nips nips2008 nips2008-241 knowledge-graph by maker-knowledge-mining

241 nips-2008-Transfer Learning by Distribution Matching for Targeted Advertising

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Author: Steffen Bickel, Christoph Sawade, Tobias Scheffer

Abstract: We address the problem of learning classiﬁers for several related tasks that may differ in their joint distribution of input and output variables. For each task, small – possibly even empty – labeled samples and large unlabeled samples are available. While the unlabeled samples reﬂect the target distribution, the labeled samples may be biased. This setting is motivated by the problem of predicting sociodemographic features for users of web portals, based on the content which they have accessed. Here, questionnaires offered to a portion of each portal’s users produce biased samples. We derive a transfer learning procedure that produces resampling weights which match the pool of all examples to the target distribution of any given task. Transfer learning enables us to make predictions even for new portals with few or no training data and improves the overall prediction accuracy. 1

Reference: text

Summary: the most important sentenses genereted by tfidf model

sentIndex sentText sentNum sentScore

1 While the unlabeled samples reﬂect the target distribution, the labeled samples may be biased. [sent-5, score-0.545]

2 This setting is motivated by the problem of predicting sociodemographic features for users of web portals, based on the content which they have accessed. [sent-6, score-0.421]

3 We derive a transfer learning procedure that produces resampling weights which match the pool of all examples to the target distribution of any given task. [sent-8, score-0.898]

4 Some of the multiple tasks will likely relate to one another, but one cannot assume that the tasks share a joint conditional distribution of the class label given the input variables. [sent-11, score-0.417]

5 A common method for learning under covariate shift (marginal shift) is to weight the biased trainp ing examples by the test-to-training ratio p test (x) to match the marginal distribution of the test data train (x) [1]. [sent-13, score-0.731]

6 Similar to the idea of learning under marginal shift by weighting the training examples, [8] devise a method for learning under joint shift of covariates and labels over multiple tasks that is based on instancespeciﬁc rescaling factors. [sent-17, score-0.704]

7 We generalize this idea to a setting where not only the joint distributions between tasks may differ but also the training and test distribution within each task. [sent-18, score-0.505]

8 Our work is motivated by the targeted advertising problem for which the goal is to predict sociodemographic features (such as gender, age, or marital status) of web users, based on their surﬁng history. [sent-19, score-0.541]

9 When sociodemographic attributes can be identiﬁed, delivering advertisements to users outside the target segment can be avoided. [sent-21, score-0.552]

10 However, for many campaigns – including campaigns for products that are not typically purchased on the web – the sole goal is to deliver the advertisement to customers in the target segment. [sent-23, score-0.491]

11 Each of several tasks z is characterized by an unknown joint distribution p test (x, y|z) = p test (x|z)p(y|x, z) over features x and labels y given the task z. [sent-29, score-0.587]

12 The joint distributions of different tasks may differ arbitrarily but usually some tasks have similar distributions. [sent-30, score-0.441]

13 The test data for task z are governed by p test (x|z). [sent-36, score-0.437]

14 The training data for task z is drawn from a joint distribution p train (x, y|z) = p train (x|z)p(y|x, z) that may differ from the test distribution in terms of the marginal distribution p train (x|z). [sent-42, score-0.938]

15 The training and test marginals may differ arbitrarily, as long as each x with positive p test (x|z) also has a positive p train (x|z). [sent-43, score-0.439]

16 The conditional distribution p(y|x, z) of test and training data is identical for a given task z, but conditionals can differ arbitrarily between tasks. [sent-45, score-0.49]

17 The entire training set over all tasks is governed by the mixed density p train (z)p train (x, y|z). [sent-46, score-0.696]

18 The feature vector x encodes the web surﬁng behavior of a user of web portal z (task). [sent-53, score-0.49]

19 For a small number of users the sociodemographic target label y (e. [sent-54, score-0.552]

20 For new portals the number of such labeled training instances is initially small. [sent-57, score-0.544]

21 The sociodemographic labels for all users of all portals are to be predicted. [sent-58, score-0.55]

22 The joint distribution p test (x, y|z) can be different between portals since they attract speciﬁc populations of users. [sent-59, score-0.386]

23 The training distribution differs from the test distribution because the response to the web surveys is not uniform with respect to the test distribution. [sent-60, score-0.552]

24 Since the completion of surveys cannot be enforced, it is intrinsically impossible to obtain labeled samples that are governed by the test distribution. [sent-61, score-0.507]

25 One reference strategy is to learn individual models for each target task z by minimizing an appropriate loss function on the portion of Lz = {(xi , yi , zi ) ∈ L : zi = z}. [sent-63, score-0.511]

