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Machine Learning for Data MiningLinear Classifiers
Andres Mendez-Vazquez
May 23, 2016
1 / 85
Outline
1 IntroductionThe Simplest FunctionsSplitting the SpaceThe Decision Surface
2 Developing an Initial SolutionGradient Descent Procedure
The Geometry of a Two-Category Linearly-Separable CaseBasic Method
Minimum Squared Error ProcedureThe Error IdeaThe Final Error EquationThe Data MatrixMulti-Class SolutionIssues with Least Squares!!!What about Numerical Stability?
2 / 85
Outline
1 IntroductionThe Simplest FunctionsSplitting the SpaceThe Decision Surface
2 Developing an Initial SolutionGradient Descent Procedure
The Geometry of a Two-Category Linearly-Separable CaseBasic Method
Minimum Squared Error ProcedureThe Error IdeaThe Final Error EquationThe Data MatrixMulti-Class SolutionIssues with Least Squares!!!What about Numerical Stability?
3 / 85
What is it?
First than anything, we have a parametric model!!!Here, we have an hyperplane as a model:
g(x) = wT x + w0 (1)
In the case of R2
We have the following function:
g (x) = w1x1 + w2x2 + w0 (2)
4 / 85
What is it?
First than anything, we have a parametric model!!!Here, we have an hyperplane as a model:
g(x) = wT x + w0 (1)
In the case of R2
We have the following function:
g (x) = w1x1 + w2x2 + w0 (2)
4 / 85
Outline
1 IntroductionThe Simplest FunctionsSplitting the SpaceThe Decision Surface
2 Developing an Initial SolutionGradient Descent Procedure
The Geometry of a Two-Category Linearly-Separable CaseBasic Method
Minimum Squared Error ProcedureThe Error IdeaThe Final Error EquationThe Data MatrixMulti-Class SolutionIssues with Least Squares!!!What about Numerical Stability?
5 / 85
Splitting The Space R2
Using a simple straight line
Class
Class
6 / 85
Outline
1 IntroductionThe Simplest FunctionsSplitting the SpaceThe Decision Surface
2 Developing an Initial SolutionGradient Descent Procedure
The Geometry of a Two-Category Linearly-Separable CaseBasic Method
Minimum Squared Error ProcedureThe Error IdeaThe Final Error EquationThe Data MatrixMulti-Class SolutionIssues with Least Squares!!!What about Numerical Stability?
7 / 85
Defining a Decision Surface
The equation g (x) = 0 defines a decision surfaceSeparating the elements in classes, ω1 and ω2.
When g (x) is linear the decision surface is an hyperplaneGiven x1 and x2 are both on the decision surface:
wT x1 + w0 = 0wT x2 + w0 = 0
Thus
wT x1 + w0 = wT x2 + w0 (3)
8 / 85
Defining a Decision Surface
The equation g (x) = 0 defines a decision surfaceSeparating the elements in classes, ω1 and ω2.
When g (x) is linear the decision surface is an hyperplaneGiven x1 and x2 are both on the decision surface:
wT x1 + w0 = 0wT x2 + w0 = 0
Thus
wT x1 + w0 = wT x2 + w0 (3)
8 / 85
Defining a Decision Surface
The equation g (x) = 0 defines a decision surfaceSeparating the elements in classes, ω1 and ω2.
When g (x) is linear the decision surface is an hyperplaneGiven x1 and x2 are both on the decision surface:
wT x1 + w0 = 0wT x2 + w0 = 0
Thus
wT x1 + w0 = wT x2 + w0 (3)
8 / 85
Defining a Decision Surface
Thus
wT (x1 − x2) = 0 (4)
Remark: Any vector in the hyperplane is perpendicular to wT i.e. wT
is normal to the hyperplane.
Something Notable
Properties
9 / 85
Defining a Decision Surface
Thus
wT (x1 − x2) = 0 (4)
Remark: Any vector in the hyperplane is perpendicular to wT i.e. wT
is normal to the hyperplane.
Something Notable
Properties
9 / 85
Defining a Decision Surface
Thus
wT (x1 − x2) = 0 (4)
Remark: Any vector in the hyperplane is perpendicular to wT i.e. wT
is normal to the hyperplane.
Something Notable
Properties
9 / 85
Therefore
The space is split in two regions (Example in R3) by the hyperplane H
10 / 85
Some Properties of the Hyperplane
Given that g (x) > 0 if x ∈ R1
11 / 85
It is moreWe can say the following
Any x ∈ R1 is on the positive side of H.Any x ∈ R2 is on the negative side of H.
In addition, g (x) can give us a way to obtain the distance from x tothe hyperplane HFirst, we express any x as follows
x = xp + rw
‖w‖
Wherexp is the normal projection of x onto H.r is the desired distance
I Positive, if x is in the positive sideI Negative, if x is in the negative side
12 / 85
It is moreWe can say the following
Any x ∈ R1 is on the positive side of H.Any x ∈ R2 is on the negative side of H.
In addition, g (x) can give us a way to obtain the distance from x tothe hyperplane HFirst, we express any x as follows
x = xp + rw
‖w‖
Wherexp is the normal projection of x onto H.r is the desired distance
I Positive, if x is in the positive sideI Negative, if x is in the negative side
12 / 85
It is moreWe can say the following
Any x ∈ R1 is on the positive side of H.Any x ∈ R2 is on the negative side of H.
In addition, g (x) can give us a way to obtain the distance from x tothe hyperplane HFirst, we express any x as follows
x = xp + rw
‖w‖
Wherexp is the normal projection of x onto H.r is the desired distance
I Positive, if x is in the positive sideI Negative, if x is in the negative side
12 / 85
It is moreWe can say the following
Any x ∈ R1 is on the positive side of H.Any x ∈ R2 is on the negative side of H.
In addition, g (x) can give us a way to obtain the distance from x tothe hyperplane HFirst, we express any x as follows
x = xp + rw
‖w‖
Wherexp is the normal projection of x onto H.r is the desired distance
I Positive, if x is in the positive sideI Negative, if x is in the negative side
12 / 85
It is moreWe can say the following
Any x ∈ R1 is on the positive side of H.Any x ∈ R2 is on the negative side of H.
In addition, g (x) can give us a way to obtain the distance from x tothe hyperplane HFirst, we express any x as follows
x = xp + rw
‖w‖
Wherexp is the normal projection of x onto H.r is the desired distance
I Positive, if x is in the positive sideI Negative, if x is in the negative side
12 / 85
It is moreWe can say the following
Any x ∈ R1 is on the positive side of H.Any x ∈ R2 is on the negative side of H.
In addition, g (x) can give us a way to obtain the distance from x tothe hyperplane HFirst, we express any x as follows
x = xp + rw
‖w‖
Wherexp is the normal projection of x onto H.r is the desired distance
I Positive, if x is in the positive sideI Negative, if x is in the negative side
12 / 85
It is moreWe can say the following
Any x ∈ R1 is on the positive side of H.Any x ∈ R2 is on the negative side of H.
In addition, g (x) can give us a way to obtain the distance from x tothe hyperplane HFirst, we express any x as follows
x = xp + rw
‖w‖
Wherexp is the normal projection of x onto H.r is the desired distance
I Positive, if x is in the positive sideI Negative, if x is in the negative side
12 / 85
We have something like this
We have then
13 / 85
NowSince g (xp) = 0We have that
g (x) = g
(xp + r
w
‖w‖
)= wT
(xp + r
w
‖w‖
)+ w0
= wT xp + w0 + rwT w
‖w‖
= g (xp) + r‖w‖2
‖w‖= r ‖w‖
Then, we have
r = g (x)‖w‖
(5)
14 / 85
NowSince g (xp) = 0We have that
g (x) = g
(xp + r
w
‖w‖
)= wT
(xp + r
w
‖w‖
)+ w0
= wT xp + w0 + rwT w
‖w‖
= g (xp) + r‖w‖2
‖w‖= r ‖w‖
Then, we have
r = g (x)‖w‖
(5)
14 / 85
NowSince g (xp) = 0We have that
g (x) = g
(xp + r
w
‖w‖
)= wT
(xp + r
w
‖w‖
)+ w0
= wT xp + w0 + rwT w
‖w‖
= g (xp) + r‖w‖2
‖w‖= r ‖w‖
Then, we have
r = g (x)‖w‖
(5)
14 / 85
NowSince g (xp) = 0We have that
g (x) = g
(xp + r
w
‖w‖
)= wT
(xp + r
w
‖w‖
)+ w0
= wT xp + w0 + rwT w
‖w‖
= g (xp) + r‖w‖2
‖w‖= r ‖w‖
Then, we have
r = g (x)‖w‖
(5)
14 / 85
NowSince g (xp) = 0We have that
g (x) = g
(xp + r
w
‖w‖
)= wT
(xp + r
w
‖w‖
)+ w0
= wT xp + w0 + rwT w
‖w‖
= g (xp) + r‖w‖2
‖w‖= r ‖w‖
Then, we have
r = g (x)‖w‖
(5)
14 / 85
NowSince g (xp) = 0We have that
g (x) = g
(xp + r
w
‖w‖
)= wT
(xp + r
w
‖w‖
)+ w0
= wT xp + w0 + rwT w
‖w‖
= g (xp) + r‖w‖2
‖w‖= r ‖w‖
Then, we have
r = g (x)‖w‖
(5)
14 / 85
In particular
The distance from the origin to H
r = g (0)‖w‖
= wT (0) + w0‖w‖
= w0‖w‖
(6)
RemarksIf w0 > 0, the origin is on the positive side of H.If w0 < 0, the origin is on the negative side of H.If w0 = 0, the hyperplane has the homogeneous form wT x andhyperplane passes through the origin.
