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Old 09-24-2013, 05:23 PM
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Default Single/Double Parameter Pareto

ASM Manual Exercise #16.21

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

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

;

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 equal to 3,000. I understand why 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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Old 09-24-2013, 05:37 PM
Abraham Weishaus Abraham Weishaus is offline
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Yes. Not hard to prove.
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Old 09-24-2013, 05:47 PM
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Thanks
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Old 09-24-2013, 05:51 PM
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Quote:
Originally Posted by V1per41 View Post
ASM Manual Exercise #16.21

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

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

;

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 equal to 3,000. I understand why 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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Old 09-25-2013, 10:22 AM
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Quote:
Originally Posted by V1per41 View Post
ASM Manual Exercise #16.21

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

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

;

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 equal to 3,000. I understand why 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.)
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Old 09-25-2013, 10:58 AM
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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 .
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Old 09-25-2013, 05:25 PM
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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 .
Just reviewed a section of ASM that dealt with it, and I think I have it down. I hope
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Old 09-26-2013, 09:54 AM
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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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Old 09-28-2013, 11:09 PM
Abraham Weishaus Abraham Weishaus is offline
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2.3.7?
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Old 09-28-2013, 11:54 PM
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2.3.7?
That's it, thank you!
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