AdaBoost. Lecturer: Authors: Center for Machine Perception Czech Technical University, Prague
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1 AdaBoost Lecturer: Jan Šochman Authors: Jan Šochman, Jiří Matas Center for Machine Perception Czech Technical University, Prague
2 Motivation Presentation 2/17 AdaBoost with trees is the best off-the-shelf classifier in the world. (Breiman 1998) Outline: AdaBoost algorithm How it works? Why it works? Online AdaBoost and other variants
3 What is AdaBoost? AdaBoost is an algorithm for constructing a strong classifier as linear combination 3/17 f(x) = T α t h t (x) t=1 of simple weak classifiers h t (x): X { 1, +1}. Terminology h t (x)... weak or basis classifier, hypothesis, feature H(x) = sign(f(x))... strong or final classifier/hypothesis Interesting properties AB is capable reducing both bias (e.g. stumps) and variance (e.g. trees) of the weak classifiers AB has good generalisation properties (maximises margin) AB output converges to the logarithm of likelihood ratio AB can be seen as a feature selector with a principled strategy (minimisation of upper bound on empirical error) AB is close to sequential decision making (it produces a sequence of gradually more complex classifiers)
4 The AdaBoost Algorithm Given: (x 1, y 1 ),..., (x m, y m ); x i X, y i { 1, +1} 4/17
5 The AdaBoost Algorithm Given: (x 1, y 1 ),..., (x m, y m ); x i X, y i { 1, +1} Initialise weights D 1 (i) = 1/m 4/17
6 The AdaBoost Algorithm Given: (x 1, y 1 ),..., (x m, y m ); x i X, y i { 1, +1} Initialise weights D 1 (i) = 1/m For t = 1,..., T : 4/17
7 The AdaBoost Algorithm Given: (x 1, y 1 ),..., (x m, y m ); x i X, y i { 1, +1} Initialise weights D 1 (i) = 1/m For t = 1,..., T : Find ht = arg min ɛ j = m D t (i) y i h j (x i ) h j H i=1 t = 1 4/17
8 The AdaBoost Algorithm Given: (x 1, y 1 ),..., (x m, y m ); x i X, y i { 1, +1} Initialise weights D 1 (i) = 1/m For t = 1,..., T : Find ht = arg min ɛ j = m D t (i) y i h j (x i ) h j H i=1 If ɛ t 1/2 then stop t = 1 4/17
9 The AdaBoost Algorithm Given: (x 1, y 1 ),..., (x m, y m ); x i X, y i { 1, +1} Initialise weights D 1 (i) = 1/m For t = 1,..., T : Find ht = arg min ɛ j = m D t (i) y i h j (x i ) h j H i=1 If ɛ t 1/2 then stop t = 1 4/17 Set αt = 12 log(1 ɛt ɛ t )
10 The AdaBoost Algorithm Given: (x 1, y 1 ),..., (x m, y m ); x i X, y i { 1, +1} Initialise weights D 1 (i) = 1/m For t = 1,..., T : Find ht = arg min ɛ j = m D t (i) y i h j (x i ) h j H i=1 If ɛ t 1/2 then stop t = 1 4/17 Set αt = Update 12 log(1 ɛt ɛ t ) D t+1 (i) = D t(i)exp( α t y i h t (x i )) Z t where Z t is normalisation factor
11 The AdaBoost Algorithm Given: (x 1, y 1 ),..., (x m, y m ); x i X, y i { 1, +1} Initialise weights D 1 (i) = 1/m For t = 1,..., T : Find ht = arg min ɛ j = m D t (i) y i h j (x i ) h j H i=1 If ɛ t 1/2 then stop t = 1 4/17 Set αt = Update 12 log(1 ɛt ɛ t ) D t+1 (i) = D t(i)exp( α t y i h t (x i )) Z t where Z t is normalisation factor Output the final classifier: ( T ) H(x) = sign α t h t (x) t=1 training error step
12 The AdaBoost Algorithm Given: (x 1, y 1 ),..., (x m, y m ); x i X, y i { 1, +1} Initialise weights D 1 (i) = 1/m For t = 1,..., T : Find ht = arg min ɛ j = m D t (i) y i h j (x i ) h j H i=1 If ɛ t 1/2 then stop t = 2 4/17 Set αt = Update 12 log(1 ɛt ɛ t ) D t+1 (i) = D t(i)exp( α t y i h t (x i )) Z t where Z t is normalisation factor Output the final classifier: ( T ) H(x) = sign α t h t (x) t=1 training error step