15 / 85
In particular
The distance from the origin to H
r = g (0)‖w‖
= wT (0) + w0‖w‖
= w0‖w‖
(6)
RemarksIf w0 > 0, the origin is on the positive side of H.If w0 < 0, the origin is on the negative side of H.If w0 = 0, the hyperplane has the homogeneous form wT x andhyperplane passes through the origin.
15 / 85
In particular
The distance from the origin to H
r = g (0)‖w‖
= wT (0) + w0‖w‖
= w0‖w‖
(6)
RemarksIf w0 > 0, the origin is on the positive side of H.If w0 < 0, the origin is on the negative side of H.If w0 = 0, the hyperplane has the homogeneous form wT x andhyperplane passes through the origin.
15 / 85
In particular
The distance from the origin to H
r = g (0)‖w‖
= wT (0) + w0‖w‖
= w0‖w‖
(6)
RemarksIf w0 > 0, the origin is on the positive side of H.If w0 < 0, the origin is on the negative side of H.If w0 = 0, the hyperplane has the homogeneous form wT x andhyperplane passes through the origin.
15 / 85
In addition...
If we do the following
g (x) = w0 +d∑
i=1wixi =
d∑i=0
wixi (7)
By making
x0 = 1 and y =
1x1...xd
=
1
x
Wherey is called an augmented feature vector.
16 / 85
In addition...
If we do the following
g (x) = w0 +d∑
i=1wixi =
d∑i=0
wixi (7)
By making
x0 = 1 and y =
1x1...xd
=
1
x
Wherey is called an augmented feature vector.
16 / 85
In addition...
If we do the following
g (x) = w0 +d∑
i=1wixi =
d∑i=0
wixi (7)
By making
x0 = 1 and y =
1x1...xd
=
1
x
Wherey is called an augmented feature vector.
16 / 85
In a similar way
We have the augmented weight vector
waug =
w0w1...wd
=
w0
w
RemarksThe addition of a constant component to x preserves all the distancerelationship between samples.The resulting y vectors, all lie in a d-dimensional subspace which isthe x-space itself.
17 / 85
In a similar way
We have the augmented weight vector
waug =
w0w1...wd
=
w0
w
RemarksThe addition of a constant component to x preserves all the distancerelationship between samples.The resulting y vectors, all lie in a d-dimensional subspace which isthe x-space itself.
17 / 85
In a similar way
We have the augmented weight vector
waug =
w0w1...wd
=
w0
w
RemarksThe addition of a constant component to x preserves all the distancerelationship between samples.The resulting y vectors, all lie in a d-dimensional subspace which isthe x-space itself.
17 / 85
More Remarks
In additionThe hyperplane decision surface H defined by wT
augy = 0 passesthrough the origin in y-space.Even though the corresponding hyperplane H can be in any positionof the x-space.
The distance from y to H is |wTaugy|
‖waug‖ or |g(x)|‖waug‖ .
Since ‖waug‖ > ‖w‖This distance is less or at least equal to the distance from x to H.
This mapping is quite usefulBecause we only need to find a weight vector waug instead of finding theweight vector w and the threshold w0.
18 / 85
More Remarks
In additionThe hyperplane decision surface H defined by wT
augy = 0 passesthrough the origin in y-space.Even though the corresponding hyperplane H can be in any positionof the x-space.
The distance from y to H is |wTaugy|
‖waug‖ or |g(x)|‖waug‖ .
Since ‖waug‖ > ‖w‖This distance is less or at least equal to the distance from x to H.
This mapping is quite usefulBecause we only need to find a weight vector waug instead of finding theweight vector w and the threshold w0.
18 / 85
More Remarks
In additionThe hyperplane decision surface H defined by wT
augy = 0 passesthrough the origin in y-space.Even though the corresponding hyperplane H can be in any positionof the x-space.
The distance from y to H is |wTaugy|
‖waug‖ or |g(x)|‖waug‖ .
Since ‖waug‖ > ‖w‖This distance is less or at least equal to the distance from x to H.
This mapping is quite usefulBecause we only need to find a weight vector waug instead of finding theweight vector w and the threshold w0.
18 / 85
More Remarks
In additionThe hyperplane decision surface H defined by wT
augy = 0 passesthrough the origin in y-space.Even though the corresponding hyperplane H can be in any positionof the x-space.
The distance from y to H is |wTaugy|
‖waug‖ or |g(x)|‖waug‖ .
Since ‖waug‖ > ‖w‖This distance is less or at least equal to the distance from x to H.
This mapping is quite usefulBecause we only need to find a weight vector waug instead of finding theweight vector w and the threshold w0.
18 / 85
More Remarks
In additionThe hyperplane decision surface H defined by wT
augy = 0 passesthrough the origin in y-space.Even though the corresponding hyperplane H can be in any positionof the x-space.
The distance from y to H is |wTaugy|
‖waug‖ or |g(x)|‖waug‖ .
Since ‖waug‖ > ‖w‖This distance is less or at least equal to the distance from x to H.
This mapping is quite usefulBecause we only need to find a weight vector waug instead of finding theweight vector w and the threshold w0.
18 / 85
Outline
1 IntroductionThe Simplest FunctionsSplitting the SpaceThe Decision Surface
2 Developing an Initial SolutionGradient Descent Procedure
The Geometry of a Two-Category Linearly-Separable CaseBasic Method
Minimum Squared Error ProcedureThe Error IdeaThe Final Error EquationThe Data MatrixMulti-Class SolutionIssues with Least Squares!!!What about Numerical Stability?
19 / 85
Outline
1 IntroductionThe Simplest FunctionsSplitting the SpaceThe Decision Surface
2 Developing an Initial SolutionGradient Descent Procedure
The Geometry of a Two-Category Linearly-Separable CaseBasic Method
Minimum Squared Error ProcedureThe Error IdeaThe Final Error EquationThe Data MatrixMulti-Class SolutionIssues with Least Squares!!!What about Numerical Stability?
20 / 85
Initial Supposition
Suppose, we haven samples x1,x2, ...,xn some labeled ω1 and some labeled ω2.
We want a vector weight w such thatwT xi > 0, if xi ∈ ω1.wT xi < 0, if xi ∈ ω2.
We suggest the following normalizationWe replace all the samples xi ∈ ω2 by their negative vectors!!!
21 / 85
Initial Supposition
Suppose, we haven samples x1,x2, ...,xn some labeled ω1 and some labeled ω2.
We want a vector weight w such thatwT xi > 0, if xi ∈ ω1.wT xi < 0, if xi ∈ ω2.
We suggest the following normalizationWe replace all the samples xi ∈ ω2 by their negative vectors!!!
21 / 85
Initial Supposition
Suppose, we haven samples x1,x2, ...,xn some labeled ω1 and some labeled ω2.
We want a vector weight w such thatwT xi > 0, if xi ∈ ω1.wT xi < 0, if xi ∈ ω2.
We suggest the following normalizationWe replace all the samples xi ∈ ω2 by their negative vectors!!!
21 / 85
Initial Supposition
Suppose, we haven samples x1,x2, ...,xn some labeled ω1 and some labeled ω2.
We want a vector weight w such thatwT xi > 0, if xi ∈ ω1.wT xi < 0, if xi ∈ ω2.
We suggest the following normalizationWe replace all the samples xi ∈ ω2 by their negative vectors!!!
21 / 85
The Usefulness of the Normalization
Once the normalization is doneWe only need for a weight vector w such that wT xi > 0 for all thesamples.
The name of this weight vectorIt is called a separating vector or solution vector.
22 / 85
The Usefulness of the Normalization
Once the normalization is doneWe only need for a weight vector w such that wT xi > 0 for all thesamples.
The name of this weight vectorIt is called a separating vector or solution vector.
22 / 85
Here, we have the solution region for w
Do not confuse this region with the decision region!!!
separat
ing plan
e
solution space
Remark: w is not unique!!! We can have different w’s solving theproblem
23 / 85
Here, we have the solution region for w
Do not confuse this region with the decision region!!!
separat
ing plan
e
solution space
Remark: w is not unique!!! We can have different w’s solving theproblem
23 / 85
Here, we have the solution region for w undernormalizationDo not confuse this region with the decision region!!!
"separat
ing" pla
ne
solution space
Remark: w is not unique!!!24 / 85
Here, we have the solution region for w undernormalizationDo not confuse this region with the decision region!!!
"separat
ing" pla
ne
solution space
Remark: w is not unique!!!24 / 85
How do we get this w?
In order to be able to do thisWe need to impose constraints to the problem.
Possible constraints!!!To find a unit-length weight vector that maximizes the minimumdistance from the samples to the separating plane.To find the minimum-length weight vector satisfying wT xi ≥ b for alli where b is a constant called the margin.Here the solution space resulting from the intersections of thehalf-spaces such that wT xi ≥ b > 0 lies within the previous solutionspace!!!
25 / 85
How do we get this w?
In order to be able to do thisWe need to impose constraints to the problem.
Possible constraints!!!To find a unit-length weight vector that maximizes the minimumdistance from the samples to the separating plane.To find the minimum-length weight vector satisfying wT xi ≥ b for alli where b is a constant called the margin.Here the solution space resulting from the intersections of thehalf-spaces such that wT xi ≥ b > 0 lies within the previous solutionspace!!!
25 / 85
How do we get this w?
In order to be able to do thisWe need to impose constraints to the problem.
Possible constraints!!!To find a unit-length weight vector that maximizes the minimumdistance from the samples to the separating plane.To find the minimum-length weight vector satisfying wT xi ≥ b for alli where b is a constant called the margin.Here the solution space resulting from the intersections of thehalf-spaces such that wT xi ≥ b > 0 lies within the previous solutionspace!!!
25 / 85
How do we get this w?
In order to be able to do thisWe need to impose constraints to the problem.
