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  #1  
Old 08-17-2003, 05:45 PM
shluffer shluffer is offline
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Default Mahler Ruin Theory question 1.1

The Tan teen Insurance Company is set up soley to jointly insure 7 independent lives.
Each life has a future lifetitme which follows a Weibull Distribution rho = 2 and theta = 30
tan teen pays 50 upon the death of the last survivor
calculate the probobility of ruin of tan teen

Mahler's answer = .56%

M<y answer = .35%

the difference is that I assumed that since assets grow continuosly at 10% per year that we need 7 deaths before 5e^(.1t) = 50, solve for t. Mahler used 5*1.1^t=50. why doest "Grow continuosly" infer continuose compounding? Or does it?

*edited to change row to rho
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Old 08-17-2003, 11:58 PM
Been There Done That Been There Done That is offline
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exp(tlog(1.1)) = 1.1^t

By the way, rho is spelled "rho".
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Old 08-18-2003, 10:31 AM
shluffer shluffer is offline
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I'm not sure I get what you are saying. If I solve for t using my equation I get a diffrent answer than if I solve for t using mahlers. The eqution you stated is the same one I used and assumes continuose compounding where the one Mahler uses does not.
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Old 08-18-2003, 12:32 PM
Been There Done That Been There Done That is offline
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I cannot understand what you wrote, and it's not just the spelling.

log(1.1) <> .1

log here is natural log
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  #5  
Old 08-18-2003, 01:33 PM
Howard Mahler Howard Mahler is offline
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Growing continuosly at 10% per year means the value at time t is (1.1^t) times the value at time zero.
If assets are 5 at time 0, they 5.5 at time 1, and 6.05 at time 2, etc.

If instead one mutliplies by Exp[.1t], one has the analog of a force of interest of .1, and if assets are 5 at time 0, they 5.526 at time 1, 6.107 at time 2, etc. Most people would describe this as growing continuosly at 10.52% per year.

This is my undestanding of the common usage of the terms. If you are relying on some place where the terms are used differently, please let me know. hmahler@mac.com

Howard Mahler
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