1
$\begingroup$

I have time series data that reflects users activity on some platform, and it has clear daily seasonal pattern. I understand that in order to fit a model such as ARMA to this data I should first detrend it and remove its seasonal component, which is commonly performed using differencing. This should leave me with a time series which is stationary and I can then use models like ARMA. Yet there is an issue I don't understand - as my data describes users activity, while the mean has a clear periodical pattern, the variance is much higher in daytime comparing to late night hours. This means that after differencing the series would still remain non-stationary.

Should the differencing eliminate the changing variance? I don't see why. If not, are there other methods to deal with data with such behavior?

$\endgroup$
2
  • $\begingroup$ Have you looked at ARCH/GARCH models? $\endgroup$ Jun 19, 2017 at 13:57
  • $\begingroup$ I wasn't aware of ARCH/GARCH models, thank you $\endgroup$
    – keren42
    Jun 19, 2017 at 14:37

1 Answer 1

1
$\begingroup$

I have time series data that reflects users activity on some platform, and it has clear daily seasonal pattern. I understand that in order to fit a model such as ARMA to this data I should first detrend it and remove its seasonal component, which is commonly performed using differencing.

Differencing is appropriate when the data has a stochastic trend (is integrated, has a unit root). It is not appropriate when the data is merely seasonal or has a deterministic trend (e.g. linear trend). By differencing in absence of a stochastic trend you will introduce a superfluous integrated MA(1) component.

Should the differencing eliminate the changing variance? I don't see why. If not, are there other methods to deal with data with such behavior?

No, differencing will not turn a time-varying variance to constant variance. But you could specify a model for the time-varying variance extra to the model for the time-varying mean (see my longer answer is this thread). An ARMA(p,q)-GARCH(s,r) model with exogenous regressors in the conditional variance equation (extra to those in the conditional mean equation) is such an example. It would look something like \begin{aligned} x_t &\sim D(\mu_t,\sigma_t^2); \\ \mu_t &= \varphi_1 \mu_{t-1} + \dotsc + \varphi_p \mu_{t-p} + (\varphi_1 + \theta_1) \varepsilon_{t-1} + \dotsc + (\varphi_m + \theta_m) \varepsilon_{t-m} \\ &+ \text{seasonal dummies or Fourier terms}; \\ \sigma_t^2 &= \omega + \alpha_1 \varepsilon_{t-1}^2 + \dotsc + \alpha_s \varepsilon_{t-s}^2 + \beta_1 \sigma_{t-1}^2 + \dotsc + \beta_r \sigma_{t-r}^2 \\ &+ \text{seasonal dummies or Fourier terms}. \\ \end{aligned} It might be that you do not need the regular GARCH terms (lagged $\varepsilon_t^2$ and lagged $\sigma_t^2$), then the conditional variance equation would collapse to $$ \sigma_t^2 = \omega + \text{seasonal dummies or Fourier terms}. $$ I do not know how to implement this directly, but there is a workaround: specify an ARMA(p,q)-GARCH(1,1) model with exogenous regressors $$ \sigma_t^2 = \omega + \alpha_1 \varepsilon_{t-1}^2 + \beta_1 \sigma_{t-1}^2 + \text{seasonal dummies or Fourier terms} $$ while fixing $\alpha_1=0$ and $\beta_1=1$. You can do this, for example, in "rugarch" package in R with functions ugarchspec and ugarchfit:

library(forecast)
library(rugarch)
p=2; q=2 # arbitrary choice, just for this example
x=rnorm(1000); x=ts(x,freq=12,start=c(1960,1)) # generated data just for this example
fourierterms=fourier(x,K=6)
spec=ugarchspec(variance.model=list(external.regressors=cbind(fourierterms)), mean.model=list(armaOrder=c(p,q), external.regressors=cbind(fourierterms)), 
fixed.pars=list(alpha1=0.0,beta1=1.0))
fit=ugarchfit(spec=spec,data=x)
$\endgroup$
2
  • $\begingroup$ Thank you for your answer. Are you familiar with equivalent package in python for GARCH models? $\endgroup$
    – keren42
    Jun 20, 2017 at 9:29
  • $\begingroup$ @keren42, no, unfortunately I only work with R. But Python is used extensively in finance, so hopefully the relevant functions exists there, too. $\endgroup$ Jun 20, 2017 at 10:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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