Possible constraints!!!To find a unit-length weight vector that maximizes the minimumdistance from the samples to the separating plane.To find the minimum-length weight vector satisfying wT xi ≥ b for alli where b is a constant called the margin.Here the solution space resulting from the intersections of thehalf-spaces such that wT xi ≥ b > 0 lies within the previous solutionspace!!!
25 / 85
We have thenA new boundary by a distance b
‖xi‖
solution region
26 / 85
Outline
1 IntroductionThe Simplest FunctionsSplitting the SpaceThe Decision Surface
2 Developing an Initial SolutionGradient Descent Procedure
The Geometry of a Two-Category Linearly-Separable CaseBasic Method
Minimum Squared Error ProcedureThe Error IdeaThe Final Error EquationThe Data MatrixMulti-Class SolutionIssues with Least Squares!!!What about Numerical Stability?
27 / 85
Gradient Descent
For this, we will define a criterion function J (w)A classic optimization
The basic procedure is as follow1 Start with a random weight vector w (1).2 Compute the gradient vector ∇J (w (1)).3 Obtain value w (2) by moving from w (1) in the direction of the
steepest descent:1 i.e. along the negative of the gradient.2 By using the following equation:
w (k + 1) = w (k)− η (k)∇J (w (k)) (8)
28 / 85
Gradient Descent
For this, we will define a criterion function J (w)A classic optimization
The basic procedure is as follow1 Start with a random weight vector w (1).2 Compute the gradient vector ∇J (w (1)).3 Obtain value w (2) by moving from w (1) in the direction of the
steepest descent:1 i.e. along the negative of the gradient.2 By using the following equation:
w (k + 1) = w (k)− η (k)∇J (w (k)) (8)
28 / 85
Gradient Descent
For this, we will define a criterion function J (w)A classic optimization
The basic procedure is as follow1 Start with a random weight vector w (1).2 Compute the gradient vector ∇J (w (1)).3 Obtain value w (2) by moving from w (1) in the direction of the
steepest descent:1 i.e. along the negative of the gradient.2 By using the following equation:
w (k + 1) = w (k)− η (k)∇J (w (k)) (8)
28 / 85
Gradient Descent
For this, we will define a criterion function J (w)A classic optimization
The basic procedure is as follow1 Start with a random weight vector w (1).2 Compute the gradient vector ∇J (w (1)).3 Obtain value w (2) by moving from w (1) in the direction of the
steepest descent:1 i.e. along the negative of the gradient.2 By using the following equation:
w (k + 1) = w (k)− η (k)∇J (w (k)) (8)
28 / 85
Gradient Descent
For this, we will define a criterion function J (w)A classic optimization
The basic procedure is as follow1 Start with a random weight vector w (1).2 Compute the gradient vector ∇J (w (1)).3 Obtain value w (2) by moving from w (1) in the direction of the
steepest descent:1 i.e. along the negative of the gradient.2 By using the following equation:
w (k + 1) = w (k)− η (k)∇J (w (k)) (8)
28 / 85
Gradient Descent
For this, we will define a criterion function J (w)A classic optimization
The basic procedure is as follow1 Start with a random weight vector w (1).2 Compute the gradient vector ∇J (w (1)).3 Obtain value w (2) by moving from w (1) in the direction of the
steepest descent:1 i.e. along the negative of the gradient.2 By using the following equation:
w (k + 1) = w (k)− η (k)∇J (w (k)) (8)
28 / 85
What is η (k)?Hereη (k) is a positive scale factor or learning rate!!!
The basic algorithm looks like thisAlgorithm 1 (Basic gradient descent)
1 begin initialize w, criterion θ, η (·), k = 02 do k = k + 13 w = w − η (k)∇J (w)4 until η (k)∇J (w) < θ
5 return w
Problem!!! How to choose the learning rate?If η (k) is too small, convergence is quite slow!!!If η (k) is too large, correction will overshot and can even diverge!!!
29 / 85
What is η (k)?Hereη (k) is a positive scale factor or learning rate!!!
The basic algorithm looks like thisAlgorithm 1 (Basic gradient descent)
1 begin initialize w, criterion θ, η (·), k = 02 do k = k + 13 w = w − η (k)∇J (w)4 until η (k)∇J (w) < θ
5 return w
Problem!!! How to choose the learning rate?If η (k) is too small, convergence is quite slow!!!If η (k) is too large, correction will overshot and can even diverge!!!
29 / 85
What is η (k)?Hereη (k) is a positive scale factor or learning rate!!!
The basic algorithm looks like thisAlgorithm 1 (Basic gradient descent)
1 begin initialize w, criterion θ, η (·), k = 02 do k = k + 13 w = w − η (k)∇J (w)4 until η (k)∇J (w) < θ
5 return w
Problem!!! How to choose the learning rate?If η (k) is too small, convergence is quite slow!!!If η (k) is too large, correction will overshot and can even diverge!!!
29 / 85
What is η (k)?Hereη (k) is a positive scale factor or learning rate!!!
The basic algorithm looks like thisAlgorithm 1 (Basic gradient descent)
1 begin initialize w, criterion θ, η (·), k = 02 do k = k + 13 w = w − η (k)∇J (w)4 until η (k)∇J (w) < θ
5 return w
Problem!!! How to choose the learning rate?If η (k) is too small, convergence is quite slow!!!If η (k) is too large, correction will overshot and can even diverge!!!
29 / 85
What is η (k)?Hereη (k) is a positive scale factor or learning rate!!!
The basic algorithm looks like thisAlgorithm 1 (Basic gradient descent)
1 begin initialize w, criterion θ, η (·), k = 02 do k = k + 13 w = w − η (k)∇J (w)4 until η (k)∇J (w) < θ
5 return w
Problem!!! How to choose the learning rate?If η (k) is too small, convergence is quite slow!!!If η (k) is too large, correction will overshot and can even diverge!!!
29 / 85
What is η (k)?Hereη (k) is a positive scale factor or learning rate!!!
The basic algorithm looks like thisAlgorithm 1 (Basic gradient descent)
1 begin initialize w, criterion θ, η (·), k = 02 do k = k + 13 w = w − η (k)∇J (w)4 until η (k)∇J (w) < θ
5 return w
Problem!!! How to choose the learning rate?If η (k) is too small, convergence is quite slow!!!If η (k) is too large, correction will overshot and can even diverge!!!
29 / 85
What is η (k)?Hereη (k) is a positive scale factor or learning rate!!!
The basic algorithm looks like thisAlgorithm 1 (Basic gradient descent)
1 begin initialize w, criterion θ, η (·), k = 02 do k = k + 13 w = w − η (k)∇J (w)4 until η (k)∇J (w) < θ
5 return w
Problem!!! How to choose the learning rate?If η (k) is too small, convergence is quite slow!!!If η (k) is too large, correction will overshot and can even diverge!!!
29 / 85
What is η (k)?Hereη (k) is a positive scale factor or learning rate!!!
The basic algorithm looks like thisAlgorithm 1 (Basic gradient descent)
1 begin initialize w, criterion θ, η (·), k = 02 do k = k + 13 w = w − η (k)∇J (w)4 until η (k)∇J (w) < θ
5 return w
Problem!!! How to choose the learning rate?If η (k) is too small, convergence is quite slow!!!If η (k) is too large, correction will overshot and can even diverge!!!
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What is η (k)?Hereη (k) is a positive scale factor or learning rate!!!
The basic algorithm looks like thisAlgorithm 1 (Basic gradient descent)
1 begin initialize w, criterion θ, η (·), k = 02 do k = k + 13 w = w − η (k)∇J (w)4 until η (k)∇J (w) < θ
5 return w
Problem!!! How to choose the learning rate?If η (k) is too small, convergence is quite slow!!!If η (k) is too large, correction will overshot and can even diverge!!!
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Using the Taylor’s second-order expansion around valuew (k)
We do the following
J (w) = J (w (k)) +∇JT (w −w (k)) + 12 (w −w (k))T H (w −w (k)) (9)
Remark: This is know as Taylor’s Second Order expansion!!!
Here, we have∇J is the vector of partial derivatives ∂J
∂wievaluated at w (k).
H is the Hessian matrix of second partial derivatives ∂2J∂wi∂wj
evaluated at w (k).
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Using the Taylor’s second-order expansion around valuew (k)
We do the following
J (w) = J (w (k)) +∇JT (w −w (k)) + 12 (w −w (k))T H (w −w (k)) (9)
Remark: This is know as Taylor’s Second Order expansion!!!
Here, we have∇J is the vector of partial derivatives ∂J
∂wievaluated at w (k).
H is the Hessian matrix of second partial derivatives ∂2J∂wi∂wj
evaluated at w (k).
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Using the Taylor’s second-order expansion around valuew (k)
We do the following
J (w) = J (w (k)) +∇JT (w −w (k)) + 12 (w −w (k))T H (w −w (k)) (9)
Remark: This is know as Taylor’s Second Order expansion!!!
Here, we have∇J is the vector of partial derivatives ∂J
∂wievaluated at w (k).
H is the Hessian matrix of second partial derivatives ∂2J∂wi∂wj
evaluated at w (k).
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Using the Taylor’s second-order expansion around valuew (k)
We do the following
J (w) = J (w (k)) +∇JT (w −w (k)) + 12 (w −w (k))T H (w −w (k)) (9)
Remark: This is know as Taylor’s Second Order expansion!!!
Here, we have∇J is the vector of partial derivatives ∂J
∂wievaluated at w (k).
H is the Hessian matrix of second partial derivatives ∂2J∂wi∂wj
evaluated at w (k).
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Then
We substitute (Eq. 8) into (Eq. 9)
w (k + 1)−w (k) = η (k)∇J (w (k)) (10)
We have then
J (w (k + 1)) ∼=J (w (k)) +∇JT (−η (k)∇J (w (k))) + ...
