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Extra resources for Mathematical Foundations of Actuarial Science. Society of Actuaries - Course 1. Casualty Actuarial - Exam 1
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13. 0; 1. Additionally, let ˛i 2 N, then as p˛;n;Qj ! 0. 80) ! 0 when n ! 1, we get Proof. Let ˛i 2 N for i D 1; 2; : : : ; N 2 N. 0; 1. 0; 1. 70), we have 1 P 0 Ä p˛;n;Qj Ä 1D see also [2]. 83) ! 0 when n ! 1, p˛;n;Qj ! 0. Â 1 Ã P 1 2 Now, let ˛i D 0. 14. Let m 2 N, f 2 Cm RN , N 49 1, x 2 RN . Assume N P C < 1, for all ˛j 2 Z , j D 1; : : : ; N W j˛j WD ˛j D m. Let @m f . 0; 1. 87) 1 ! 0, as n ! 88) given that kf˛ k1 < 1, for all ˛ W j˛j D Qj, Qj D 1; : : : ; m. Furthermore, for ˛j 2 N, as n !
If ˛i is odd, ; if ˛i is even. see also [2]. Assume that ˛i is even. 98) Then 1 X ˛i i 2˛O i C 2˛O n Á ˇ i D1 for all i when ˇ > Ä 1 X ˛i 2˛ˇ O i D i D1 ˛i C1 . 99), we get 2N ˛ˇ O K 3n ; n i ! 100), for ˛O 2 N and ˇ > n ! 1, q˛;n;Qj ! 0. 100) we have as n !
52), we get uQ; n Ä R3;˛i àà ÂN Á k k2 r Q 2˛O 2˛O 1C j i j˛i i C n 1D 1 ND 1  N iD1 à  Ãnr  N iD1 Á ˇÃ 1 1 P P Q Q 2˛O k k 2N ˛ˇ O ˛i 2 N 2˛O 1C 2 ::: D n i i C n n iD1 iD1 1 D1 N D1   à r Á ˇÃ 1 N 1 P P Q i 2N ˛ˇ O ˛i 2˛O 2˛O 1 C 2NCr ::: C Ä Nr n n i i n iD1 1 D1 N D1 ! 0; 1. 0; 1, ˛i 2 N, ˇ > For ˛i D 0, we observe that M2;0 D 2 1CrC˛i , 2˛O 1 P 2˛ˇ O n i D1 Ä2  1C i Ãr 2˛O i n  i 1C 1 P 2˛ˇ O n and i D 1; : : : ; N. 0; 1, and ˇ > rC2 . 58), the proof is done. 9. A. Anastassiou and M.