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Boosting and other Expert Fusion Strategies. References Chapter 9.5 Duda Hart & Stock Leo Breiman Boosting Bagging Arcing Presentation.

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Presentation on theme: "Boosting and other Expert Fusion Strategies. References Chapter 9.5 Duda Hart & Stock Leo Breiman Boosting Bagging Arcing Presentation."— Presentation transcript:

1 Boosting and other Expert Fusion Strategies

2 References http://www.boosting.org Chapter 9.5 Duda Hart & Stock Leo Breiman Boosting Bagging Arcing Presentation adapted from: Rishi Sinha, Robin Dhamankar

3 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

4 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

5 Online Experts Selection N strategies (experts) At time t: –Learner A chooses a distribution over N experts –Let p t (i) be the probability of i-th expert –  p t (i) = 1 and for a loss vector l t Loss at time t:  p t (i) l t (i) Assume bounded loss, l t (i) in [0,1]

6 Experts Algorithm: Greedy For each expert define its cumulative loss: Greedy: At time t choose the expert with minimum loss, namely, arg min L i t

7 Greedy Analysis Theorem: Let L G T be the loss of Greedy at time T, then Proof in notes. Weakness: Relies on a single expert for every observation

8 Better Multiple Experts Algorithms Would like to bound Better Bound: Hedge Algorithm Utilizes all experts for each observation

9 Multiple Experts Algorithm: Hedge Maintain weight vector at time t: w t Probabilities p t (k) = w t (k) /  w t (j) Initialization w 1 (i) = 1/N Updates:  w t+1 (k) = w t (k) U b (l t (k))  where b in [0,1] and  b r < U b (r) < 1-(1-b)r

10 Hedge Analysis Lemma: For any sequence of losses Proof (Mansour’s scribe) Corollary:

11 Hedge: Properties Bounding the weights Similarly for a subset of experts.

12 Hedge: Performance Let k be with minimal loss Therefore

13 Hedge: Optimizing b For b=1/2 we have Better selection of b:

14 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  (c t ) m  Efficiency.

15 Occam Razor Theorem A: ( ,  )-Occam algorithm for C using H D distribution over inputs X c t in C the target function Sample size: with probability 1-  A(S)=h has error(h) < 

16 Occam Razor Theorem Use the bound for finite hypothesis class. Effective hypothesis class size 2 size(h) size(h) < n  m  Sample size:

17 Weak and Strong Learning

18 PAC Learning model (Strong Learning) There exists a distribution D over domain X Examples: –use c for target function (rather than c t ) Goal: –With high probability (1-  ) –find h in H such that –error(h,c ) <  –  arbitrarily small, thus STRONG LEARNING

19 Weak Learning Model Goal: error(h,c) < ½ -  (Slightly above chance) The parameter  is small –constant Intuitively: A much easier task Question: –Assume C is weak learnable, –C is PAC (strong) learnable

20 Majority Algorithm Hypothesis: h M (x)= MAJ[ h 1 (x),..., h T (x) ] size(h M ) < T size(h t ) Using Occam Razor

21 Majority: outline Sample m example Start with a distribution 1/m per example. Modify the distribution and get h t Hypothesis is the majority Terminate when perfect classification – of the sample

22 Majority: Algorithm Use the Hedge algorithm. The “experts” will be associate with points. Loss would be a correct classification. –l t (i)= 1 - | h t (x i ) – c(x i ) | Setting b= 1-  h M (x) = MAJORITY( h i (x)) Q: How do we set T?

23 Majority: Analysis Consider the set of errors S S={i | h M (x i )  c(x i ) } For every i in S: L i / T < ½ (Proof!) From Hedge properties:

24 MAJORITY: Correctness Error Probability: Number of Rounds: Terminate when error less than 1/m

25 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

26 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)

27 Boosting W(x) is the distribution of weights over N training observations ∑ W(x i )=1 Initially assign uniform weights W 0 (x) = 1/N for all x, step k=0 At each iteration k : –Find best weak classifier C k (x) using weights W k (x) –With error rate ε k and based on a loss function: weight α k the classifier C k ‘s weight in the final hypothesis For each x i, update weights based on ε k to get W k+1 (x i ) C FINAL (x) =sign [ ∑ α i C i (x) ]

28 Boosting (Algorithm)

29 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

30 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

31 Boosting As Additive Model AdaBoost for Classification: L(y, f (x)) = exp(-y ∙ f (x)) - the exponential loss function

32 Boosting As Additive Model First assume that β is constant, and minimize w.r.t. G:

33 Boosting As Additive Model err m : 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.

34 Boosting As Additive Model Now that we have found G, we minimize w.r.t. β:

35 Boosting (Recall) W(x) is the distribution of weights over the N training observations ∑ W(x i )=1 Initially assign uniform weights W 0 (x) = 1/N for all x, step k=0 At each iteration k : –Find best weak classifier C k (x) using weights W k (x) –With error rate ε k and based on a loss function: weight α k the classifier C k ‘s weight in the final hypothesis For each x i, update weights based on ε k to get W k+1 (x i ) C FINAL (x) =sign [ ∑ α i C i (x) ]