12 (−η (k)∇J (w (k)))T H (−η (k)∇J (w (k)))
Finally, we have
J (w (k + 1)) ∼= J (w (k))− η (k) ‖∇J‖2 + 12η
2 (k)∇JT H∇J (11)
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Then
We substitute (Eq. 8) into (Eq. 9)
w (k + 1)−w (k) = η (k)∇J (w (k)) (10)
We have then
J (w (k + 1)) ∼=J (w (k)) +∇JT (−η (k)∇J (w (k))) + ...
12 (−η (k)∇J (w (k)))T H (−η (k)∇J (w (k)))
Finally, we have
J (w (k + 1)) ∼= J (w (k))− η (k) ‖∇J‖2 + 12η
2 (k)∇JT H∇J (11)
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Then
We substitute (Eq. 8) into (Eq. 9)
w (k + 1)−w (k) = η (k)∇J (w (k)) (10)
We have then
J (w (k + 1)) ∼=J (w (k)) +∇JT (−η (k)∇J (w (k))) + ...
12 (−η (k)∇J (w (k)))T H (−η (k)∇J (w (k)))
Finally, we have
J (w (k + 1)) ∼= J (w (k))− η (k) ‖∇J‖2 + 12η
2 (k)∇JT H∇J (11)
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Derive with respect to η (k) and make the result equal tozero
We have then
−‖∇J‖2 + η (k)∇JT H∇J = 0 (12)
Finally
η (k) = ‖∇J‖2
∇JT H∇J(13)
Remark This is the optimal step size!!!
Problem!!!Calculating H can be quite expansive!!!
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Derive with respect to η (k) and make the result equal tozero
We have then
−‖∇J‖2 + η (k)∇JT H∇J = 0 (12)
Finally
η (k) = ‖∇J‖2
∇JT H∇J(13)
Remark This is the optimal step size!!!
Problem!!!Calculating H can be quite expansive!!!
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Derive with respect to η (k) and make the result equal tozero
We have then
−‖∇J‖2 + η (k)∇JT H∇J = 0 (12)
Finally
η (k) = ‖∇J‖2
∇JT H∇J(13)
Remark This is the optimal step size!!!
Problem!!!Calculating H can be quite expansive!!!
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We can have an adaptive linear search!!!
We can use the idea of having everything fixed, but η (k)Then, we can have the following functionf (η (k)) = w (k)− η (k)∇J (w (k))
We can optimized using linear search methods
Linear Search MethodsBacktracking linear searchBisection methodGolden ratioEtc.
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We can have an adaptive linear search!!!
We can use the idea of having everything fixed, but η (k)Then, we can have the following functionf (η (k)) = w (k)− η (k)∇J (w (k))
We can optimized using linear search methods
Linear Search MethodsBacktracking linear searchBisection methodGolden ratioEtc.
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We can have an adaptive linear search!!!
We can use the idea of having everything fixed, but η (k)Then, we can have the following functionf (η (k)) = w (k)− η (k)∇J (w (k))
We can optimized using linear search methods
Linear Search MethodsBacktracking linear searchBisection methodGolden ratioEtc.
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We can have an adaptive linear search!!!
We can use the idea of having everything fixed, but η (k)Then, we can have the following functionf (η (k)) = w (k)− η (k)∇J (w (k))
We can optimized using linear search methods
Linear Search MethodsBacktracking linear searchBisection methodGolden ratioEtc.
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We can have an adaptive linear search!!!
We can use the idea of having everything fixed, but η (k)Then, we can have the following functionf (η (k)) = w (k)− η (k)∇J (w (k))
We can optimized using linear search methods
Linear Search MethodsBacktracking linear searchBisection methodGolden ratioEtc.
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We can have an adaptive linear search!!!
We can use the idea of having everything fixed, but η (k)Then, we can have the following functionf (η (k)) = w (k)− η (k)∇J (w (k))
We can optimized using linear search methods
Linear Search MethodsBacktracking linear searchBisection methodGolden ratioEtc.
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Example: Golden Ratio
Imagine that you have a linear function f : L→ R
Where: Chose a and b such that a+ba = a
b (The Golden Ratio).
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Example: Golden Ratio
Imagine that you have a linear function f : L→ R
Where: Chose a and b such that a+ba = a
b (The Golden Ratio).
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The process is as follow
Given f1, f2, f3, wheref1 = f (x1)f2 = f (x2)f3 = f (x3)
We have thenif f2 is smaller than f1 and f3 then the minimum lies in [x1, x3]
Now, we generate x4 with f4 = f (x4)In the largest subinterval!!! [x2, x3]
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The process is as follow
Given f1, f2, f3, wheref1 = f (x1)f2 = f (x2)f3 = f (x3)
We have thenif f2 is smaller than f1 and f3 then the minimum lies in [x1, x3]
Now, we generate x4 with f4 = f (x4)In the largest subinterval!!! [x2, x3]
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The process is as follow
Given f1, f2, f3, wheref1 = f (x1)f2 = f (x2)f3 = f (x3)
We have thenif f2 is smaller than f1 and f3 then the minimum lies in [x1, x3]
Now, we generate x4 with f4 = f (x4)In the largest subinterval!!! [x2, x3]
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Finally
Two casesIf f4a > f2 then the minimum lies between x1 and x4 and the newtriplet is x1, x2 and x4.If f4b < f2 then the minimum lies between x2 and x3 and the newtriplet is x2, x4 and x3.
ThenRepeat the procedure!!!
For more, please read the paper“SEQUENTIAL MINIMAX SEARCH FOR A MAXIMUM” by J. Kiefer
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Finally
Two casesIf f4a > f2 then the minimum lies between x1 and x4 and the newtriplet is x1, x2 and x4.If f4b < f2 then the minimum lies between x2 and x3 and the newtriplet is x2, x4 and x3.
ThenRepeat the procedure!!!
For more, please read the paper“SEQUENTIAL MINIMAX SEARCH FOR A MAXIMUM” by J. Kiefer
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Finally
Two casesIf f4a > f2 then the minimum lies between x1 and x4 and the newtriplet is x1, x2 and x4.If f4b < f2 then the minimum lies between x2 and x3 and the newtriplet is x2, x4 and x3.
ThenRepeat the procedure!!!
For more, please read the paper“SEQUENTIAL MINIMAX SEARCH FOR A MAXIMUM” by J. Kiefer
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Finally
Two casesIf f4a > f2 then the minimum lies between x1 and x4 and the newtriplet is x1, x2 and x4.If f4b < f2 then the minimum lies between x2 and x3 and the newtriplet is x2, x4 and x3.
ThenRepeat the procedure!!!
For more, please read the paper“SEQUENTIAL MINIMAX SEARCH FOR A MAXIMUM” by J. Kiefer
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We have another method...
Derive the second Taylor expansion by w
J (w) = J (w (k)) +∇JT (w −w (k)) + 12 (w −w (k))T H (w −w (k))
We get
∇J + Hw −Hw (k) = 0 (14)
Thus
Hw = Hw (k)−∇JH−1Hw = H−1Hw (k)−H−1∇J
w = w (k)−H−1∇J
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We have another method...
Derive the second Taylor expansion by w
J (w) = J (w (k)) +∇JT (w −w (k)) + 12 (w −w (k))T H (w −w (k))
We get
∇J + Hw −Hw (k) = 0 (14)
Thus
Hw = Hw (k)−∇JH−1Hw = H−1Hw (k)−H−1∇J
w = w (k)−H−1∇J
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We have another method...
Derive the second Taylor expansion by w
J (w) = J (w (k)) +∇JT (w −w (k)) + 12 (w −w (k))T H (w −w (k))
We get
∇J + Hw −Hw (k) = 0 (14)
Thus
Hw = Hw (k)−∇JH−1Hw = H−1Hw (k)−H−1∇J
w = w (k)−H−1∇J
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The Newton-Raphson Algorithm
We have the following algorithmAlgorithm 2 (Newton descent)
1 begin initialize w, criterion θ2 do k = k + 13 w = w −H−1∇J (w)4 until H−1∇J (w) < θ
5 return w
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The Newton-Raphson Algorithm
We have the following algorithmAlgorithm 2 (Newton descent)
1 begin initialize w, criterion θ2 do k = k + 13 w = w −H−1∇J (w)4 until H−1∇J (w) < θ
5 return w
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The Newton-Raphson Algorithm
We have the following algorithmAlgorithm 2 (Newton descent)
1 begin initialize w, criterion θ2 do k = k + 13 w = w −H−1∇J (w)4 until H−1∇J (w) < θ
5 return w
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The Newton-Raphson Algorithm
We have the following algorithmAlgorithm 2 (Newton descent)
1 begin initialize w, criterion θ2 do k = k + 13 w = w −H−1∇J (w)4 until H−1∇J (w) < θ
5 return w
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The Newton-Raphson Algorithm
We have the following algorithmAlgorithm 2 (Newton descent)
1 begin initialize w, criterion θ2 do k = k + 13 w = w −H−1∇J (w)4 until H−1∇J (w) < θ
5 return w
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The Newton-Raphson Algorithm
We have the following algorithmAlgorithm 2 (Newton descent)
1 begin initialize w, criterion θ2 do k = k + 13 w = w −H−1∇J (w)4 until H−1∇J (w) < θ
5 return w
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The Newton-Raphson Algorithm
We have the following algorithmAlgorithm 2 (Newton descent)
1 begin initialize w, criterion θ2 do k = k + 13 w = w −H−1∇J (w)4 until H−1∇J (w) < θ
5 return w
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Outline
1 IntroductionThe Simplest FunctionsSplitting the SpaceThe Decision Surface
2 Developing an Initial SolutionGradient Descent Procedure
The Geometry of a Two-Category Linearly-Separable CaseBasic Method
Minimum Squared Error ProcedureThe Error IdeaThe Final Error EquationThe Data MatrixMulti-Class SolutionIssues with Least Squares!!!What about Numerical Stability?
