Privacy Preserving OLAP

1. Privacy Preserving OLAP Rakesh Agrawal, IBM Almaden Ramakrishnan Srikant, IBM Almaden Dilys Thomas, Stanford University

2. Horizontally Partitioned Personal Information Client C1 Original Row r1 Perturbed p1 Table T for analysis at server Client C2 Original Row r2 Perturbed p2 EXAMPLE: What number of children in this county go to college? Client Cn Original Row rn Perturbed pn

3. Vertically Partitioned Enterprise Information EXAMPLE: What fraction of United customers to New York fly Virgin Atlantic to travel to London? Original Relation D1 Perturbed Relation D’1 Perturbed Joined Relation D’ Original Relation D2 Perturbed Relation D’2

4. Talk Outline • Motivation • Problem Definition • Query Reconstruction • Privacy Guarantees • Experiments

5. Privacy Preserving OLAP Compute select count(*) from T where P1and P2 and P3 and …. Pk i.e. COUNTT( P1and P2 and P3 and …. Pk ) We need to provide error bounds to analyst. provide privacy guarantees to data sources. scale to larger # of attributes.

6. Uniform Retention ReplacementPerturbation 5 4 3 BIAS=0.2 1 3 HEADS: RETAIN TAILS: REPLACE U.A.R FROM [1-5]

7. Retention Replacement Perturbation • Done for each column • The replacing pdf need not be uniform • Different columns can have different biases for retention

8. Talk Outline • Motivation • Problem Definition • Query Reconstruction Inversion method Single attribute Multiple attributes Iterative method • Privacy Guarantees • Experiments

9. Single Attribute Example What is the fraction of people in this building with age 30-50? • Assume age between 0-100 • Whenever a person enters the building flips a coin of bias p=0.2 for heads say. • Heads -- report true age • Tails -- random number uniform in 0-100 reported • Totally 100 randomized numbers collected. • Of these 22 are 30-50. • How many among the original are 30-50?

10. Analysis 20 Retained 80 Perturbed Out of 100 : 80 perturbed (0.8 fraction), 20 retained (0.2 fraction)

11. Analysis Contd. 16 Perturbed, Age[30-50] 20 Retained 64 Perturbed , NOT Age[30-50] 20% of the 80 randomized rows, i.e. 16 of them satisfy Age[30-50]. The remaining 64 don’t.

12. Analysis Contd. 6 Retained, Age[30-50] 16 Perturbed, Age[30-50] Since there were 22 randomized rows in [30-50]. 22-16=6 of them come from the 20 retained rows. 14 Retained , NOT Age[30-50] 64 Perturbed, NOT Age[30-50]

13. Scaling up 30 ? Thus 30 people had age 30-50 in expectation.

14. Multiple Attributes (k=2)

15. Architecture

16. Formally : Select count(*) from R where P p = retention probability (0.2 in example) 1-p = probability that an element is replaced by replacing p.d.f. b = probability that an element from the replacing p.d.f. satisfies predicate P ( in example) a = 1-b

17. Transition matrix = i.e. Solve xA=y A00 = probability that original element satisfies : P and after perturbation satisfies : P p = probability it was retained (1-p)a = probability it was perturbed and satisfies : P A00 = (1-p)a+p

18. Multiple Attributes For k attributes, • x, y are vectors of size 2k • x=y A-1 Where A=A1 ­A2­ .. ­ Ak [Tensor Product]

19. Error Bounds • In our example, we want to say when estimated answer is 30, the actual answer lies in [28-32] with probability greater than 0.9 • Given T !a T’ , with n rows f(T) is (n,e,d) reconstructible by g(T’) if |f(T) – g(T’)| < max (e, e f(T)) with probability greater than (1- d). f(T) =2,  =0.1 in above example

20. Results Fraction, f, of rows in [low,high] in the original table estimated by matrix inversion on the table obtained after uniform perturbation is a (n, ,  ) estimator for f if n > 4 log(2/)(p )-2 , by Chernoff bounds Vector, x, obtained by matrix inversion is the MLE (maximum likelihood estimator), by using Lagrangian Multiplier method and showing that the Hessian is negative

