Classification.. continued

Prediction and Classification

• Last week we discussed the classification problem..– Used the Naïve Bayes Method

• Today..we will dive into more details..

• But first how do we evaluate classifier

Abstract Binary Classification Problem

• Given n data samples where xi is a

data vector and yi is label {-1,1}.

• Aim is to learn a function

• Such that f is “accurate” on unseen data.

• [ill-specified as defined]

Algorithms to Learn Classifier

• We can use an algorithm A to learn the function f: X Y

• Then we write f as fA

• One example of A is Naïve Bayes.

• Other examples {Logistic Regression, Neural Networks, Support Vector Machines, Decision Trees, Random Forests,….}

Training vs. Test Data

• In practice to take care of the “unseen” part…we split the data into training and test sets

• We learn fA on the training set using an algorithm A

• The learned function fA is then evaluated on the test set.

Example

• Suppose we learn a function F on training set.

• Our test set consists of four data points (z1,1),(z2,-1),(z3,1),(z4,-1).

• We apply F on the four data points (without labels) and we get F(z1)=1, F(z2)=1,F(z3)=-1 and F(z4) = -1.

• Then F correctly classified z1 and z4 but incorrectly classified z2 and z3.

Confusion Matrix

Actual Label (1) Actual Label (-1)

Predicted Label (1) True Positive (N1) False Positive (N2)

Predicted Label (-1) False Negatives (N3) True Negatives (N4)

Label 1 is called Positive, Label -1 is called Negative

Let the number of test samples be N

N = N1 + N2 + N3 + N4.

True Positive Rate (TPR) = N1/(N1+N3)

True Negative Rate (TNR) = N4/(N4+N2)

False Positive Rate (FPR) = N2/(N2+N4)

False Negative Rate (FNR) = N3/(N1+N3)

Accuracy = (N1+N4)/(N1+N2+N3+N4)

Precision = N1/(N1+N2) Recall = N1/(N1+N3)

Example

Actual Label (1) Actual Label (-1)

Predicted Label (1) 10 3

Predicted Label (-1) 2 20

TPR = 5/6; TNR = 20/23; FPR = 3/23; FNR = 2/12;

Accuracy = 30/35

Precision = 10/13 and Recall = 10/12

ROC (Receiver Operating Characteristic) Curves

• Generally a learning algorithm A will return a real number…but what we want is a label {1 or -1}

• We can apply a threshold..TA 0.7 0.6 0.5 0.2 0.1 0.09 0.08 0.02 0.01

T=0.1 1 1 1 1 1 -1 -1 -1 -1

True Label

1 1 -1 -1 1 1 -1 -1 -1

A 0.7 0.6 0.5 0.2 0.1 0.09 0.08 0.02 0.01

T=0.2 1 1 1 1 -1 -1 -1 -1 -1

True Label

1 1 -1 -1 1 1 -1 -1 -1

TPR = 3/4FPR = 2/5

TPR = 2/4FPR = 2/5

ROC Curve

• An ROC Curve is the plot where the x-axis is FPR, the y-axis is the TPR and for each threshold t, the point on the plot represents the pair (FPR(t), TPR(t))

• Lets Look at the Wikipedia ROC Entry

Discussion..

• If F: Symptoms {Disease, No-Disease}– Higher Recall or Precision ?– What is the relative cost of a mis-diagnosis (and

which way)

• If F: Banner Ad {Click, No-Click}– Higher Precision means more revenue?

Random Variables

• A r.v. is a numerical quantity associated with events in an experiment.

• Suppose we roll two dice. Let X = k be the sum of the two faces.

• X can take values ranging from {2….12}.• P(X=12) = 1/36. Why ?

– Event associated with X=12 is {(6,6)}• P(X=7) = 6/36 = 1/6

– Associated Event: {(1,6),(6,1),(2,5),(5,2),(3,4),(4,3)}

Random Variable

• A random variable X can take values in a set which is:– discrete and finite.

• Lets toss a coin and X = 1 if it’s a head and X=0 if it’s a tail. X is random variable

– discrete and infinite (countable)• Let X be the number of accidents in Sydney in a day.. Then

X = 0,1,2,…..

– Infinite (uncountable)• Let X be the height of a Sydney-sider.

– X = 150, 150.11,150.112,……

Random Variable Properties

• Let X be a discrete valued random variable taking values in a set S.

The Expected (average) Value of X, E(X) is

• The Variance is

Examples

• Let X be a random variable which takes values 1 with probability p and 0 with probability 1-p. Then

Examples

• Let X be a random variable which denotes the number of “spam emails” in a batch of n emails. Assuming the probability of spam email is p.

• X={0,1,2,3,4,5}

X is a r.v. which follows a binomial distribution with parameters (n,p)… X ~ Binomial(n,p)– E(X) = np ; Var(X) = np(1-p)

Examples

• Let X be a random variable which denotes the number of tcp packets that arrive in a unit time. Then X can be modeled to follow a Poisson distribution..

• E(X) = Var(X) = λ

Continuous Distribution

• Ofcourse the most common continuous distribution is the Normal/Gaussian distribution… denoted

•

How to use r.v. for classification

• To use r.v. in classification…we have to make an assumption.– For example..Sepal Length follows a Normal Distribution.– Is this a good/reasonable assumption.

• Then we use data to estimate the parameters of the distribution..– The parameters of a Normal distribution are the mean and the

variance (square of standard deviation).– For the moment we can just use Matlab/program to do that…– Once we have the parameters we can use the distribution to

estimate the “probability” of Sepal Length taking a new value..

Fitting Distributions..Examples• 0,1,0,1,0,0

– Assume data from a Binomial distribution with 6 trials and 2 successes• In Matlab:>> binofit(2,6) = 0.3333

• 10,20,5,3,3,100– Assume data is from a Poisson distribution– X=[10 20 5 3 3 100]; – Poissfit(X); – Ans: 23.50

• What is happening ? We are just taking sample averages. The more data we have the more reliable these estimates become..

• Suppose we take Sepal Length…data vector x>> [mean,std] = normfit(x);>> ans: mean = 5.8, std=0.81

Return to the Iris Example

• We will redo the Iris Classification Example..but now will use “continuous” values for the attributes…