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Initial Setup
ImportantWe get away from our initial normalization of the samples!!!
Now, we are going to use the method know asMinimum Squared Error
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Initial Setup
ImportantWe get away from our initial normalization of the samples!!!
Now, we are going to use the method know asMinimum Squared Error
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Now, assume the following
Imagine that your problem has two classes ω1 and ω2 in R2
1 They are linearly separable!!!2 You require to label them.
We have a problem!!!Which is the problem?
We do not know the hyperplane!!!Thus, what distance each point has to the hyperplane?
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Now, assume the following
Imagine that your problem has two classes ω1 and ω2 in R2
1 They are linearly separable!!!2 You require to label them.
We have a problem!!!Which is the problem?
We do not know the hyperplane!!!Thus, what distance each point has to the hyperplane?
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Now, assume the following
Imagine that your problem has two classes ω1 and ω2 in R2
1 They are linearly separable!!!2 You require to label them.
We have a problem!!!Which is the problem?
We do not know the hyperplane!!!Thus, what distance each point has to the hyperplane?
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Now, assume the following
Imagine that your problem has two classes ω1 and ω2 in R2
1 They are linearly separable!!!2 You require to label them.
We have a problem!!!Which is the problem?
We do not know the hyperplane!!!Thus, what distance each point has to the hyperplane?
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A Simple Solution For Our Quandary
Label the Classesω1 =⇒ +1ω2 =⇒ −1
We produce the following labels1 if x ∈ ω1 then yideal = gideal (x) = +1.2 if x ∈ ω2 then yideal = gideal (x) = −1.
Remark: We have a problem with this labels!!!
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A Simple Solution For Our Quandary
Label the Classesω1 =⇒ +1ω2 =⇒ −1
We produce the following labels1 if x ∈ ω1 then yideal = gideal (x) = +1.2 if x ∈ ω2 then yideal = gideal (x) = −1.
Remark: We have a problem with this labels!!!
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A Simple Solution For Our Quandary
Label the Classesω1 =⇒ +1ω2 =⇒ −1
We produce the following labels1 if x ∈ ω1 then yideal = gideal (x) = +1.2 if x ∈ ω2 then yideal = gideal (x) = −1.
Remark: We have a problem with this labels!!!
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A Simple Solution For Our Quandary
Label the Classesω1 =⇒ +1ω2 =⇒ −1
We produce the following labels1 if x ∈ ω1 then yideal = gideal (x) = +1.2 if x ∈ ω2 then yideal = gideal (x) = −1.
Remark: We have a problem with this labels!!!
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Outline
1 IntroductionThe Simplest FunctionsSplitting the SpaceThe Decision Surface
2 Developing an Initial SolutionGradient Descent Procedure
The Geometry of a Two-Category Linearly-Separable CaseBasic Method
Minimum Squared Error ProcedureThe Error IdeaThe Final Error EquationThe Data MatrixMulti-Class SolutionIssues with Least Squares!!!What about Numerical Stability?
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Now, What?
Assume true function f is given by
ynoise = gnoise (x) = wT x + w0 + ε (15)
Where the εIt has a ε ∼ N
(µ, σ2)
Thus, we can do the following
ynoise = gnoise (x) = gideal (x) + ε (16)
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Now, What?
Assume true function f is given by
ynoise = gnoise (x) = wT x + w0 + ε (15)
Where the εIt has a ε ∼ N
(µ, σ2)
Thus, we can do the following
ynoise = gnoise (x) = gideal (x) + ε (16)
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Now, What?
Assume true function f is given by
ynoise = gnoise (x) = wT x + w0 + ε (15)
Where the εIt has a ε ∼ N
(µ, σ2)
Thus, we can do the following
ynoise = gnoise (x) = gideal (x) + ε (16)
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Thus, we have
What to do?
ε = ynoise − gideal (x) (17)
Graphically
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Thus, we haveWhat to do?
ε = ynoise − gideal (x) (17)
Graphically
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Outline
1 IntroductionThe Simplest FunctionsSplitting the SpaceThe Decision Surface
2 Developing an Initial SolutionGradient Descent Procedure
The Geometry of a Two-Category Linearly-Separable CaseBasic Method
Minimum Squared Error ProcedureThe Error IdeaThe Final Error EquationThe Data MatrixMulti-Class SolutionIssues with Least Squares!!!What about Numerical Stability?
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Sum Over All ErrorsWe can do the following
J (w) =N∑
i=1ε2i =
N∑i=1
(yi − gideal (x))2 (18)
Remark: Know as least squares (Fitting the vertical offset!!!)
GeneralizeIf
The dimensionality of each sample (data point) is d,You can extend each vector sample to be xT = (1,x′),We have:
N∑i=1
(yi − xT w
)2= (y −Xw)T (y −Xw) = ‖y −Xw‖22 (19)
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Sum Over All ErrorsWe can do the following
J (w) =N∑
i=1ε2i =
N∑i=1
(yi − gideal (x))2 (18)
Remark: Know as least squares (Fitting the vertical offset!!!)
GeneralizeIf
The dimensionality of each sample (data point) is d,You can extend each vector sample to be xT = (1,x′),We have:
N∑i=1
(yi − xT w
)2= (y −Xw)T (y −Xw) = ‖y −Xw‖22 (19)
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Sum Over All ErrorsWe can do the following
J (w) =N∑
i=1ε2i =
N∑i=1
(yi − gideal (x))2 (18)
Remark: Know as least squares (Fitting the vertical offset!!!)
GeneralizeIf
The dimensionality of each sample (data point) is d,You can extend each vector sample to be xT = (1,x′),We have:
N∑i=1
(yi − xT w
)2= (y −Xw)T (y −Xw) = ‖y −Xw‖22 (19)
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Sum Over All ErrorsWe can do the following
J (w) =N∑
i=1ε2i =
N∑i=1
(yi − gideal (x))2 (18)
Remark: Know as least squares (Fitting the vertical offset!!!)
GeneralizeIf
The dimensionality of each sample (data point) is d,You can extend each vector sample to be xT = (1,x′),We have:
N∑i=1
(yi − xT w
)2= (y −Xw)T (y −Xw) = ‖y −Xw‖22 (19)
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Outline
1 IntroductionThe Simplest FunctionsSplitting the SpaceThe Decision Surface
2 Developing an Initial SolutionGradient Descent Procedure
The Geometry of a Two-Category Linearly-Separable CaseBasic Method
Minimum Squared Error ProcedureThe Error IdeaThe Final Error EquationThe Data MatrixMulti-Class SolutionIssues with Least Squares!!!What about Numerical Stability?
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What is X
It is the Data Matrix
X =
1 (x1)1 · · · (x1)j · · · (x1)d...
......
1 (xi)1 (xi)j (xi)d...
......
1 (xN )1 · · · (xN )j · · · (xN )d
(20)
We know the followingdxTAx
dx= Ax+ATx,
dAx
dx= A (21)
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What is X
It is the Data Matrix
X =
1 (x1)1 · · · (x1)j · · · (x1)d...
......
1 (xi)1 (xi)j (xi)d...
......
1 (xN )1 · · · (xN )j · · · (xN )d
(20)
We know the followingdxTAx
dx= Ax+ATx,
dAx
dx= A (21)
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Note about other representations
We could have xT = (x1, x2, ..., xd, 1) thus
X =
(x1)1 · · · (x1)j · · · (x1)d 1...
......
(xi)1 (xi)j (xi)d 1...
......
(xN )1 · · · (xN )j · · · (xN )d 1
(22)
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We can expand our quadratic formula!!!
Thus
(y −Xw)T (y −Xw) = yT y−wT XT y−yT Xw + wT XT Xw (23)
Making Possible to have by deriving with respect to w and assumingthat XT X is invertible
w =(XT X
)−1XT y (24)
Note:XT X is always positive semi-definite. If it is also invertible, it ispositive definite.
Thus, How we get the discriminant function?Any Ideas?
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We can expand our quadratic formula!!!
Thus
(y −Xw)T (y −Xw) = yT y−wT XT y−yT Xw + wT XT Xw (23)
Making Possible to have by deriving with respect to w and assumingthat XT X is invertible
w =(XT X
)−1XT y (24)
Note:XT X is always positive semi-definite. If it is also invertible, it ispositive definite.
Thus, How we get the discriminant function?Any Ideas?
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We can expand our quadratic formula!!!
Thus
(y −Xw)T (y −Xw) = yT y−wT XT y−yT Xw + wT XT Xw (23)
Making Possible to have by deriving with respect to w and assumingthat XT X is invertible
w =(XT X
)−1XT y (24)
Note:XT X is always positive semi-definite. If it is also invertible, it ispositive definite.
Thus, How we get the discriminant function?Any Ideas?
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The Final Discriminant Function
Very Simple!!!
g(x) = xT w = xT(XT X
)−1XT y (25)
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Pseudo-inverse of a Matrix
DefinitionSuppose that A ∈ Rm×n and rank (A) = m. We call the matrix
A+ =(ATA
)−1AT
the pseudo inverse of A.
ReasonA+ inverts A on its image
What?If w ∈ image (A), then there is some v ∈ Rn such that w = Av. Hence:
A+w = A+Av =(ATA
)−1ATAv
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Pseudo-inverse of a Matrix
DefinitionSuppose that A ∈ Rm×n and rank (A) = m. We call the matrix
A+ =(ATA
)−1AT
the pseudo inverse of A.
ReasonA+ inverts A on its image
What?If w ∈ image (A), then there is some v ∈ Rn such that w = Av. Hence:
A+w = A+Av =(ATA
)−1ATAv
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Pseudo-inverse of a Matrix
DefinitionSuppose that A ∈ Rm×n and rank (A) = m. We call the matrix
A+ =(ATA
)−1AT
the pseudo inverse of A.
