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Evaluation �and Credibility

How much should we believe in what was learned?

Lectures in DM

Academic year 2020/21

Jamolbek Mattiev

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Outline

Introduction
Classification with Train, Test, �and Validation sets

Handling Unbalanced Data
Parameter Tuning

Cross-validation
Comparing Data Mining Schemes

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Introduction

How predictive is the model we learned?
Error on the training data is not a good indicator of performance on future data

Q: Why?
A: Because new data will probably not be exactly the same as the training data!

Overfitting – fitting the training data too precisely – usually leads to poor results on new data

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Evaluation issues

Possible evaluation measures:

Classification Accuracy
Total cost/benefit – when different errors involve different costs
Lift and ROC curves
Error in numeric predictions

How reliable are the predicted results?

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Classifier error rate

Natural performance measure for classification problems: error rate

Success: instance’s class is predicted correctly
Error: instance’s class is predicted incorrectly
Error rate: proportion of errors made over the whole set of instances

Training set error rate: is way too optimistic!

you can find patterns even in random data

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Evaluation on “LARGE” data, 1

If many (thousands) of examples are available, including several hundred examples from each class, then how can we evaluate our classifier method?

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Evaluation on “LARGE” data, 2

A simple evaluation is sufficient

Randomly split data into training and test sets (usually 2/3 for train, 1/3 for test)

Build a classifier using the train set and evaluate it using the test set.

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Classification Step 1: �Split data into train and test sets

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Results Known

+

-

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THE PAST

Data

Training set

Testing set

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Classification Step 2: �Build a model on a training set

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Training set

Results Known

+

-

+

THE PAST

Data

Model Builder

Testing set

Evaluation

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Classification Step 3:�Evaluate on test set (Re-train?)

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Data

Predictions

Y

N

Results Known

Training set

Testing set

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-

+

Model Builder

Evaluate

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-

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-

Evaluation

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Unbalanced data

Sometimes, classes have very unequal frequency

Attrition prediction: 97% stay, 3% attrite (in a month)
medical diagnosis: 90% healthy, 10% disease
eCommerce: 99% don’t buy, 1% buy
Security: >99.99% of Americans are not terrorists

Similar situation with multiple classes
Majority class classifier can be 97% correct, �but useless

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Handling unbalanced data – how?

If we have two classes that are very unbalanced, then how can we evaluate our classifier method?

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Balancing unbalanced data, 1

With two classes, a good approach is to build BALANCED train and test sets, and train model on a balanced set

randomly select desired number of minority class instances
add equal number of randomly selected �majority class

How do we generalize “balancing” to multiple classes?

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Balancing unbalanced data, 2

Generalize “balancing” to multiple classes

Ensure that each class is represented with approximately equal proportions in train and test

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A note on parameter tuning

It is important that the test data is not used �in any way to create the classifier
Some learning schemes operate in two stages:

Stage 1: builds the basic structure
Stage 2: optimizes parameter settings

The test data can’t be used for parameter tuning!
Proper procedure uses three sets: training data, validation data, and test data

Validation data is used to optimize parameters

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Making the most of the data

Once evaluation is complete, all the data can be used to build the final classifier
Generally, the larger the training data the better the classifier (but returns diminish)
The larger the test data the more accurate the error estimate

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Classification: Train, Validation, Test split

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Data

Predictions

Y

N

Results Known

Training set

Validation set

+

-

+

Model Builder

Evaluate

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-

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-

Final Model

Final Test Set

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Final Evaluation

Model

Builder

Evaluation

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*Predicting performance

Assume the estimated error rate is 25%. How close is this to the true error rate?

Depends on the amount of test data

Prediction is just like tossing a biased (!) coin

“Head” is a “success”, “tail” is an “error”

In statistics, a succession of independent events like this is called a Bernoulli process
Statistical theory provides us with confidence intervals for the true underlying proportion!

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*Confidence intervals

We can say: p lies within a certain specified interval with a certain specified confidence
Example: S=750 successes in N=1000 trials

Estimated success rate: 75%
How close is this to true success rate p?

