[R] How does ccf() really work?

[Spencer Graves]

	  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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