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1 A Model for an Intrusion Prior Related to Example 4.2: n=10,N=100 (Slides 25-27) Jim Lynch NISS/SAMSI & University of South Carolina

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2 Framework Y=(y 1,…, y N ) is the Population X*=(x* 1,…, x* n ) is the Released Data The Released Data, X*, will be a random sample, X=(x 1,…, x n ), of Y after it has been masked

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3 The Random Sample X Let P be a random permutation of size n from N={1,2,…,N} where is uniformly distributed over these permutations. For P=(j(1),…,j(n)), X =(x 1,…, x n )=y P =( y j(1),…, y j(n) ) So f(X|Y)=1/ N P n for X =y P

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4 The Masked Released Data X* For X =(x 1,…, x n )=y P =( y j(1),…, y j(n) ), x* i = x i m j(i) = y j(i) m j(i) where y j(i) and m j(i) are independent with m j(i) ~ g j(i),P,PThus, f(X*|Y,P) = P g j(i) (x* i /x i )/x i = f(X*|X,P) where P denotes the product over i=1,…,n with P=(j(1),…,j(n)),PP,PSo, f(X*|Y)= P f(X*|Y,P) f(P|Y)= P f(X*|Y,P)/ N P n

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5 The Prior To construct a prior we need to know the parameter space. Two choices for this, Y and (Y,P). I think it is easier to work with (Y,P).,P,P –Note that since f(X*|X,P)= f(X*|Y,P) in the second case, we can consider the reduced parameter space (X,P). A prior on (X,P) –The intruder is interested in the smallest y WLOG label the smallest as y 1 –Doesnt care about inferring anything else about the population –The intruder uses a supra model to model Y y 2,…, y N where are iid f 2 =f y 1 has density f 1 =Nf(1-F) N-1 This model doesnt seem that unreasonable but, given y 1, perhaps we should have used that y 2,…, y N are iid f 2 (x)=f(x)/(1-F( y 1 )) with x> y 1

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6 The prior (X,P) If 1 is not in P, (X,P)= f 2 (x i )/ N P n If 1 is in P and j(i)=1, (X,P)=f 1 (x i ) i f 2 (x k )/ N P n where i is over k not equal to i

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7 The posterior (X,P|X*) If 1 is not in P, (X,P|X*) P g j(i) (x* i /x i )/x i f 2 (x i )/ N P n If 1 is in P = (j(1),…,j(n)) and j(i)=1, (X,P|X*) [f 1 (x i )g 1 (x* i /x i )/x i ] i f 2 (x k )g j(k) (x* k /x k )/x k ]/ N P n where i is over k not equal to i Note that the posterior involves [f 1 (x i )g 1 (x* i /x i )/x i ] i f 2 (x k )g j(k) (x* k /x k )/x k ]. Last time we seemed to integrate this out to do a posterior calculation. In any event the posterior is complicated but calculable –Use this to calculate P(j(i)=1|X*)=p i1 for i=1,…,n –Calculate (P|X*) and (X|P,X*)= (X,P|X*)/ (P|X*) If 1 is in P and j(i)=1(so, y 1 =x i ), (X|P,X*) [f 1 (x i )g 1 (x* i /x i )/x i ] i f 2 (x k )g j(k) (x* k /x k )/x k ]/ N-1 P n-1 Integrating this out over x k, k not equal to i, gives, (dy 1 |P,X*) = [f 1 (x i )g 1 (x* i /x i )/x i ]C(P,X*) Now, (dy 1 |X*) = (dy 1 |P,X*) (P|X*) = [f 1 (x i )g 1 (x* i /x i )/x i ]C(P,X*) (P|X*) where the is the sum over those P where 1 is in P. Note that (dy 1 |X*) is a subdistribution. Integrating y 1 out gives the probability that y 1 has been released and equals [h 1 (x* i )]C(P,X*) (P|X*)= p i1 where h 1 is the multiplicative convolution of f 1 and g

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8 The posterior (X,P|X*) –If 1 is not in P, (X,P|X*) P g j(i) (x* i /x i )/x i f 2 (x i )/ N P n –If 1 is in P = (j(1),…,j(n)) and j(i)=1, (X,P|X*) [f 1 (x i )g 1 (x* i /x i )/x i ] i f 2 (x k )g j(k) (x* k /x k )/x k ]/ N P n where i is over k not equal to i On slide 25 g i =g for all i. –If 1 is not in P, (X,P|X*) g(x* i /x i )/x i f 2 (x i )/ N P n –If 1 is in P = (j(1),…,j(n)) and j(i)=1, (X,P|X*) [f 1 (x i )g(x* i /x i )/x i ] i f 2 (x k )g(x* k /x k )/x k ]/ N P n where i is over k not equal to i. –Integrating out over X gives (where h i is the multiplicative convolution of f i and g) (P|X*) h 2 (x* i )/ N P n if 1 is not in P (P|X*) h 1 (x* i ) i h 2 (x* k )/ N P n if 1 is in P = (j(1),…,j(n)) and j(i)=1 Thus, (j(i)=1|X*) = [ N-1 P n-1 h 1 (x* i ) i h 2 (x* k )/ N P n ]x{[ N-1 P n-1 h 1 (x* i ) i h 2 (x* k )/ N P n ]+[ N-1 P n h 2 (x* i )/ N P n ]} -1 = [h 1 (x* i )/h 2 (x* i )]x{[h 1 (x* i )/h 2 (x* i )]+(N-n)} -1 For Lamberts example, N=100 and n=10. So, (j(i)=1|X*) =[h 1 (x* i )/h 2 (x* i )]x{[h 1 (x* i )/h 2 (x* i )]+90]} -1 which agrees with Lamberts answer in Section of her paper.

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