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#1
02-02-2007, 11:47 AM
 Kabaka Member SOA Join Date: Aug 2006 Location: O Canada Studying for NOTHING! :) Favorite beer: Root Posts: 2,185
ASM Q7.16

Help me understand this one. It sounds easy, but it's Friday.

Losses follow a single parameter Pareto distribution with alpha = 3 and theta = 500.

Determine the deductible d needed to achieve a loss elimination ratio of 20%.
Spoiler:
Here is my solution. Why is it incorrect?

Solve E[X^d] / E[X] = 0.2 for d with parameters for single Pareto. My result d = 322.75

Correct solution:
Spoiler:

"For single parameter Pareto, E[X] = (alpha) (theta) / (alpha - 1) = 750.

If the deductible, d, is less than 500, E[X^d]=d (since the probability that a loss is less than 500 is 0, so the random variable P(X<=0) = 0). But d = 150 eliminates 20% of the losses."

#2
02-02-2007, 11:50 AM
 Jski Member Join Date: Feb 2003 Favorite beer: Schlitz Posts: 410
Not too bad

The LER is E(X^d) / E(X), right? So use the tables and the parameters given and plug / chug. Am I missing something?
#3
02-02-2007, 11:52 AM
 Kazodev Member SOA Join Date: May 2004 Posts: 3,393

He's saying that's what he did, but the solution seems to state that d = 150 is the correct solution since 150/750 = .2?
#4
02-02-2007, 11:55 AM
 Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 26,452

Are you recognizing that the formula for E(X^d) is different for d > theta and for d < theta in the single parameter Pareto.
#5
02-02-2007, 12:03 PM
 Kabaka Member SOA Join Date: Aug 2006 Location: O Canada Studying for NOTHING! :) Favorite beer: Root Posts: 2,185

Quote:
 Originally Posted by Gandalf Are you recognizing that the formula for E(X^d) is different for d > theta and for d < theta in the single parameter Pareto.
Right, I see now that it says in the tables that the formula for E[X^d] is valid for d>= theta, or in this case for d>=500.

I still need some help here.
#6
02-02-2007, 12:10 PM
 Kabaka Member SOA Join Date: Aug 2006 Location: O Canada Studying for NOTHING! :) Favorite beer: Root Posts: 2,185

So I think what the solution is saying, is that since d<500, we can't use the formula in the tables and we have to use another method (that I can't find explained anywhere). How do we know that d<500? Is it just because after finding my solution of 322.75, I check it with the F(x) formula to find that F(x) < 0 (ie. invalid)?
#7
02-02-2007, 12:30 PM
 Abraham Weishaus Member SOA AAA Join Date: Oct 2001 Posts: 6,195

Quote:
 Originally Posted by Kabaka we have to use another method (that I can't find explained anywhere).
But I thought the explanation in the manual was quite clear. If the loss size is guaranteed to be at least 500, and I give you the minimum of the loss size and 300, what is the expected value of what I give you?
#8
02-02-2007, 01:15 PM
 Kabaka Member SOA Join Date: Aug 2006 Location: O Canada Studying for NOTHING! :) Favorite beer: Root Posts: 2,185

Quote:
 Originally Posted by Abraham Weishaus But I thought the explanation in the manual was quite clear. If the loss size is guaranteed to be at least 500, and I give you the minimum of the loss size and 300, what is the expected value of what I give you?
I get 300.

.......

Ahhhhhhh, I get it. The solution in #17 helps to clarify for me. It says, "We can't use the trick of the last problem though since 0.8 (750) > theta..."

So we DO use the LER formula 0.2 = E[X^d] / E[X] to solve for E[X^d]. Then we have E[X^d] = 0.2 (750) = 150. But E[X^d] = d for the reason you state above. Ok, all good. Thanks for the prodding.
#9
02-02-2007, 01:21 PM
 DudeMan Member Join Date: Jan 2006 Location: teh Po' Studying for shts&ggles Favorite beer: Ginger Beer Posts: 11,532

Quote:
 Originally Posted by Kabaka So I think what the solution is saying, is that since d<500, we can't use the formula in the tables and we have to use another method (that I can't find explained anywhere). How do we know that d<500? Is it just because after finding my solution of 322.75, I check it with the F(x) formula to find that F(x) < 0 (ie. invalid)?
Kabaka.....I believe in a single parameter pareto, the value theta (500 in this case) is the minimum possible value of a claim. A deductible of 150 effects EVERY claim since S(500)=1.....E(X)=750......E(X^150)=150*S(150)=150 .....150/750=0.2
#10
02-02-2007, 04:16 PM
 beck Member Join Date: Oct 2006 Posts: 349

any of you guys get 7.25 ??

i dont get where 10(1-F(2000)) comes from...

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