Hi Quantitative Finance Stack Exchange,

It's my first go at GARCH models so give me a chance with my phrasing. I'm looking for an answer to a general question.

First, I understand that you can have a forecasting model to forecast returns and a GARCH model to forecast volatility. Let's proceed with the simplest example:

Forecasting returns:

$$\hat{y_t}=\alpha\cdot y_{t-1} + \epsilon_t$$



Now, I've developed my trading strategy and let's say I found that it works, namely buy when $\hat{y_t} > 0.0020\%$. My question is this. What is the standard way of looking at how GARCH compliments my strategy, if at all?

The way I see it is that both predicts different things. One predicts $\hat{y_t}$ and another predicts $\hat{\sigma^2_{t}}$. Therefore, GARCH is only readily implementable if you somehow found a way to incorporate volatility in your strategy. If my existing strategy $\hat{y_t} > 0.0020\%$ works fine, there isn't a need for GARCH correct?

Thank you for your help, Donny

  • 1
    $\begingroup$ As explained in MathsQuant525's answer, complementing your raw AR(1) model with a GARCH layer allows you to enrich your forecasts: you move from a simple conditional mean model, to a conditional mean + variance model. Because you are using a simple trading rule (or signal generation step) "predicted return > threshold" this might not be clear. But you could use something more "complicated" like mean-variance optimisation à la Markowitz. $\endgroup$
    – Quantuple
    Nov 14, 2016 at 10:20
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    $\begingroup$ For a very simple example: having a high expected return makes it attractive to buy the asset, but a high expected variance might tell you to take a smaller position than usual. This is an example of the kind of more complicated strategy that Quantuple may be referring to. $\endgroup$
    – nbbo2
    Nov 14, 2016 at 13:36
  • $\begingroup$ Have you considered accepting the answer? See how this is supposed to work. $\endgroup$ Dec 13, 2020 at 18:15

3 Answers 3


Your random innovations in the returns model depend on the volatility model. In this setting, we have $\epsilon_t ~ N(0,\hat{\sigma_t}^2)$. The effect of this at a very layman level is that when the volatility is higher, the random innovations are more likely to take larger values, which increases the probability of the returns taking larger values. This is exactly what we wants, as it should increase the jumps between consecutive returns, hence making the volatility of the predicted returns series higher.

I think you’re forgetting that there is a dependence on $\sigma_t$ hidden inside the random innovation $\epsilon_t$.


One aspect of modelling that has been overlooked in the comments and answers so far is that including a time-varying conditional variance in your model will not only (1) give you time-varying conditional variances but also (2) affect the estimates of the conditional model.

For example, if a certain ARMA-GARCH model approximates the data better than a pure ARMA model with constant conditional variance, then it makes sense to model the data as ARMA-GARCH not only (1) to have better forecasts of volatility but also (2) because neglecting the GARCH part will negatively affect the estimates of the ARMA parameters, making them inefficient and likely even inconsistent.


Your mean forecast ($y$) already incorporates the GARCH component, (ie $\sigma$).

However, by being focus on the mean you loose some informations because a forecast mean of 0.02% with a volatility of 0.004% is not similar, in term of risk, of the same forecast mean (0.02%) with a lower volatility forecast (ex :0.002%) - I assume Gaussian errors.

The central tendency (the mean forecast) does not appropriately summarizes your forecast, you should also consider the density forecast. As an example, you can use an interval of confidence (ex: $y> \mu + 1.96 \sigma$) instead of a fix threshold (ex: $y> \mu$) to better capture your risk. $\mu$ is your mean threshold level (0.002% in your example) and $\sigma$ the forecasted volatility. 1.96 is a parameter that depend of your risk aversion and of the error term distribution.

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    $\begingroup$ You do not need to assume normality where you do, the claim holds regardless. $\endgroup$ Nov 15, 2016 at 7:11
  • $\begingroup$ I'm not so sure because for a more complex error distribution , others higher moments parameters play also a role. $\endgroup$
    – Malick
    Nov 15, 2016 at 11:48
  • $\begingroup$ Assuming the same error distribution (except for the variance) across the alternative cases should be enough. If the distributions are all Normal, all Student's-$t$ or the like, your claim holds. $\endgroup$ Nov 15, 2016 at 12:48
  • $\begingroup$ Yes and that's why I have said that I assume Gaussian errors, my claim doesn't holds regardless (cf your first comment) $\endgroup$
    – Malick
    Nov 15, 2016 at 14:32
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    $\begingroup$ OK, I understand. Your assumption is too strict, but indeed you cannot seamlesly switch between distributions in your comparison. However, normally the distribution does not change within the model (except for its location and scale as modelled by, say, ARMA-GARCH), so that you are always comparing apples to apples in practice as long as you stick to the same model that produces the forecasts. $\endgroup$ Nov 15, 2016 at 15:01

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