26 In addition, it minimizes the loss with respect to p train (x, y|z); the minimum of this optimization problem will not generally coincide with the minimal loss on p test (x, y|z). [sent-65, score-0.438]

27 The training sample may deviate arbitrarily from the target distribution p test (x, y|z). [sent-67, score-0.579]

28 3 Transfer Learning by Distribution Matching In learning a classiﬁer ft (x) for target task t, we seek to minimize the loss function with respect to p test (x, y|t) = p(x, y|t, s = −1). [sent-70, score-0.522]

29 Our approach is to create a task-speciﬁc resampling weight rt (x, y) for each element of the pool of examples. [sent-73, score-0.724]

30 The resampling weights match the pool distribution to the target distribution p(x, y|t, s = −1). [sent-74, score-0.783]

31 The resampled pool is governed by the correct target distribution, but is larger than the labeled sample of the target task. [sent-75, score-1.078]

32 The expected weighted loss with respect to the mixture distribution that governs the pool equals the loss with respect to the target distribution p(x, y|t, s = −1). [sent-77, score-0.811]

33 Equation 5 rearranges some terms and Equation 6 is the expected loss over the distribution of all tasks weighted by rt (x, y). [sent-82, score-0.752]

34 This amounts to minimizing the expected loss with respect to the target distribution p(x, y|t, s = −1). [sent-84, score-0.382]

35 The resampling weights of Equation 2 have an intuitive interpretation: The ﬁrst fraction accounts for the difference in the joint distributions across tasks, and the second fraction accounts for the covariate shift within the target task. [sent-85, score-0.691]

36 Equation 5 leaves us with the problem of estimating the product of two density ratios rt (x, y) = p(x,y|t,s=1) p(x|t,s=−1) . [sent-86, score-0.421]

37 p(x,y|t,s=1) 1 in terms of a conditional We reformulate the ﬁrst density ratio rt (x, y) = z p(z|s=1)p(x,y|z,s=1) model p(t|x, y, s = 1). [sent-91, score-0.463]

38 This conditional has the following intuitive meaning: Given that an instance (x, y) has been drawn at random from the pool distribution z p(z|s = 1)p(x, y|z, s = 1) over all tasks (including target task t); the probability that (x, y) originates from p(x, y|t, s = 1) is p(t|x, y, s = 1). [sent-92, score-0.818]

39 The right hand side of Equation 8 can be evaluated based on a model p(t|x, y, s = 1) that discriminates labeled instances of the target task against labeled instances of the pool of examples for all non-target tasks. [sent-97, score-1.106]

40 2 Similar to the ﬁrst density ratio, the second density ratio rt (x) = p(x|t,s=−1) can be expressed p(x|t,s=1) using a conditional model p(s = 1|x, t). [sent-98, score-0.518]

41 2 Estimation of Discriminative Density Ratios 1 For estimation of rt (x, y) we model p(t|x, y, s = 1) of Equation 8 with a logistic regression model p(t|x, y, s = 1, ut ) = 1 1 + exp(−uT Φ(x, y)) t over model parameters ut using a problem-speciﬁc feature mapping Φ(x, y). [sent-104, score-0.801]

42 Empirically, we observe that a separate binary logistic regression model (as described above) for each task yields more accurate results with the drawback of a slightly increased overall training time. [sent-110, score-0.373]

43 Optimization Problem 1 For task t: over parameters ut , maximize log p(t|x, y, s = 1, ut ) + (x,y)∈Lt log(1 − p(t|x, y, s = 1, ut )) − (x,y)∈L\Lt uT ut t . [sent-111, score-0.728]

44 The estimated vector ut leads to the ﬁrst part of the weighting factor rt (x, y) ∝ p(t|x, y, s = 1, ut ) according to Equation 8. [sent-113, score-0.701]

45 rt (x, y) is normalized so that the weighted ˆ1 ˆ1 1 ˆ1 empirical distribution over the pool L sums to one, |L| (x,y)∈L rt (x, y) = 1. [sent-114, score-1.082]

46 1 2 According to Equation 10 density ratio rt (x) = p(x|t,s=−1) ∝ p(s=1|x,t) − 1 can be inferred p(x|t,s=1) from p(s = 1|x, t) which is the likelihood that a given x for task t originates from the training distribution, as opposed to from the test distribution. [sent-115, score-0.821]

47 A model of p(s = 1|x, t) can be obtained by discriminating a sample governed by p(x|t, s = 1) against a sample governed by p(x|t, s = −1) using a probabilistic classiﬁer. [sent-116, score-0.378]