21. Talk Outline • Motivation • Problem Definition • Query Reconstruction Inversion method Iterative method • Privacy Guarantees • Experiments

22. Iterative Algorithm [AS00] Initialize: x0=y Iterate: xpT+1 = Sq=0t yq (apqxpT / (Sr=0t arq xrT)) [ By Application of Bayes Rule] Stop Condition: Two consecutive x iterates do not differ much

23. Iterative Algorithm • RESULT [AA01]: The Iterative Algorithm gives the MLE with the additional constraint that 0 < xi , 8 0 < i < 2k-1

24. Talk Outline • Motivation • Problem Definition • Query Reconstruction • Privacy Guarantees • Experiments

25. Privacy Guarantees Say initially know with probability < 0.3 that Alice’s age > 25 After seeing perturbed value can say that with probability > 0.95 Then we say there is a (0.3,0.95) privacy breach

26. Privacy Guarantees (1,2) Privacy breach (s,1,2) Privacy breach • Let X, Y be random variables where X = original value, Y= perturbed value. Let Q, S be subsets of their domains • Apriori Probability P[ X 2 Q] = Pq·r1 • Posteriori Probability P[X 2 Q | Y 2 ] ¸r2 where 0 < r1 < r2 < 1 and P[Y 2 ] > 0 Q S S Q Where pq/mq < s, i.e. Q is a rare set (mq = probability of Q under replacing pdf)

27. (s, r1, r2) vs (r1,r2) metric • Provides more privacy to rare sets e.g. : in market basket data, medicines are rarer than bread, so we provide more privacy for medicines than for bread • For multiple columns, s expresses correlations • Works for retention replacement perturbation on numeric attributes

28. (s,1,2) Guarantees The median value of s is 1 There is no (s,1,2) privacy breach for s < f(1,2,p) for retention replacement perturbation on single as well as multiple columns

29. Application to Classification[AS00] • For the first split to compute split criterion/gini index Count(age[0-30] and class-var=‘-’) Count(age[0-30] and class-var=‘+’) Count(: age[0-30] and class-var=‘-’) Count(: age[0-30] and class-var=‘+’)

30. Talk Outline • Motivation • Problem Definition • Query Reconstruction • Privacy Guarantees • Experiments

31. Experiments • Real data: Census data from the UCI Machine Learning Repository having 32000 rows • Synthetic data: Generated multiple columns of Zipfian data, number of rows varied between 1000 and 1000000 • Error metric: l1 norm of difference between x and y. Eg for 1-dim queries |x1 – y1| + | x0 – y0|

32. Inversion vs Iterative Reconstruction 2 attributes: Census Data 3 attributes: Census Data Iterative algorithm (MLE on constrained space) outperforms Inversion (global MLE)

33. Privacy Obtained Privacy as a function of retention probability on 3 attributes of census data

34. Error vs Number of Columns: Census Data Inversion Algorithm Iterative Algorithm Error increases exponentially with increase in number of columns

35. Error as a function of number of Rows Error has square root n dependence on number of rows

36. Conclusion Possible to run OLAP on data across multiple servers so that probabilistically approximate answers are obtained and data privacy is maintained The techniques have been tested experimentally on real and synthetic data. More experiments in the paper. PRIVACY PRESERVING OLAP is PRACTICAL

37. References • [AS00] Agrawal, Srikant: Privacy Preserving Data Mining • [AA01] Agarwal, Aggarwal: On the Quantification of… • [W65] Randomized Response.. • [EGS] Evfimievski, Gehrke, Srikant: Limiting Privacy Breaches.. Others in the paper..

38. Error vs Number of Columns:Iterative Algorithm: Zipf Data The error in the iterative algorithm flattens out as its maximum value is bounded by 2

39. Supported by Privacy Group at Stanford: Rajeev and Hector