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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.

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)[1] 50.60668 # the sample mean is closearima(x, order = c(1, 0, 0))Coefficients: ar1 intercept <--here's the problem0.8971 50.6304 <--or here, one of these has to changes.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: ar1mean0.8971 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(x, order=1, demean=F, intercept=T)Note thatCoefficients: 1 0.9052 <--estimate of φIntercept: 4.806 (2.167) <--yes, it IS the intercept as you know and love it

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. 0.0793arima(diff(x), order = c(1, 0, 0)) #(2)Coefficients: ar1 interceptarima(x, order = c(1, 1, 0), xreg=1:length(x)) #(1+)<-- remember, this is the mean of diff(x)-0.0031 1.1163and NOT the intercepts.e. 0.1002 0.0897Coefficients: ar1 1:length(x)<-- this is the intercept of the model-0.0031 1.1169for 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.

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 2fit1=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:

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

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 33 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'tregressx on lag(x,1) because that would beprogress:)##-- WRONG:cor(x,xb1) # correlation[1] 1 ... is one?lm(x~xb1) # regressionCoefficients: (Intercept) xb1 6.112e-17 1.000e+00do 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]) # regressionCoefficients: (Intercept) y[, 2] 0.5022 0.7315##-- OR: y=ts.intersect(x,xb1) # tie them together first this way lm(y[,1]~y[,2]) # regressionCoefficients: (Intercept) y[, 2] 0.5022 0.7315cor(y[,1],y[,2]) # correlation[1] 0.842086By 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.

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 hereand 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.