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  #1  
Old 05-02-2008, 04:48 PM
njtp08 njtp08 is offline
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Question SOA sample #102

What's the meaning of ? Also I don't understand the way they found E(Dividend)?
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Old 05-02-2008, 05:09 PM
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Quote:
Originally Posted by njtp08 View Post
What's the meaning of ? Also I don't understand the way they found E(Dividend)?
is the probability that retained repair costs, for the two factories combined, are k. Since the retained repair costs at a single factory are 0 or 1 (insurance would pay any excess over 1), the combined costs are 0, 1 or 2.

E(dividend) = sum of [(dividend given repairs costs are k) * probability repair costs are k)]
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Old 10-24-2008, 10:39 PM
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Can anyone explain to me why I can't calculate retained major repair costs using 2*E[X^1]? thanks a lot
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Old 10-24-2008, 10:48 PM
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"Retained major repair costs" = Major repair costs - costs paid by insurer.

So if X (major repair costs, before applying deductible) = 1, then

Retained major repair costs = 1-1 = 0
but 1^1 = 1.

They are also different if X = 2 or X = 3.
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Old 10-24-2008, 11:00 PM
uncafe uncafe is offline
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If major repair costs before applying deductible =1, then retained major repair costs should be 1, right? Because insurer don't need to pay anything in this case with d=1.

Last edited by uncafe; 10-24-2008 at 11:37 PM..
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Old 10-26-2008, 05:15 PM
blahbla blahbla is offline
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Quote:
Originally Posted by uncafe View Post
If major repair costs before applying deductible =1, then retained major repair costs should be 1, right? Because insurer don't need to pay anything in this case with d=1.
there are two factories so there are two insurance contracts with ordinary deductible d, does that help?
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Old 10-08-2017, 08:05 PM
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Quote:
Originally Posted by Gandalf View Post
"Retained major repair costs" = Major repair costs - costs paid by insurer.

So if X (major repair costs, before applying deductible) = 1, then

Retained major repair costs = 1-1 = 0
but 1^1 = 1.

They are also different if X = 2 or X = 3.
I still don't understand this. If the repair cost is 1, shouldn't Retained repair cost = 1- 0 = 1?

Why does the insurer pay 1 if the problem states that the insurer pays the cost in excess of the ordinary deductible of 1?

Also, shouldn't E(X) = E(retained cost) + E(cost paid by insurer)

so 1 = 0.6 + 0.4 ?
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Old 10-08-2017, 08:34 PM
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Gee thanks. Remind me of stupid things I wrote back on 2008? Isnīt there a statute of limitations?

I should have just said that retained major repair costs are a random variable, not the expected value of a random variable. Calculating E(X^1) does not help you solve E(Dividend). I should have stopped without giving specific examples of retained repair costs, especially not wrong specific examples.
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Old 10-08-2017, 08:53 PM
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Isn't retained repair cost K ^ 1?

So for example,

k=0, retained repair cost = 0, Probability=0.4
k=1, retained repair cost = 1 Probability=0.3
k=2, retained repair cost = 1 Probability=0.2
k=3, retained repair cost = 1 Probability=0.1

What can't we say the expected retained repair cost is E(K ^ 1)=0.6 for each factory since

E(K)= 1 = E(K ^ 1) + E((K-1)+) and ((K-1)+)=0.4?

This is super confusing to me

Last edited by Ren_St; 10-08-2017 at 09:07 PM..
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Old 10-08-2017, 10:04 PM
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Why do you want the expected retained repair costs? You cannot use it to calculate the expected dividend, at least without further adjustment for retained claims = 2.

I.e, it is not true that Expected Dividend = Expected revenues - Expected insurance premiums - expected expenses - Expected retained repair costs. If you think itīs most efficient or understandable to calculate Expected retained repair costs and the adjustment term, go ahead. It would give you the right answer, which would match the SOA answer.
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