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#11
03-07-2012, 07:50 PM
 phd_actuary CAS SOA Join Date: Feb 2012 Location: TAMPA Studying for EXAM FM. College: University of south Florida Posts: 7

Now that is a helpful post. Need some answer checking now.
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#12
10-03-2012, 10:41 AM
 student823 Member SOA Join Date: Dec 2011 Posts: 260

I have a question about Sample Question 207. I understand the solution that the SOA gives, but it's not how I tried to do the problem and seems inconsistent with how we usually solve payment per payment problems. I used the formula for e(d), the mean excess loss, which is the expected payment per payment. Here is my solution. Please tell me what I'm doing wrong or why this doesn't work. Note that E[X ^ 4] is not the 4th raw moment, but the limited expected value.

e(4) = (E[X]-E[X ^ 4])/S(4)
The SOA and I both agree that S(4) = 0.84
E[X] = the integral of 0.02x^2 from 0 to 10, which is 6.666667
E[X ^ 4] = (the integral of 0.02x^2 from 0 to 4)*F(4) + 4*S(4) = .4266667*.16 + 4*.84 = 3.42826667
So then e(4) = (6.666667 - 3.428256)/0.84 = 3.855238488

I am calculating the INSURER'S expected payment per payment. That is what they are asking for, right?
#13
10-03-2012, 10:55 AM
 Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 31,301

Quote:
 Originally Posted by student823 E[X ^ 4] = (the integral of 0.02x^2 from 0 to 4)*F(4) + 4*S(4)
Where did you get that formula from?
#14
10-03-2012, 11:00 AM
 Celtics4Life Member SOA AAA Join Date: Jul 2011 Studying for Nothing. Favorite beer: Straight Edge Posts: 832

Quote:
 Originally Posted by student823 e(4) = (E[X]-E[X ^ 4])/S(4) The SOA and I both agree that S(4) = 0.84 E[X] = the integral of 0.02x^2 from 0 to 10, which is 6.666667 E[X ^ 4] = (the integral of 0.02x^2 from 0 to 4)*F(4) + 4*S(4) = .4266667*.16 + 4*.84 = 3.42826667 So then e(4) = (6.666667 - 3.428256)/0.84 = 3.855238488
The bolded part is not needed. The integration you perform of .02x^2 from 0 to 4 doesn't need to be multiplied by F(4), but rather doesn't need to be multiplied by anything.

Btw, your question may be moved I think this thread was mainly to point people to where other questions are answered.
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Last edited by Celtics4Life; 10-03-2012 at 11:01 AM.. Reason: typo
#15
10-03-2012, 11:19 AM
 student823 Member SOA Join Date: Dec 2011 Posts: 260

Quote:
 Originally Posted by Celtics4Life The bolded part is not needed. The integration you perform of .02x^2 from 0 to 4 doesn't need to be multiplied by F(4), but rather doesn't need to be multiplied by anything. Btw, your question may be moved I think this thread was mainly to point people to where other questions are answered.
Celtics4Life, thanks so much. That makes perfect sense.
#16
12-28-2012, 06:38 AM
 abglim Join Date: Jan 2010 College: UCL Alumni Posts: 4
You have done a great job to collect all these. Appreciate wholeheartedly

Thanks so much
#17
04-29-2013, 09:45 PM
 toshane Join Date: Oct 2007 Posts: 16
SOA 289 - question 196

Could you please post a detailed explanation for SOA 289 - question 196?

I do not understand why in the official answer, there is no f(x) for loss of 300 (with policy limit of 20,000) and how do you deal with loss >10,000 with policy limit of 10,000? thanks in advance.
#18
04-29-2013, 10:47 PM
 Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 31,301

Quote:
 Originally Posted by toshane Could you please post a detailed explanation for SOA 289 - question 196? I do not understand why in the official answer, there is no f(x) for loss of 300 (with policy limit of 20,000)
There is an f(300)^4 in the official solution
Quote:
 and how do you deal with loss >10,000 with policy limit of 10,000? thanks in advance.
There is (1-F(10,000))^6 in the official solution.
#19
05-12-2013, 04:01 PM
 toshane Join Date: Oct 2007 Posts: 16
#260

Stupid question, In #260, the answer used f(5) to calculate p(x=5 l theta=8). Why? Thanks.
#20
06-13-2013, 11:56 AM
 ActuaryActually SOA Join Date: Apr 2013 Posts: 27

Thank you very much!
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