Some R Time Series Issues

There are a few items related to the analysis of time series with R that will have you scratching your head. The issues mentioned below are meant to help get you past the sticky points. Most of the issues are related to the stats package, which is essentially a base package in that it is included with R, and loaded automatically when you start R. If you use other time series packages that have scripts with the same or similar names, then these issues might not apply. Then again, they just may apply because mistakes tend to propagate. The final issue (Issue 7) has to do with auto.arima from the forecast package... automated model fitting can lead to some strange models.

Issue 0: how will R end?

A recent issue with the package dplyr and lag from the stats package (essestially part of the base of R) came to my attention. You can read about it in how will R end... not with a bang but a package. The bottom line is, if you load dplyr, then you should know what it breaks. Stay tuned.

Issue 1: when is the intercept the mean?

When fitting ARIMA models, R calls the estimate of the mean, the estimate of the intercept. This is ok if there's no AR term, but not if there is an AR term.

For example, if x(t) = α + φ*x(t-1) + w(t) is stationary, then taking expectations, μ = α + φ*μ or α = μ*(1-φ). So, the intercept, α, is not the mean, μ, unless φ=0. In general, the mean and the intercept are the same only when there is no AR term.

Here's a numerical example:

# generate an AR(1) with mean 50
set.seed(66)      # so you can reproduce these results
x = arima.sim(list(order=c(1,0,0), ar=.9), n=100) + 50   
  [1] 50.60668   # the sample mean is close
arima(x, order = c(1, 0, 0))  
           ar1  intercept  <--  here's the problem
        0.8971    50.6304  <--  or here, one of these has to change
  s.e.  0.0409     0.8365

The result is telling you the estimated model is
  x(t) = 50.6304 + .8971*x(t-1) + w(t)
whereas, it should be telling you the estimated model is
  x(t)−50.6304 = .8971*[x(t-1)−50.6304] + w(t)
  x(t) = 5.21 + .8971*x(t-1) + w(t).   Note that 5.21 = 50.6304*(1-.8971).

The easy thing to do is simply change "intercept" to "mean":

           ar1       mean
        0.8971    50.6304
  s.e.  0.0409     0.8365

So WTF were they thinking? I don't know, but I'll guess... if you write an AR(1) model as
x(t) = μ + φ*[x(t-1) - μ] + w(t)
then in someone's twisted mind or altered state, μ might look like an intercept.

Issue 2: your arima is drifting

When fitting ARIMA models with R, a constant term is NOT included in the model if there is any differencing. The best R will do by default is fit a mean if there is no differencing [type ?arima for details]. What's wrong with this? Well (with a time series in x), for example:

arima(x, order = c(1, 1, 0))          # (1)

will not produce the same result as

arima(diff(x), order = c(1, 0, 0))    # (2)

because in (1), R will fit the model [with ∇x(s) = x(s)-x(s-1)]
  ∇x(t)= φ*∇x(t-1) + w(t)    (no constant)
whereas in (2), R will fit the model
  ∇x(t) = α + φ*∇x(t-1) + w(t).    (constant)

If there's drift (i.e., α is NOT zero), the two fits can be extremely different and using (1) will lead to an incorrect fit and consequently bad forecasts (see Issue 3 below).

If α is NOT zero, then what you have to do to correct (1) is use xreg as follows:

arima(x, order = c(1, 1, 0), xreg=1:length(x))    # (1+)

Why does this work? In symbols, xreg = t and consequently, R will replace x(t) with y(t) = x(t)-β*t; that is, it will fit the model
  ∇y(t)= φ*∇y(t-1) + w(t),
  ∇[x(t) - β*t] = φ*∇[x(t-1) - β*(t-1)] + w(t).
  ∇x(t) = α + φ*∇x(t-1) + w(t) where α = β*(1-φ).

