# Time Series Analysis and Its Applications: With R Examples

## Second Edition

This is the site for the second edition of the text and is no longer maintained. Follow this link if you're looking for the site of the third edition
.

# 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 (and remedies) mentioned below are meant to help get you past the sticky points.

It's unlikely that the problems mentioned here will be fixed, so I wrote some short R scripts to help overcome these problems. You can find the code along with instructions and examples by going to EXAMPLES.

## 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, suppose x(t) = α + φ*x(t-1) + w(t) is stationary. Then taking expectations we have 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
mean(x)
 50.60668   # the sample mean is close
arima(x, order = c(1, 0, 0))
Coefficients:
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)
or
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":

```  Coefficients:
ar1       mean
0.8971    50.6304
s.e.  0.0409     0.8365
```

I should mention that I reported this flub to the R folks, but I was told that it is a matter of opinion. But, if you use ar.ols, you get anotheR opinion:
```ar.ols(x, order=1, demean=F, intercept=T)
Coefficients:
1
0.9052               <-- estimate of φ
Intercept: 4.806 (2.167) <-- yes, it IS the intercept as you know and love it
```
Note that arima() uses MLE, whereas ar.ols() uses OLS to fit the model, and hence the differences in the estimates. One thing is certain, the use of the term intercept in R is open to interpretation, which is not exactly an optimal situation.

## ISSUE 2: Why does arima fit different models for different orders? And what does xregdo?

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)
whereas in (2), R will fit the model
∇x(t) = α + φ*∇x(t-1) + w(t).

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 x(t)-β*t; that is, it will fit the model
∇[x(t) - β*t] = φ*∇[x(t-1) - β*(t-1)] + w(t).
Simplifying,
∇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)

Coefficients:
ar1
0.6031
s.e.  0.0793

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

Coefficients:
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+)

Coefficients:
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)
or
∇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: Why does predict.Arima give strange forecasts?

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 Chapter 3. In what you'll see below, the first method gives the wrong results and the second method gives the correct results. I think it's just best to stay away from the first method. If you use sarima and sarima.for, then you'll avoid these problems.

``` u=read.table("/mydata/globtemp2.dat")  # read the data
gtemp=ts(u[,2],start=1880,freq=1)      # yearly temp in col 2
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))
par(mfrow=c(2,1))
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: tsdiag.Arima gives 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. And YES, at least one R core developer knows this.

## ISSUE 5: When is lead, lag, and why is yesterday perfectly correlated with today?

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

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

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

Note that the default of the command lag(x) is lag(x,1). So, in R, if you have a series x(t), then
y(t) = lag{x(t)} = x(t+1), and NOT x(t-1).
This seems awkward, and it's not typical of other programs. But, that's the way Splus does it, so why not R? As long as you know the convention, you'll be ok ...

... well, then I started wondering how this plays out in other things. So, I started playing around with some 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

`ccf(soi,rec)`

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,1) because that would be progress :)

##-- WRONG:
cor(x,xb1)  # correlation
 1     ... is one?
lm(x~xb1)   # regression
Coefficients:
(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
Coefficients:
(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
Coefficients:
(Intercept)       y[, 2]
0.5022       0.7315
cor(y[,1],y[,2]) # correlation
 0.842086

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

## ISSUE 6: Why do you want to see the zero lag value of the ACF, and what's this mess?

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)
par(mfcol=c(2,1))
acf(MA1,20)
pacf(MA1,20)```
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:

```par(mfrow=c(2,1))
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. So I wrote an R function called acf2 that will do everything at once and save you some time and typing. You can get acf2 on the web page for the text under R CODE (Ch 1-5) - use the blue bar on top of this page to get there.