48 The labeled pool L 1 over all training examples weighted by rt (x, y) can serve as a sample governed by p(x|t, s = 1); the labels y can be ignored for this step. [sent-118, score-1.295]

49 We model p(s = 1|x, vt ) of Equation 10 with a regularized logistic regression on target variable s with parameters vt (Optimization Problem 2). [sent-120, score-0.653]

50 Labeled examples L are weighted by the estimated ﬁrst factor rt (x, y) using the outcome of Optimization Problem 1. [sent-121, score-0.547]

51 ˆ1 Optimization Problem 2 For task t: over parameters vt , maximize rt (x, y) log p(s = 1|x, vt ) + ˆ1 log p(s = −1|x, vt ) − x∈Tt (x,y)∈L T vt vt . [sent-122, score-1.079]

52 rt (x) is normalized so that the ﬁnal weighted empirical distribution ˆ2 1 ˆ1 over the pool sums to one, |L| (x,y)∈L rt (x, y)ˆt (x) = 1. [sent-124, score-1.082]

53 3 Weighted Empirical Loss and Target Model The learning procedure ﬁrst determines resampling weights rt (x, y) = rt (x, y)ˆt (x) by solving ˆ ˆ1 r2 Optimization Problems 1 and 2. [sent-126, score-0.953]

54 These weights can now be used to reweight the labeled pool over all tasks and train the target model for task t. [sent-127, score-1.085]

55 E(x,y)∼L rt (x, y)ˆt (x) (f (x, t), y) + ˆ1 r2 T wt wt 2 2σw (11) Optimization Problem 3 minimizes Equation 11, the weighted regularized loss over the training 2 data using a standard Gaussian log-prior with variance σw on the parameters wt . [sent-130, score-0.899]

56 Optimization Problem 3 For task t: over parameters wt , minimize 1 |L| rt (x, y)ˆt (x) (f (x, wt ), y) + ˆ1 r2 (x,y)∈L T wt wt . [sent-132, score-0.758]

57 2 2σw In order to train target models for all tasks, instances of Optimization Problems 1 to 3 are solved for each task. [sent-133, score-0.383]

58 4 Targeted Advertising We study the beneﬁt of distribution matching and other reference methods on targeted advertising for four web portals. [sent-134, score-0.673]

59 A small proportion of users is asked to ﬁll out a web questionnaire that collects sociodemographic user proﬁles. [sent-140, score-0.577]

60 The completion of the questionnaire cannot be enforced and it is therefore not possible to obtain labeled data that is governed by the test distribution of all users that surf the target portal. [sent-143, score-0.988]

61 We add the correct proportion (25%) of users who have taken the survey, and thereby construct a sample that is governed by an approximation of the test distribution. [sent-146, score-0.43]

62 Table 1: Portal statistics: number of accepted, partially rejected, and test examples (mix of all partial reject (=75%) and 25% accept); ratio of male users in training (accept) and test set. [sent-149, score-0.729]

63 portal family TV channel news 1 news 2 # accept 8073 8848 3051 2247 # partial reject 2035 1192 149 143 # test 2713 1589 199 191 % male training 53. [sent-150, score-0.774]

64 0% We compare distribution matching on labeled and unlabeled data (Optimization Problems 1 to 3) and distribution matching only on labeled data by setting rt (x) = 1 in Optimization Problem 3 to ˆ2 the following reference models. [sent-158, score-1.357]

65 The ﬁrst baseline is a one-size-ﬁts-all model that directly trains a logistic regression on L (setting rt (x, y)ˆt (x) = 1 in Optimization Problem 3). [sent-159, score-0.561]

66 The second baseline ˆ1 r2 is a logistic regression trained only on Lt , the training examples of the target task. [sent-160, score-0.596]

67 Training only on the reweighted target task data and correcting for marginal shift with respect to the unlabeled test data is the third baseline [4]. [sent-161, score-0.735]

68 Training a logistic regression with their feature mapping over training examples from all tasks is equivalent to a joint MAP estimation of all model parameters and the mean of the Gaussian prior. [sent-164, score-0.552]

69 We evaluate the methods using all training examples from non-target tasks and different numbers of training examples of the target task. [sent-165, score-0.815]

70 From all available accept examples of the target task we randomly select a certain number (0-1600) of training examples. [sent-166, score-0.631]

71 From the remaining accept examples of the target task we randomly select an appropriate number and add them to all partial reject examples of the target task so that the evaluation set has the right proportions as described above. [sent-167, score-1.057]

