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  #51  
Old 11-12-2001, 04:21 PM
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Quote:
On 2001-11-09 21:40, phdmom wrote:
thing,

I put E on #32 also (same reasoning as you). But the key says D...any thoughts?
Any answers to #31??? Why can't d be right?
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  #52  
Old 11-12-2001, 04:47 PM
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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 twice-usual 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.

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  #53  
Old 11-12-2001, 04:57 PM
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#39

The increasing annuity cancels inflation, so a whole life annuity-due would be worth 1000e-dot<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
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  #54  
Old 11-12-2001, 05:07 PM
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Will it be possible to argue that #31 can be D considered the near infinity to supernova situation??
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  #55  
Old 11-12-2001, 05:20 PM
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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...
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  #56  
Old 11-12-2001, 06:47 PM
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#15

This is a (P-P)/P problem. (Batten helps)

(Px - P^(1)x:n)/Px:n^(1) = kVx
(WL - Term)/Endowment = kVx

so (.09-Term)/.00864 = .563
therefore Term = .0851

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  #57  
Old 11-12-2001, 06:54 PM
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#16

Note: All continuous, not discrete
A50 - P40 * a50
or
1-a60 / a50
among others

Since De Moivre's Law
A40 = a(w-x)/(w-x) => a60/60
A40=.3233
A50=.3742
so
a40=13.869
a50=12.826

Answer = .0752

No need to integrate!

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  #58  
Old 11-12-2001, 07:04 PM
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#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 2001-11-13 09:57 ]</font>
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  #59  
Old 11-13-2001, 08:53 AM
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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;Non-Smokers</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;Non-Smokers</sub>
= .3*(.06/(.06+2*.08)) + .7*(.03/(.03+2*.08))
= .192344

Var(a<sub>T(x)</sub>) = &delta;<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

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  #60  
Old 11-13-2001, 09:11 AM
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Well done all, especially you Math Guy.
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