Download A Bayesian predictive approach to determining the number of by Dey D. K., Kuo L., Sahu S. K. PDF

By Dey D. K., Kuo L., Sahu S. K.

This paper describes a Bayesian method of mix modelling and a mode in keeping with predictive distribution to figure out the variety of elements within the combinations. The implementation is finished by using the Gibbs sampler. the strategy is defined in the course of the combos of standard and gamma distributions. research is gifted in a single simulated and one actual facts instance. The Bayesian effects are then in comparison with the possibility procedure for the 2 examples.

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1 (Compatible conditional probability matrices). , P as described above) which has the columns and rows, respectively, of A and B as its conditional distributions. , N = {(i, j) : aij > 0}). As remarked earlier, compatibility of A and B can only occur if N A = B N . We will always assume that each row and each column of A (and B) contains at least one positive element (otherwise we would redefine the list of possible values to leave out the zero rows and/or columns). 1: A and B are compatible iff they have identical incidence sets and if there exist vectors u and v for which A cij = aij /bij = ui vj , ∀(i, j) ∈ N A .

7 (Cone generated by a matrix). Let A be a real matrix of dimension m × n. The polyhedral convex cone generated by A, denoted by π(A), is the set of all vectors in IR m which can be expressed as linear combinations of the columns of A with nonnegative coefficients. 8 (Dual or polar cone). Let π be a cone in IR m . The dual or polar cone of π is the set Ω(π) = {u ∈ IR m : v ′ u ≤ 0, ∀v ∈ π} . 2 Review and Extensions of Compatibility Results 27 In the case in which π is the cone generated by A, the dual cone admits a simple description involving the matrix A.

40) 32 2. Exact and Near Compatibility with elements dij = aij + bij . 40), it is necessary to make some assumptions about the incidence set of the matrix D. For example, D must not be block diagonal. A reasonable requirement that we will assume is that some power of D, perhaps D itself, have all elements strictly positive. 40). Then P ∗ must satisfy the following system of equations: p∗ij p∗ij + ∗ = dij , ∗ pi. j i = 1, 2, . . , I, j = 1, 2, . . , J. 41) using a simple iterative algorithm: (n) (n+1) pi,j (n) dij /[1/pi.

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