Spearman's Rank Correlation

Date: 02/17/99 at 17:45:10
From: Bobby
Subject: Spearman's Rank Correlation

I would like to know where the 6 in the formula of Spearman's rank 
correlation originated. How did it turn out to be 6? I have tried 
consulting people and encyclopedias but without answers. 

Thank you very much.

Date: 02/17/99 at 18:19:49
From: Doctor Pat
Subject: Re: Spearman's Rank Correlation

I am not sure, but I believe that the 6 comes from the sum of squares 
of integers in the denominator. Remember that the Spearman rho is (if 
there are no ties) equivalent to the Pearson's r when the ranks are 
treated as x,y coordinates. The denominator has (n-1)sx*sy. The sums 
of the squares of the x variables, since they are ranks, must be 
consecutive integers. Similar method and reasoning works for the y 
values. When we find the variances by summing these squares we get a 
six in each denominator, and taking the square root of 6*6 gives us a 
6 outside the radical. To simplify the mathematics, multiply numerator 
and denominator by 6 and it shows up at the top...

Good luck!

- Doctor Pat, The Math Forum
  http://mathforum.org/dr.math/   

Date: 02/18/99 at 16:17:28
From: Doctor Pat
Subject: Re: Spearman's Rank Correlation

After I sent you the first response, I posted a note on a statistics 
list. Here is a MUCH better response to your question that I received 
from a correspondant of mine who is an excellent statistician.
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The Spearman rank correlation coefficient is based upon the sum 
(differences^2) in the two rankings. Consider the extreme situations 
of the rankings. If there are N ranks and the two rankings are
identical in every place, then sum(differences^2) = 0. If they are 
the reverse of each other, for example, if one is 1, 2, 3, 4, 5 and 
the other is 5, 4, 3, 2, 1, then the sum(differences^2) = N(N^2 - 1)/3. 
The closer to 0 the sum(differences^2) is, the more alike the rankings. 
The closer to N(N^2 - 1)/3 the sum(differences^2), the more disparate 
the rankings. But we want to have a common scale to judge by, not one 
that depends upon N. Moreover, we would like the scale to go from -1 
to 1, so we want to find a transformation that takes 0 to 1 and 
N(N^2 - 1)/3 to -1. Write the equation of the line passing through the 
points (0, 1) and (N(N^2 - 1)/3, -1). This line has a slope of 
-6/(N(N^2 - 1)) and an intercept of 1.   

- Doctor Pat, The Math Forum
  http://mathforum.org/dr.math/