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Does the autocorrelation function have any meaning with a non-stationary time series?

The time series is generally assumed to be stationary before autocorrelation is used for Box and Jenkins modeling purposes.

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if your series is non stationnary, the ACF will decline very slowly, to the point of being useless (it essentially a constant). What do you mean by 'have any meaning' ? –  user603 Sep 13 '10 at 17:02
    
If the time series is not stationary, often the 1st difference of the series will be stationary (for example, financial time series). –  John Salvatier Dec 3 '10 at 22:18

2 Answers 2

up vote 10 down vote accepted

@whuber gave a nice answer. I would just add, that you can simulate this very easily in R:

op <- par(mfrow = c(2,2), mar = .5 + c(0,0,0,0))

N <- 500
# Simulate a Gaussian noise process
y1 <- rnorm(N)
# Turn it into integrated noise (a random walk)
y2 <- cumsum(y1)

plot(ts(y1), xlab="", ylab="", main="", axes=F); box()
plot(ts(y2), xlab="", ylab="", main="", axes=F); box()
acf(y1, xlab="", ylab="", main="", axes=F); box()
acf(y2, xlab="", ylab="", main="", axes=F); box()

par(op)

Which ends up looking somewhat like this:

alt text

So you can easily see that the ACF function trails off slowly to zero in the case of a non-stationary series. The rate of decline is some measure of the trend, as @whuber mentioned, although this isn't the best tool to use for that kind of analysis.

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+1 (Sorry about the belated vote): nice illustrations. –  whuber Feb 24 '11 at 15:13

In its alternative form as a variogram, the rate at which the function grows with large lags is roughly the square of the average trend. This can sometimes be a useful way to decide whether you have adequately removed any trends.

(You can think of the variogram as the squared correlation multiplied by an appropriate variance and flipped upside down.)

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