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Transcript of Boosting and other Expert Fusion Strategies. References Chapter 9.5 Duda Hart & Stock Leo Breiman...
Boosting and other Expert Fusion Strategies
References
• http://www.boosting.org• Chapter 9.5 Duda Hart & Stock• Leo Breiman Boosting Bagging
Arcing• Presentation adapted from:
Rishi Sinha, Robin Dhamankar
Types of Multiple Experts
• Single expert on full observation space
• Single expert for sub regions of observation space (Trees)
• Multiple experts on full observation space
• Multiple experts on sub regions of observation space
Types of Multiple Experts Training
• Use full observation space for each expert
• Use different observation features for each expert
• Use different observations for each expert
• Combine the above
Online Experts Selection
• N strategies (experts)
• At time t:– Learner A chooses a distribution over N
experts
– Let pt(i) be the probability of i-th expert
– pt(i) = 1 and for a loss vector lt
Loss at time t: pt(i) lt(i)
• Assume bounded loss, lt(i) in [0,1]
Experts Algorithm: Greedy
• For each expert define its cumulative loss:
• Greedy: At time t choose the expert with minimum loss, namely, arg min Li
t
t
j
ji
t
ilL
1
Greedy Analysis
• Theorem: Let LGT be the loss of
Greedy at time T, then
• Proof in notes.• Weakness: Relies on a single
expert for every observation
)1min( Ti
i
TG LNL
Better Multiple Experts Algorithms
• Would like to bound
• Better Bound: Hedge AlgorithmUtilizes all experts for each observation
Ti
i
TA LL min
Multiple Experts Algorithm: Hedge
• Maintain weight vector at time t: wt
• Probabilities pt(k) = wt(k) / wt(j)• Initialization w1(i) = 1/N• Updates: wt+1(k) = wt(k) Ub(lt(k)) where b in [0,1] and br < Ub (r) < 1-(1-b)r
Hedge Analysis
• Lemma: For any sequence of losses
• Proof (Mansour’s scribe)• Corollary:
H
N
j
T Lbjw )1())(ln(1
1
b
jw
H
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T
L
1
))(ln(1
1
Hedge: Properties
• Bounding the weights
• Similarly for a subset of experts.
Ti
T
t
t
Lil
T
t
tb
T
biwbiw
ilUiwiw
)()(
))(()()(
1)(
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Hedge: Performance
• Let k be with minimal loss
• Therefore
TkLT
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1
1
bbLN
b
bN
H
Tk
TkL
L
1)/1ln()ln(
1
)1
ln(
Hedge: Optimizing b
• For b=1/2 we have
• Better selection of b:
)2ln(2)ln(2 TkH LNL
)ln(ln2min NNLLL iiH
Occam Razor
• Finding the shortest consistent hypothesis.
• Definition: ()-Occam algorithm– >0 and <1– Input: a sample S of size m– Output: hypothesis h– for every (x,b) in S: h(x)=b– size(h) < size(ct) m
• Efficiency.
Occam Razor Theorem
• A: (,)-Occam algorithm for C using H• D distribution over inputs X• ct in C the target function• Sample size:
• with probability 1- A(S)=h has error(h) <
)1(121
ln1
2
n
m
Occam Razor Theorem
• Use the bound for finite hypothesis class.• Effective hypothesis class size 2size(h)
• size(h) < n m
• Sample size:
1ln
12lnln
1 2 mn
mmn
Weak and Strong Learning
PAC Learning model (Strong Learning)
• There exists a distribution D over domain X
• Examples: <x, c(x)>– use c for target function (rather than ct)
• Goal: – With high probability (1-)– find h in H such that – error(h,c ) < – arbitrarily small, thus STRONG LEARNING
Weak Learning Model
• Goal: error(h,c) < ½ - (Slightly above chance)• The parameter is small
– constantIntuitively: A much easier task
• Question:– Assume C is weak learnable, – C is PAC (strong) learnable
Majority Algorithm
• Hypothesis: hM(x)= MAJ[ h1(x), ... , hT(x) ]
• size(hM) < T size(ht)
• Using Occam Razor
Majority: outline
• Sample m example• Start with a distribution 1/m per
example.
• Modify the distribution and get ht
• Hypothesis is the majority • Terminate when perfect classification
– of the sample
Majority: Algorithm
• Use the Hedge algorithm.• The “experts” will be associate with
points.• Loss would be a correct classification.
– lt(i)= 1 - | ht(xi) – c(xi) |
• Setting b= 1- • hM(x) = MAJORITY( hi(x))
• Q: How do we set T?
Majority: Analysis
• Consider the set of errors SS={i | hM(xi)c(xi) }
• For every i in S:Li / T < ½ (Proof!)