13 The AdaBoost Algorithm Given: (x 1, y 1 ),..., (x m, y m ); x i X, y i { 1, +1} Initialise weights D 1 (i) = 1/m For t = 1,..., T : Find ht = arg min ɛ j = m D t (i) y i h j (x i ) h j H i=1 If ɛ t 1/2 then stop t = 3 4/17 Set αt = Update 12 log(1 ɛt ɛ t ) D t+1 (i) = D t(i)exp( α t y i h t (x i )) Z t where Z t is normalisation factor Output the final classifier: ( T ) H(x) = sign α t h t (x) t=1 training error step
14 The AdaBoost Algorithm Given: (x 1, y 1 ),..., (x m, y m ); x i X, y i { 1, +1} Initialise weights D 1 (i) = 1/m For t = 1,..., T : Find ht = arg min ɛ j = m D t (i) y i h j (x i ) h j H i=1 If ɛ t 1/2 then stop t = 4 4/17 Set αt = Update 12 log(1 ɛt ɛ t ) D t+1 (i) = D t(i)exp( α t y i h t (x i )) Z t where Z t is normalisation factor Output the final classifier: ( T ) H(x) = sign α t h t (x) t=1 training error step
15 The AdaBoost Algorithm Given: (x 1, y 1 ),..., (x m, y m ); x i X, y i { 1, +1} Initialise weights D 1 (i) = 1/m For t = 1,..., T : Find ht = arg min ɛ j = m D t (i) y i h j (x i ) h j H i=1 If ɛ t 1/2 then stop t = 5 4/17 Set αt = Update 12 log(1 ɛt ɛ t ) D t+1 (i) = D t(i)exp( α t y i h t (x i )) Z t where Z t is normalisation factor Output the final classifier: ( T ) H(x) = sign α t h t (x) t=1 training error step
16 The AdaBoost Algorithm Given: (x 1, y 1 ),..., (x m, y m ); x i X, y i { 1, +1} Initialise weights D 1 (i) = 1/m For t = 1,..., T : Find ht = arg min ɛ j = m D t (i) y i h j (x i ) h j H i=1 If ɛ t 1/2 then stop t = 6 4/17 Set αt = Update 12 log(1 ɛt ɛ t ) D t+1 (i) = D t(i)exp( α t y i h t (x i )) Z t where Z t is normalisation factor Output the final classifier: ( T ) H(x) = sign α t h t (x) t=1 training error step
17 The AdaBoost Algorithm Given: (x 1, y 1 ),..., (x m, y m ); x i X, y i { 1, +1} Initialise weights D 1 (i) = 1/m For t = 1,..., T : Find ht = arg min ɛ j = m D t (i) y i h j (x i ) h j H i=1 If ɛ t 1/2 then stop t = 7 4/17 Set αt = Update 12 log(1 ɛt ɛ t ) D t+1 (i) = D t(i)exp( α t y i h t (x i )) Z t where Z t is normalisation factor Output the final classifier: ( T ) H(x) = sign α t h t (x) t=1 training error step
18 The AdaBoost Algorithm Given: (x 1, y 1 ),..., (x m, y m ); x i X, y i { 1, +1} Initialise weights D 1 (i) = 1/m For t = 1,..., T : Find ht = arg min ɛ j = m D t (i) y i h j (x i ) h j H i=1 If ɛ t 1/2 then stop t = 40 4/17 Set αt = Update 12 log(1 ɛt ɛ t ) D t+1 (i) = D t(i)exp( α t y i h t (x i )) Z t where Z t is normalisation factor Output the final classifier: ( T ) H(x) = sign α t h t (x) t=1 training error step
19 Reweighting 5/17 Effect on the training set D t+1 (i) = D t(i)exp( α t y i h t (x i )) exp( α t y i h t (x i )) Z t { < 1, yi = h t (x i ) > 1, y i h t (x i ) Increase (decrease) weight of wrongly (correctly) classified examples The weight is the upper bound on the error of a given example All information about previously selected features is captured in D t
20 Reweighting 5/17 Effect on the training set D t+1 (i) = D t(i)exp( α t y i h t (x i )) exp( α t y i h t (x i )) Z t { < 1, yi = h t (x i ) > 1, y i h t (x i ) Increase (decrease) weight of wrongly (correctly) classified examples The weight is the upper bound on the error of a given example All information about previously selected features is captured in D t
21 Reweighting 5/17 Effect on the training set D t+1 (i) = D t(i)exp( α t y i h t (x i )) exp( α t y i h t (x i )) Z t { < 1, yi = h t (x i ) > 1, y i h t (x i ) Increase (decrease) weight of wrongly (correctly) classified examples The weight is the upper bound on the error of a given example All information about previously selected features is captured in D t
