[R] How does ccf() really work?
Spencer Graves
spencer.graves at pdf.com
Fri Apr 21 17:35:50 CEST 2006
The standard estimate of cross correlation uses the same denominator
for all lags AND ignores the reduction in the number of observations.
Consider the following:
a. <- a-mean(a)
b. <- b-mean(b)
SSa <- sum(a.^2)
SSb <- sum(b.^2)
SaSb <- sqrt(SSa*SSb)
sum(a.*b.)/SaSb
# 0.618 = cor(a, b)
sum(a.[-1]*b.[-5])/SaSb
# 0.568 = cc lag 1
sum(a.[1]*b.[5])/SaSb
# -0.065
sum(a.[1:2]*b.[4:5])/SaSb
# -0.289
These numbers match the results you reported below. If I'm not
mistaken, this also matches the definition of the cross correlation
function in the original Box and Jenkins book [or the more recent Box,
Jenkins, Reinsel], "Time Series Analysis, Forecasting and Control". The
rationale, as I recall, is to reduce the false alarm rate by biasing
estimates with larger lags toward zero, thereby compensating slightly
for their increased random variability.
hope this helps.
spencer graves
p.s. Thanks for including such a simple, self-contained example. Posts
that don't include examples like this are typically much more difficult
to understand, which in turn increases the chances that a response will
not help the questionner.
Robert Lundqvist wrote:
> I can't understand the results from cross-correlation function ccf()
> even though it should be simple.
> Here's my short example:
> *********
> a<-rnorm(5);b<-rnorm(5)
> a;b
> [1] 1.4429135 0.8470067 1.2263730 -1.8159190 -0.6997260
> [1] -0.4227674 0.8602645 -0.6810602 -1.4858726 -0.7008563
>
> cc<-ccf(a,b,lag.max=4,type="correlation")
>
> cc
> Autocorrelations of series 'X', by lag
>
> -4 -3 -2 -1 0 1 2 3 4
> -0.056 -0.289 -0.232 0.199 0.618 0.568 -0.517 -0.280 -0.012
> **********
> With lag 4 and vectors of length 5 there should as far as I can see
> only be 2 pairs of observations. The correlation would then be 1.
> Guess I am missing something really simple here. Anyone who could
> explain what is happening?
>
> Robert
>
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