ReasonA+ inverts A on its image
What?If w ∈ image (A), then there is some v ∈ Rn such that w = Av. Hence:
A+w = A+Av =(ATA
)−1ATAv
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What lives where?
We haveX ∈ RN×(d+1)
Image (X) = span{
Xcol1 , ..., Xcol
d+1
}xi ∈ Rd
w ∈ Rd+1
Xcoli ,y ∈ RN
Basically y, the list of desired inputs the is being protected into
span{
Xcol1 , ..., Xcol
d+1
}(26)
by the projection operator X(XT X
)−1XT .
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What lives where?
We haveX ∈ RN×(d+1)
Image (X) = span{
Xcol1 , ..., Xcol
d+1
}xi ∈ Rd
w ∈ Rd+1
Xcoli ,y ∈ RN
Basically y, the list of desired inputs the is being protected into
span{
Xcol1 , ..., Xcol
d+1
}(26)
by the projection operator X(XT X
)−1XT .
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What lives where?
We haveX ∈ RN×(d+1)
Image (X) = span{
Xcol1 , ..., Xcol
d+1
}xi ∈ Rd
w ∈ Rd+1
Xcoli ,y ∈ RN
Basically y, the list of desired inputs the is being protected into
span{
Xcol1 , ..., Xcol
d+1
}(26)
by the projection operator X(XT X
)−1XT .
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What lives where?
We haveX ∈ RN×(d+1)
Image (X) = span{
Xcol1 , ..., Xcol
d+1
}xi ∈ Rd
w ∈ Rd+1
Xcoli ,y ∈ RN
Basically y, the list of desired inputs the is being protected into
span{
Xcol1 , ..., Xcol
d+1
}(26)
by the projection operator X(XT X
)−1XT .
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What lives where?
We haveX ∈ RN×(d+1)
Image (X) = span{
Xcol1 , ..., Xcol
d+1
}xi ∈ Rd
w ∈ Rd+1
Xcoli ,y ∈ RN
Basically y, the list of desired inputs the is being protected into
span{
Xcol1 , ..., Xcol
d+1
}(26)
by the projection operator X(XT X
)−1XT .
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What lives where?
We haveX ∈ RN×(d+1)
Image (X) = span{
Xcol1 , ..., Xcol
d+1
}xi ∈ Rd
w ∈ Rd+1
Xcoli ,y ∈ RN
Basically y, the list of desired inputs the is being protected into
span{
Xcol1 , ..., Xcol
d+1
}(26)
by the projection operator X(XT X
)−1XT .
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Geometric Interpretation
We have1 The image of the mapping w to Xw is a linear subspace of RN .2 As w runs through all points Rd+1, the function value Xw runs
through all points in the image spaceimage (X) = span
{Xcol
1 , ..., Xcold+1
}.
3 Each w defines one point Xw =∑d
j=0wjXcolj .
4 w is the point which minimizes the distance d (y, image (X)).
55 / 85
Geometric Interpretation
We have1 The image of the mapping w to Xw is a linear subspace of RN .2 As w runs through all points Rd+1, the function value Xw runs
through all points in the image spaceimage (X) = span
{Xcol
1 , ..., Xcold+1
}.
3 Each w defines one point Xw =∑d
j=0wjXcolj .
4 w is the point which minimizes the distance d (y, image (X)).
55 / 85
Geometric Interpretation
We have1 The image of the mapping w to Xw is a linear subspace of RN .2 As w runs through all points Rd+1, the function value Xw runs
through all points in the image spaceimage (X) = span
{Xcol
1 , ..., Xcold+1
}.
3 Each w defines one point Xw =∑d
j=0wjXcolj .
4 w is the point which minimizes the distance d (y, image (X)).
55 / 85
Geometric Interpretation
We have1 The image of the mapping w to Xw is a linear subspace of RN .2 As w runs through all points Rd+1, the function value Xw runs
through all points in the image spaceimage (X) = span
{Xcol
1 , ..., Xcold+1
}.
3 Each w defines one point Xw =∑d
j=0wjXcolj .
4 w is the point which minimizes the distance d (y, image (X)).
55 / 85
Geometrically
Ahhhh!!!
56 / 85
Outline
1 IntroductionThe Simplest FunctionsSplitting the SpaceThe Decision Surface
2 Developing an Initial SolutionGradient Descent Procedure
The Geometry of a Two-Category Linearly-Separable CaseBasic Method
Minimum Squared Error ProcedureThe Error IdeaThe Final Error EquationThe Data MatrixMulti-Class SolutionIssues with Least Squares!!!What about Numerical Stability?
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Multi-Class Solution
What to do?1 We might reduce the problem to c− 1 two-class problems.2 We might use c(c−1)
2 linear discriminants, one for every pair of classes.
However
58 / 85
Multi-Class Solution
What to do?1 We might reduce the problem to c− 1 two-class problems.2 We might use c(c−1)
2 linear discriminants, one for every pair of classes.
However
58 / 85
Multi-Class Solution
What to do?1 We might reduce the problem to c− 1 two-class problems.2 We might use c(c−1)
2 linear discriminants, one for every pair of classes.
However
58 / 85
What to do?
Define c linear discriminant functions
gi (x) = wT x + wi0 for i = 1, ..., c (27)
This is known as a linear machineRule: if gk (x) > gj (x) for all j 6= k =⇒ x ∈ ωk
Nice Properties (It can be proved!!!)1 Decision Regions are Singly Connected.2 Decision Regions are Convex.
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What to do?
Define c linear discriminant functions
gi (x) = wT x + wi0 for i = 1, ..., c (27)
This is known as a linear machineRule: if gk (x) > gj (x) for all j 6= k =⇒ x ∈ ωk
Nice Properties (It can be proved!!!)1 Decision Regions are Singly Connected.2 Decision Regions are Convex.
59 / 85
What to do?
Define c linear discriminant functions
gi (x) = wT x + wi0 for i = 1, ..., c (27)
This is known as a linear machineRule: if gk (x) > gj (x) for all j 6= k =⇒ x ∈ ωk
Nice Properties (It can be proved!!!)1 Decision Regions are Singly Connected.2 Decision Regions are Convex.
59 / 85
Proof of Properties
Proof
Actually quite simpleGiven
y = λxA + (1− λ) xB
with λ ∈ (0, 1).
60 / 85
Proof of Properties
Proof
Actually quite simpleGiven
y = λxA + (1− λ) xB
with λ ∈ (0, 1).
60 / 85
Proof of Properties
We know that
gk (y) = wT (λxA + (1− λ) xB) + w0
= λwT xA + λw0 + (1− λ) wT xB + (1− λ)w0
= λgk (xA) + (1− λ) gk (xA)> λgj (xA) + (1− λ) gj (xA)> gj (λxA + (1− λ) xB)> gj (y)
For all j 6= k
Or...y belongs to an area k defined by the rule!!!This area is Convex and Singly Connected because the definition ofy.
61 / 85
Proof of Properties
We know that
gk (y) = wT (λxA + (1− λ) xB) + w0
= λwT xA + λw0 + (1− λ) wT xB + (1− λ)w0
= λgk (xA) + (1− λ) gk (xA)> λgj (xA) + (1− λ) gj (xA)> gj (λxA + (1− λ) xB)> gj (y)
For all j 6= k
Or...y belongs to an area k defined by the rule!!!This area is Convex and Singly Connected because the definition ofy.
61 / 85
Proof of Properties
We know that
gk (y) = wT (λxA + (1− λ) xB) + w0
= λwT xA + λw0 + (1− λ) wT xB + (1− λ)w0
= λgk (xA) + (1− λ) gk (xA)> λgj (xA) + (1− λ) gj (xA)> gj (λxA + (1− λ) xB)> gj (y)
For all j 6= k
Or...y belongs to an area k defined by the rule!!!This area is Convex and Singly Connected because the definition ofy.
61 / 85
Proof of Properties
We know that
gk (y) = wT (λxA + (1− λ) xB) + w0
= λwT xA + λw0 + (1− λ) wT xB + (1− λ)w0
= λgk (xA) + (1− λ) gk (xA)> λgj (xA) + (1− λ) gj (xA)> gj (λxA + (1− λ) xB)> gj (y)
For all j 6= k
Or...y belongs to an area k defined by the rule!!!This area is Convex and Singly Connected because the definition ofy.
61 / 85
Proof of Properties
We know that
gk (y) = wT (λxA + (1− λ) xB) + w0
= λwT xA + λw0 + (1− λ) wT xB + (1− λ)w0
= λgk (xA) + (1− λ) gk (xA)> λgj (xA) + (1− λ) gj (xA)> gj (λxA + (1− λ) xB)> gj (y)
For all j 6= k
Or...y belongs to an area k defined by the rule!!!This area is Convex and Singly Connected because the definition ofy.
61 / 85
Proof of Properties
We know that
gk (y) = wT (λxA + (1− λ) xB) + w0
= λwT xA + λw0 + (1− λ) wT xB + (1− λ)w0
= λgk (xA) + (1− λ) gk (xA)> λgj (xA) + (1− λ) gj (xA)> gj (λxA + (1− λ) xB)> gj (y)
For all j 6= k
Or...y belongs to an area k defined by the rule!!!This area is Convex and Singly Connected because the definition ofy.
61 / 85
Proof of Properties
We know that
gk (y) = wT (λxA + (1− λ) xB) + w0
= λwT xA + λw0 + (1− λ) wT xB + (1− λ)w0
= λgk (xA) + (1− λ) gk (xA)> λgj (xA) + (1− λ) gj (xA)> gj (λxA + (1− λ) xB)> gj (y)
For all j 6= k
Or...y belongs to an area k defined by the rule!!!This area is Convex and Singly Connected because the definition ofy.
