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2.5. A closed-form expression for f d 59In fact, as soon as we know that SX d is a sum of exponentials multiplied by multivariatepolynomials, we know which β i ’s can appear in the multivariate polynomials. In<strong>de</strong>ed, as it isa fraction of f d , we know that SX d is finite and even boun<strong>de</strong>d between 0 and 1 for every tupleof β i ’s. But SX d would explo<strong>de</strong> as β i goes to infinity w<strong>here</strong>as the other ones are fixed if this β iappeared in a multivariate polynomial and not in the exponential.We can now prove the final step towards Proposition 2.5.1. We claim that for I ⊂ {1, . . . , d},SI d that we <strong>de</strong>fine as ∑SI d ={X|X i=2 if i∈I,X i≠2 if i∉I}already has an appropriate form, whence Proposition 2.5.1 becausef d (β 1 , . . . , β d ) =∑SI d .I⊂{1,...,d}For I, J ⊂ {1, . . . , d} such that I ∩ J = ∅, we <strong>de</strong>fine X(I, J) as the only vector in {0, 1, 2} dsuch that⎧⎨ 2 if i ∈ I,X i = 1 if i ∈ J,⎩0 otherwise.We <strong>de</strong>note SX(I,J) d simply by Sd I,J so thatS d I = ∑J⊂I c S d I,J .S XWe <strong>de</strong>fine in the same way T d I,J and T d Iand so on when I ≠ ∅.Proposition 2.5.15. The function SId is a symmetric function in the β i’s such that i ∉ I, aswell as in the β i ’s such that i ∈ I.For I = ∅, we haveS∅ d = 1 ∑3 d∑ 2 #J 4 − j∈J βj :J⊂{1,...,d}and for {d} ⊂ I = {j 0 + j 1 + 1, . . . , d}, we haveS d I = 2j23 j2 d∏i=j 0+j 1+1((1 − 4 −βi ) S d−j2∅− Ξ d I).∑For {d} ⊂ I = {j 0 + j 1 + 1, . . . , d}, Ξ d I is a sum for J ⊂ Ic of terms of the form 4 − j∈J βjmultiplied by a multivariate polynomial of <strong>de</strong>gree in β j exactly #I if j ∈ J, 0 otherwise, and oftotal <strong>de</strong>gree min(d − 1, #I#J).Proof. This assertion does not <strong>de</strong>pend on the exact value of I, but only on its cardinality #I,even if the value of SI d does: one has to permute the β i’s to <strong>de</strong>duce one from another. Hence, wecan assume that I = {j 0 + j 1 + 1, . . . , d}. The symmetry of SI d in each subset of variables followsfrom its <strong>de</strong>finition. The proof goes by induction on j 2 = #I.Suppose first that j 2 = 0, i.e. I = ∅. We go by induction on d. For d = 1,S 1 ∅ = S1 (0) + S1 (1) = f 1(β 1 ) = 2 3 4−β1 + 1 3 .

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