22 Upper Bound Theorem Theorem: The following upper bound holds on the training error of H 6/17 Proof: By unravelling the update rule 1 m {i : H(x i) y i } T t=1 D T +1 (i) = D t(i)exp( α t y i h t (x i )) Z t = exp( t α ty i h t (x i )) m t Z t Z t = exp( y if(x i )) m t Z t If H(x i ) y i then y i f(x i ) 0 implying that exp( y i f(x i )) > 1, thus 1 m H(x i ) y i exp( y i f(x i )) H(x i ) y i 1 exp( y i f(x i )) m i i err = i ( t Z t )D T +1 (i) = t 0 Z t yf(x)
23 Consequences of the Theorem Instead of minimising the training error, its upper bound can be minimised 7/17 This can be done by minimising Zt in each training round by: Choosing optimal h t, and Finding optimal α t AdaBoost can be proved to maximise margin AdaBoost iteratively fits an additive logistic regression model
24 Choosing α t We attempt to minimise Z t = i D t(i)exp( α t y i h t (x i )): 8/17 m dz dα = D(i)y i h(x i )e y iα i h(x i ) = 0 i=1 D(i)e α + D(i)e α = 0 i:y i =h(x i ) i:y i h(x i ) e α (1 ɛ) + e α ɛ = 0 α = 1 2 log 1 ɛ ɛ The minimisator of the upper bound is α t = 1 2 log 1 ɛ t ɛ t
25 Weak classifier examples Choosing h t 9/17 Decision tree (or stump), Perceptron H infinite Selecting the best one from given finite set H Justification of the weighted error minimisation Having α t = 1 2 log 1 ɛ t ɛ t Z t = = m D t (i)e y iα i h t (x i ) i=1 i:y i =h t (x i ) D t (i)e α t + i:y i h t (x i ) D t (i)e α t = (1 ɛ t )e α t + ɛ t e α t = 2 ɛ t (1 ɛ t ) Z t is minimised by selecting h t with minimal weighted error ɛ t
26 Generalisation (Schapire & Singer 1999) Maximising margins in AdaBoost 10/17 P (x,y) S [yf(x) θ] 2 T T t=1 ɛ 1 θ t (1 ɛ t ) 1+θ where f(x) = Choosing ht(x) with minimal ɛt in each step one minimises the margin α h(x) α 1 Margin in SVM use the L2 norm instead: ( α h(x))/ α 2 Upper bounds based on margin With probability 1 δ over the random choice of the training set S P (x,y) D [yf(x) 0] P (x,y) S [yf(x) θ] + O ( 1 d log 2 (m/d) m θ 2 + log(1/δ) ) 1/2 where D is a distribution over X {+1, 1}, and d is pseudodimension of H. Problem: The upper bound is very loose. In practice AdaBoost works much better.
27 Convergence (Friedman et al. 1998) Proposition 1 The discrete AdaBoost algorithm minimises J(f(x)) = E(e yf(x) ) by adaptive Newton updates 11/17 Lemma J(f(x)) is minimised at Hence f(x) = T α t h t (x) = 1 2 t=1 log P (y = 1 x) P (y = 1 x) P (y = 1 x) = e f(x) e f(x) + e f(x) and P (y = 1 x) = e f(x) e f(x) + e f(x) Additive logistic regression model T a t (x) = log t=1 P (y = 1 x) P (y = 1 x) Proposition 2 By minimising J(f(x)) the discrete AdaBoost fits (up to a factor 2) an additive logistic regression model
28 The Algorithm Recapitulation Given: (x 1, y 1 ),..., (x m, y m ); x i X, y i { 1, +1} 12/17
29 The Algorithm Recapitulation Given: (x 1, y 1 ),..., (x m, y m ); x i X, y i { 1, +1} Initialise weights D 1 (i) = 1/m 12/17
30 The Algorithm Recapitulation Given: (x 1, y 1 ),..., (x m, y m ); x i X, y i { 1, +1} Initialise weights D 1 (i) = 1/m For t = 1,..., T : 12/17
31 The Algorithm Recapitulation Given: (x 1, y 1 ),..., (x m, y m ); x i X, y i { 1, +1} Initialise weights D 1 (i) = 1/m For t = 1,..., T : Find ht = arg min ɛ j = m D t (i) y i h j (x i ) h j H i=1 t = 1 12/17
32 The Algorithm Recapitulation Given: (x 1, y 1 ),..., (x m, y m ); x i X, y i { 1, +1} Initialise weights D 1 (i) = 1/m For t = 1,..., T : Find ht = arg min ɛ j = m D t (i) y i h j (x i ) h j H i=1 If ɛ t 1/2 then stop t = 1 12/17
