I have a data set that contains outliers (big orders) i need to forecast this series taking the outliers into consideration. I already know what the top 11 big orders are so i dont need to detect them first. I have tried a few ways to deal with this 1) forecast the data 10 times each time replacing the biggest outlier with the next biggest until the last set is run with them all replaced and then compare results 2) forecast the data another 10 times removing the outliers in each until they are all removed in the last set. Both of these work but they dont consistently give accurate forecasts. I was wondering if anyone knew another way to approach this?

One way i was thinking was running a weighted ARIMA and work it so that less/minimal weight is put on those specific data points. Is this possible?

I just want to point out that removing the known outliers does not delete that point completely, only minimizes it as there are other deals that happened in that quarter

One of my data sets is the following:

data <- matrix(c("08Q1",    "08Q2", "08Q3", "08Q4", "09Q1", "09Q2", "09Q3", "09Q4", "10Q1", "10Q2", "10Q3", "10Q4", "11Q1", "11Q2", "11Q3", "11Q4", "12Q1", "12Q2", "12Q3", "12Q4", "13Q1", "13Q2", "13Q3", "13Q4", "14Q1", "14Q2", "14Q3", "14Q4",155782698,   159463653.4,    172741125.6,    204547180,  126049319.8,    138648461.5,    135678842.1,    242568446.1,    177019289.3,    200397120.6,    182516217.1,    306143365.6,    222890269.2,    239062450.2,    229124263.2,    370575382.9,    257757410.5,    256125841.6,    231879306.6,    419580274,  268211059,  276378232.1,    261739468.7,    429127062.8,    254776725.6,    329429882.8,    264012891.6,    496745973.9),ncol=2,byrow=FALSE)

the known outliers in this series are:

outliers <- matrix(c("14Q4","14Q2","12Q1","13Q1","14Q2","11Q1","11Q4","14Q2","13Q4","14Q4","13Q1",20193525.68,18319234.7,12896323.62,12718744.01,12353002.09,11936190.13,11356476.28,11351192.31,10101527.85,9723641.25,9643214.018),ncol=2,byrow=FALSE)

please do not say about seasonality as this is only one type of data set, i have many ones without seaonality and i need the code to work for both types.

Edit by javlacalle: This is a plot of the observed data and the time points defined in the first column of outliers.

original data and outliers

  • $\begingroup$ What are the values in the second column of the matrix outliers? Why are the points "13Q1", "14Q2" and "14Q4" duplicated in the first column of this matrix? The series does not seem to have as many outliers as you detected. Maybe you are interested in forecasting a smooth pattern of the series? $\endgroup$
    – javlacalle
    Commented Apr 27, 2015 at 18:39
  • 2
    $\begingroup$ Plotting your data on logarithmic scale makes the supposed outliers even less convincing. For whatever reason, you have peaks every 4th quarter. So you have trend, seasonality, some irregularity. That's par for the course, and nothing pathological. Even @IrishStat's analysis, which often produces models more complicated than would suit many other researchers, comes close to saying that. $\endgroup$
    – Nick Cox
    Commented Apr 28, 2015 at 9:34
  • 2
    $\begingroup$ Further, your request not to mention seasonality misses the point that the "outliers" in this example are interpretable largely as a reflection of seasonality. That is, seasonality really is germane. $\endgroup$
    – Nick Cox
    Commented Apr 28, 2015 at 11:16
  • 1
    $\begingroup$ A simple solution, why don't you simply subtract those big orders from your data and run the forecast model? $\endgroup$
    – forecaster
    Commented Apr 28, 2015 at 16:50
  • 1
    $\begingroup$ I respect your practical constraints, but it is no longer clear what the question is. Your question as posted above concerns a specific dataset, but your comments downplay that dataset entirely and ask for a much more general strategy, roughly: I may have outliers; I may have seasonality; I want a general robust method for forecasting. The thread seems in permanent tension between your posing a very specific question and your seemingly wanting a much more general answer. $\endgroup$
    – Nick Cox
    Commented Apr 30, 2015 at 11:29

3 Answers 3


The OP insists in dealing with the points that are reported in the question as outliers without considering them as part of a possible seasonal pattern. Below I first give an idea to treat these points separately. In the second part of the answer I propose an alternative approach in the lines of the answer given by @Irishstat, which is a more appropriate analysis of the data.

The effect of these observations can be weighted by means of regression on seasonal dummies (variables that take the value 1 at the time points related to the outliers and 0 otherwise). Then, an ARIMA model for the residuals of the regression could be fitted and used to obtain forecasts.