If you want to see the differences, generate a random walk with drift and try to fit an ARIMA(1,1,0) model to it. Here's how:

set.seed(1)           # so you can reproduce the results
v = rnorm(100,1,1)    # v contains 100 iid N(1,1) variates
x = cumsum(v)         # x is a random walk with drift = 1 
plot.ts(x)            # pretty picture...

arima(x, order = c(1, 1, 0))         #(1)

  s.e.  0.0793

arima(diff(x), order = c(1, 0, 0))   #(2)

           ar1  intercept  <-- remember, this is the mean of diff(x)
       -0.0031     1.1163      and NOT the intercept
  s.e.  0.1002     0.0897

arima(x, order = c(1, 1, 0), xreg=1:length(x))    #(1+)

            ar1  1:length(x)  <-- this is the intercept of the model
        -0.0031      1.1169      for diff(x)... got a headache?                
  s.e.   0.1002      0.0897

Let me explain what's going on here. The model generating the data is
x(t) = 1 + x(t-1) + w(t)
where w(t) is N(0,1) noise. Another way to write this is
[x(t)-x(t-1)] = 1 + 0*[x(t-1)-x(t-2)] + w(t)
∇x(t) = 1 + 0*∇x(t-1) + w(t)
so, if you fit an AR(1) to ∇x(t), the estimates should be, approximately, ar1 = 0 and intercept = 1.

Note that (1) gives the WRONG answer because it's forcing the regression to go through the origin. But, (2) and (1+) give the correct answers expressed in two different ways.

Issue 3: you call that a forecast?

If you want to get predictions from an ARIMA(p,d,q) fit when there is differencing (i.e., d > 0), then the previous issue continues to be a problem. Here's an example using the global temperature data from Edition 2 of the text. In what you'll see below, the first method gives the wrong results and the second method gives the correct results. If you use sarima and sarima.for, then you'll avoid these problems.

fit1 = arima(gtemp, order=c(1,1,1))   
fore1 = predict(fit1, 15)   
nobs = length(gtemp)                
fit2 = arima(gtemp, order=c(1,1,1), xreg=1:nobs)  
fore2 = predict(fit2, 15, newxreg=(nobs+1):(nobs+15))
ts.plot(gtemp,fore1$pred, col=1:2, main="WRONG")
ts.plot(gtemp,fore2$pred, col=1:2, main="RIGHT")

Here's the graphic:


Issue 4: the wrong p-values

If you use tsdiag() for diagnostics after an ARIMA fit, you will get a graphic that looks like this:


The p-values shown for the Ljung-Box statistic plot are incorrect because the degrees of freedom used to calculate the p-values are lag instead of lag - (p+q). That is, the procedure being used does NOT take into account the fact that the residuals are from a fitted model.

Issue 5: lead from behind

You have to be careful when working with lagged components of a time series. Note that lag(x) is a FORWARD shift and lag(x,-1) is a BACKWARD shift. Try a small example:

x = ts(1:5)
cbind(x, lag(x), lag(x,-1))

Time Series:
Start = 0 
End = 6 
Frequency = 1 
   x lag(x) lag(x, -1)
0 NA     1         NA
1  1     2         NA
2  2     3          1
3  3     4          2 <-- here, x is 3, lag(x) is 4, lag(x,-1) is 2
4  4     5          3               
5  5    NA          4
6 NA    NA          5

In other words, if you have a series x(t), then
y(t) = lag{x(t)} = x(t+1), and NOT x(t-1).
In fact, this is reasonable in that y(t) actually does "lag" x(t) by one time period. But, it seems awkward, and it's not typical of other programs. As long as you know the convention, you'll be ok ...

... well, then I started wondering how this plays out in other things. If you do a lag plot

x = rnorm(500)

you get x(t) [vertical axis] vs x(t+1) [horizontal axis], but your brain is saying it's x(t-1) on the horizontal.


So, I started playing around with some other commands. In what you'll see next, I'm using two simultaneously measured series presented in the text called soi and rec... it doesn't matter what they are for this demonstration. First, I entered the command

acf(cbind(soi, rec))

and I got:


Before you scroll down, try to figure out what the graphs are giving you (in particular, the off-diagonal plots ... and yes they're CCFs, but what's the lead-lag relationship in each plot???) ...
... here you go:


The jpg is messy, but you'll get the point... the writing is mine. When you see something like 'rec "leads" here', it means rec comes in time before soi, and so on. Anyway, to be consistent, shouldn't the graph in the 2nd row, 1st column be corr{rec(t+Lag}, soi(t)} for positive values of Lag ... or ... shouldn't the title be soi & rec?? oops.