33 The Algorithm Recapitulation Given: (x 1, y 1 ),..., (x m, y m ); x i X, y i { 1, +1} Initialise weights D 1 (i) = 1/m For t = 1,..., T : Find ht = arg min ɛ j = m D t (i) y i h j (x i ) h j H i=1 If ɛ t 1/2 then stop t = 1 12/17 Set αt = 12 log(1 ɛt ɛ t )
34 The Algorithm Recapitulation Given: (x 1, y 1 ),..., (x m, y m ); x i X, y i { 1, +1} Initialise weights D 1 (i) = 1/m For t = 1,..., T : Find ht = arg min ɛ j = m D t (i) y i h j (x i ) h j H i=1 If ɛ t 1/2 then stop t = 1 12/17 Set αt = Update 12 log(1 ɛt ɛ t ) D t+1 (i) = D t(i)exp( α t y i h t (x i )) Z t
35 The Algorithm Recapitulation Given: (x 1, y 1 ),..., (x m, y m ); x i X, y i { 1, +1} Initialise weights D 1 (i) = 1/m For t = 1,..., T : Find ht = arg min ɛ j = m D t (i) y i h j (x i ) h j H i=1 If ɛ t 1/2 then stop t = 1 12/17 Set αt = Update 12 log(1 ɛt ɛ t ) D t+1 (i) = D t(i)exp( α t y i h t (x i )) Z t Output the final classifier: ( T ) H(x) = sign α t h t (x) t=1 training error step
36 The Algorithm Recapitulation Given: (x 1, y 1 ),..., (x m, y m ); x i X, y i { 1, +1} Initialise weights D 1 (i) = 1/m For t = 1,..., T : Find ht = arg min ɛ j = m D t (i) y i h j (x i ) h j H i=1 If ɛ t 1/2 then stop t = 2 12/17 Set αt = Update 12 log(1 ɛt ɛ t ) D t+1 (i) = D t(i)exp( α t y i h t (x i )) Z t Output the final classifier: ( T ) H(x) = sign α t h t (x) t=1 training error step
37 The Algorithm Recapitulation Given: (x 1, y 1 ),..., (x m, y m ); x i X, y i { 1, +1} Initialise weights D 1 (i) = 1/m For t = 1,..., T : Find ht = arg min ɛ j = m D t (i) y i h j (x i ) h j H i=1 If ɛ t 1/2 then stop t = 3 12/17 Set αt = Update 12 log(1 ɛt ɛ t ) D t+1 (i) = D t(i)exp( α t y i h t (x i )) Z t Output the final classifier: ( T ) H(x) = sign α t h t (x) t=1 training error step
38 The Algorithm Recapitulation Given: (x 1, y 1 ),..., (x m, y m ); x i X, y i { 1, +1} Initialise weights D 1 (i) = 1/m For t = 1,..., T : Find ht = arg min ɛ j = m D t (i) y i h j (x i ) h j H i=1 If ɛ t 1/2 then stop t = 4 12/17 Set αt = Update 12 log(1 ɛt ɛ t ) D t+1 (i) = D t(i)exp( α t y i h t (x i )) Z t Output the final classifier: ( T ) H(x) = sign α t h t (x) t=1 training error step
39 The Algorithm Recapitulation Given: (x 1, y 1 ),..., (x m, y m ); x i X, y i { 1, +1} Initialise weights D 1 (i) = 1/m For t = 1,..., T : Find ht = arg min ɛ j = m D t (i) y i h j (x i ) h j H i=1 If ɛ t 1/2 then stop t = 5 12/17 Set αt = Update 12 log(1 ɛt ɛ t ) D t+1 (i) = D t(i)exp( α t y i h t (x i )) Z t Output the final classifier: ( T ) H(x) = sign α t h t (x) t=1 training error step
40 The Algorithm Recapitulation Given: (x 1, y 1 ),..., (x m, y m ); x i X, y i { 1, +1} Initialise weights D 1 (i) = 1/m For t = 1,..., T : Find ht = arg min ɛ j = m D t (i) y i h j (x i ) h j H i=1 If ɛ t 1/2 then stop t = 6 12/17 Set αt = Update 12 log(1 ɛt ɛ t ) D t+1 (i) = D t(i)exp( α t y i h t (x i )) Z t Output the final classifier: ( T ) H(x) = sign α t h t (x) t=1 training error step
41 The Algorithm Recapitulation Given: (x 1, y 1 ),..., (x m, y m ); x i X, y i { 1, +1} Initialise weights D 1 (i) = 1/m For t = 1,..., T : Find ht = arg min ɛ j = m D t (i) y i h j (x i ) h j H i=1 If ɛ t 1/2 then stop t = 7 12/17 Set αt = Update 12 log(1 ɛt ɛ t ) D t+1 (i) = D t(i)exp( α t y i h t (x i )) Z t Output the final classifier: ( T ) H(x) = sign α t h t (x) t=1 training error step
