchange.com/questions/26024/moving-average-model-error-terms). But in our GARCH(p,q) model the residulas $\varepsilon_t$ seem to be calculated by $\varepsilon_t = y_t - \mu$ and with $y_t$ I mean the return in period $t$ and with $\mu$ the mean of all returns. – Billy Klein Jan 19 '21 at 12:36
  • Sorry, it is a bit hard to follow. Do you have any issues with ARMA alone or GARCH alone? Both are models for an observed time series $Y_t$. ARMA models the conditional mean as time varying in a particular way. GARCH models the conditional variance as time varying in a particular way. When the cond. mean of $Y_t$ follows ARMA, the cond. variance of $Y_t$ can be constant, follow GARCH or whatever. When the cond. variance of $Y_t$ follows GARCH, the cond. mean of $Y_t$ can be constant, follow ARMA or whatever. Estimation of each model is another story. – Richard Hardy Jan 19 '21 at 16:41
  • I think I understood it now after doing more research and reading some of your posts. I will talk about financial data in the following, so basically returns it it what I try to model. So, when you direcly apply a GARCH model, we assume that the mean is constant, and that is why we see people calculating the residuals by $\varepsilon_t=r_t-\mu$ and with $r_t$ I mean the return. But for our ARMA-GARCH model we apply the ARMA model first to get residuals with a constant mean which we then feed into our GARCH model, correct? – Billy Klein Jan 20 '21 at 10:51
  • After that, we have our $\sigma_t^2$ which is basically a "new series" with an almost constant variance. From there, we can calculate our new residuals by $\hat{\varepsilon_t} = \sqrt{\sigma_t^2}Z_t$ with $Z_t$ being white noise. And for our ARMA-GARCH model we apply these new $\hat{\varepsilon_t}$ then? – Billy Klein Jan 20 '21 at 11:03
  • This is not how I tend to put it, but I cannot see a serious error in your line of thought. Perhaps you can continue thinking as you do without any negative consequences. – Richard Hardy Jan 20 '21 at 11:53
  • Thanks a lot! You seem to be an expert in this field. Do you personally recommend any other specific models for modeling financial data and forecasting? – Billy Klein Jan 20 '21 at 12:19
  • ARMA(1,0)-GARCH(1,1) is a decent benchmark for many cases. You could experiment with other modifications of GARCH, there are a number of them. If you have data on realized variances available, [realized GARCH](http://econ.au.dk/fileadmin/site_files/filer_oekonomi/subsites/creates/Diverse_2016/PhD_High-Frequency/3RealizedGARCH.pdf) tends to be good for variance modelling. It depends on what exactly you will be using your model for. – Richard Hardy Jan 20 '21 at 12:34
  • I really appreciate your time and effort. I will work through the lecture you just posted! – Billy Klein Jan 20 '21 at 13:27
  • Thanks, I am glad if I can help. Just note that realized GARCH works only if you have additional data on (i.e. an additional time series of) realized variances. – Richard Hardy Jan 20 '21 at 15:50
  • Hey Richard, I still have a question about the AR-GARCH Model, I think there lies my mistake of thinking. We use AR / ARMA Models to model the conditional mean, and given $E[Y_{t+1}]$ we get the expected value for tomorrow's return. With a GARCH model alone we assume that the mean is constant, and here we model the conditional variance. We get a forecast of tomorrows variance. When using an AR-GARCH Model now, the expected Value of tomorrow's return doesn't change, but we have a new value for our variance. Is that correct? Because I always assumed we'd get a new value for the return aswell. – Billy Klein Jan 21 '21 at 13:31
  • Sorry, I am quite busy with teaching at my main job, so I am only commenting here (usually on the questions that I find interesting or easy) when I have time. Perhaps you should read the linked thread carefully one more time and ponder upon it. – Richard Hardy Jan 21 '21 at 13:34
  • I wrote the following above: *When the cond. variance of $Y_t$ follows GARCH, the cond. mean of $Y_t$ can be constant, follow ARMA or whatever.* GARCH just specifies how the cond. variance is modelled, it does not restrict the other aspects of the model such as the cond. mean equation. See the section **ARMA-GARCH** in [this](https://stats.stackexchange.com/questions/41509/what-is-the-difference-between-garch-and-arma/231512#231512) answer. – Richard Hardy Jan 21 '21 at 13:54
  • I think I understand this, but my question is about tomorrow's Expected Value. Let $Y_t=c+\phi Y_{t-1}+\varepsilon_t$ be the AR(1) model of a given times series, and $\hat{Y}_t=c+\phi\hat{Y}_{t-1}+\hat{\varepsilon}_t$ be the AR(1)-GARCH(1,1) model for the same time series, with $\hat{\varepsilon}_t=\sigma_tZ_t$ and $\sigma_t^2=\omega+\alpha\hat{\varepsilon}_{t-1}+\beta\sigma_{t-1}^2$. Is it correct, that $E[Y_{t+1}] = E[\hat{Y}_{t+1}]$? Only the variance $\sigma_{t+1}^2$ would be different here compared to a GARCH(1,1) model alone. – Billy Klein Jan 21 '21 at 14:16
  • Yes, that is correct. (You forgot to square epsilon in the cond. variance equation, but that is a minor thing.) – Richard Hardy Jan 21 '21 at 14:22
  • Thank you so much, now it all makes sense! Will continue my studies and hopefully learn more from you in the future :) – Billy Klein Jan 21 '21 at 14:46

1 Answers1

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Consider a dependent variable $Y_t$ which is an original time series or a transformation (such as log-returns) thereof.

  • ARMA models its conditional mean as time varying in a particular way.
    When the cond. mean of $Y_t$ follows ARMA, the cond. variance of $Y_t$ can be constant, follow GARCH or whatever.
    (See this answer for details.)
  • GARCH models its conditional variance as time varying in a particular way.
    When the cond. variance of $Y_t$ follows GARCH, the cond. mean of $Y_t$ can be constant, follow ARMA or whatever.
    (See this answer for details.)

Consider a basic example of ARMA(1,1)-GARCH(1,1): \begin{aligned} y_t &= c+\varphi_1 y_{t-1}+\varepsilon_t+\theta_1\varepsilon_{t-1}, \\ \varepsilon_t &= \sigma_t z_t, \\ \sigma^2_t &= \omega+\alpha_1\varepsilon_{t-1}^2+\beta_1\sigma_{t-1}^2, \\ z_t &\sim i.i.d(0,1). \end{aligned} Given the estimated model, a one-step-ahead estimate/forecast of the conditional mean and variance of $Y_t$ are the following: \begin{aligned} \mathbb{E_t}(y_{t+1}) &= c+\hat\varphi_1 y_{t}+\hat\theta_1\hat\varepsilon_{t}, \\ \text{Var}_t(y_{t+1}) &= c+\hat\alpha_1\hat\varepsilon_{t}^2+\hat\beta_1\hat\sigma_{t}^2. \end{aligned}

Richard Hardy
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