36 AdaBoost W(x) is the distribution of weights over the N training points ∑ W(x i )=1 Initially assign uniform weights W 0 (x) = 1/N for all x. At each iteration k : –Find best weak classifier C k (x) using weights W k (x) –Compute ε k the error rate as ε k = [ ∑ W(x i ) ∙ I(y i ≠ C k (x i )) ] / [ ∑ W(x i )] –weight α k the classifier C k ‘s weight in the final hypothesis Set α k = log ((1 – ε k )/ε k ) –For each x i, W k+1 (x i ) = W k (x i ) ∙ exp[α k ∙ I(y i ≠ C k (x i ))] C FINAL (x) =sign [ ∑ α i C i (x) ]

37 AdaBoost(Example) Original Training set : Equal Weights to all training samples Taken from “A Tutorial on Boosting” by Yoav Freund and Rob Schapire

38 AdaBoost (Example) ROUND 1

39 AdaBoost (Example) ROUND 2

40 AdaBoost (Example) ROUND 3

41 AdaBoost (Example)

42 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).

43 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.

44 Boosting performance Steep decrease in error

45 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 ?

46 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.

47 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

48 Adaboost Algorithm:

49 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

50 Boosting Trees : Basics Formally a tree is The parameters found by minimizing the empirical risk. Finding: –  j given R j : typically mean of y i in R j –R j : Is tough but solutions exist.

51 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

52 Numerical Optimization Loss Function is So the problem boils down to finding Which in optimization procedures are solved as

53 Numerical Optimization Methods Steepest Descent Loss on Training Data converges to 0.

54 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

55 Algorithm

56 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

57 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.

58 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.

59

60 Controlling M (Regularization) After each iteration the training risk L(f M ). As M  , L(f M )  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.

61 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.

62

63

64 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.

65 Expert: Goal Match the loss of best expert. Loss: –L A –L i Can we hope to do better?

66 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.

67 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. RockPaperScissor s Rock1/210 Paper01/21 Scissor s 101/2

68 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”

69 Adaboost Line 1: Given input space X and training examples x 1,…x m 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 D t Line 5: Get a weak hypothesis h t which maps the input space to the label space. The error of this hypothesis is ε t Line 6: α t = (1/2)ln((1- ε t )/ ε t ) Line 7: D t (instance i )=(1/ Z t )(D t (instance i )x{e - αt }if the hypothesis correctly matched the instance to the label x{e αt } otherwise

70 Adaboost Final hypothesis: H(x) = sign(sum(α t h t (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

71 AdaBoost: Dynamic Boosting Better bounds on the error No need to “know”  Each round a different b –as a function of the error

72 AdaBoost: Input Sample of size m: A distribution D over examples –We will use D(x i )=1/m Weak learning algorithm A constant T (number of iterations)

73 AdaBoost: Algorithm Initialization: w 1 (i) = D(x i ) For t = 1 to T DO –p t (i) = w t (i) /  w t (j) –Call Weak Learner with p t –Receive h t –Compute the error  t of h t on p t –Set b t =  t /(1-  t ) –w t+1 (i) = w t (i) (b t ) e, where e=1-|h t (x i )-c(x i )| Output

74 AdaBoost: Analysis Theorem: –Given  1,...,  T –the error  of h A is bounded by

75 AdaBoost: Proof Let l t (i) = 1-|h t (x i )-c(x i )| By definition: p t l t = 1 –  t Upper bounding the sum of weights –From the Hedge Analysis. Error occurs only if

76 AdaBoost Analysis (cont.) Bounding the weight of a point Bounding the sum of weights Final bound as function of b t Optimizing b t : –b t =  t / (1 –  t )

77 AdaBoost: Fixed bias Assume  t = 1/2 -  We bound:

78 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

79 Set Cover - Definition Input: S 1, …, S t and S i  U Output: S i1, …, S ik and  j S jk =U Question: Are there k sets that cover U? NP-complete

80 Set Cover Greedy algorithm j=0 ; U j =U; C=  While U j   –Let S i be arg max |S i  U j | –Add S i to C –Let U j+1 = U j – S i –j = j+1

81 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 U j Some S in C’ covers U j /k elements of U j Analysis of U j : |U j+1 |  |U j | - |U j |/k Solving the recursion. Number of sets j < k ln |U|

82 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

83 Building an Occam algorithm Positive examples: –Search for a small subset of LIT –Which classifies S + correctly –For a literal z build T z ={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))

84 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.

85 Boosting Trees Presented by Rishi Sinha

86 Occam Razor

87 Occam algorithm and compression A B S (x i,b i ) x 1, …, x m

88 compression Option 1: –A sends B the values b 1, …, b m –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)

89 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 for Bayesian Independent Component Analysis” ICA chapter by Rizwan A. Choudrey


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