42 The Algorithm Recapitulation Given: (x 1, y 1 ),..., (x m, y m ); x i X, y i { 1, +1} Initialise weights D 1 (i) = 1/m For t = 1,..., T : Find ht = arg min ɛ j = m D t (i) y i h j (x i ) h j H i=1 If ɛ t 1/2 then stop t = 40 12/17 Set αt = Update 12 log(1 ɛt ɛ t ) D t+1 (i) = D t(i)exp( α t y i h t (x i )) Z t Output the final classifier: ( T ) H(x) = sign α t h t (x) t=1 training error step
43 Freund & Schapire 1995 AdaBoost Variants 13/17 Discrete (h : X {0, 1}) Multiclass AdaBoost.M1 (h : X {0, 1,..., k}) Multiclass AdaBoost.M2 (h : X [0, 1]k) Real valued AdaBoost.R (Y = [0, 1], h : X [0, 1]) Schapire & Singer 1999 Confidence rated prediction (h : X R, two-class) Multilabel AdaBoost.MR, AdaBoost.MH (different formulation of minimised loss) Oza 2001 Online AdaBoost Many other modifications since then: cascaded AB, WaldBoost, probabilistic boosting tree,...
44 Online AdaBoost 14/17 Given: Offline Given: Online Set of labeled training samples X = {(x 1, y 1 ),..., (x m, y m ) y = ±1} Weight distribution over X D 0 = 1/m For t = 1,..., T For t = 1,..., T Train a weak classifier using samples and weight distribution Calculate error ɛt h t (x) = L(X, D t 1 ) Calculate coeficient αt from ɛt Update weight distribution Dt Output: F (x) = sign( T t=1 α th t (x)) Output: F (x) = sign( T t=1 α th t (x))
45 Given: Offline Online AdaBoost Set of labeled training samples X = {(x 1, y 1 ),..., (x m, y m ) y = ±1} Weight distribution over X D 0 = 1/m For t = 1,..., T Given: Online One labeled training sample (x, y) y = ±1 Strong classifier to update For t = 1,..., T 14/17 Train a weak classifier using samples and weight distribution Calculate error ɛt h t (x) = L(X, D t 1 ) Calculate coeficient αt from ɛt Update weight distribution Dt Output: F (x) = sign( T t=1 α th t (x)) Output: F (x) = sign( T t=1 α th t (x))
46 Given: Offline Online AdaBoost Set of labeled training samples X = {(x 1, y 1 ),..., (x m, y m ) y = ±1} Weight distribution over X D 0 = 1/m For t = 1,..., T Given: Online One labeled training sample (x, y) y = ±1 Strong classifier to update Initial importance λ = 1 For t = 1,..., T 14/17 Train a weak classifier using samples and weight distribution Calculate error ɛt h t (x) = L(X, D t 1 ) Calculate coeficient αt from ɛt Update weight distribution Dt Output: F (x) = sign( T t=1 α th t (x)) Output: F (x) = sign( T t=1 α th t (x))
47 Given: Offline Online AdaBoost Set of labeled training samples X = {(x 1, y 1 ),..., (x m, y m ) y = ±1} Weight distribution over X D 0 = 1/m For t = 1,..., T Train a weak classifier using samples and weight distribution Calculate error ɛt h t (x) = L(X, D t 1 ) Calculate coeficient αt from ɛt Given: Online One labeled training sample (x, y) y = ±1 Strong classifier to update Initial importance λ = 1 For t = 1,..., T 14/17 Update the weak classifier using the sample and the importance h t (x) = L(h t, (x, y), λ) Update weight distribution Dt Output: F (x) = sign( T t=1 α th t (x)) Output: F (x) = sign( T t=1 α th t (x))
48 Given: Offline Online AdaBoost Set of labeled training samples X = {(x 1, y 1 ),..., (x m, y m ) y = ±1} Weight distribution over X D 0 = 1/m For t = 1,..., T Train a weak classifier using samples and weight distribution h t (x) = L(X, D t 1 ) Given: Online One labeled training sample (x, y) y = ±1 Strong classifier to update Initial importance λ = 1 For t = 1,..., T 14/17 Update the weak classifier using the sample and the importance h t (x) = L(h t, (x, y), λ) Calculate error ɛt Calculate coeficient αt from ɛt Update weight distribution Dt Output: F (x) = sign( T t=1 α th t (x)) Update error estimation ɛt Update weight αt based on ɛt Output: F (x) = sign( T t=1 α th t (x))
