42 Mr. C. J. Hargreave on the Valuation of Lije Contingencies 



Value of annuity for life of survivor expectant on the death of 

 one of the two is Aj + Ag— 2AjA2; and value of present perpe- 

 tuity, payment of which is suspended from death of first to death 

 of secona, is 1 — (A, + Aj) 4- 2A1A2. 



Probability that both are living is 5 <^"(1) or s^ or a^a^] value 



of annuity on joint lives is Aj A^ ; value of perpetuity expectant 

 on death of one of the two is 1 — AjA^. 



Now let us take the contingencies from the lives of three 

 persons, Ai, Ag, and Ay. Here 



1 



-2^r(i)= 



>3- 



Probability that all are dead is 1— ^i+^g'^'^ai value of per- 

 petuity expectant on death of last survivor is 



1- (Ai + A2+ A3) + (A,A2 + AiA3 + A2A3) -AjAgAg; 



and, taking the complement, we have value of annuity for the 

 lives and life of the survivor 



Aj + A2 + A3- (A, A2 + A1A3 + A2A3) + Ai A2A3. 



Probability that one only is living is 51—2^2 + ^%; ^^^ value 

 of annuity for life of last survivor expectant upon the decease of 

 both the others is 



Ai + Ag + A3~2(Ai A2 + A1A3 + A2A3) + 3A1A2A3. 



Probability that one or none is living is 1 — ^g + 2^3 ; therefore 

 value of perpetuity expectant on the death of two of the three is 

 1 — (AiA2+AiA3H-A2A3)+2AiA2A3; and value of present an- 

 nuity payable so long as two or more of the three are living is 

 Ai Ag + Ai A3 + A2A3 — 2Ai A2A3. 



Probability that exactly two are living is ^2 — 3% ; and value 

 of annuity for joint life of the survivors expectant upon the 

 decease of the first of the three is 



AiA2 + AiA3 + A2A3— 3A1A2A3. 



Probability that two or one are living is s^—s^-, so that value 

 of annuity for life of the two survivors and the survivor of them 

 expectant on the decease of the first of the three is 



A1 + A2 + A3— (AiA2 + AiA3-f A2A3). 



Probability that all are living is s^; so that value of annuity 

 for the joint lives of the three is AjAgAg; and the complemen- 

 tary perpetuity is 1— AjAgAg. 



The above formulae thus solve every possible contingency that 



