Graduate Do Residuals Always Sum to Zero in Regression Analysis?

  • Thread starter Thread starter FallenApple
  • Start date Start date
  • Tags Tags
    Zero
Click For Summary
In regression analysis, when comparing two groups using different parameterizations, the design matrix can either include or exclude a column of ones, affecting the residuals. The discussion centers on whether residuals sum to zero in both models, noting that if the models are equivalent, the parameter estimates should also be equivalent. It is suggested that while residuals may not necessarily sum to zero when minimizing sum-squared-errors, they will sum to zero when using Maximum Likelihood estimation with the appropriate regressors. The conversation highlights uncertainty about the existence of a theorem regarding residuals summing to zero, with some participants agreeing on the lack of such a theorem. The conclusion emphasizes that the presence of an intercept typically leads to residuals summing to zero.
FallenApple
Messages
564
Reaction score
61
Ok so say I'm comparing two groups. I can do it this way ##Y_{i}=b_{1}*I(G1)+b_{2}*I(G2)+e_{i}## where I(G1) is 1 if in group 1 and 0 if not. I(G2) is 1 if in group 2 and 0 if not. In that case, my design matrix will not have a column of ones.

However, if I reparameterise to ##Y_{i}=b_{0}+b_{2}*I(G2)+e_{i}## since I know I(G2) and I(G1) has to sum to 1. I will get a design matrix with ones in the first column. I think there is a theorem that says that the residuals sum to 0 if this is the case.

Now, does this mean that the residuals sum to zero for the first parameterization as well? After all, the two models should be equivalent.
 
Physics news on Phys.org
The two models are equivalent, since ##G1=1-G2##, ##b_0=b_1## and ##b'_2=b_2-b_1## where ##b'_2## is the coefficient of ##I(G_2)## in the second model.

Given the models are equivalent, I imagine that the parameter estimates will be equivalent. Conceivably that may differ by estimation method. The method for OLS is Maximum Likelihood and I'm pretty sure that would give identical estimates, but one would need to work through the equations for the estimates, substituting the equivalences in the preceding paragraph, to be sure.

If the parameter estimates are equivalent then the residuals will be identical since the linear estimators will be identical, so if the residuals sum to zero for the first model they will do that for the second as well.
 
  • Like
Likes FallenApple and FactChecker
If you are minimizing the sum-squared-errors for your parameter estimates, I don't think that the residuals have to sum to zero.
 
  • Like
Likes FallenApple
I agree with FactChecker. I am not aware of any theorem about residuals summing to zero. If one is using Maximum-Likelihood estimation of the coefficients (ie the usual, simplest way) then the sum of the products of residuals ##\varepsilon_i## with regressors ##I(G2_i)## will be zero, that is, ##\sum_i \varepsilon_i I(G2_i)=0##. Could that be the theorem you had in mind?
 
  • Like
Likes FallenApple
andrewkirk said:
I agree with FactChecker. I am not aware of any theorem about residuals summing to zero. If one is using Maximum-Likelihood estimation of the coefficients (ie the usual, simplest way) then the sum of the products of residuals ##\varepsilon_i## with regressors ##I(G2_i)## will be zero, that is, ##\sum_i \varepsilon_i I(G2_i)=0##. Could that be the theorem you had in mind?

I thought that if there is an intercept or that there could be a transformation to the intercept, the residuals sum to 0.
 
The standard _A " operator" maps a Null Hypothesis Ho into a decision set { Do not reject:=1 and reject :=0}. In this sense ( HA)_A , makes no sense. Since H0, HA aren't exhaustive, can we find an alternative operator, _A' , so that ( H_A)_A' makes sense? Isn't Pearson Neyman related to this? Hope I'm making sense. Edit: I was motivated by a superficial similarity of the idea with double transposition of matrices M, with ## (M^{T})^{T}=M##, and just wanted to see if it made sense to talk...

Similar threads

  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
3
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
4
Views
8K