# How to make time series data with changing variance stationary?

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?

• Have you looked at ARCH/GARCH models? – Michael Chernick Jun 19 '17 at 13:57
• I wasn't aware of ARCH/GARCH models, thank you – keren42 Jun 19 '17 at 14:37

## 1 Answer

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)

• Thank you for your answer. Are you familiar with equivalent package in python for GARCH models? – keren42 Jun 20 '17 at 9:29
• @keren42, no, unfortunately I only work with R. But Python is used extensively in finance, so hopefully the relevant functions exists there, too. – Richard Hardy Jun 20 '17 at 10:05