4-2-5-KO-Exercise
2
Krzys’ Ostaszewski: http://www.krzysio.net Author of the ASM Manual (the “Been There Done That!” manual) for Course P/1 http://smartURL.it/krzysioP or http://smartURL.it/krzysioPe Instructor for Course P/1 seminar: http://www.math.ilstu.edu/actuary/prepcourses.html If you find these exercises valuable, please consider buying the manual or attending our seminar, and if you can’t, please consider making a donation to the Actuarial Program at Illinois State University: https://www.math.ilstu.edu/actuary/giving/ Donations will be used for scholarships for actuarial students. Donations are tax- deductible to the extent allowed by law. Questions about these exercises? E-mail: [email protected] Dr. Ostaszewski’s online exercise posted April 2, 2005 In a block of car insurance business you are considering, there is a 50% chance that a claim will be made during the upcoming year. Once a claim is submitted, the claim size has the Pareto distribution with parameters ! = 3 and ! = 1000, for which the mean is given by the formula ! " # 1 , and the second moment is given by the formula 2 ! 2 " # 2 ( ) " # 1 ( ) . Only one claim will happen during the year. Determine the variance of the unconditional distribution of the claim size. A. 62500 B. 437500 C. 500000 D. 750000 E. 1000000 Solution. Let Y be the random claim size. Its probability distribution is mixed: 50% in the Pareto distribution with parameters ! = 3 and ! = 1000 and 50% in a point-mass at 0. Its mean is simply half of the Pareto distribution with parameters ! = 3 and ! = 1000, i.e., half of ! " # 1 , or 1 2 ! 1000 2 = 250. The second moment is also half of the second moment of that Pareto distribution, i.e., half of 2 ! 2 " # 2 ( ) " # 1 ( ) , or 1 2 ! 2 ! 1000 2 1 ! 2 = 500000. The variance is therefore 500000 ! 250 2 = 437500. That’s Answer B. You could also do this by defining X = 0 when there is no claim, and X = 1 when there is a claim, with Pr X = 0 ( ) = Pr X = 1 ( ) = 1 2 , so that X is a Bernoulli Trial with p = 1 2 . Then we see that Y X = 0 ( ) is degenerate distribution equal to 0 with probability 1, while Y X = 1 ( ) is Pareto with ! = 3 and ! = 1000. Based on this EY X = x ( ) = 0 x = 0, ! " # 1 = 1000 3 # 1 = 500 x = 1, $ % & ' & so that EY X = x ( ) = 500 x, and
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Krzys’ Ostaszewski: http://www.krzysio.net Author of the ASM Manual (the “Been There Done That!” manual) for Course P/1http://smartURL.it/krzysioP or http://smartURL.it/krzysioPeInstructor for Course P/1 seminar: http://www.math.ilstu.edu/actuary/prepcourses.html If you find these exercises valuable, please consider buying the manual or attending our seminar, and if you can’t, please consider making a donation to the Actuarial Program at Illinois State University: https://www.math.ilstu.edu/actuary/giving/ Donations will be used for scholarships for actuarial students. Donations are tax-deductible to the extent allowed by law. Questions about these exercises? E-mail: [email protected]
Dr. Ostaszewski’s online exercise posted April 2, 2005 In a block of car insurance business you are considering, there is a 50% chance that a claim will be made during the upcoming year. Once a claim is submitted, the claim size has the Pareto distribution with parameters ! = 3 and ! = 1000, for which the mean is
given by the formula !" #1
, and the second moment is given by the formula
2! 2
" # 2( ) " #1( ) . Only one claim will happen during the year. Determine the variance of
the unconditional distribution of the claim size.
A. 62500 B. 437500 C. 500000 D. 750000 E. 1000000
Solution. Let Y be the random claim size. Its probability distribution is mixed: 50% in the Pareto distribution with parameters ! = 3 and ! = 1000 and 50% in a point-mass at 0. Its meanis simply half of the Pareto distribution with parameters ! = 3 and ! = 1000, i.e., half of!
" #1, or 1
2!10002
= 250. The second moment is also half of the second moment of that
Pareto distribution, i.e., half of 2! 2
" # 2( ) " #1( ) , or 12!2 !10002
1 !2= 500000. The variance is
therefore 500000 ! 2502 = 437500. That’s Answer B. You could also do this by definingX = 0 when there is no claim, and X = 1 when there is a claim, with
Pr X = 0( ) = Pr X = 1( ) = 12
, so that X is a Bernoulli Trial with p = 12
. Then we see that
Y X = 0( ) is degenerate distribution equal to 0 with probability 1, while Y X = 1( ) isPareto with ! = 3 and ! = 1000. Based on this
E Y X = x( ) =0 x = 0,!
" #1= 1000
3#1= 500 x = 1,
$
%&
'&
so that E Y X = x( ) = 500x, and
Var Y X = x( ) =0 x = 0,
2! 2
" # 2( ) " #1( ) #!
" #1$%&
'()
2
= 2 *10002
3# 2( ) 3#1( ) # 5002 = 750000 x = 1,
+
,-
.--
so that Var Y X = x( ) = 750000x. Therefore
E Var Y X( )( ) = E 750,000X( ) = 750,000 !E X( ) = 750,000 ! 12= 375,000,
Var E Y X( )( ) = Var 500X( ) = 5002 !Var X( ) = 5002 !0.5 !0.5 = 62,500, Var Y( ) = E Var Y X( )( ) +Var E Y X( )( ) = 375,000 + 62,500 = 437,500. Answer B, again. © Copyright 2005 by Krzysztof Ostaszewski. All rights reserved. Reproduction in whole or in part without express written permission from the author is strictly prohibited.
