# Data Mining Chapter 5 Credibility: Evaluating What’s Been Learned

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Data Mining Chapter 5 Credibility: Evaluating What’s Been Learned
Kirk Scott

The Lontara Script, and Example of an Abugida

Comparison of various abugidas descended from Brahmi script
Comparison of various abugidas descended from Brahmi script. May Śiva protect those who take delight in the language of the gods. (Kalidasa)

Bopomofo system of phonetic notation for the transcription of spoken Chinese

Oracle bone script: (from left)

5.0 On to the Topic at Hand…

One thing you’d like to do is evaluate the performance of a data mining algorithm
More specifically, you’d like to evaluate results obtained by applying a data mining algorithm to a certain data set For example, can you estimate what percent of new instances it will classify correctly?

Problems include: The performance level on the training set is not a good indicator of performance on other data sets Data may be scarce, so that enough data for separate training and test data sets may be hard to come by A related question is how to compare the performance of 2 different algorithms

Parameters for evaluation and comparison:
Are you predicting classification? Or are you predicting the probability that an instance falls into a classification? Or are you doing numeric prediction? What does performance evaluation mean for things other than prediction, like association rules?

Can you define the cost of a misclassification or other “failure” of machine learning?
There are false positives and false negatives Different kinds of misclassifications will have different costs in practice

For evaluation of one algorithm, a large amount of data makes estimating performance easy For smaller amounts of data, a technique called cross-validation is commonly used Comparing the performance of different data mining algorithms relies on statistics

5.1 Training and Testing

For tasks like prediction, a natural performance measure is the error rate
This is the number of incorrect predictions out of all made How to estimate this?

A rule set, a tree, etc. may be imperfect
It may not classify all of the training set instances correctly But it has been derived from the training set In effect, it is optimized for the training set Its error rate on an independent test set is likely to be much higher

The error rate on the training set is known as the resubstitution error
You put the same data into the classifier that was used to create it The resubstitution error rate may be of interest But it is not a good estimate of the “true” error rate

In a sense, you can say that the classifier, by definition, is overfitted on the training set
If the training set is a very large sample, its error rate could be closer to the true rate Even so, as time passes and conditions change, new instances might not fall into quite the same distribution as the training instances and will have a higher rate of misclassification

Not surprisingly, the way to estimate the error rate is with a test set
You’d like both the training set and the test set to be representative samples of all possible instances It is important that the training set and the test set be independent Any test data should have played no role in the training of the classifier

There is actually a third set to consider
The validation set It may serve either of these two purposes: Selecting one of several possible data mining algorithms Optimizing the one selected

All three sets should be independent
Test only with the test set Train only with the training set Validate only with the validation set In general, error rate estimation is done on the test set After all decisions are made, it is permissible to combine all test sets and retrain on this superset for better results

5.2 Predicting Performance

This section can be dealt with quickly
Its main conclusions are based on statistical concepts, which are presented in a box You may be familiar with the derivations from statistics class They are beyond the scope of this course

These are the basic ideas:
Think in terms of a success rate rather than an error rate Based on a sample of n instances, you have an observed success rate in the test data set

Statistically, you can derive a confidence interval around the observed rate that depends on the sample size Doing this provides more complete knowledge about the real success rate, whatever it might be, based on the observed success rate

5.3 Cross-Validation

Cross-validation is the general term for techniques that can be used to select training/testing data sets and estimate performance when the overall data set is limited in size In simple terms, this might be a reasonable rule of thumb: Hold out 1/3 of the data for testing, for example, and use the rest for training (and validation)

Statistical Problems with the Sample
Even if you do a proper random sample, poorly matched test and training sets can lead to poor error estimates Suppose that all of the instances of one classification fall into the test set and none fall into the training set

The rules resulting from training will not be able to classify those test set instances
The error rate will be high (justifiably) But a different selection of training set and test set would give rules that covered that classification And when the rules are evaluated using the test set, a more realistic error rate would result

Stratification Stratification can be used to better match the test set and the training set You don’t simply obtain the test set by random sampling You randomly sample from each of the classifications in the overall data set in proportion to their occurrence there This means all classifications are represented in both the test and training sets