61 / 85
Proof of Properties
We know that
gk (y) = wT (λxA + (1− λ) xB) + w0
= λwT xA + λw0 + (1− λ) wT xB + (1− λ)w0
= λgk (xA) + (1− λ) gk (xA)> λgj (xA) + (1− λ) gj (xA)> gj (λxA + (1− λ) xB)> gj (y)
For all j 6= k
Or...y belongs to an area k defined by the rule!!!This area is Convex and Singly Connected because the definition ofy.
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However!!!
No so nice properties!!!It limits the power of classification for multi-objective function.
62 / 85
How do we train this Linear Machine?
We know that each ωk class is described bygk (x) = wT
k x + w0 where k = 1, ..., c
We then design a single machine
g (x) = W T x (28)
63 / 85
How do we train this Linear Machine?
We know that each ωk class is described bygk (x) = wT
k x + w0 where k = 1, ..., c
We then design a single machine
g (x) = W T x (28)
63 / 85
Where
We have the following
W T =
1 w11 w12 · · · w1d
1 w21 w22 · · · w2d
1 w31 w32 · · · w3d...
......
...1 wc1 wc2 · · · wcd
(29)
What about the labels?OK, we know how to do with 2 classes, What about many classes?
64 / 85
Where
We have the following
W T =
1 w11 w12 · · · w1d
1 w21 w22 · · · w2d
1 w31 w32 · · · w3d...
......
...1 wc1 wc2 · · · wcd
(29)
What about the labels?OK, we know how to do with 2 classes, What about many classes?
64 / 85
How do we train this Linear Machine?
Use a vector ti with dimensionality c to identify each element at eachclassWe have then the following dataset
{xi, ti} for i = 1, 2, ..., N
We build the following Matrix of Vectors
T =
tT
1tT
2...
tTN
(30)
65 / 85
How do we train this Linear Machine?
Use a vector ti with dimensionality c to identify each element at eachclassWe have then the following dataset
{xi, ti} for i = 1, 2, ..., N
We build the following Matrix of Vectors
T =
tT
1tT
2...
tTN
(30)
65 / 85
Thus, we create the following Matrix
A Matrix containing all the required information
XW − T (31)
Where we have the following vector[xT
i w1,xTi w2,x
Ti w3, ...,x
Ti wc
](32)
Remark: It is the vector result of multiplication of row i of X againstW on XW .
That is compared to the vector tTi on T by using the subtraction of
vectors
εi =[xT
i w1,xTi w2,x
Ti w3, ...,x
Ti wc
]− tT
i (33)
66 / 85
Thus, we create the following Matrix
A Matrix containing all the required information
XW − T (31)
Where we have the following vector[xT
i w1,xTi w2,x
Ti w3, ...,x
Ti wc
](32)
Remark: It is the vector result of multiplication of row i of X againstW on XW .
That is compared to the vector tTi on T by using the subtraction of
vectors
εi =[xT
i w1,xTi w2,x
Ti w3, ...,x
Ti wc
]− tT
i (33)
66 / 85
Thus, we create the following Matrix
A Matrix containing all the required information
XW − T (31)
Where we have the following vector[xT
i w1,xTi w2,x
Ti w3, ...,x
Ti wc
](32)
Remark: It is the vector result of multiplication of row i of X againstW on XW .
That is compared to the vector tTi on T by using the subtraction of
vectors
εi =[xT
i w1,xTi w2,x
Ti w3, ...,x
Ti wc
]− tT
i (33)
66 / 85
What do we want?
We want the quadratic error12ε
2i
This specific quadratic errors are at the diagonal of the matrix
(XW − T )T (XW − T )
We can use the trace function to generate the desired total error of
J (·) = 12
N∑i=1
ε2i (34)
67 / 85
What do we want?
We want the quadratic error12ε
2i
This specific quadratic errors are at the diagonal of the matrix
(XW − T )T (XW − T )
We can use the trace function to generate the desired total error of
J (·) = 12
N∑i=1
ε2i (34)
67 / 85
What do we want?
We want the quadratic error12ε
2i
This specific quadratic errors are at the diagonal of the matrix
(XW − T )T (XW − T )
We can use the trace function to generate the desired total error of
J (·) = 12
N∑i=1
ε2i (34)
67 / 85
Then
The trace allows to express the total error
J (W ) = 12Trace
{(XW − T )T (XW − T )
}(35)
Thus, we have by the same derivative method
W =(XT X
)XT T = X+T (36)
68 / 85
Then
The trace allows to express the total error
J (W ) = 12Trace
{(XW − T )T (XW − T )
}(35)
Thus, we have by the same derivative method
W =(XT X
)XT T = X+T (36)
68 / 85
How we train this Linear Machine?
Thus, we obtain the discriminant
g (x) = W T x = T T(X+
)x (37)
69 / 85
Outline
1 IntroductionThe Simplest FunctionsSplitting the SpaceThe Decision Surface
2 Developing an Initial SolutionGradient Descent Procedure
The Geometry of a Two-Category Linearly-Separable CaseBasic Method
Minimum Squared Error ProcedureThe Error IdeaThe Final Error EquationThe Data MatrixMulti-Class SolutionIssues with Least Squares!!!What about Numerical Stability?
70 / 85
Issues with Least SquaresRobustness
1 Least squares works only if X has full column rank, i.e. if XT X isinvertible.
2 If XT X almost not invertible, least squares is numerically unstable.1 Statistical consequence: High variance of predictions.
Not suited for high-dimensional data1 Modern problems: Many dimensions/features/predictors (possibly
thousands).2 Only a few of these may be important:
1 It needs some form of feature selection.2 Possible some type of regularization
Why?1 Treats all dimensions equally2 Relevant dimensions are averaged with irrelevant ones
71 / 85
Issues with Least SquaresRobustness
1 Least squares works only if X has full column rank, i.e. if XT X isinvertible.
2 If XT X almost not invertible, least squares is numerically unstable.1 Statistical consequence: High variance of predictions.
Not suited for high-dimensional data1 Modern problems: Many dimensions/features/predictors (possibly
thousands).2 Only a few of these may be important:
1 It needs some form of feature selection.2 Possible some type of regularization
Why?1 Treats all dimensions equally2 Relevant dimensions are averaged with irrelevant ones
71 / 85
Issues with Least SquaresRobustness
1 Least squares works only if X has full column rank, i.e. if XT X isinvertible.
2 If XT X almost not invertible, least squares is numerically unstable.1 Statistical consequence: High variance of predictions.
Not suited for high-dimensional data1 Modern problems: Many dimensions/features/predictors (possibly
thousands).2 Only a few of these may be important:
1 It needs some form of feature selection.2 Possible some type of regularization
Why?1 Treats all dimensions equally2 Relevant dimensions are averaged with irrelevant ones
71 / 85
Issues with Least SquaresRobustness
1 Least squares works only if X has full column rank, i.e. if XT X isinvertible.
2 If XT X almost not invertible, least squares is numerically unstable.1 Statistical consequence: High variance of predictions.
Not suited for high-dimensional data1 Modern problems: Many dimensions/features/predictors (possibly
thousands).2 Only a few of these may be important:
1 It needs some form of feature selection.2 Possible some type of regularization
Why?1 Treats all dimensions equally2 Relevant dimensions are averaged with irrelevant ones
71 / 85
Issues with Least SquaresRobustness
1 Least squares works only if X has full column rank, i.e. if XT X isinvertible.
2 If XT X almost not invertible, least squares is numerically unstable.1 Statistical consequence: High variance of predictions.
Not suited for high-dimensional data1 Modern problems: Many dimensions/features/predictors (possibly
thousands).2 Only a few of these may be important:
1 It needs some form of feature selection.2 Possible some type of regularization
Why?1 Treats all dimensions equally2 Relevant dimensions are averaged with irrelevant ones
71 / 85
Issues with Least SquaresRobustness
1 Least squares works only if X has full column rank, i.e. if XT X isinvertible.
2 If XT X almost not invertible, least squares is numerically unstable.1 Statistical consequence: High variance of predictions.
Not suited for high-dimensional data1 Modern problems: Many dimensions/features/predictors (possibly
thousands).2 Only a few of these may be important:
1 It needs some form of feature selection.2 Possible some type of regularization
Why?1 Treats all dimensions equally2 Relevant dimensions are averaged with irrelevant ones
71 / 85
Issues with Least SquaresRobustness
1 Least squares works only if X has full column rank, i.e. if XT X isinvertible.
2 If XT X almost not invertible, least squares is numerically unstable.1 Statistical consequence: High variance of predictions.
Not suited for high-dimensional data1 Modern problems: Many dimensions/features/predictors (possibly
thousands).2 Only a few of these may be important:
1 It needs some form of feature selection.2 Possible some type of regularization
Why?1 Treats all dimensions equally2 Relevant dimensions are averaged with irrelevant ones
71 / 85
Issues with Least SquaresRobustness
1 Least squares works only if X has full column rank, i.e. if XT X isinvertible.
2 If XT X almost not invertible, least squares is numerically unstable.1 Statistical consequence: High variance of predictions.
Not suited for high-dimensional data1 Modern problems: Many dimensions/features/predictors (possibly
thousands).2 Only a few of these may be important:
1 It needs some form of feature selection.2 Possible some type of regularization
Why?1 Treats all dimensions equally2 Relevant dimensions are averaged with irrelevant ones
71 / 85
Issues with Least SquaresRobustness
1 Least squares works only if X has full column rank, i.e. if XT X isinvertible.
2 If XT X almost not invertible, least squares is numerically unstable.1 Statistical consequence: High variance of predictions.
