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pirates
03-28-2002, 11:00 AM
Question 9.6. The question asks for the median of the portion of the 1995 loss distribution above 200. What if it had asked for the 40th percentile of the portion of the 1995 loss above 200?

would it be (1-0.4(0.6162)) or (1-0.6(0.6162))

Please refer to the solution in the manual for this.

My second question is 11.7 (page 69)
i don't quite understand the solution in the manual. If all losses increased by 10% how come 1-Fz(1000) is less than 1- Fx(1000). if losses increase, how can the portion above 1000 decrease?

Thank you for your help!

New at pd
03-28-2002, 11:42 AM
On 2002-03-28 11:00, pirates wrote:
Question 9.6. The question asks for the median of the portion of the 1995 loss distribution above 200. What if it had asked for the 40th percentile of the portion of the 1995 loss above 200?

would it be (1-0.4(0.6162)) or (1-0.6(0.6162))

Please refer to the solution in the manual for this.

My second question is 11.7 (page 69)
i don't quite understand the solution in the manual. If all losses increased by 10% how come 1-Fz(1000) is less than 1- Fx(1000). if losses increase, how can the portion above 1000 decrease?

Thank you for your help!

What are the questions? I don't have my ASM in front of me.

MathGuy
03-28-2002, 12:42 PM
The key is this: If we want to find the x value for the median value above 200, we need to find the point which coresponds to the halfway point of the area after 200, as shown in this image:
http://www.villagephotos.com/pubimage.asp?id_=183136
So, the area is .3838 + .5(.6162) = 1 - .6162/2.

In your example, the answer would be:
http://www.villagephotos.com/pubimage.asp?id_=183135
The math is: y = .3838 + .4(.6162) = (1 - .6162) + .4(.6162) = 1 - .6(.6162).

MathGuy
03-28-2002, 12:55 PM
The solution to 11.7 must be in error, as it computes 1 - &Phi;(-0.0628) = 0.4749, which implies that &Phi;(-0.0628) = .5251, but &Phi;(x) > .5 if x > 0 and < .5 if x < 0.

The fact that the final solution computes (1 - F<sub>x</sub>(1000))/(1 - F<sub>z</sub>(1000)) - 1, should also be a hint, as this would indicate a decrease in claims above 1000, not an increase.

pirates
03-28-2002, 02:02 PM
Thank you so much! That helps a lot. So you are in agreement that 11.7 is incorrect?

Abraham Weishaus
03-31-2002, 08:34 PM
This has been incorporated in my updated errata list, which should be posted at www.studymanuals.com shortly. In the meantime, I will send it to anybody who requests it. Send requests to abe_weishaus@glic.com (underscore between e and w).

M.
04-01-2002, 04:53 PM
I tried your email address, but received a delivery error message. I would very much like to receive your updated errata.

Thanks!