Why can't we add all the individual Pearson's $r$'s in a multiple regression and calculate $R^2$ based on this sum? Is there an easy mathematical explanation to this as $r^2$ is squared and don't add up some way?

  • 2
    $\begingroup$ When there are just two variables and $r$ is negative, you have no hope of creating a nonnegative $R^2$ from its sum, do you? Instead of inventing a formula and asking why it doesn't work, why not study the theory and the formulas that do work? That is a much faster and surer way towards understanding. $\endgroup$
    – whuber
    Oct 30, 2014 at 21:58
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    $\begingroup$ What is $R^2$ intended to measure? What would be the purpose of adding the correlations? What would it measure? $\endgroup$
    – Glen_b
    Oct 30, 2014 at 22:07
  • $\begingroup$ You're asking the same question at stats.stackexchange.com/questions/122109/… Please don't duplicate questions. $\endgroup$
    – Nick Cox
    Oct 30, 2014 at 22:31
  • $\begingroup$ I don't see this as the same question, @NickCox. It is an extension / follow-on question though. $\endgroup$ Oct 30, 2014 at 22:36
  • $\begingroup$ My point is that we don't need this question too. By all the other question can edited or extended. $\endgroup$
    – Nick Cox
    Oct 30, 2014 at 22:38

1 Answer 1


It is hard to know how to answer this. I guess I would just say that you can't add up the univariate Pearson's $r$'s and get $R$ or $R^2$. Bear in mind that $r$ is bound by $[-1,\ 1]$, and that $R^2$ is bound by $[0,\ 1]$. So it may be easy to see that simple addition would yield sums that violate those rules. Here is an example you can play with yourself:

## here are 3 simple variables x1, x2, & y:
x1 = c(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20)
x2 = c(2,1,4,3,6,5,8,7,10,9,12,11,14,13,16,15,18,17,20,19)
y  = c(3,1,2,6,4,5,9,7,8,12,10,11,15,13,14,18,16,17,20,19)

## here are their Pearson's correlations:
cor(x1,y)                 # r = 0.9714286
cor(x2,y)                 # r = 0.9593985
cor(x1,x2)                # r = 0.9849624

## here I fit a multiple regression model & get R, & R^2:
model = lm(y~x1+x2)
cor(y, fitted(model))     # R   = 0.9715432
summary(model)$r.squared  # R^2 = 0.9438961

## we can square R to get R^2:  
0.9715432^2               # = 0.9438962

## adding r(x1,y) + r(x2,y) does not equal either R or R^2, though:
0.9714286 + 0.9593985     # = 1.930827
  • $\begingroup$ great great great - I got it! $\endgroup$ Oct 30, 2014 at 22:59

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