need help understanding this problem. The answer did not explain how they got the results well. In particular, how E[Y^2|X > d] is equal to 2(10)^2 in the next step. Thanks!
need help understanding this problem. The answer did not explain how they got the results well. In particular, how E[Y^2|X > d] is equal to 2(10)^2 in the next step. Thanks!
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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I need help understanding this problem. The answer did not explain how they got the results well. In particular, how E[Y^2|X > d] is equal to 2(10)^2 in the next step. Thanks!
![[SOA 288] Losses, X, under an insurance policy are exponentially
distributed with mean 10. For each loss, the claim payment Y is equal to
the amount of the loss in excess of a deductible d > 0.
Calculate Var(Y).
X No answer selected
A
B
с
E
100 - d
(100 - d)²
100e-d/100
100 (2e-d/10 - e-d/5)
(10 - d)³ (2e-d/10 - e-d/5)
Because X is exponential, we can use the memoryless property, which
says that the claim payment is exponential with mean 10 whenever the
loss is greater than d, and 0 otherwise. We can then break things into
those 2 cases:
Var[Y] = E[Y2] - E[Y]²
= E[Y² | X > d] · P[X > d] +0 • P[X <d]
-(E[YX> d] P[X > d] + 0·P[X < [d])²
=
= 2(10)²e¯d/¹⁰ + 0 − (10e-d/10 + 0)²
= 200e-d/10
- 100e-2d/10
e-d/5
=100 (2e-d/10](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5d456ce1-cbfb-470f-8ef9-05bd7d57f044%2Fd8774e02-549d-42be-a20f-49000eef121c%2Fqisoxfg_processed.png&w=3840&q=75)
Transcribed Image Text:[SOA 288] Losses, X, under an insurance policy are exponentially
distributed with mean 10. For each loss, the claim payment Y is equal to
the amount of the loss in excess of a deductible d > 0.
Calculate Var(Y).
X No answer selected
A
B
с
E
100 - d
(100 - d)²
100e-d/100
100 (2e-d/10 - e-d/5)
(10 - d)³ (2e-d/10 - e-d/5)
Because X is exponential, we can use the memoryless property, which
says that the claim payment is exponential with mean 10 whenever the
loss is greater than d, and 0 otherwise. We can then break things into
those 2 cases:
Var[Y] = E[Y2] - E[Y]²
= E[Y² | X > d] · P[X > d] +0 • P[X <d]
-(E[YX> d] P[X > d] + 0·P[X < [d])²
=
= 2(10)²e¯d/¹⁰ + 0 − (10e-d/10 + 0)²
= 200e-d/10
- 100e-2d/10
e-d/5
=100 (2e-d/10
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