• From Hedge properties:
2/)())(ln( 2 TxD
MSi iL
MAJORITY: Correctness
• Error Probability:
• Number of Rounds:
• Terminate when error less than 1/m
Si ixD )(
2/2 Te
2
ln2
m
T
Bagging• Generate a random sample from training set by selecting
elements with replacement.
• Repeat this sampling procedure, getting a sequence of k “independent” training sets
• A corresponding sequence of classifiers C1,C2,…,Ck is constructed for each of these training sets, by using the same classification algorithm
• To classify an unknown sample X, let each classifier predict.
• The Bagged Classifier C* then combines the predictions of the individual classifiers to generate the final outcome. (sometimes combination is simple voting)
Taken from Lecture slides for Data Mining Concepts and Techniques by Jiawei Han and M Kamber
Boosting
• Also Ensemble Method. =>The final prediction is a combination of the prediction of several predictors.
• What is different?– Its iterative. – Boosting: Successive classifiers depends upon its
predecessors. Previous methods : Individual classifiers were “independent”
– Training Examples may have unequal weights.– Look at errors from previous classifier step to decide how
to focus on next iteration over data– Set weights to focus more on ‘hard’ examples. (the ones
on which we committed mistakes in the previous iterations)
Boosting
• W(x) is the distribution of weights over N training observations ∑ W(xi)=1
• Initially assign uniform weights W0(x) = 1/N for
all x, step k=0
• At each iteration k :– Find best weak classifier Ck(x) using weights Wk(x)
– With error rate εk and based on a loss function:
• weight αk the classifier Ck‘s weight in the final hypothesis
• For each xi , update weights based on εk to get Wk+1(xi )
• CFINAL(x) =sign [ ∑ αi Ci (x) ]
Boosting (Algorithm)
Boosting As Additive Model
• The final prediction in boosting f(x) can be expressed as an additive expansion of individual classifiers
• The process is iterative and can be expressed as follows.
• Typically we would try to minimize a loss function on the training examples
);()(1
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Boosting As Additive Model
• Simple case: Squared-error loss
• Forward stage-wise modeling amounts to just fitting the residuals from previous iteration.
• Squared-error loss not robust for classification
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Boosting As Additive Model
• AdaBoost for Classification: L(y, f (x)) = exp(-y ∙ f (x)) - the exponential loss function
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Boosting As Additive Model
First assume that β is constant, and minimize w.r.t. G:
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Boosting As Additive Model
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errm : It is the training error on the weighted samples
The last equation tells us that in each iteration we must find a classifier that minimizes the training error on the weighted samples.
Boosting As Additive Model
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Now that we have found G, we minimize w.r.t. β:
Boosting (Recall)
• W(x) is the distribution of weights over the N training observations ∑ W(xi)=1
• Initially assign uniform weights W0(x) = 1/N for
all x, step k=0
• At each iteration k :– Find best weak classifier Ck(x) using weights Wk(x)
– With error rate εk and based on a loss function:
• weight αk the classifier Ck‘s weight in the final hypothesis
• For each xi , update weights based on εk to get Wk+1(xi )
• CFINAL(x) =sign [ ∑ αi Ci (x) ]
AdaBoost
• W(x) is the distribution of weights over the N training points ∑ W(xi)=1
• Initially assign uniform weights W0(x) = 1/N for all x.
• At each iteration k :– Find best weak classifier Ck(x) using weights Wk(x)
– Compute εk the error rate as
εk= [ ∑ W(xi ) ∙ I(yi ≠ Ck(xi )) ] / [ ∑ W(xi )]
– weight αk the classifier Ck‘s weight in the final hypothesis Set αk = log ((1 – εk )/εk )
– For each xi , Wk+1(xi ) = Wk(xi ) ∙ exp[αk ∙ I(yi ≠ Ck(xi ))]
• CFINAL(x) =sign [ ∑ αi Ci (x) ]
AdaBoost(Example)
Original Training set : Equal Weights to all training samples
Taken from “A Tutorial on Boosting” by Yoav Freund and Rob Schapire
AdaBoost (Example)
ROUND 1
AdaBoost (Example)
ROUND 2
AdaBoost (Example)
ROUND 3
AdaBoost (Example)
AdaBoost (Characteristics)• Why exponential loss function?
– Computational• Simple modular re-weighting• Derivative easy so determining optimal
parameters is relatively easy
– Statistical • In a two label case it determines one half
the log odds of P(Y=1|x) => We can use the sign as the classification rule
• Accuracy depends upon number of iterations ( How sensitive.. we will see soon).
Boosting performance
Decision stumps are very simple rules of thumb that test condition on a single attribute.
Decision stumps formed the individual classifiers whose predictions were combined to generate the final prediction.