Repeated Holdout With additional computation, you can improve on the error estimate obtained from a stratified sample alone n times, randomly hold out 1/3 of the instances for training, possibly using stratification Average the error rates over the n times This is called repeated holdout

Cross-Validation The idea of multiple samples can be generalized
Partition the data into n “folds” (partitions) Do the following n times: Train on (n – 1) of the partitions Test on the remaining 1 Average the error rates over the n times

Stratified 10-Fold Cross-Validation
The final refinement of cross validation is to make the partitions so that all classifications are roughly in proportion The standard rule of thumb is to use 10 partitions Why? The answer boils down to this essentially: Human beings have 10 fingers…

In general, it has been observed that 10 partitions lead to reasonable results
10 partitions is not computationally out of the question

The final refinement presented in the book is this:
If you want really good error estimates, do 10-fold cross validation 10 times with different stratified partitions and average the results At this point, estimating error rates has become a computationally intensive task

5.4 Other Estimates

10 times 10-fold stratified cross-validation is the current standard for estimating error rates
There are other methods, including these two: Leave-One-Out Cross-Validation The Bootstrap

Leave-One-Out Cross-Validation
This is basically n-fold cross validation taken to the max For a data set with n instances, you hold out one for testing and train on the remaining (n – 1) You do this for each of the instances and then average the results

It’s deterministic: There’s no sampling In a sense, you maximize the information you can squeeze out of the data set

It’s computationally intensive By definition, a holdout of 1 can’t be stratified By definition, the classification of a single instance will not conform to the distribution of classifications in the remaining instances This can lead to poor error rate estimates

The Bootstrap This is another technique that ultimately relies on statistics (presented in a box) The question underlying the development and use of the bootstrap technique is this: Is there a way of estimating the error rate that is especially well-suited to small data sets?

The basic statistical idea is this:
Do not just randomly select a test set from the data set Instead, select a training set (the larger of the two sets) using sampling with replacement

Sampling with replacement in this way will lead to these results:
Some of the instances will not be selected for the training set These instances will be the test set By definition, you expect duplicates in the training set

Having duplicates in the training set will mean that it is a poorer match to the test set
The error estimate is skewed to the high side This is corrected by combining this estimate with the resubstitution error rate, which is naturally skewed to the low side

This is the statistically based formula for the overall error rate:
.632 * test set error rate * training set error rate To improve the estimate, randomly sample with replacement multiple times and average the results

Like the leave-out-one technique, there are cases where this technique does not give good results
This depends on the distribution of the classifications in the overall data set Sampling with replacement in some cases can lead to the component error rates being particularly skewed in such a way that they don’t compensate for each other

5.5 Comparing Data Mining Schemes

This section can be dealt with quickly because it is largely based on statistical derivations presented in a box The fundamental idea is this: Given a data set (or data sets) and two different data mining algorithms, you’d like to choose which one is best

For researchers in the field, this is the broader question:
Across all possible data sets in a given problem domain (however that may be defined) which algorithm is superior overall? We’re really only interested in the simpler, applied question

At heart, to compare two algorithms, you compare their estimated error or success rates
In a previous section it was noted that you can get a confidence interval for an estimated success rate In a situation like this, you have two probabilistic quantities you want to compare

The paired t-test is an established statistical technique for comparing two such quantities and determining if they are significantly different

5.6 Predicting Probabilities

The discussion so far had to do with evaluating schemes that do simple classification
Either an instance is predicted to be in a certain classification or it isn’t Numerically, you could say a successful classification was a 1 and an unsuccessful classification was a 0 In this situation, evaluation boiled down to counting the number of successes

Some classification schemes are finer tuned Instead of just predicting a classification, they produce the following: For each instance examined, a probability that that instance falls into each of the classes

The set of k probabilities can be thought of as a vector of length k containing elements pi
Each pi represents the probability of one of the classifications for that instance The pi sum to 1

The following discussion will be based on the case of one instance, not initially considering the fact that a data set will contain many instances How do you judge the goodness of the vector of pi‘s that a data mining algorithm produces?

