Now, using (7) one sees that (9) is equivalent to ��N2 ? (?g(k)/?��N) = kg(k), k = 1,2,��. It follows that ��N2 ? (?��N/?��N) = ��k=1�ަ�N2 ? (?g(k)/?��N) http://www.selleckchem.com/products/17-AAG(Geldanamycin).html = ��k=1��kg(k) = ��N. Proof of Theorem 5 ��Let MY(t) = exp CY(t) be the moment generating function of Y. Expressed in terms of the mean scaled severity Z = Y/�� one has MY(t) = MZ(��t). The relationship (7) for the cgf CN(t) yields the series expansion:C(t)=CN(ln?MZ(��t))=��k=1��g(k)?MZk(��t)?��N.(15)By Theorem 2 one has X��C-�� if and only if the equation ��2 ? (?C/?��) ? ((?C/?t) ? ��) = 0 is satisfied. With the series representation for C(t) and the assumption ��2 ? (?��N/?��) = ��, this equation is equivalent to the following condition (use that MZ(t) does not depend on =MZ(��t)?��2?��k=1��?g(k)?��?MZk?1(��t).
(16)Now,??��):(��?��2t)?MZ��(��t)?��k=1��kg(k)?MZk?1(��t) by Lemma 7 and (10), one has the identity (use the differential chain rule)��N??g(k)?��=?��N?��?kg(k),(17)which, together with ��2 ? (?��N/?��) = ��, implies that��2??g(k)?��=��2��N??��N?��?kg(k)=�̦�N?kg(k).(18)Inserted into the above expression one obtains the ordinary differential equation:��N?(1?��2t)?MZ��(t)=MZ(t),(19)whose unique solution is MZ(t) = (1 ? ��2t)?��. Since ��2�� = ��Y/��, one sees thatCY(t)=ln?MZ(��t)=��?ln?��(��?��Yt)(20)is the cgf of a gamma-distributed random variable. The proof is complete.The proof uses the so-called natural parameterization (p, ��N, ��Y, ��) of the compound gamma distribution. It is interesting to obtain explicit parameters orthogonal to the means of N, Y, and X.
By the assumption N��C-�� one has ��N = ��N(p, ��N)��N, and since Y is gamma distributed, one has Y��C-�� with ��Y��. It remains to construct a parameter vector orthogonal to the mean of X such that��=��N(p,��N)?��Y��(��N,��,��),(21)where �� = ��(p, ��N, ��Y, ��) must be determined. This task can be solved in a unified way for a lot of counting distributions (see [18, Section 4]). To illustrate the method, it suffices to consider here a single example.Example 9 (compound negative binomial gamma distribution) ��Let N ~ NB(��N, p), ��N > 0, p (0,1), be a negative binomial random variable. Its cumulant pgf (6) reads G(s) = ?��N ? ln (1 ? p ? s), ��N = ?��N ? ln (1 ? p). One has the following identity (see [18, equation (4.7)]):p?G(s)?p=s?G��(s),(22)which implies for s = 1 that p(?��N/?p) = ��N. Together, this shows that (10) is satisfied.
Therefore, one has N��C-�� and ��N = ��N ? p/(1 ? p)��N. Now, by Theorem 5 the compound negative binomial will be a compound negative binomial gamma if X��C-�� and ��2 ? (?��N/?��) = �� is satisfied. Written in terms of the Brefeldin_A parameter �� = (��N��2)?1, �� = ��/��, the latter equation is equivalent to the condition (1/��N)?(?��N/?��) = ��/��. With ?��N/?�� = (?��N/?p)?(?p/?��) = (��N/p)?(?p/?��) one obtains the differential equation (1/p)?(?p/?��) = ��/��, which has the solution p = �� ? �̦� for some ��.
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