By Magnus Ekdahl.

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**Sample text**

Let Ti be the i’th tree. v’s that are in the i’th tree. Let vi denote the subset of a sample x that is included in the i’th tree, with Vi as the set all possible values of vi . For the purpose of ﬁnding the point with maximal probability the distributive law will be useful. 1 When G is a forest arg max Pξ (x|G) = x∈X arg max PV1 (v1 |G) v1 ∈V1 ... arg max PVt (vt |G) vt ∈Vt (53) February 13, 2006 (13:19) 47 Proof This follows since Pξ (x|G) = PV1 (v1 |G) · . . 1, equation (52)) So if we can compute arg maxvi ∈Vi PVi (vi |G) eﬀectively we can also compute arg maxx∈X Pξ (x|G) eﬀectively.

Samples in x(n) will be independent given the BN, but not necessarily identically distributed unless they are in the same class. 1, which derives the complete SC for both classes and the samples given the class models. 1. 2 we will discuss the computational complexity of ﬁnding the SC optimal model for a Bayesian Network. 1 SC for a given number of classes, k In this subsection we will explain how SC is calculated for a given classiﬁcation of size k. Letting cl be the class of sample l the prevalence of class index j is λj = Pς (cl ).

6 1 − Pξ (y) Eξ(n) − log Pˆξ(n) (ξ (n) ) n · 2 log 2 (36) Proof We start by proving that 1 − Pξ (y) 1. 1. February 13, 2006 (13:19) (37) 34 2. We assume that the assumption holds for n = j, and show that it holds for n = j + 1. ,xj xj+1 Pξj+1 |ξ(j) (xj+1 |x1 , . . , xj )Pξ(j) (x1 , . . , xj ) · log2 (Pξj+1 |ξ(j) (xj+1 |x1 , . . , xj )Pξ(j) (x1 , . . ,xj Pξ(j) (x1 , . . , xj ) xj+1 Pξj+1 |ξ(j) (xj+1 |x1 , . . , xj ) · log2 (Pξj+1 |ξ(j) (xj+1 |x1 , . . ,xj Pξ(j) (x1 , . . , xj ) log2 (Pξ(j) (x1 , .