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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)
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
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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