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Old 12-16-2015, 05:46 AM
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Default About a kind of Pareto model

I am reading an old book "Foundations of Casualty Actuarial
Science (Fourth Edition)" by the CAS published in 2001, in order to learn some reinsruance pricing approaches.

In Chapter 7 "Reinsurance", there is a model called Censored Pareto model, as shown in the picture attached.

However, it can be seen that F(1) is not equal to 1 as there is no definition for x>1. So it is not a reasonable probability model, is it?

In addition, when it computed E[x;1], it utilized the limited expected value equation. But the latter only applies to a Pareto distribution which is not censored. So the result seems to have something wrong.

Anyone holds interest in that question? Thanks in advance.
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Old 12-16-2015, 09:42 AM
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F(1) does equal 1 since you are given that 1-F(1) = 0.

There is a point mass at 1 which corresponds to the probability that the uncensored distribution is bigger than 1.
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Old 12-16-2015, 09:54 AM
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Okay, so I agree that it is presented funny, but I don't have the full context, so I may be missing something. However, I'm pretty sure the misunderstanding is associated with this line:

Quote:
where the claim size X is expressed as a percent of MPL
So MPL being 10,000,000, x is 1 when the claim size is 10,000,000, .5 when claim size is 5,000,000 and so on. Is it possible that the claim size could be 20,000,000? From the perspective of the reinsurer, Absolutely. So if you run the integrand over the distribution for all of x it will evaluate to 1 as expected.

However, from the perspective of the insurer, reinsurance kicks in once x exceeds 1. There will never be a claim size of 120% of MPL, for example. Since severity is capped at 100%, the "normal" pareto formula for E[X;1] will correctly calculate the expected fraction of MPL for the severity.

-Riley
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Old 12-16-2015, 12:50 PM
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You know, I don't regret missing that game. I told the boys I had to go see about a Pareto model.
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Old 12-18-2015, 07:35 AM
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The integral x*d(F(x)) from 0 to 1 is not equal to 1. But there is a stacked probability at x=1, Pr(x=1)=(0.1/1.1)^2=1/121. So the integral from 0 to 1 plus Pr(x=1) is equal to 1.
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Old 04-20-2016, 02:40 PM
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You know, I don't regret missing that game. I told the boys I had to go see about a Pareto model.
that bastard stole my line
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Old 04-21-2016, 02:45 PM
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You know, I don't regret missing that game. I told the boys I had to go see about a Pareto model.
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that bastard stole my line
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