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I have a two class problem at hand, 160 data samples which were classified with a linear support vector machine. The obtained classification accuracy (test accuracy) is 71% (this is the average over 70 folds). I want to calculate now the p-value for this result, i.e. the probability that this result is purely due to chance. However I have not found a clear (and for my level understandable) description of how to do that, and am not sure if I need more information about the dataset to be able to perform such a test. Any help appreciated

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It is very unusual to perform a significance test on a classifier (also it is very unusual to use a 70 fold on a 160 dataset - the most common is 5 or 10 folds. For the number of folds you used you could have chosen a Leave-one-out procedure)

The issue is the null hypothesis. You probably want to know if your classifier is significantly better than a random classifier - one that did not really learned anything from the data.

Let us assume that the dataset is binary (only two classes, + and -) where p+ is the proportion of positive classes in the dataset. Let us assume the classifier that randomly answers + with 50% probability. The chance that a data will be + is p+. Finally since the classifier output is independent of the data value itself, the probability that the classifier will be correct on a + prediction is 0.5*p+. Similarly, the probability of being right on a - prediction is 0.5*p-.

If p+ is 0.5, than the classifier will be right 0.5 of the time. And that is the null hypothesis for the situation where p+=0.5.

But if p+=0.9, a classifier that guesses + with 0.5 probability will still have a

 0.5*0.9+0.5*0.1 = 0.5 

probability of being right. But a "smarter" random classifier, that makes a + guess with 0.9 probability, will have an accuracy of

0.9*0.9+0.1*0.1 = 0.82

probability of being right, which is the maximum probability for a random classifier.

Thus, the null hypothesis for a daaset with p+ proportion of positives is an accuracy of

acc_null = p+^2 + p-^2

So you need to collect the p+ and p- of your dataset and compute the acc_null.

The question now is whether your 71% accuracy is significantly different than acc_null. Than can only be answered if you know the number of times your classified was right, and you know it. Of the 160 data points, the classifier was correct 0.71*160 = 133.6 = 134 times.

Thus you need a binomial test to figure out the probability that a random process that generates a "correct" or a 1 or a "success" with probability acc_null would have generated 134 "correct" ou "success" of 160 tries. This is the p-value you are looking for.

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  • $\begingroup$ Unfortunately, this answer is wrong and it leads to biassed p-values. It assumes that each sample is independent, which is not the case for cross-validation. This was shown for example in Noirhomme 2014 sciencedirect.com/science/article/pii/S2213158214000485 . The correct way to do this, is to use a permutation test and refit the model each time on shuffled labels, thus creating correct null-distribution. $\endgroup$ – rep_ho Jul 5 '18 at 9:06
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To follow up on Jaques' great explanation of the matter with a practical solution: To just calculate a test of whether a SVM is a better classifier for your data than a null model: you could load them into Weka (http://www.cs.waikato.ac.nz/ml/weka/index.html). This software provides a module called "experimenter", which allows you to easily compare classification accuracy of different algorithms across different data sets. One would just add the data set, then add SVM (the algorithm is called SMO in Weka) and the null model (called ZeroR there) as algorithms. As evaluation method one would select the default value (10fold cross validation), and "run the experiment". Comparison can be done using any of a comprehensive list of possible evaluation measures (accuracy, AUC, F-measure, ...).

To get aquainted with the software one would like to have a look on their online course "data mining with weka", the material of which is available on youtube (https://www.youtube.com/playlist?list=PLm4W7_iX_v4NqPUjceOGd-OKNVO4c_cPD)

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