72 We use a logistic loss as the target loss of distribution matching in Optimization Problem 3 and all reference methods. [sent-169, score-0.783]

73 • The inner loop temporarily tunes σw by cross-validation on rescaled L¬t merged with the rescaled current training folds of Lt . [sent-178, score-0.407]

74 In each loop Optimization Problem 3 is solved with ﬁxed rt (x) = 1. [sent-180, score-0.433]

75 Optimization Problem 3 is solved for each outer loop with the temporary σw and with rt (x) = 1. [sent-182, score-0.511]

76 Test data Tt of the target task as well as the weighted pool L (weighted by rt (x, y), ˆ1 based on previously tuned σu ) are split into ten folds. [sent-187, score-1.085]

77 With the nine training folds of the test data and the nine training folds of the weighted pool L, Optimization Problem 2 is solved. [sent-188, score-0.838]

78 data distribution matching on labeled data hierarchical Bayes one-size-fits-all on pool of labeled data training only on lab. [sent-192, score-0.962]

79 56 0 25 50 100 200 400 800 1600 0 25 50 100 200 400 800 1600 training examples for target portal training examples for target portal news 1 news 2 0. [sent-203, score-1.414]

80 72 0 25 50 100 200 400 800 1600 training examples for target portal 0 25 50 100 200 400 800 1600 training examples for target portal Figure 1: Accuracy over different number of training examples for target portal. [sent-209, score-1.751]

81 Error bars indicate the standard error of the differences to distribution matching on labeled data. [sent-210, score-0.407]

82 σv is chosen to maximize the log-likelihood rt (x, y) log p(s = 1|x, vt ) + ˆ1 (x,y)∈Ltune log p(s = −1|x, vt ) tune x∈Tt on the tuning folds of test data and weighted pool (denoted by Ltune and Tttune ) over all ten cross-validation loops. [sent-211, score-1.158]

83 We follow [1] and smooth the estimated weights by rt (x)λ before including ˆ2 them into Optimization Problem 3. [sent-214, score-0.43]

84 Without looking at the test data of the target task we tune η on the non-target tasks so that the accuracy of the distribution matching method is maximized. [sent-216, score-0.848]

85 Finally, σw is tuned by cross-validation on L rescaled by rt (x, y)ˆt (x) (based on the previously ˆ1 r2 tuned parameters σu and σv ). [sent-220, score-0.545]

86 Figure 1 displays the accuracies over different numbers of labeled data for the four different target portals. [sent-222, score-0.487]

87 The error bars are the standard errors of the differences to the distribution matching method on labeled data (solid blue line). [sent-223, score-0.407]

88 For the “family” and “TV channel” portals the distribution matching method on labeled and unlabeled data outperforms all other methods in almost all cases. [sent-224, score-0.708]

89 The distribution matching method on labeled data outperforms the baselines trained only on the data of the target task for all portals and all data set sizes and it is at least as good as the one-size-ﬁts-all model in almost all cases. [sent-225, score-0.963]

90 The hierarchical Bayesian method yields low accuracies for smaller numbers of training examples but becomes comparable to the distribution matching method when training set sizes of the target portal increase. [sent-226, score-1.059]

91 The simple covariate shift model that trains only on labeled and unlabeled data of the target task does not improve over the iid model that only trains on the labeled data of the target task. [sent-227, score-1.382]

92 This indicates that the marginal shift between training and test distributions is small, or could indicate that the approximation of the reject distribution which we use in our experimentation is not sufﬁciently close. [sent-228, score-0.509]

93 Either reason also explains why accounting for the marginal shift in the distribution matching method does not always improve over distribution matching using only labeled data. [sent-229, score-0.77]

94 Transfer learning by distribution matching passes all examples for all tasks to the underlying logistic regressions. [sent-230, score-0.553]

95 For example, the single task baseline trains only one logistic regression on the examples of the target task. [sent-232, score-0.642]

96 This led to an algorithm that discriminatively estimates these resampling weights by training two simple conditional models. [sent-235, score-0.37]

97 After weighting the pooled examples over all tasks the target model for a speciﬁc task can be trained. [sent-236, score-0.65]

98 In our empirical study on targeted advertising, we found that distribution matching using labeled data outperforms all reference methods in almost all cases; the differences are particularly large for small sample sizes. [sent-237, score-0.66]

99 Distribution matching with labeled and unlabeled data outperforms the reference methods and distribution matching with only labeled data in two out of four portals. [sent-238, score-0.933]

100 Even with no labeled data of the target task the performance of the distribution matching method is comparable to training on 1600 examples of the target task for all portals. [sent-239, score-1.327]

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