49 Given: Offline Online AdaBoost Set of labeled training samples X = {(x 1, y 1 ),..., (x m, y m ) y = ±1} Weight distribution over X D 0 = 1/m For t = 1,..., T Train a weak classifier using samples and weight distribution h t (x) = L(X, D t 1 ) Given: Online One labeled training sample (x, y) y = ±1 Strong classifier to update Initial importance λ = 1 For t = 1,..., T 14/17 Update the weak classifier using the sample and the importance h t (x) = L(h t, (x, y), λ) Calculate error ɛt Calculate coeficient αt from ɛt Update weight distribution Dt Output: F (x) = sign( T t=1 α th t (x)) Update error estimation ɛt Update weight αt based on ɛt Update importance weight λ Output: F (x) = sign( T t=1 α th t (x))
50 Given: Offline Online AdaBoost Set of labeled training samples X = {(x 1, y 1 ),..., (x m, y m ) y = ±1} Weight distribution over X D 0 = 1/m For t = 1,..., T Train a weak classifier using samples and weight distribution h t (x) = L(X, D t 1 ) Given: Online One labeled training sample (x, y) y = ±1 Strong classifier to update Initial importance λ = 1 For t = 1,..., T 14/17 Update the weak classifier using the sample and the importance h t (x) = L(h t, (x, y), λ) Calculate error ɛt Calculate coeficient αt from ɛt Update weight distribution Dt Output: F (x) = sign( T t=1 α th t (x)) Update error estimation ɛt Update weight αt based on ɛt Update importance weight λ Output: F (x) = sign( T t=1 α th t (x))
51 Online AdaBoost 15/17 Converges to offline results given the same training set and the number of iterations N N. Oza and S. Russell. Online Bagging and Boosting. Artificial Inteligence and Statistics, 2001.
52 Advantages Pros and Cons of AdaBoost 16/17 Very simple to implement General learning scheme - can be used for various learning tasks Feature selection on very large sets of features Good generalisation Seems not to overfit in practice (probably due to margin maximisation) Disadvantages Suboptimal solution (greedy learning)
53 Selected references 17/17 Y. Freund, R.E. Schapire. A Decision-theoretic Generalization of On-line Learning and an Application to Boosting. Journal of Computer and System Sciences R.E. Schapire, Y. Freund, P. Bartlett, W.S. Lee. Boosting the Margin: A New Explanation for the Effectiveness of Voting Methods. The Annals of Statistics, 1998 R.E. Schapire, Y. Singer. Improved Boosting Algorithms Using Confidence-rated Predictions. Machine Learning J. Friedman, T. Hastie, R. Tibshirani. Additive Logistic Regression: a Statistical View of Boosting. Technical report N.C. Oza. Online Ensemble Learning. PhD thesis
54 Selected references 17/17 Y. Freund, R.E. Schapire. A Decision-theoretic Generalization of On-line Learning and an Application to Boosting. Journal of Computer and System Sciences R.E. Schapire, Y. Freund, P. Bartlett, W.S. Lee. Boosting the Margin: A New Explanation for the Effectiveness of Voting Methods. The Annals of Statistics, 1998 R.E. Schapire, Y. Singer. Improved Boosting Algorithms Using Confidence-rated Predictions. Machine Learning J. Friedman, T. Hastie, R. Tibshirani. Additive Logistic Regression: a Statistical View of Boosting. Technical report N.C. Oza. Online Ensemble Learning. PhD thesis Thank you for attention
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Learning Ensembles. 293S T. Yang. UCSB, 2017.
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