Evaluation is based on what is known as a loss function
Loss is a measure of the probabilities against the actual classification found A good prediction of probabilities should have a low loss value In other words, the difference is small between what is observed and the predicted probabilities

Consider the loss for one instance
Out of k possible classifications, the instance actually falls into one of them, say the ith This fact can be represented by a vector a with a 1 in position ai and 0 in all others

The data mining algorithm finds a vector of length k containing the probabilities that an instance falls into one of the classifications This fact can be represented by a vector p containing values pi

Now for a single instance, consider this sum over the different classes and their probabilities:

Ignore the squaring for a moment
Excluding the ith case, this is a sum of the probabilities that didn’t come true For a good prediction, you’d like this to be small

For the ith case you have pi – ai = pi – 1
For a good scheme you’d like the probability of predicting the correct classification to be as near to 1 as possible In other words, you’d like the difference between the probability and 1 to be as small as possible

Putting the two parts back together again, you’d like this sum to be small:
As for the squaring, this is basically just statistical standard operating procedure Squaring makes the function amenable to calculus techniques

The final outcome is this
The goal is to minimize the mean square error (MSE) Because of the squaring this is called the quadratic loss function Recall that this discussion was for one instance Globally you want to minimize the MSE over all instances

Informational Loss Function
The informational loss function is another way of evaluating a probabilistic predictor Like the discussion of the quadratic loss function, this will be discussed in terms of a single instance For a whole data set you would have to consider the losses over all instances

The book explains that the informational loss function is related to the information function for tree formation It is also related to logistic regression Suppose that a particular instance is classified into the ith category For that instance, this is the loss function: - log2 pi

I will give a simplified information theoretic explanation at the end
In the meantime, it is possible to understand what the function does with a few basic definitions and graphs Remember the following: The sum of all of the pi = 1 Each individual pi < 1

Now observe that the information loss function involves only pi, the probability value for the category that the instance actually classified into The information loss function does not include any information about the probabilities of the categories that the instance didn’t classify into

So, restrict your attention to the ith category, the one the instance classified into
A measure of a good predictor would be how close the prediction probability, pi, is to the value 1 (perfect prediction) for that instance

Here is the standard log function
log2 pi Its graph is shown on the following overhead

Here is the informational loss function
- log2 pi Its graph is shown on the following overhead We are only interested in that part of the graph where 0 < pi < 1, since that is the range of possible values for pi

The goal is to minimize the loss function
What does that mean, visually? It means pushing pi as close to one as possible When pi = 1 the information loss function takes on the value 0 An information loss function value of 0 is the optimal result

In vague information theoretic terms, you might say this:
The greater the probability of your prediction, the more (valid) information you apparently had to base it on Or, the less information you would have needed in order to make it optimal As pi  1, -log2 pi  0

Conversely, the less the probability of your prediction the less (valid) information you apparently had to base it on Or the more information you would have needed to make it optimal As pi  0, -log2 pi  ∞

The book observes that if you assign a probability of 0 to an event and that event happens, the informational loss function “cost” is punishing—namely, infinite A prediction that is flat wrong would require an unmeasurable amount of information to correct

This leads to another very informal insight into the use of the logarithm in the function, and what that means If you view the graph from right to left instead of right to left, as the probability of a prediction gets smaller and smaller, the information needed in order to make it better grows “exponentially”

In practice, because of the potential for an infinite informational loss function value, no event is given a probability of 0

Consider the graph again
Assume that information was measured in bits rather than decimal numbers A more theoretical statement about the relative costs would go like this: The informational loss function represents the relative number of bits of information that would be needed to effect a change in the probability of a prediction

Discussion Which is better, the quadratic loss function or the informational loss function? It’s a question of preference The quadratic loss function is based on the probability of cases that didn’t occur as well as cases that occurred The information loss function is based only on the probability of the case that occurred

The quadratic loss function is bounded (a finite sum of squares of differences)
The informational loss function is unbounded Statisticians would probably find the quadratic loss function (MSE criterion) familiar and useful

Information theorists would probably find the informational loss function familiar and useful
The book points out that in theory you can measure the information contained in a structural representation in bits Advanced analyses can be based on formulas that combine the information in a representation and the bits involved in measuring the error rate (bits still needed)

The End