Weka, SVM technique output - svm

Question

I am working with SVM at Weka
I have some data and I try SVM (I tried different values of C) technique to analize the data. But the output has totally confused me, that is why I hope some help.
This is the output for a polynomial kernel of degree 1:
Scheme:weka.classifiers.meta.CVParameterSelection -P "C 0.001 10.0 5.0" -X 10 -S 1 -W weka.classifiers.functions.SMO -- -C 0.7 -L 0.001 -P 1.0E-12 -N 0 -V -1 -W 1 -K "weka.classifiers.functions.supportVector.PolyKernel -C 250007 -E 1.0"
Classifier Options: -C 7.5003 -L 0.001 -P 1.0E-12 -N 0 -V -1 -W 1 -K "weka.classifiers.functions.supportVector.PolyKernel -C 250007 -E 1.0"
=== Summary ===
Correctly Classified Instances 83 79.0476 %
Incorrectly Classified Instances 22 20.9524 %
Kappa statistic 0.6555
Mean absolute error 0.0487
Root mean squared error 0.1549
Relative absolute error 91.5633 %
Root relative squared error 100.2828 %
Total Number of Instances 105
This is the output for a polynomial kernel of degree 2:
Scheme:weka.classifiers.meta.CVParameterSelection -P "C 0.001 10.0 5.0" -X 10 -S 1 -W weka.classifiers.functions.SMO -- -C 0.7 -L 0.001 -P 1.0E-12 -N 0 -V -1 -W 1 -K "weka.classifiers.functions.supportVector.PolyKernel -C 250007 -E 2.0"
Classifier Options: -C 2.5008 -L 0.001 -P 1.0E-12 -N 0 -V -1 -W 1 -K "weka.classifiers.functions.supportVector.PolyKernel -C 250007 -E 2.0"
=== Summary ===
Correctly Classified Instances 87 82.8571 %
Incorrectly Classified Instances 18 17.1429 %
Kappa statistic 0.7236
Mean absolute error 0.0486
Root mean squared error 0.1547
Relative absolute error 91.4748 %
Root relative squared error 100.1845 %
Total Number of Instances 105
This is the output for a gaussian kernel and gamma value of 1.0:
Scheme:weka.classifiers.meta.CVParameterSelection -P "C 0.001 10.0 5.0" -X 10 -S 1 -W weka.classifiers.functions.SMO -- -C 0.7 -L 0.001 -P 1.0E-12 -N 0 -V -1 -W 1 -K "weka.classifiers.functions.supportVector.RBFKernel -C 250007 -G 1.0"
Classifier Options: -C 2.5008 -L 0.001 -P 1.0E-12 -N 0 -V -1 -W 1 -K "weka.classifiers.functions.supportVector.RBFKernel -C 250007 -G 1.0"
=== Summary ===
Correctly Classified Instances 87 82.8571 %
Incorrectly Classified Instances 18 17.1429 %
Kappa statistic 0.721
Mean absolute error 0.0486
Root mean squared error 0.1547
Relative absolute error 91.4571 %
Root relative squared error 100.1651 %
Total Number of Instances 105
These are my questions:
Why the classifiers at “Classifier Options: -Cxxxxx” have different values of C? I think it is related to the values of C that I am checking with “C 0.001 10.0 5.0”, but I may be wrong and I may be getting wrong conclusions. I think that with “C 0.001 10.0 5.0” it tries the values 0.001, 0.01, 0.1, 1.0 and 10.0. If that is true why the values of C are: 7.5003 and 2.5008??
The percentage of the Correctly Classified Instances is very similar in all of the different runs, I don’t understand why… What does it mean? If it were 12% I would think SVM was not a proper technique (it wasn't linearly separable) but with 80% (no or little variations after changing the parameters) I don’t know what to think…
If I check the data in the “Visualize” of Weka the data doesn’t seem linearly separable but due to the conversion of a bigger number of dimensions when I use SVM I don’t think it is possible to get conclusions from the visualize.

Answer 1

1) Yes you use CVParameterSelection which trys different parameters in your case from 0.001 to 10 in 5 steps. The steps would be 0.001 + k * (10-0.001)/4) because your first step is allready defined as 0.001. If you round the following values (for k= 0/1/2/3/4) you see that they fit.
Step 1) 0.001
Step 2) 2.50075
Step 3) 5.005
Step 4) 7.50025
Step 5) 10.0
2) If you would have 12% accuracy there would be something really strange. Random classification (assuming you have balanced data) would result in 50%. I cannot tell you if 82% is a good result because i don't know about your data.
3) To your question about linar separability. Thats exactly why the SVM is so good. SVM with the right Kernel transforms your data into a higher featurespace to get rid of the non-separability. An RBF kernel can transform your data even into an infinite-dimensional feature space. Thats why a perfect linar separation is allways possible. At this point overfitting can occur. To avoid that and reach a good generalisation you have your complexity parameter C.
Here you can read more in a good stackexchange post about the last point
https://stats.stackexchange.com/questions/80398/how-can-svm-find-an-infinite-feature-space-where-linear-separation-is-always-p