It may be more efficient to estimate jointly the coefficients for the dummies and those of the ARIMA model, but I did not get a satisfactory result so I decided to split it in two steps as show below.

x <- ts(as.numeric(data[,2]), frequency = 4, start = c(2008, 1))
outliers <- c(2011.00, 2011.75, 2012.00, 2013.00, 2013.75, 2014.25, 2014.75)
# create dummies
dummies <- matrix(0, nrow = length(x), ncol = length(outliers)) 
for (i in seq_along(outliers))
  dummies[which(time(x) == outliers[i]),i] <- 1
# estimate the weights for these dummies and store the residuals
fitaux <- lm(x ~ dummies)
resid <- residuals(fitaux)
# fit an ARIMA model to the residuals and display forecasts
fit <- auto.arima(resid, ic = "bic")
fcast <- forecast(fit, 8)
# full code of the plot shown below is not posted to save space

forecasts of first approach

There is high uncertainty in the forecasts (wide lower and upper bounds). Although not shown, the residuals do not show autocorrelation but there is some sign of overdifferencing. The choice of the ARIMA model should be explored further, but I think this gives you the idea.

As mentioned in the comments above, I don't think the above approach is appropriate. I would do and analysis in the lines of the answer given by Irishstat. The R package tsoutliers follows the approach proposed in Chen and Liu (1993) to detect outliers in time series (e.g. additive outlies, level shifts). This is what I get:

fit2 <- tso(x, args.tsmethod=list(ic="bic"))
# ARIMA(0,0,0)(0,1,0)[4] with drift         
# Coefficients:
#         drift        LS4
#       8810020  -64443697
# s.e.  1289215   14293608
# sigma^2 estimated as 5.529e+14:  log likelihood=-366.02
# AIC=738.04   AICc=739.24   BIC=741.57
# Outliers:
#   type ind    time   coefhat  tstat
# 1   LS   4 2008:04 -64443697 -4.509
# type plot(fit2) to see the shape of the detected outlier(s)
# refit the model with the series adjusted for outliers
# (this will save arrangements to display forecasts
# the same model as in fit2$fit is chosen
fit2 <- auto.arima(fit2$yadj, ic="bic")
plot(forecast(fit2, 8))

forecasts based on second approach

The series is relatively clean from outliers. None of the outliers initially proposed in the question were detected. Similarly to the results shown by Irishstat, the forecasts look now more reliable, since they reflect the overall dynamics of the data.

  • $\begingroup$ +1 for great visualization and analysis. This puts everything is perspective on outliers. $\endgroup$
    – forecaster
    Commented Apr 28, 2015 at 18:11

If you start with a bad/insufficient model then one can incorrectly find many outliers. A good model will capture systematic patterns in the data and good diagnostic analysis can suggest periods in time where the model was inadequate suggesting required enhancements. Your 28 quarterly values graphed here enter image description here suggest the following model including two outlier adjustments enter image description here The two outliers (low values) can be seen here in the actual/cleansed plot enter image description here . The model generated the following residuals enter image description here which are free of any apparent auto-correlative structure enter image description here . The forecasts reflect the adjustment for the two anomalies enter image description here

  • $\begingroup$ Thankyou for your response. I am unsure of how to do this in R though. $\endgroup$ Commented Apr 27, 2015 at 7:59
  • $\begingroup$ @Summer-JadeGleek'away I suggest you look at IrishStat's user page and visit the website mentioned therein. He is probably using software called Autobox. That would be a place to start. Reproducing results in R might take a lot of study. $\endgroup$ Commented Apr 27, 2015 at 21:47
  • $\begingroup$ @Summer-JadeGleek'away See also here: stats.stackexchange.com/questions/32742/… $\endgroup$ Commented Apr 27, 2015 at 22:07
  • 1
    $\begingroup$ The downvote here seems unfair and I've reversed it. Not using R is not good grounds for a downvote and that is the only negative comment made. $\endgroup$
    – Nick Cox
    Commented Apr 28, 2015 at 11:18
  • $\begingroup$ @NickCox I did not downvote, but I would prefer to know the name of the procedure used and/or a citation for the procedure. $\endgroup$ Commented Apr 28, 2015 at 13:10

To me it's essentially the choice between pushing the uncertainty of "big" shocks to the forecasts or to remove them from the forecast. For instance, you could identify the outliers and put dummies in those dates. Then your dummies will catch the uncertainty, and since you don't forecast dummies, these uncertainties will be trapped in the dummies. You seem to expect that big sales may happen in the future, but in the approach with outliers your forecast will not reflect this possibility.

The second option is to leave the outliers in. In this case your error variance will increase, to the extent the big shocks are not correlated with your explanatory variables. Hence, when you forecast your confidence bands will widen reflecting the uncertainty you couldn't capture in your explanatory variables. After all you're not sure if the big shock would pop up at any time in future. the issue with this approach is that outliers often introduce bias in estimates, so it's not a clean packaging of uncertainty into the error term.

Whether you go with one option or another is a matter of preference and circumstances.


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