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#51




Quote:

#52




To HOSS re: 31
Under the stated conditions, ruin is possible in year 1 if there are twice as many claims as usual in year 1. That's possible, so ruin is possible. Suppose the initial surplus was 60 instead of 6. It would take 10 consecutive years of having twice as many claims as usual, but ruin could still occur. Set the initial surplus as high as you like, and I can tell you how many consecutive years of twiceusual claims to ruin. You mentioned (in another thread) that the surplus could be set to "near infinite", but the time frame is actually infinite. You're thinking just because if we had a billion surplus that ruin wouldn't happen before the sun goes nova, that ruin is impossible. 
#53




#39
The increasing annuity cancels inflation, so a whole life annuitydue would be worth 1000edot<sub>x</sub> = 11050. Take away 3000 for the deferal period to get the present value of the deferred annuity is 8050. We need P such that P + .99P/1.04 + (.99)(.98)P/1.04^2 = 8050. This solves to P = 2825.6247..., B 
#54




Will it be possible to argue that #31 can be D considered the near infinity to supernova situation??

#55




by my count, the only 4 questions which have not yet been addressed (in this thread or on another) are 15, 16, 17, and 38. But personally, the obsessive need to know I can actually solve these problems (given enough time) is beginning to burn off...

#56




#15
This is a (PP)/P problem. (Batten helps) (Px  P^(1)x:n)/Px:n^(1) = kVx (WL  Term)/Endowment = kVx so (.09Term)/.00864 = .563 therefore Term = .0851 
#57




#16
Note: All continuous, not discrete A50  P40 * a50 or 1a60 / a50 among others Since De Moivre's Law A40 = a(wx)/(wx) => a60/60 A40=.3233 A50=.3742 so a40=13.869 a50=12.826 Answer = .0752 No need to integrate! 
#58




#38
use formula p'(j)x=p(t)x^[q(j)/q(t)x] p(t)64=(1.025)*(1.035)*(1.2)=.7527 q^(1)64=.02204 2q^(1)64=q^(1)64+p^(t)64*q^(1)65 2q^(1)64=.02204+.7527*(.02716)=.04248 <font size=1>[ This Message was edited by: dinosaur on 20011113 09:57 ]</font> 
#59




Drum roll please...
17. Just use weighted averages to determine A<sub>x</sub> and <sup>2</sup>A<sub>x</sub>: A<sub>x</sub> = .3*A<sub>x;Smokers</sub> + .7*A<sub>x;NonSmokers</sub> = .3*(.06/(.06+.08)) + .7*(.03/(.03+.08)) = .31948 <sup>2</sup>A<sub>x</sub> = .3*<sup>2</sup>A<sub>x;Smokers</sub> + .7*<sup>2</sup>A<sub>x;NonSmokers</sub> = .3*(.06/(.06+2*.08)) + .7*(.03/(.03+2*.08)) = .192344 Var(a<sub>T(x)</sub>) = δ<sup>2</sup>(<sup>2</sup>A<sub>x</sub>  A<sub>x</sub><sup>2</sup>) = .08<sup>2</sup>(.192344  .31948<sup>2</sup>) = 14.105 (D) And that's that 
#60




Well done all, especially you Math Guy.

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