The misclassification rate of the Boosting algorithm was plotted against the number of iterations performed.
Boosting performance
Steep decrease in error
Boosting performance
• Pondering over how many iterations would be sufficient….
• Observations– First few ( about 50) iterations increase
the accuracy substantially.. Seen by the steep decrease in misclassification rate.
– As iterations increase training error decreases ? and generalization error decreases ?
Can Boosting do well if?
• Limited training data?– Probably not ..
• Many missing values ?• Noise in the data ?• Individual classifiers not very
accurate ?– It cud if the individual classifiers have
considerable mutual disagreement.
Adaboost
• “Probably one of the three most influential ideas in machine learning in the last decade, along with Kernel methods and Variational approximations.”
• Original idea came from Valiant• Motivation: We want to improve the
performance of a weak learning algorithm
Adaboost• Algorithm:
Boosting Trees Outline
• Basics of boosting trees.• A numerical optimization problem• Control the model complexity,
generalization– Size of trees– Number of Iterations – Regularization
• Interpret the final model– Single variable– Correlation of variables
Boosting Trees : Basics
• Formally a tree is
• The parameters found by minimizing the empirical risk.
• Finding: j given R j : typically mean of yi in Rj
– Rj : Is tough but solutions exist.
JjjR 1},{
)(),(1
j
J
jj RxIxT
J
j Rxji
ji
yL1
),(minargˆ
Basics Continued …
• Approximate criterion for optimizing
• Boosted tree model is sum of such trees induced in a forward stage wise manner
• In case of binary classification and exponential loss functions this reduces to Ada Boost
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minarg~
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Numerical Optimization
• Loss Function is
• So the problem boils down to finding
• Which in optimization procedures are solved as
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Numerical Optimization Methods
• Steepest Descent
• Loss on Training Data converges to 0.
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Generalization
• Gradient Boosting– We want the algorithm to generalize. – Gradient on the other hand is defined only
on the training data points.– So fit the tree T to the negative gradient
values by least squares.
• MART – Multiple additive regression trees
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Tuning the Parameters
• The parameters that can be tuned are– The size of constituent trees J.– The number of boosting iterations M.– Shrinkage– Penalized Regression
Right-sized trees
• The optimal for one step might not be the optimal for the algorithm– Using very large tree (such as C4.5) as
weak learner to fit the residue assumes each tree is the last one in the expansion. Usually degrade performance and increase computation
• Solution : restrict the value of J to be the same for all trees.
Right sized trees.
• For trees the higher order interactions effects present in large trees suffer inaccuracies.
• J is the factor that helps control the higher order interactions. Thus we would like to keep J low.
• In practice the value of 4 J 8 is seen to have worked the best.
Controlling M (Regularization)
• After each iteration the training risk L(fM).
• As M , L(fM) 0• But this would risk over fitting the
training data.• To avoid this monitor prediction risk
on a validation sample.• Other methods in the Next Chapter.
Shrinkage
• Scale the contribution of each tree by a factor 0 < < 1 to control the learning rate.
and M control the prediction risk on training data.
• Are not independent of each other.
J
jjmjmmm RxIxff
1
^
1 )(*)(
Experts: Motivation
• Given a set of experts– No prior information– No consistent behavior– Goal: Predict as the best expert
• Model– online model– Input: historical results.
Expert: Goal
• Match the loss of best expert.• Loss:
– LA
– Li
• Can we hope to do better?
Example: Guessing letters
• Setting:– Alphabet of k letters
• Loss:– 1 incorrect guess– 0 correct guess
• Experts:– Each expert guesses a certain letter always.
• Game: guess the most popular letter online.
Example 2: Rock-Paper-Scissors
• Two player game.• Each player chooses: Rock, Paper, or Scissors.• Loss Matrix:
• Goal: Play as best as we can given the opponent.
Rock Paper Scissors
Rock 1/2 1 0
Paper 0 1/2 1
Scissors
1 0 1/2
Example 3: Placing a point
• Action: choosing a point d.• Loss (give the true location y): ||d-y||.• Experts: One for each point.• Important: Loss is Convex
• Goal: Find a “center”
||||)1(||||))1(( 2121 ydydydd
Adaboost
• Line 1: Given input space X and training examples x1,…xm and label space Y = {-1,1}
• Line 2: Initialize a distribution D to 1/m where m is the number of instances in the input space.
