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  #31  
Old 06-20-2016, 10:20 PM
newleaf9413 newleaf9413 is offline
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Unhappy Question #274

The answered provided by SOA is as follows:

S(x) = e^-H(x)

H(x) = 0.5534 (I got this part.)

Nelson-Aalen Estimate = 1/50 + 3/49 + 5/k + 7/12 = 0.5534

---> k = 36.

Why the last term is NOT 7/21? Did I miss something?

Please let me know at your earliest convenience thanks.


L
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  #32  
Old 06-20-2016, 10:28 PM
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Quote:
Originally Posted by newleaf9413 View Post
The answered provided by SOA is as follows:

S(x) = e^-H(x)

H(x) = 0.5534 (I got this part.)

Nelson-Aalen Estimate = 1/50 + 3/49 + 5/k + 7/12 = 0.5534

---> k = 36.

Why the last term is NOT 7/21? Did I miss something?

Please let me know at your earliest convenience thanks.


L
Hi newleaf, that's just a typo in the solution. The last term is actually 7/21. You'll notice if you plug in k = 35, and correct the last term, the equation sums correctly to .5534
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  #33  
Old 06-20-2016, 10:46 PM
newleaf9413 newleaf9413 is offline
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Thank you so much. I'm just trying to remember whatever I can. Wednesday is the day.
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  #34  
Old 07-22-2016, 09:57 AM
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I am struggling with question 70. I was able to get EPV by doing [.2(.5)+.8(1) + .2(.8) + .8(1] /2 . But i cannot get VHM. (2.3-1.8)^2 * .5*.5 which i think is intuitive.
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  #35  
Old 08-13-2016, 09:03 PM
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SOA Question 2:
The number of claims has a Poisson Dist.
Claims have a Pareto dist with parameters theta = .5 and alpha =6. The number of claims and claim sizes are independent.
The observed pure premium should be within 2% of the expected pure premium 90% of the time.

Calculate the expected number of claims needed for full credibility.


What key words in this question indicate that we are looking for number of claims needed for full credibility of aggregate losses?
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  #36  
Old 08-13-2016, 09:35 PM
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pure premium
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  #37  
Old 08-13-2016, 09:38 PM
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SOA Question 4:
How is a * 3^(-a-1) * a * 6^(-a-1) * a * 14^(-a-1) * (25^-a)^2
proportional to:
a^3 * (3*6*14*625)^-a

I understand how we get a^3, but I am not sure for the second piece how the exponent for all numbers can be -a when for three of those values, the exponent was (-a-1)?
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Old 08-13-2016, 09:45 PM
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I think you have a type in your last term.
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  #39  
Old 08-13-2016, 10:19 PM
Actuarialsuck Actuarialsuck is offline
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Quote:
Originally Posted by mathematicalhype View Post
SOA Question 4:
How is a * 3^(-a-1) * a * 6^(-a-1) * a * 14^(-a-1) * (25^-a)^2
proportional to:
a^3 * (3*6*14*625)^-a

I understand how we get a^3, but I am not sure for the second piece how the exponent for all numbers can be -a when for three of those values, the exponent was (-a-1)?
So you can see how the a^3 is in there right?

So now we have to deal with 3^{-a-1} * 6^{-a-1} * 14^{-a-1} * 25^{-2a} = 3^{-1} * 3^{-a} * 6^{-a} * 6^{-1} * 14^{-a} * 14^{-1} * 625^{-a}

Everything that is not raised to the a'th (or -a'th) power is a constant so hence we can say it's proportional to and you are left with (3*6*14*625)^{-a}
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  #40  
Old 08-21-2016, 03:19 PM
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Quote:
Originally Posted by Actuarialsuck View Post
So you can see how the a^3 is in there right?

So now we have to deal with 3^{-a-1} * 6^{-a-1} * 14^{-a-1} * 25^{-2a} = 3^{-1} * 3^{-a} * 6^{-a} * 6^{-1} * 14^{-a} * 14^{-1} * 625^{-a}

Everything that is not raised to the a'th (or -a'th) power is a constant so hence we can say it's proportional to and you are left with (3*6*14*625)^{-a}
Thank you!
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