Download e-book for iPad: Abstract inference by Grenander U.

By Grenander U.

ISBN-10: 0471082678

ISBN-13: 9780471082675

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6 (A difficult case). 36) 1/2 0 1/2 30 2. 3. 6. w1 6 –6 –3 0 w2 12 –6 –6 0 w3 12 –6 –3 0 v1 30 –18 –12 1   1/3 2/3 0 B= 0 1/3 2/3  . 37) We cannot resolve this case using cross-product ratio matrices (there are no positive 2 × 2 submatrices). 2 may be used without difficulty. 38)  τ =   , τ ≥ 0,  1/3 −1/3   0 0     0 1/6   −1/6 0     0 0   0 0     0 −1/3 1/3 0 1 1 1 1 which by removing redundant equations can be written as:     1/6 0 −1/6 0 0   1/3 −1/6 0  τ =   , τ ≥ 0.

66) i=1 j=1 which can also be written as I J 2 2 2 (pij − (bij pi. j − bij pi. j − bij pi. ) /2 . j aij + pi. bij )/2, I i = 1, 2, . . , I, j = 1, 2, . . 68) J subject to i=1 j=1 pij = 1. j ’s). 68) over j for each fixed i, and over i for each fixed j, we are led to the following system of I + J linear equations:   J 1 pi. j aij + pi.  , i = 1, 2, . . j = 1 pi. j + 2 i=1 j = 1, 2, . . , J. 6 Related Discrepancy Measures 39 If we define an (I + J)-dimensional stochastic vector 1 (p1. , p2. , .

32) 3/4 1/2 B= 1/6 5/6 3/5 2/5 . 1. 4. 2. 5. 23) becomes     1/8 −3/20 0 5/12 −1/5   0     −1/8 3/20 τ =    0  , τ ≥ 0. 34) At this stage, if desired, we can remove redundant rows and linearly dependent rows before proceeding to computing the generators of the dual cone. For example, we can remove Equations 3 and 4, because they are exactly the same (sign changed) as Equations 1 and 2, respectively, to obtain the new system:     1/8 −3/20 0  5/12 −1/5  τ =  0  , τ ≥ 0. 2.

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Abstract inference by Grenander U.


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