• Line 3: for( int t=0;t<T;t++)• Line 4: Train weak learning algorithm using Dt
• Line 5: Get a weak hypothesis ht which maps the input space to the label space. The error of this hypothesis is εt
• Line 6: αt = (1/2)ln((1- εt)/ εt)• Line 7: Dt(instancei)=(1/ Zt)(Dt(instancei)x{e- αt }if the
hypothesis correctly matched the instance to the label• x{eαt } otherwise
Adaboost
• Final hypothesis: H(x) = sign(sum(αt ht (x)))• Main ideas:
– Adaboost forces the weak learner to focus on incorrectly classified instances
– Training error decreases exponentially– Does boosting overfit?
• Baum showed Generalization error = O(sqrt(Td/m))• Schapire showed error = O(sqrt(d/mθ))• Does Generalization error depend on T or not? The jury
is still out.– No overfit mechanism
AdaBoost: Dynamic Boosting
• Better bounds on the error• No need to “know” • Each round a different b
– as a function of the error
AdaBoost: Input
• Sample of size m: < xi,c(xi) >
• A distribution D over examples – We will use D(xi)=1/m
• Weak learning algorithm• A constant T (number of iterations)
AdaBoost: Algorithm
• Initialization: w1(i) = D(xi)• For t = 1 to T DO
– pt(i) = wt(i) / wt(j)– Call Weak Learner with pt
– Receive ht
– Compute the error t of ht on pt
– Set bt= t/(1-t)– wt+1(i) = wt(i) (bt)e, where e=1-|ht(xi)-c(xi)|
• Output
T
t t
T
ttt
A bxhbIxh1
1
1log21)()1(log)(
AdaBoost: Analysis
• Theorem: – Given 1, ... , T
– the error of hA is bounded by
T
ttt
T
1
)1(2
AdaBoost: Proof
• Let lt(i) = 1-|ht(xi)-c(xi)|
• By definition: pt lt = 1 –t
• Upper bounding the sum of weights– From the Hedge Analysis.
• Error occurs only if
T
t t
T
ttt bxcxhb
11
1log21 |)()(|)1(log
AdaBoost Analysis (cont.)
• Bounding the weight of a point• Bounding the sum of weights
• Final bound as function of bt
• Optimizing bt:
– bt= t / (1 – t)
AdaBoost: Fixed bias
• Assume t= 1/2 - • We bound:
TT e222/2 )41(
Learning OR with few attributes
• Target function: OR of k literals• Goal: learn in time:
– polynomial in k and log n– and constant
• ELIM makes “slow” progress – disqualifies one literal per round– May remain with O(n) literals
Set Cover - Definition
• Input: S1 , … , St and Si U
• Output: Si1, … , Sik and j Sjk=U
• Question: Are there k sets that cover U?
• NP-complete
Set Cover Greedy algorithm
• j=0 ; Uj=U; C=
• While Uj – Let Si be arg max |Si Uj|
– Add Si to C
– Let Uj+1 = Uj – Si
– j = j+1
Set Cover: Greedy Analysis
• At termination, C is a cover.• Assume there is a cover C’ of size k.
• C’ is a cover for every Uj
• Some S in C’ covers Uj/k elements of Uj
• Analysis of Uj: |Uj+1| |Uj| - |Uj|/k
• Solving the recursion.• Number of sets j < k ln |U|
Building an Occam algorithm
• Given a sample S of size m– Run ELIM on S – Let LIT be the set of literals– There exists k literals in LIT that
classify correctly all S
• Negative examples: – any subset of LIT classifies theme
correctly
Building an Occam algorithm
• Positive examples: – Search for a small subset of LIT – Which classifies S+ correctly– For a literal z build Tz={x | z satisfies x}– There are k sets that cover S+
– Find k ln m sets that cover S+
• Output h = the OR of the k ln m literals • Size (h) < k ln m log 2n• Sample size m =O( k log n log (k log n))
Application : Data mining
• Challenges in real world data mining problems– Data has large number of observations and large number of
variables on each observation.– Inputs are a mixture of various different kinds of variables– Missing values, outliers and variables with skewed
distribution.– Results to be obtained fast and they should be
interpretable.
• So off-shelf techniques are difficult to come up with.• Boosting Decision Trees ( AdaBoost or MART) come
close to an off-shelf technique for Data Mining.
Boosting TreesPresented by Rishi Sinha
Occam Razor
Occam algorithm and compression
A BS(xi,bi)
x1, … , xm
compression
• Option 1:
– A sends B the values b1 , … , bm
– m bits of information• Option 2:
– A sends B the hypothesis h– Occam: large enough m has size(h) < m
• Option 3 (MDL):– A sends B a hypothesis h and “corrections”– complexity: size(h) + size(errors)
Independent Component Analysis (ICA)
• This is the first ICA paper– My source for this explanation of ICA
is “Independent Component Analysis: A Tutorial” by Aapo Hyvarinen and “Variational Methods forBayesian Independent Component Analysis” ICA chapter by Rizwan A. Choudrey