Discuss About Application of Coefficient of determination in Neural Networks

Hi All
when we are running a ANN code , I have noticed that the MSE and Coefficient of determination is calculated before the neural model , as a Linear Model, but It is never used afterward to be compared to the results and error of the neural model to judge the performance here is the link that is is discussed about.
my question is where we use the R^2 finally ? if it's before the double loop, it has no application somehow !

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Greg Heath
Greg Heath 2015년 3월 14일
What is being done is to increase the order of the models to better understand and quantify the final model.
The 0th order net is just a constant output, regardless of the input. The constant output that yields minimum MSE is the rowwise mean. The MSE, MSE00, is just the mean of the rowwise variances.
The 1st order net is a linear model that can be implemented with no hidden layers. However, the weights can be obtained more easily with the typical backslash division approach. The MSE is denoted MSE0 and the fraction of the target variance that is not modeled by the net is just the normalized MSE0, NMSE0 = MSE0/MSE00. The fraction that is modeled is the coefficient of determination or Rsquared0 = 1- NMSE0 (see Wilikpedia).
Higher order nets can be implemented with a single hidden layer. The measures of performance are the normalized mean-squared-error, NMSE = MSE/MSE00 and coefficient od determination or Rsquared = 1- NMSE , the fraction of target variance that is represented by the model.
Hope this helps.
Thank you for formally accepting my answer
Greg

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Thank you Very much Professor
one question is that
in the example I mentioned , in the linear model you got :
R2a = 1-MSE0a/MSE00a % 0.2524
and in neural :
disp( ' Nepochs R2 R2a ' )
disp ([ Nepochs R2 R2a ] )
% Nepochs R2 R2a
% 6. 0.0874 -2.1942
% 36 0.9078 0.6772 % Best
% 2 0.2427 -1.6505
% 4 0.4008 -1.0971
% 15 0.2492 -1.6278
% 26 0.2492 -1.6278
% 7 0.2492 -1.6278
% 9 0.3424 -1.3016
% 11 0.0906 -2.1828
% 16 0.9078 0.6772 % Best
on what basis you have selected the Best ones ? and how have you compared it to the linear model ?
MSE00 = mean(var(t'),1) Didn't work for me ,
I entered : MSE00 = mean(var(t'),2); and it became a scalar ,
I tried it and this is what I got, can R^2 be always 1 ??? :
Nepochs R2 R2a
1.0e+03 *
0 0 0
1.4850 0.0010 0.0010
0 0 0
0 0 0
0 0 0
0.1280 0.0010 0.0010
0 0 0
0 0 0
0 0 0
0.3480 0.0010 0.0010
0 0 0
0 0 0
0 0 0
2.0000 0.0010 0.0010
0 0 0
0 0 0
0 0 0
0.6150 0.0010 0.0010
0 0 0
0 0 0
0 0 0
1.1550 0.0010 0.0010
0 0 0
0 0 0
0 0 0
1.3040 0.0010 0.0010
0 0 0
0 0 0
0 0 0
0.3350 0.0010 0.0010
0 0 0
0 0 0
0 0 0
0.3390 0.0010 0.0010
0 0 0
0 0 0
0 0 0
0.6190 0.0010 0.0010
0 0 0
0 0 0
0 0 0
0.1690 0.0010 0.0010
0 0 0
0 0 0
0 0 0
0.5610 0.0010 0.0010
0 0 0
0 0 0
0 0 0
0.6520 0.0010 0.0010
0 0 0
0 0 0
0 0 0
0.5900 0.0010 0.0010
0 0 0
0 0 0
0 0 0
0.8060 0.0010 0.0010
0 0 0
0 0 0
0 0 0
0.1480 0.0010 0.0010
0 0 0
0 0 0
0 0 0
0.7440 0.0010 0.0010
0 0 0
0 0 0
0 0 0
0.3270 0.0010 0.0010
0 0 0
0 0 0
0 0 0
0.6240 0.0010 0.0010
0 0 0
0 0 0
0 0 0
0.6700 0.0010 0.0010

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