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Dr. Krzysztof Ostaszewski, FSA, FSAS, CERA, CFA, MAAA http://www.krzysio.net E-mail: [email protected] Errata for the Solutions Manual for Bowers' et al Actuarial Mathematics, 2007 Edition Posted November 2, 2010 Solutions to 6.4b and 6.4c should be: b. Let P be the premium rate sought. We have L = e- T - PaT and this is a decreasing function of T. The 50-th percentile of L is found at the value corresponding to the 50-th ln 2 ln 2 = 34.66 years. Based on this percentile of T , which is ln2 times the mean, or µ 0.02

0=e

so that

-0.06

ln 2 0.02

0.02 1- e -P 0.06

-0.06

ln 2

P=

e

-0.06

ln 2 0.02 ln 2

0.02 1- e 0.06 c. Let P be the premium rate sought. With zero force of interest e- T = 1 and the prospective loss function at policy duration 0 becomes

-0.06

1 2 -3 8 = 6 0.008571. = -3 = 7 100 7 100 1- 2 8 6 0.06

L = 1 - P dt = 1 - PT .

0

T

Then

1 1 0.50 = Pr L > 0 = Pr 1 - PT > 0 = Pr > T = Pr T < . P P

(

)

(

)

This implies that

1 ln 2 1 µ 0.02 , or P = = 0.028854. is the median of T, i.e., = P µ P ln 2 ln 2

Posted September 19, 2009 In the solution of Exercise 3.4b the last formula should be: 11 11 7 11 15 44 45 1 s (1) = - + - = - + =- + = > 0. 6 4 8 6 8 24 24 24 instead of 11 11 7 11 37 176 111 65 1 s = - + - =- + =- + =- 2 6 8 32 6 32 96 96 96.

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