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Short-Term Actuarial Math Old Exam C Forum

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  #11  
Old 03-07-2012, 07:50 PM
phd_actuary phd_actuary is offline
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Now that is a helpful post. Need some answer checking now.
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  #12  
Old 10-03-2012, 10:41 AM
student823 student823 is offline
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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?
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  #13  
Old 10-03-2012, 10:55 AM
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Gandalf Gandalf is offline
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Quote:
Originally Posted by student823 View Post
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?
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  #14  
Old 10-03-2012, 11:00 AM
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Celtics4Life Celtics4Life is offline
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Quote:
Originally Posted by student823 View Post

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
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  #15  
Old 10-03-2012, 11:19 AM
student823 student823 is offline
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Quote:
Originally Posted by Celtics4Life View Post
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.
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  #16  
Old 12-28-2012, 06:38 AM
abglim abglim is offline
 
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Default You have done a great job to collect all these. Appreciate wholeheartedly

Thanks so much
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  #17  
Old 04-29-2013, 09:45 PM
toshane toshane is offline
 
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Default 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.
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  #18  
Old 04-29-2013, 10:47 PM
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Originally Posted by toshane View Post
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.
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  #19  
Old 05-12-2013, 04:01 PM
toshane toshane is offline
 
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Default #260

Stupid question, In #260, the answer used f(5) to calculate p(x=5 l theta=8). Why? Thanks.
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  #20  
Old 06-13-2013, 11:56 AM
ActuaryActually ActuaryActually is offline
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Thank you very much!
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