Post on 02-Jan-2016
description
Linear Classifier
Team teaching
2
Linear Methods for Classification
Lecture Notes for CMPUT 466/551
Nilanjan Ray
3
Linear Classification
• What is meant by linear classification?– The decision boundaries in the in the feature
(input) space is linear• Should the regions be contiguous?
R1 R2
R3R4
X1
X2
Piecewise linear decision boundaries in 2D input space
4
Linear Classification…
• There is a discriminant function k(x) for each
class k
• Classification rule:
• In higher dimensional space the decision boundaries are piecewise hyperplanar
• Remember that 0-1 loss function led to the classification rule:
• So, can serve as k(x)
)}(maxarg:{ xkxR jj
k
)}|(maxarg:{ xXjGPkxRj
k
)|( XkGP
5
Linear Classification…
• All we require here is the class boundaries {x:k(x) = j(x)} be linear for every (k, j) pair
• One can achieve this if k(x) themselves are linear or any monotone transform of k(x) is linear– An example:
xxXGP
xXGP
xxXGP
x
xxXGP
T
T
T
T
0
0
0
0
])|2(
)|1(log[
)exp(1
1)|2(
)exp(1
)exp()|1(
Linear
So that
6
Linear Discriminant Analysis
K
lll
kk
xf
xfxXkG
1
)(
)()|Pr(
Essentially minimum error Bayes’ classifier
Assumes that the conditional class densities are (multivariate) Gaussian
Assumes equal covariance for every class
Posterior probability
k is the prior probability for class k
fk(x) is class conditional density or likelihood density
Application ofBayes rule
))()(2
1exp(
||)2(
1)( 1
2/12/ kT
kpk xxxf
ΣΣ
7
LDA…
)2
1(log)
2
1(log
loglog)|Pr(
)|Pr(log
1111l
Tll
Tlk
Tkk
Tk
l
k
l
k
xx
f
f
xXlG
xXkG
ΣΣΣΣ
)(xl)(xk
)(maxarg)(ˆ xxG kk
)|Pr(maxarg)(ˆ xXkGxGk
Classification rule:
is equivalent to:
The good old Bayes classifier!
8
LDA…
kkg ik Nxi
/ˆ
NNkk /ˆ
)/()ˆ)(ˆ(ˆ1
KNxxK
k g
Tkiki
i
Σ
Training data utilized to estimate
Prior probabilities:
Means:
Covariance matrix:
When are we going to use the training data?
Nixg ii :1),,( Total N input-output pairs Nk number of pairs in class k Total number of classes: K
9
LDA: Example
LDA was able to avoid masking here
Study case
• Factory “ABC” produces very expensive and high quality chip rings that their qualities are measured in term of curvature and diameter. Result of quality control by experts is given in the table below.
Curvature Diameter Quality Control Result
2.95 6.63 Passed
2.53 7.79 Passed
3.57 5.65 Passed
3.57 5.45 Passed
3.16 4.46 Not passed
2.58 6.22 Not passed
2.16 3.52 Not passed
• As a consultant to the factory, you get a task to set up the criteria for automatic quality control. Then, the manager of the factory also wants to test your criteria upon new type of chip rings that even the human experts are argued to each other. The new chip rings have curvature 2.81 and diameter 5.46.
• Can you solve this problem by employing Discriminant Analysis?
Solutions
• When we plot the features, we can see that the data is linearly separable. We can draw a line to separate the two groups. The problem is to find the line and to rotate the features in such a way to maximize the distance between groups and to minimize distance within group.
• X = features (or independent variables) of all data. Each row (denoted by ) represents one object; each column stands for one feature.
• Y = group of the object (or dependent variable) of all data. Each row represents one object and it has only one column.
x= y=
€
2.95
2.35
3.57
3.16
2.58
2.16
3.27
6.63
7.79
5.65
5.47
4.46
6.22
3.52
⎡
⎣
⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥
€
1
1
1
1
2
2
2
⎡
⎣
⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥
• Xk = data of row k, for example x3 =
• g=number of gropus in y, in our example, g=2• Xi = features data for group i . Each row
represents one object; each column stands for one feature. We separate x into several groups based on the number of category in y.
€
3.57 5.65[ ]
x1= x2=
€
2.95
2.53
3.57
3.16
6.63
7.79
5.65
5.47
⎡
⎣
⎢ ⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥ ⎥
€
2.58
2.16
3.27
4.46
6.22
3.52
⎡
⎣
⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥
• μi = mean of features in group i, which is average of xi
• μ1 = , μ2 =
• μ = global mean vector, that is mean of the whole data set.
• In this example, μ =
€
2.67 4.73[ ]
€
3.05 6.38[ ]
€
2.88 5.676[ ]
• = mean corrected data, that is the features data for group i, xi , minus the global mean vector μ
= =
€
−0.305
−0.732
0.386
−1.218
0.547
−2.155
⎡
⎣
⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥
€
0.060
−0.357
0.679
0.269
0.951
2.109
−0.025
−0.209
⎡
⎣
⎢ ⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥ ⎥
€
x i0
€
x10
€
x20
Covariance matrix of group i =
C1 = C2 =
€
c i =(x i
0)T x i0
ni
€
0.166
−0.192
−0.192
1.349
⎡
⎣ ⎢
⎤
⎦ ⎥
€
0.259
−0.286
−0.286
2.142
⎡
⎣ ⎢
⎤
⎦ ⎥
= pooled within group covariance matrix. It is calculated for each entry in the matrix. In our example, 4/7*0.166 + 3/7*0.259=0.206 , 4/7*(-0.192)+3/7*(-0.286)=-0.233 and 4/7*1.349+3/7*2.142=1.689 , therefore
€
C(r,s) =1
nnic i(r,s)
i=1
g
∑
C =
The inverse of covariance matrix is :
C-1 = €
0.206
−0.233
−0.233
1.689
⎡
⎣ ⎢
⎤
⎦ ⎥
€
5.745
0.791
0.791
0.701
⎡
⎣ ⎢
⎤
⎦ ⎥
• P = prior probability vector (each row represent prior probability of group ). If we do not know the prior probability, we just assume it is equal to total sample of each group divided by the total samples, that is
p = =
€
0.571
0.429
⎡
⎣ ⎢
⎤
⎦ ⎥
€
4 /7
3/7
⎡
⎣ ⎢
⎤
⎦ ⎥
• discriminant function
• We should assign object k to group i that has maximum fi
€
f i = μ iC−1xk
T −1
2μC−1μ i
T + ln(pi)
LDA
Tugas
• Gunakan excel/matlab/tools lain untuk mengklasifikasi data set breast tissue secara :
• Naïve Bayes• LDA
Presentasikan minggu depan