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MSE result is too high

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NeverPerfecT
NeverPerfecT on 28 Mar 2018
Commented: NeverPerfecT on 29 Mar 2018
I have a 122x499 dataset, consisting of -1, 0, and 1 value. I made a neural network using "nntool" which uses Feedforward backprop, 2 hidden layer, the first layer having 100 hidden neuron and the second as output layer. I then trained the network using these parameters:
rng('default')
network1.divideFcn = 'dividerand';
network1.divideParam.trainRatio = 0.7;
network1.divideParam.valRatio = 0.15;
network1.divideParam.testRatio = 0.15;
network1.trainParam.epochs = 10000;
network1.trainParam.lr = 0.01;
network1.trainParam.mc = 0.9;
% Train the Network
[net,tr] = train(network1,input,output);
% Test the Network
y = net(input);
e = gsubtract(output,y);
mse = perform(net,output,y)
accuracy = 1-(0.5*mean(abs(e./(abs(output)+abs(y)))))
Now, for some reason I get MSE value of 1.4188.. As far as I know MSE only goes up to 1 right? The same thing happens when I use 200 neuron, MSE value went up to 1.5594.. But when I increased the neuron to 500, MSE goes down to 0.5852.. Did I do something wrong here? Is my MSE equation wrong? Please help!

Answers (1)

Greg Heath
Greg Heath on 28 Mar 2018
Edited: Greg Heath on 28 Mar 2018
1. GUESS: fitnet ?
2. Unnecessary assignments of default values to
divideFcn and ratios.
3. Erroneous use of "output" instead of "target"
4. Inadvisable normalization. Normalize using the
mse from the simple case where the target
is constant and the corresponding mse is the
variance:
e = target - output;
NMSE = mse(e)/mean(var(target',1))
% NMSE = mse(e)/var(target) for 1-dim
which yields
Rsquare = 1 - NMSE % R^2
See Wikipedia for an extensive discussion of Rsquare.
Hope this helps.
Thank you for formally accepting my answer
Greg
  1 Comment
NeverPerfecT
NeverPerfecT on 29 Mar 2018
Thanks for the answer Greg. Just a few uncertainties I'd like to ask..
So in this case Rsquared would be the accuracy of the network and NMSE is the correct MSE equation right?

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