Answer: with 80% confidence p∈[73.2,76.7]

Another example: S=75 and N=100

Estimated success rate: 75%
With 80% confidence p∈[69.1,80.1]

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*Mean and variance (also Mod 7)

Mean and variance for a Bernoulli trial:�p, p(1–p)
Expected success rate f=S/N
Mean and variance for f : p, p (1–p)/N
For large enough N, f follows a Normal distribution
c% confidence interval [–z ≤ X ≤ z] for random variable with 0 mean is given by:

With a symmetric distribution:

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*Confidence limits

Confidence limits for the normal distribution with 0 mean and a variance of 1:

Thus:

To use this we have to reduce our random variable f to have 0 mean and unit variance

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Pr[X ≥ z]	z
0.1%	3.09
0.5%	2.58
1%	2.33
5%	1.65
10%	1.28
20%	0.84
40%	0.25

–1 0 1 1.65

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*Transforming f

Transformed value for f :��(i.e. subtract the mean and divide by the standard deviation)�
Resulting equation:

Solving for p :

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*Examples

f = 75%, N = 1000, c = 80% (so that z = 1.28):

f = 75%, N = 100, c = 80% (so that z = 1.28):

Note that normal distribution assumption is only valid for large N (i.e. N > 100)
f = 75%, N = 10, c = 80% (so that z = 1.28):

(should be taken with a grain of salt)

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Evaluation on “small” data, 1

The holdout method reserves a certain amount for testing and uses the remainder for training

Usually: one third for testing, the rest for training

For “unbalanced” datasets, samples might not be representative

Few or none instances of some classes

Stratified sample: �advanced version of balancing the data

Make sure that each class is represented with approximately equal proportions in both subsets

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Evaluation on “small” data, 2

What if we have a small data set?
The chosen 2/3 for training may not be representative.
The chosen 1/3 for testing may not be representative.

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Repeated holdout method, 1

Holdout estimate can be made more reliable by repeating the process with different subsamples

In each iteration, a certain proportion is randomly selected for training �(possibly with stratification)
The error rates on the different iterations are averaged to yield an overall error rate

This is called the repeated holdout method

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Repeated holdout method, 2

Still not optimum: the different test sets overlap.
Can we prevent overlapping?

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Cross-validation

Cross-validation avoids overlapping test sets

First step: data is split into k subsets of equal size
Second step: each subset in turn is used for testing and the remainder for training

This is called k-fold cross-validation
Often the subsets are stratified before the cross-validation is performed
The error estimates are averaged to yield an overall error estimate

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Cross-validation example:

Break up data into groups of the same size

Hold aside one group for testing and use the rest to build the model�
Test

Repeat

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Leave-One-Out cross-validation

Leave-One-Out:�a particular form of cross-validation:

Set number of folds to number of training instances
I.e., for n training instances, build classifier n times

Makes best use of the data
Involves no random subsampling
Very computationally expensive(exception: NN)

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Leave-One-Out-CV and stratification

Disadvantage of Leave-One-Out-CV: �stratification is not possible

It guarantees a non-stratified sample because there is only one instance in the test set!

Extreme example: �random dataset split equally into two classes

Best inducer predicts majority class
50% accuracy on fresh data
Leave-One-Out-CV estimate is 100% error!

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*The bootstrap

CV uses sampling without replacement

The same instance, once selected, can not be selected again for a particular training/test set

The bootstrap uses sampling with replacement to form the training set

Sample a dataset of n instances n times with replacement to form a new dataset�of n instances
Use this data as the training set
Use the instances from the original�dataset that don’t occur in the new�training set for testing

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*The 0.632 bootstrap

Also called the 0.632 bootstrap

A particular instance has a probability of 1–1/n of not being picked
Thus its probability of ending up in the test data is:

This means the training data will contain approximately 63.2% of the instances

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*Estimating error�with the bootstrap

The error estimate on the test data will be very pessimistic

Trained on just ~63% of the instances

Therefore, combine it with the resubstitution error:

The resubstitution error gets less weight than the error on the test data
Repeat process several times with different replacement samples; average the results

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*More on the bootstrap

Probably the best way of estimating performance for very small datasets
However, it has some problems

Consider the random dataset from above
A perfect memorizer will achieve� 0% resubstitution error and� ~50% error on test data
Bootstrap estimate for this classifier:

True expected error: 50%

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Comparing data mining schemes

Frequent situation: we want to know which one of two learning schemes performs better

Note: this is domain dependent!

