# How is the auto-correlation of vectors defined?

Suppose $$v$$ is an $$n$$-ary vector with entries from the set $$\{0,1\}$$ (i.e. a vector of ones and zeros).

A paper I am reading defines the "auto-correlation sequences" $$v*v$$ where $$*$$ denotes the correlation operator.

1) What is an auto-correlation sequence of a vector?

2) What is the correlation operator? (I'm assuming it can be applied to two distinct vectors too)

My first guess was that to auto-correlate a vector you try all the possible rotational permutations of the vector and measure the cosine of the angle between each permuted vector with the original. However, Mathematica's CorrelationFunction on $$\{1,0\}$$ with $$lag=0$$ returns 1 and with $$lag=1$$ returns $$-\frac{1}{2}$$, which shoots down my theory since I would expect orthogonal vectors to have $$0$$ correlation. So what is Mathematica doing here?

The sample correlation of vectors $$(X_1, \dots, X_n)$$ and $$(Y_1, \dots, Y_n)$$ is

$$\rho_{(X,Y)} = \frac{\frac{1}{n-1}\sum_{i=1}^n (X_i - \bar X)(Y_i - \bar Y) }{S_XS_Y},$$ where $$\bar X, \bar Y$$ are the respective sample means and $$S_XS_Y$$ are the respective sample standard deviations.

Roughly speaking, the sample autocorrelation of lag $$\ell$$ of a vector $$(X_1, \dots X_n)$$ is the sample correlation of the vector $$(X_1, \dots, X_{n-\ell})$$ and and the lagged vector $$(X_\ell, X_{\ell + 1}, \dots, X_n).$$

Various refinements are used in specific applications. Perhaps the one you are looking for is of the following form:

$$\rho_\ell = \frac{\sum_{i=1}^{n-\ell} (X_1 - \bar X)(X_\ell - \bar X) }{(n-1)S_X^2},$$ Notice that $$\bar X$$ and $$S_X^2$$ are based on the entire sequence. Also, when $$\ell=0,$$ we have $$\rho_\ell = 1.$$ See Wikipedia at the last bullet under Estimation.

As I recall, this is used in the R function acf:

set.seed(1115)
x = round(rnorm(10,200,15))-20*(1:10); x
[1] 176 169 127  99  96  92  45  70  10  12
acf(x)
acf(x, plot=F)

Autocorrelations of series ‘x’, by lag

0      1      2      3      4      5      6      7      8      9
1.000  0.625  0.299  0.138 -0.060 -0.177 -0.304 -0.363 -0.434 -0.223


• Thanks! Very clear and concise.
– Mike
Commented Nov 16, 2018 at 14:43