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#1
09-24-2013, 05:23 PM
 V1per41 Member SOA Join Date: Apr 2010 Location: Colorado Studying for FAP College: Purdue University Alumni Favorite beer: Shipyard Smashed Pumpkin Posts: 2,137
Single/Double Parameter Pareto

ASM Manual Exercise #16.21

-----------------------------------------------

On an insurance coverage, loss size has the following distribution:

$F(x) = 1 - (\frac{2000}{x})^3$ ;$x \geq 2000$

The number of claims has a negative binomial distribution with mean 0.3, variance 0.6. Claim counts and loss sizes are independent. A deductible of 3000 is applied to each claim. Calculate the variance of aggregate payments.

-----------------------------------------------

The formula above is that of a single-parameter Pareto distribution, but the book calculates E(x) and Var(x) based on a Two-parameter Pareto with $\theta$ equal to 3,000. I understand why $\theta$ should be increased to 3,000. My question is, is that in doing so does this always convert a single-paramter into a two-parameter Pareto?
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#2
09-24-2013, 05:37 PM
 Abraham Weishaus Member SOA AAA Join Date: Oct 2001 Posts: 6,754

Yes. Not hard to prove.
#3
09-24-2013, 05:47 PM
 V1per41 Member SOA Join Date: Apr 2010 Location: Colorado Studying for FAP College: Purdue University Alumni Favorite beer: Shipyard Smashed Pumpkin Posts: 2,137

Thanks
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#4
09-24-2013, 05:51 PM
 Jim Daniel Member SOA Join Date: Jan 2002 Location: Davis, CA College: Wabash College 1962 (!) Posts: 2,199

Quote:
 Originally Posted by V1per41 ASM Manual Exercise #16.21 ----------------------------------------------- On an insurance coverage, loss size has the following distribution: $F(x) = 1 - (\frac{2000}{x})^3$ ;$x \geq 2000$ The number of claims has a negative binomial distribution with mean 0.3, variance 0.6. Claim counts and loss sizes are independent. A deductible of 3000 is applied to each claim. Calculate the variance of aggregate payments. ----------------------------------------------- The formula above is that of a single-parameter Pareto distribution, but the book calculates E(x) and Var(x) based on a Two-parameter Pareto with $\theta$ equal to 3,000. I understand why $\theta$ should be increased to 3,000. My question is, is that in doing so does this always convert a single-paramter into a two-parameter Pareto?
A 1-parameter Pareto X with parameters alpha and theta has the same distribution as theta+Y where Y is a 2-parameter Pareto with parameters alpha and theta. So X-d | X>d is just Y+theta-d | Y+theta > d which is the amount per payment with severity Y and deductible d-theta. That amount per payment random variable is therefore 2-Pareto with parameters alpha'=alpha and theta'=theta+(d-theta)=d.
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#5
09-25-2013, 10:22 AM
 Statatak Member SOA Join Date: Jul 2012 Location: Pacific Ocean Studying for GI Intro to Ratemaking Favorite beer: Game Day Posts: 10,917

Quote:
 Originally Posted by V1per41 ASM Manual Exercise #16.21 ----------------------------------------------- On an insurance coverage, loss size has the following distribution: $F(x) = 1 - (\frac{2000}{x})^3$ ;$x \geq 2000$ The number of claims has a negative binomial distribution with mean 0.3, variance 0.6. Claim counts and loss sizes are independent. A deductible of 3000 is applied to each claim. Calculate the variance of aggregate payments. ----------------------------------------------- The formula above is that of a single-parameter Pareto distribution, but the book calculates E(x) and Var(x) based on a Two-parameter Pareto with $\theta$ equal to 3,000. I understand why $\theta$ should be increased to 3,000. My question is, is that in doing so does this always convert a single-paramter into a two-parameter Pareto?
This is the kind of problem I struggle on. I can't solve it.

I just don't know how to calculate second moments of expected loss with a deductible. It's killing me. Can anyone walk me through this problem? (I don't have the same ASM version as OP.)
#6
09-25-2013, 10:58 AM
 V1per41 Member SOA Join Date: Apr 2010 Location: Colorado Studying for FAP College: Purdue University Alumni Favorite beer: Shipyard Smashed Pumpkin Posts: 2,137

That's basically what my struggle was initially. However, like it's been mentioned above, because of the deductible this becomes a two-parameter pareto with alpha staying the same and theta becoming 3,000. The formula sheets tell you how to get the second raw moment, and therefore the variance.

I've seen similar problems where the distribution is exponential. From there you just use the memoryless property of exponential distributions to get $E(Y^L)$.
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#7
09-25-2013, 05:25 PM
 Statatak Member SOA Join Date: Jul 2012 Location: Pacific Ocean Studying for GI Intro to Ratemaking Favorite beer: Game Day Posts: 10,917

Quote:
 Originally Posted by V1per41 That's basically what my struggle was initially. However, like it's been mentioned above, because of the deductible this becomes a two-parameter pareto with alpha staying the same and theta becoming 3,000. The formula sheets tell you how to get the second raw moment, and therefore the variance. I've seen similar problems where the distribution is exponential. From there you just use the memoryless property of exponential distributions to get $E(Y^L)$.
Just reviewed a section of ASM that dealt with it, and I think I have it down. I hope
#8
09-26-2013, 09:54 AM
 OldOne Member Join Date: Nov 2008 College: ASU MBA Posts: 82

Quote:
 Originally Posted by Statatak Just reviewed a section of ASM that dealt with it, and I think I have it down. I hope
Do you recall what section that was? I was looking for this topic in my version of ASM yesterday & couldn't find it.
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#9
09-28-2013, 11:09 PM
 Abraham Weishaus Member SOA AAA Join Date: Oct 2001 Posts: 6,754

2.3.7?
#10
09-28-2013, 11:54 PM
 OldOne Member Join Date: Nov 2008 College: ASU MBA Posts: 82

Quote:
 Originally Posted by Abraham Weishaus 2.3.7?
That's it, thank you!
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