Obvious way: compare 10-fold CV estimates

Problem: variance in estimate

Variance can be reduced using repeated CV
However, we still don’t know whether the results are reliable

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Significance tests

Significance tests tell us how confident we can be that there really is a difference
Null hypothesis: there is no “real” difference
Alternative hypothesis: there is a difference
A significance test measures how much evidence there is in favor of rejecting the null hypothesis
Let’s say we are using 10 times 10-fold CV
Then we want to know whether the two means of the 10 CV estimates are significantly different

Student’s paired t-test tells us whether the means of two samples are significantly different

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*Paired t-test

Student’s t-test tells whether the means of two samples are significantly different
Take individual samples from the set of all possible cross-validation estimates
Use a paired t-test because the individual samples are paired

The same CV is applied twice

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William Gosset

Born: 1876 in Canterbury; Died: 1937 in Beaconsfield, England.�Obtained a post as a chemist in the Guinness brewery in Dublin in 1899. Invented the t-test to handle small samples for quality control in brewing. Wrote under the name "Student".

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*Distribution of the means

x₁ x₂ … x_k and y₁ y₂ … y_k are the 2k samples for a k-fold CV
m_x and m_y are the means
With enough samples, the mean of a set of independent samples is normally distributed

Estimated variances of the means are σ_x²/k and σ_y²/k

If μ_x and μ_y are the true means then��are approximately normally distributed with�mean 0 and variance 1

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*Student’s distribution

With small samples (k < 100) the mean follows Student’s distribution with k–1 degrees of freedom

Confidence limits:

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Pr[X ≥ z]	z
0.1%	4.30
0.5%	3.25
1%	2.82
5%	1.83
10%	1.38
20%	0.88

Pr[X ≥ z]	z
0.1%	3.09
0.5%	2.58
1%	2.33
5%	1.65
10%	1.28
20%	0.84

9 degrees of freedom normal distribution

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*Distribution of the differences

Let m_d= m_x– m_y
The difference of the means (m_d) also has a Student’s distribution with k–1 degrees of freedom
Let σ_d² be the variance of the difference
The standardized version of m_d is called the t-statistic:

We use t to perform the t-test

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*Performing the test

Fix a significance level α

If a difference is significant at the α% level,�there is a (100-α)% chance that there really is a difference

Divide the significance level by two because the test is two-tailed

I.e. the true difference can be +ve or –ve

Look up the value for z that corresponds to α/2
If t ≤ –z or t ≥ z then the difference is significant

I.e. the null hypothesis can be rejected

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Unpaired observations

If the CV estimates are from different randomizations, they are no longer paired

(or maybe we used k-fold CV for one scheme, and j-fold CV for the other one)

Then we have to use an un paired t-test with min(k, j) – 1 degrees of freedom
The t-statistic becomes:

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*Interpreting the result

All our cross-validation estimates are based on the same dataset
Hence the test only tells us whether a complete �k-fold CV for this dataset would show a difference

Complete k-fold CV generates all possible partitions of the data into k folds and averages the results

Ideally, should use a different dataset sample for each of the k-fold CV estimates used in the test to judge performance across different training sets

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T-statistic many uses

Looking ahead, we will come back to use of T-statistic for gene filtering in later modules

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*Predicting probabilities

Performance measure so far: success rate
Also called 0-1 loss function:

Most classifiers produces class probabilities
Depending on the application, we might want to check the accuracy of the probability estimates
0-1 loss is not the right thing to use in those cases

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*Quadratic loss function

p₁ … p_k are probability estimates for an instance�
c is the index of the instance’s actual class�
a₁ … a_k= 0, except for a_c which is 1�
Quadratic loss is:

Want to minimize

Can show that this is minimized when p_j= p_j^*, the true probabilities

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*Informational loss function

The informational loss function is –log(p_c),�where c is the index of the instance’s actual class
Number of bits required to communicate the actual class
Let p₁^* … p_k^*be the true class probabilities
Then the expected value for the loss function is:

Justification: minimized when p_j= p_j^*
Difficulty: zero-frequency problem

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*Discussion

Which loss function to choose?

Both encourage honesty
Quadratic loss function takes into account all class probability estimates for an instance
Informational loss focuses only on the probability estimate for the actual class
Quadratic loss is bounded:� it can never exceed 2
Informational loss can be infinite

Informational loss is related to �MDL principle

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Evaluation Summary:

Use Train, Test, Validation sets for �“LARGE” data
Balance “un-balanced” data
Use Cross-validation for small data
Don’t use test data for parameter tuning – use separate validation data

Most Important: Avoid Overfitting

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