Now, try this


and you get


What you're seeing is corr{soi(t+Lag), rec(t)} versus Lag. So on the positive side of Lag, rec leads, and on the negative side of Lag, soi leads.

We're not done with this yet. If you want to do a regression of x on lag(x,-1), for example, you have to "tie" them together first. Here's an example.

x = arima.sim(list(order=c(1,0,0), ar=.9), n=20)  # 20 obs from an AR(1)
xb1 = lag(x,-1)
## you wouldn't regress x on lag(x) because that would be progress :) 
##-- WRONG:
cor(x,xb1)  # correlation
   [1] 1     ... is one?  
lm(x~xb1)   # regression   
  (Intercept)          xb1  
    6.112e-17    1.000e+00
do it the WRONG way and you'll think x(t)=x(t-1)
##-- RIGHT:
y=cbind(x,xb1)   # tie them together first
lm(y[,1]~y[,2])  # regression 
  (Intercept)       y[, 2]  
       0.5022       0.7315  

##-- OR:
y=ts.intersect(x,xb1)  # tie them together first this way
lm(y[,1]~y[,2])  # regression 
  (Intercept)       y[, 2]                   
       0.5022       0.7315              
cor(y[,1],y[,2]) # correlation
  [1] 0.842086

By the way, (Intercept) is used correctly here.

R does warn you about this (type ?lm and scroll down to "Using time series"), so consider this a heads-up, rather than an issue. See our little tutorial for more info on this.

Issue 6: u-g-l-y, you ain't got no alibi

When you're trying to fit an ARMA model to data, one of the first things you do is look at the ACF and PACF of the data. Let's try this for a simulated MA(1) process. Here's how:

MA1 = arima.sim(list(order=c(0,0,1), ma=.5), n=100)

and here's the output:


What's wrong with this picture? First, the two graphs are on different scales. The ACF axis goes from -.2 to 1, whereas the PACF axis goes from -.2 to .4. Also, the lag axis on the ACF plot starts at 0 (the 0 lag ACF is always 1 so you have to ignore it or put your thumb over it), whereas the lag axis on the PACF plot starts at 1.

So, instead of getting a nice picture by default, you get a messy picture. Ok, the remedy is as follows:

acf(MA1,20,xlim=c(1,20))   # set the x-axis limits to start at 1 then
                           # look at the graph and note the y-axis limits
pacf(MA1,20,ylim=c(-.2,1)) # then use those limits here

and here's the output:


Looks nice, but who wants to get carpal tunnel syndrome sooner than necessary? Not me... and hence acf2 was born.

Issue 7: don't use auto.arima

A student recently asked me if she could use auto.arima from the forecast package. My experience is that these automated procedures don't work very well (Shumway had a version in the original Windows version of astsa that I found tended to overfit models). I decided to try it on the AirPassengers data set from Box & Jenkins, who found that a seasonal ARIMA (0,1,1) × (0,1,1)12 was appropriate for the logged data. That example is in our text and we confirm B&Js model. Here is the run using auto.arima and assuming that you looked at the data first and decided you should log it first.

require(forecast)    # download it first if you don't have it
  Series: log(AirPassengers) 
              ma1     sar1     sar2     sma1     sma2
          -0.3632  -0.4852  -0.0933  -0.0204  -0.1803
    s.e.   0.0949   0.1561   0.1107   0.1700   0.1117

  sigma^2 estimated as 0.001258:  log likelihood=251.52
  AIC=-409.06   AICc=-408.39   BIC=-391.81
  Warning message:
  In auto.arima(log(AirPassengers)) :
   Unable to fit final model using maximum likelihood. AIC value approximated

Notice that the model is overly complex and 3 of the 5 regression parameters are not significant. Also you get a warning that it can't fit the model using MLE, but it reports a likelihood value(?). These facts should be a heads up that something is wrong, but you don't get any alternatives.