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Howard Mahler
02-22-2005, 03:00 PM
F.Y.I.

Howard:

Gwendolyn Anderson has addressed your concerns. The updated information is posted under "2005 Syllabus Update" (see http://www.casact.org/admissions/syllabus/2005/update.htm).

Regards,
Tom
-- Thomas Downey
Casualty Actuarial Society

Howard Mahler wrote:

In the second edition of Anderson, Insurance to Value (April 2004).
In Example 7 at page 22, it should read: In one in four cases, the damage will amount to 50% of the property value.
In equations 17 at page 19, it would be worthwhile to note that the conditional expected value = Ff.
Howard Mahler

BassFreq
02-23-2005, 01:41 PM
If no one else is going to say it, then I will......

Thank you!

Tommy Vercetti
02-10-2006, 01:41 PM
about question 54 in Mahler's study note,
The solution treat the 80% coinsurance provision as the insured would pay 20% of the expense excess the deductible.

However, page 18 of the note treat the 80% coinsurance provision as the insured would pay 80% of the expense excess the deductible.

Which one is correct??

also, questions 45, I got 0.44, which is E, not B.

Howard Mahler
02-10-2006, 09:52 PM
My solution agrees with that of the CAS for this exam Q.
As stated, in the comment on question 54, the definitions in Personal Insurance and Anderson differ:

Comment: Intended to be answered from page 12.10 of Personal Insurance. See also pages 1 and 2 of Anderson, where a 20% coinsurance clause means the insured retains 20%, while Personal Insurance describes the same provision as an 80% coinsurance clause, since the insurer pays 80%.

Q.45 is a CAS exam question on property insurance.
Again I believe that my solution agrees with that of the CAS.

Howard

P.S. Please send me any future questions by Email.

about question 54 in Mahler's study note,
The solution treat the 80% coinsurance provision as the insured would pay 20% of the expense excess the deductible.

However, page 18 of the note treat the 80% coinsurance provision as the insured would pay 80% of the expense excess the deductible.

Which one is correct??

also, questions 45, I got 0.44, which is E, not B.

pythag0ras
02-15-2006, 11:44 AM
Shouldn't the conditional expected value E(I|F<L<infinity) be equal to F rather than Ff? L>F means that a nonzero loss has occurred, so since a nonzero loss is assumed in the condition I think it is redundant and incorrect to multiply by f. By that same logic I think equations 16.a, 16.b, 17.a, and 17.b should all not have f in the numerator since the conditions in those conditional expected values also all assume a nonzero loss L. Equation 18.a and all of the equations after that point are correct to include f as a factor, though, since the equations are not conditional expected values so a nonzero loss is not a given. Am I missing something or on to something?

shluffer
02-15-2006, 02:24 PM
FYI

P.S. Please send me any future questions by Email.

pythag0ras
02-15-2006, 04:03 PM
My post wasn't really a question for Mr. Mahler but for the board in general: is the E(I|F<L<infinity) = Ff equation on p. 19 of the Anderson text correct?

SouthOfSanity
02-23-2006, 12:30 PM
Shouldn't the conditional expected value E(I|F<L<infinity) be equal to F rather than Ff? L>F means that a nonzero loss has occurred, so since a nonzero loss is assumed in the condition I think it is redundant and incorrect to multiply by f. By that same logic I think equations 16.a, 16.b, 17.a, and 17.b should all not have f in the numerator since the conditions in those conditional expected values also all assume a nonzero loss L. Equation 18.a and all of the equations after that point are correct to include f as a factor, though, since the equations are not conditional expected values so a nonzero loss is not a given. Am I missing something or on to something?

The equations in the paper are correct. Working based on equation 16, f is multiplied with s(L) to get the unconditional probability of L. Using more conventional probability notations, f = Pr(L>0) and s(L) = Pr(L=l|L>0).
f*s(L) is therefore equivalent to Pr(L>0)*Pr(L=l|L>0), which by basic probability axioms would equal Pr(L=l).
To get the [conditional] expected indemnity payments, you would take the unconditional expected value divided by the probability of the condition. That is why you would need the unconditional probability, e.g. f*s(L), in the numerator of the formulas.