I doubt the answer to Course 3,Fall 2004 #4:

Here is the problem: For a population which contains equal numbers of males and females at birth

(i) for male, force of mortality u1(x) = 0.1 ;

(ii) for female, force of mortality u2(x) = 0.08.

Calculate q(60)=?

My answer is:

P(60)= {P(60)|[it is male]} X Pr[it is male] + {P(60) | [it is female]} X Pr[it is female] = {exp( - 0.1X60 )} X 0.5 + {exp ( -0.08X60)}X0.5=0.914 ; Then q(60)=1-p(60) = 0.086

While the "right" answer shows:

S(61)={S(61)|[it is male]} X Pr[it is male] + {S(61) | [it is female]} X Pr[it is female] = {exp( - 0.1X61 )} X 0.5 + {exp ( -0.08X61)}X0.5=0.00492

S(60)={S(60)|[it is male]} X Pr[it is male] + {S(60) | [it is female]} X Pr[it is female] = {exp( - 0.1X60 )} X 0.5 + {exp ( -0.08X60)}X0.5=0.005354

then q(60)=1-P(60)=0.081

I am very confused why my method is not right? Does it really make difference if I condition directly on P(60) instead of condition on S(60)?

I would greatly appreciate if anybody could give me clues.

Here is the problem: For a population which contains equal numbers of males and females at birth

(i) for male, force of mortality u1(x) = 0.1 ;

(ii) for female, force of mortality u2(x) = 0.08.

Calculate q(60)=?

My answer is:

P(60)= {P(60)|[it is male]} X Pr[it is male] + {P(60) | [it is female]} X Pr[it is female] = {exp( - 0.1X60 )} X 0.5 + {exp ( -0.08X60)}X0.5=0.914 ; Then q(60)=1-p(60) = 0.086

While the "right" answer shows:

S(61)={S(61)|[it is male]} X Pr[it is male] + {S(61) | [it is female]} X Pr[it is female] = {exp( - 0.1X61 )} X 0.5 + {exp ( -0.08X61)}X0.5=0.00492

S(60)={S(60)|[it is male]} X Pr[it is male] + {S(60) | [it is female]} X Pr[it is female] = {exp( - 0.1X60 )} X 0.5 + {exp ( -0.08X60)}X0.5=0.005354

then q(60)=1-P(60)=0.081

I am very confused why my method is not right? Does it really make difference if I condition directly on P(60) instead of condition on S(60)?

I would greatly appreciate if anybody could give me clues.

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