Not suited for high-dimensional data1 Modern problems: Many dimensions/features/predictors (possibly
thousands).2 Only a few of these may be important:
1 It needs some form of feature selection.2 Possible some type of regularization
Why?1 Treats all dimensions equally2 Relevant dimensions are averaged with irrelevant ones
71 / 85
Issues with Least Squares
Problem with OutliersNo Outliers Outliers
72 / 85
Issues with Least Squares
What about the Linear Machine?Please, run the algorithm and tell me...
73 / 85
What to Do About Numerical Stability?
RegularityA matrix which is not invertible is also called a singular matrix. A matrixwhich is invertible (not singular) is called regular.
In computationsIntuitions:
1 A singular matrix maps an entire linear subspace into a single point.2 If a matrix maps points far away from each other to points very close
to each other, it almost behaves like a singular matrix.
Mapping is related to the eigenvalues!!!Large positive eigenvalues ⇒ the mapping is large!!!Small positive eigenvalues ⇒ the mapping is small!!!
74 / 85
What to Do About Numerical Stability?
RegularityA matrix which is not invertible is also called a singular matrix. A matrixwhich is invertible (not singular) is called regular.
In computationsIntuitions:
1 A singular matrix maps an entire linear subspace into a single point.2 If a matrix maps points far away from each other to points very close
to each other, it almost behaves like a singular matrix.
Mapping is related to the eigenvalues!!!Large positive eigenvalues ⇒ the mapping is large!!!Small positive eigenvalues ⇒ the mapping is small!!!
74 / 85
What to Do About Numerical Stability?
RegularityA matrix which is not invertible is also called a singular matrix. A matrixwhich is invertible (not singular) is called regular.
In computationsIntuitions:
1 A singular matrix maps an entire linear subspace into a single point.2 If a matrix maps points far away from each other to points very close
to each other, it almost behaves like a singular matrix.
Mapping is related to the eigenvalues!!!Large positive eigenvalues ⇒ the mapping is large!!!Small positive eigenvalues ⇒ the mapping is small!!!
74 / 85
What to Do About Numerical Stability?
RegularityA matrix which is not invertible is also called a singular matrix. A matrixwhich is invertible (not singular) is called regular.
In computationsIntuitions:
1 A singular matrix maps an entire linear subspace into a single point.2 If a matrix maps points far away from each other to points very close
to each other, it almost behaves like a singular matrix.
Mapping is related to the eigenvalues!!!Large positive eigenvalues ⇒ the mapping is large!!!Small positive eigenvalues ⇒ the mapping is small!!!
74 / 85
What to Do About Numerical Stability?
RegularityA matrix which is not invertible is also called a singular matrix. A matrixwhich is invertible (not singular) is called regular.
In computationsIntuitions:
1 A singular matrix maps an entire linear subspace into a single point.2 If a matrix maps points far away from each other to points very close
to each other, it almost behaves like a singular matrix.
Mapping is related to the eigenvalues!!!Large positive eigenvalues ⇒ the mapping is large!!!Small positive eigenvalues ⇒ the mapping is small!!!
74 / 85
Outline
1 IntroductionThe Simplest FunctionsSplitting the SpaceThe Decision Surface
2 Developing an Initial SolutionGradient Descent Procedure
The Geometry of a Two-Category Linearly-Separable CaseBasic Method
Minimum Squared Error ProcedureThe Error IdeaThe Final Error EquationThe Data MatrixMulti-Class SolutionIssues with Least Squares!!!What about Numerical Stability?
75 / 85
What to Do About Numerical Stability?
All this comes from the following statementA positive semi-definite matrix A is singular ⇐⇒ smallest eigenvalue is 0
Consequence for StatisticsIf a statistical prediction involves the inverse of an almost-singular matrix,the predictions become unreliable (high variance).
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What to Do About Numerical Stability?
All this comes from the following statementA positive semi-definite matrix A is singular ⇐⇒ smallest eigenvalue is 0
Consequence for StatisticsIf a statistical prediction involves the inverse of an almost-singular matrix,the predictions become unreliable (high variance).
76 / 85
What can be done?Ridge RegressionRidge regression is a modification of least squares. It tries to make leastsquares more robust if XT X is almost singular.
The solution
wRidge =(XT X + λI
)−1XT y (38)
where λ is a tunning parameter
Thus, we can do the following given that XT X is positive definiteAssume that ξ1, ξ2, ..., ξd+1 are eigenvectors of XT X with eigenvaluesλ1, λ2, ..., λd+1: (
XT X + λI)ξi = (λi + λ) ξi (39)
i.e. λi + λ is an eigenvalue for(XT X + λI
)77 / 85
What can be done?Ridge RegressionRidge regression is a modification of least squares. It tries to make leastsquares more robust if XT X is almost singular.
The solution
wRidge =(XT X + λI
)−1XT y (38)
where λ is a tunning parameter
Thus, we can do the following given that XT X is positive definiteAssume that ξ1, ξ2, ..., ξd+1 are eigenvectors of XT X with eigenvaluesλ1, λ2, ..., λd+1: (
XT X + λI)ξi = (λi + λ) ξi (39)
i.e. λi + λ is an eigenvalue for(XT X + λI
)77 / 85
What can be done?Ridge RegressionRidge regression is a modification of least squares. It tries to make leastsquares more robust if XT X is almost singular.
The solution
wRidge =(XT X + λI
)−1XT y (38)
where λ is a tunning parameter
Thus, we can do the following given that XT X is positive definiteAssume that ξ1, ξ2, ..., ξd+1 are eigenvectors of XT X with eigenvaluesλ1, λ2, ..., λd+1: (
XT X + λI)ξi = (λi + λ) ξi (39)
i.e. λi + λ is an eigenvalue for(XT X + λI
)77 / 85
What does this mean?
Something NotableYou can control the singularity by detecting the smallest eigenvalue.
ThusWe add an appropriate tunning value λ.
78 / 85
What does this mean?
Something NotableYou can control the singularity by detecting the smallest eigenvalue.
ThusWe add an appropriate tunning value λ.
78 / 85
Thus, what we need to do?
Process1 Find the eigenvalues of XT X
2 If all of them are bigger than zero we are fine!!!3 Find the smallest one, then tune if necessary.4 Build wRidge =
(XT X + λI
)−1XT y.
79 / 85
Thus, what we need to do?
Process1 Find the eigenvalues of XT X
2 If all of them are bigger than zero we are fine!!!3 Find the smallest one, then tune if necessary.4 Build wRidge =
(XT X + λI
)−1XT y.
79 / 85
Thus, what we need to do?
Process1 Find the eigenvalues of XT X
2 If all of them are bigger than zero we are fine!!!3 Find the smallest one, then tune if necessary.4 Build wRidge =
(XT X + λI
)−1XT y.
79 / 85
Thus, what we need to do?
Process1 Find the eigenvalues of XT X
2 If all of them are bigger than zero we are fine!!!3 Find the smallest one, then tune if necessary.4 Build wRidge =
(XT X + λI
)−1XT y.
79 / 85
What about Thousands of Features?
There is a technique for thatLeast Absolute Shrinkage and Selection Operator (LASSO) invented byRobert Tibshirani that uses L1 =
∑di=1 |wi|.
The Least Squared Error takes the form ofN∑
i=1
(yi − xT w
)2+
d∑i=1|wi| (40)
However
You have other regularizations as L2 =√∑d
i=1 |wi|2
80 / 85
What about Thousands of Features?
There is a technique for thatLeast Absolute Shrinkage and Selection Operator (LASSO) invented byRobert Tibshirani that uses L1 =
∑di=1 |wi|.
The Least Squared Error takes the form ofN∑
i=1
(yi − xT w
)2+
d∑i=1|wi| (40)
However
You have other regularizations as L2 =√∑d
i=1 |wi|2
80 / 85
What about Thousands of Features?
There is a technique for thatLeast Absolute Shrinkage and Selection Operator (LASSO) invented byRobert Tibshirani that uses L1 =
∑di=1 |wi|.
The Least Squared Error takes the form ofN∑
i=1
(yi − xT w
)2+
d∑i=1|wi| (40)
However
You have other regularizations as L2 =√∑d
i=1 |wi|2
80 / 85
Graphically
The first area correspond to the L1 regularization and the second one?
81 / 85
Graphically
Yes the circle defined as L2 =√∑d
i=1 |wi|2
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The seminal paper by Robert Tibshirani
An initial study of this regularization can be seen in“Regression Shrinkage and Selection via the LASSO” by Robert Tibshirani
- 1996
83 / 85
This out the scope of this class
However, it is worth noticing that the most efficient method forsolving LASSO problems is“Pathwise Coordinate Optimization” By Jerome Friedman, Trevor Hastie,
Holger Ho and Robert Tibshirani
NeverthelessIt will be a great seminar paper!!!
84 / 85
This out the scope of this class
However, it is worth noticing that the most efficient method forsolving LASSO problems is“Pathwise Coordinate Optimization” By Jerome Friedman, Trevor Hastie,
Holger Ho and Robert Tibshirani
NeverthelessIt will be a great seminar paper!!!
84 / 85
ExercisesDuda and HartChapter 5
1, 3, 4, 7, 13, 17
BishopChapter 4
4.1, 4.4, 4.7,
TheodoridisChapter 3 - Problems
Using python 3.6Chapter 3 - Computer Experiments
Using python 3.1Using python and Newton 3.2
85 / 85
ExercisesDuda and HartChapter 5
1, 3, 4, 7, 13, 17
BishopChapter 4
4.1, 4.4, 4.7,
TheodoridisChapter 3 - Problems
Using python 3.6Chapter 3 - Computer Experiments
Using python 3.1Using python and Newton 3.2
85 / 85
ExercisesDuda and HartChapter 5
1, 3, 4, 7, 13, 17
BishopChapter 4
4.1, 4.4, 4.7,
TheodoridisChapter 3 - Problems
Using python 3.6Chapter 3 - Computer Experiments
Using python 3.1Using python and Newton 3.2
85 / 85