Estimation, Evaluation, and Selection of Actuarial Models

More documents

Recommendations

Info

108 APPENDIX A. SOLUTIONS TO EXERCISES The derivative is l 0 = 14 w − 16 1 − w + 4 1 − 0.4w . Set the derivative equal to zero and clear the denominators to produce the equation 0 = 14(1 − w)(1 − 0.4w) − 16w(1 − 0.4w)+4w(1 − w) 0 = 14− 31.6w +8w 2 and the solution is w = q 35 =0.508 (the other root is greater than one and so cannot be the solution). Exercise 34 The survival function is and the density function is The likelihood function is S(t) = f(t) =−S 0 (t) = ½ e −λ 1 t , 0 ≤ t0.5) = 1 − 0.5w. For the one death, the contribution is Pr(T ≤ 1) = w. Then L = (1− w) 8 (1 − 0.5w)w ln L = 8ln(1− w)+ln(1− 0.5w)+lnw d ln L = − 8 dw 1 − w − 0.5 1 − 0.5w + 1 w . Setting the derivative equal to zero and clearing the denominators gives 0=−8w(1 − 0.5w) − 0.5w(1 − w)+(1− w)(1 − 0.5w).
109 The only root of this quadratic that is less than one is w =0.10557 = ˆq x . Exercise 36 For the two lives that died, the contribution to the likelihood function is f(10) while for the eight lives that were censored, the contribution is S(10). Wehave Therefore, ˆk =25. Exercise 37 We have f(t) = −S 0 (t) = 0.5 k L = f(10) 2 S(10) 8 = µ 1 − t −0.5 k µ 0.5 2 µ 1 − 10 k k ln L = 3ln(k − 10) − 5lnk d ln L 3 = dk k − 10 − 5 k =0 0 = 3k − 5(k − 10) = 50 − 2k. −1 µ 1 − 10 k L = f(1100)f(3200)f(3300)f(3500)f(3900)[S(4000)] 495 4 ∝ (k − 10)3 k 5 = θ −1 e −1100/θ θ −1 e −3200/θ θ −1 e −3300/θ θ −1 e −3500/θ θ −1 e −3900/θ [e −4000/θ ] 495 = θ −5 e −1,995,000/θ 1, 995, 000 θ = − 5 1, 995, 000 + θ θ 2 =0 ln L = −5lnθ − d ln L dθ and the solution is ˆθ =1, 995, 000/5 = 399, 000. Exercise 38 For maximum likelihood, the contributions to the likelihood function are (where q denotes the constant value of the time to death density function) Then, Event Contribution Survive to 36 Pr(T >1) = 1 − q Censored at 35.6 Pr(T >0.6) = 1 − 0.6q Die prior to 35.6 Pr(T ≤ 0.6) = 0.6q Die after 35.6 Pr(0.6
Page 1:
Estimation, Evaluation, and Selecti
Page 4 and 5:
iv CONTENTS 4.5.2 Anderson-Darlingt
Page 6 and 7:
2 CHAPTER 1. INTRODUCTION Exercises
Page 8 and 9:
4 CHAPTER 2. MODEL ESTIMATION Throu
Page 10 and 11:
6 CHAPTER 2. MODEL ESTIMATION 2.2 E
Page 12 and 13:
8 CHAPTER 2. MODEL ESTIMATION and t
Page 14 and 15:
10 CHAPTER 2. MODEL ESTIMATION of t
Page 16 and 17:
12 CHAPTER 2. MODEL ESTIMATION Exer
Page 18 and 19:
14 CHAPTER 2. MODEL ESTIMATION i d
Page 20 and 21:
16 CHAPTER 2. MODEL ESTIMATION 8. C
Page 22 and 23:
Page 24 and 25:
20 CHAPTER 2. MODEL ESTIMATION In e
Page 26 and 27:
22 CHAPTER 2. MODEL ESTIMATION Exam
Page 28 and 29:
24 CHAPTER 2. MODEL ESTIMATION That
Page 30 and 31:
26 CHAPTER 2. MODEL ESTIMATION Unle
Page 32 and 33:
28 CHAPTER 2. MODEL ESTIMATION like
Page 34 and 35:
30 CHAPTER 2. MODEL ESTIMATION For
Page 36 and 37:
32 CHAPTER 2. MODEL ESTIMATION like
Page 38 and 39:
34 CHAPTER 2. MODEL ESTIMATION wher
Page 40 and 41:
Page 42 and 43:
38 CHAPTER 3. SAMPLING PROPERTIES O
Page 44 and 45:
Page 46 and 47:
Page 48 and 49:
Page 50 and 51:
Page 52 and 53:
Page 54 and 55:
Page 56 and 57:
Page 58 and 59:
Page 60 and 61:
Page 62 and 63: 58 CHAPTER 3. SAMPLING PROPERTIES O
Page 64 and 65: 60 CHAPTER 3. SAMPLING PROPERTIES O
Page 66 and 67: 62 CHAPTER 4. MODEL EVALUATION AND
Page 90 and 91: 86 CHAPTER 5. MODELS WITH COVARIATE
Page 100 and 101: 96 APPENDIX A. SOLUTIONS TO EXERCIS
Page 102 and 103: 98 APPENDIX A. SOLUTIONS TO EXERCIS
Page 104 and 105: 100 APPENDIX A. SOLUTIONS TO EXERCI
Page 136 and 137: 132 APPENDIX B. USING MICROSOFT EXC
Page 142: 138 APPENDIX B. USING MICROSOFT EXC
show all

Estimation, Evaluation, and Selection of Actuarial Models

Create successful ePaper yourself

Delete template?

Save as template?