a <- c(10,11)
b <- c(2,3)

The covariance (C) betweena and b =

sum ( (a-mean(a)) * (b-mean(b)) ) # equal to  cov(a,b) in this case as n=2

The sample covariance (Csamp) betweena and b =

sum ( (a-mean(a)) * (b-mean(b)) ) / n-1 # equal to  cov(a,b)

Now lets say the sum of square differences for a = SSa. This means the Variance(V) for a = SSa/n and the sample variance(Vsamp) is SSa/n-1. The sample standard deviation (SDsamp) for a would be sqrt(SSa/n-1)

The correlation coeffictent equations as I have found it are:

(1) C  /  sqrt(SSa) * sqrt(SSb)

# or

(2)  C  /  SDsamp_a * SDsamp_b

# (1) uses sqrt(SSa), not V or V samp and (2) uses SDsamp  not V and does not use Csamp

I am confused over how to calculate the correlation coefficient (CF) and the sample correlation coefficient (CFsamp). The two formulas above do not seem the same to me. can someone explain how to calculate CF and CFsamp?

  • $\begingroup$ From help("cov"): "The denominator n - 1 is used which gives an unbiased estimator of the (co)variance for i.i.d. observations." So, cov(a,b) actually calculates the sample covariance. $\endgroup$ – Roland Nov 22 '13 at 14:25
  • $\begingroup$ @Roland yes I saw that thanks but in this case C and Csamp are not different as n=2, ill edit to make clearer $\endgroup$ – user1320502 Nov 22 '13 at 14:32
  • $\begingroup$ You need to divide the covariance by n if you don't estimate the means. So, C is the covariance multiplied by n. And if you consider that 2-1=1 there is no difference between SSa and SDsamp_a in your example. $\endgroup$ – Roland Nov 22 '13 at 14:42
  • $\begingroup$ "divide the covariance by n if you don't estimate the means (i.e., your C)" not sure what you mean here sorry? $\endgroup$ – user1320502 Nov 22 '13 at 14:48
  • $\begingroup$ The population covariance is your C divided by n. $\endgroup$ – Roland Nov 22 '13 at 14:49

The pearson product correlation coefficient is defined as:


In order to estimate it from the sample, you put in the sample estimates of covariance and standard deviation:


Fortunately, $n-1$ is cancelled out and you get


Due to this last step it doesn't matter for the correlation coefficient if you divide by $n$ or by $(n-1)$ when calculating the covariance and standard deviations.

  • $\begingroup$ thanks, the papers I had added to my confusion by not being explicit between Sample and population estimates. However realising the cancellation of n or n-1 respectively makes everything clear. $\endgroup$ – user1320502 Nov 22 '13 at 14:44
  • 1
    $\begingroup$ Sorry, not able to comment. Very clear answer. I think the second $\sqrt{\Sigma (X_i - \bar X)^2}$ was meant to be $\sqrt{\Sigma (Y_i - \bar Y)^2}$. $\endgroup$ – pentandrous Sep 23 '15 at 20:57
  • $\begingroup$ @pentandrous Thanks. TeX is nice, but it's easy to get lost in the formula syntax. $\endgroup$ – Roland Sep 24 '15 at 7:24

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