求教P Sample Questions #147

147. The amount of a claim that a car insurance company pays out follows an exponential distribution. By imposing a deductible of d, the insurance company reduces the expected claim payment by 10%. Calculate the percentage reduction on the variance of the claim payment.
(A) 1%
(B) 5%
(C) 10%
(D) 20%
(E) 25%

147. Key: A
Let X denote the amount of a claim before application of the deductible. Let Y denote the amount of a claim payment after application of the deductible. Let
k be the mean of X, which because X is exponential, implies that k^2 is the variance of X and E(X^2)=2k^2. By the memoryless property of the exponential distribution, the conditional distribution of the portion of a claim above the deductible given that the claim exceeds the deductible is an exponential distribution with mean k. Given that E(Y)=0.9k, this implies that the probability of a claim exceeding the deductible is 0.9 and thus E(Y^2)=0.9*2k^2=1.8k^2.

我不太理解为什么 E(Y^2)=0.9*2k^2=1.8k^2。麻烦大家赐教~

ahixyz 发表于 2010-7-21 10:44
147. The amount of a claim that a car insurance company pays out follows an exponential distribution. By imposing a deductible of d, the insurance company reduces the expected claim payment by 10%. Calculate the percentage reduction on the variance of the claim payment.
(A) 1%
(B) 5%
(C) 10%
(D) 20%
(E) 25%

147. Key: A
Let X denote the amount of a claim before application of the deductible. Let Y denote the amount of a claim payment after application of the deductible. Let
k be the mean of X, which because X is exponential, implies that k^2 is the variance of X and E(X^2)=2k^2. By the memoryless property of the exponential distribution, the conditional distribution of the portion of a claim above the deductible given that the claim exceeds the deductible is an exponential distribution with mean k. Given that E(Y)=0.9k, this implies that the probability of a claim exceeding the deductible is 0.9 and thus E(Y^2)=0.9*2k^2=1.8k^2.

我不太理解为什么 E(Y^2)=0.9*2k^2=1.8k^2。麻烦大家赐教~

Keep in mind:
Y = 0 if X<=d
Y = X-d if X > d

Then:
E(Y^2)
= Pr(X>d) E(Y^2| X>d) + Pr(X<=d) * E(Y^2| X<=d)
=Pr(X>d) E(Y^2| X>d) + 0
= 0.9 E((X-d)^2| X>d)
= 0.9 E(X^2)
= 0.9 * 2k^2
= 1.8 k^2

楼主 辛苦了 啊

我瞎说一个啊，你看对不对。。。
E(x)=∫(0-无穷)P(X>x)dx
E(x^2)=∫(0-无穷)P(X^2>x)d(x^2)
又因为exponential dist.的无记忆性,得出E(x^2)=2[E(x)]^2，这是成比例的
所以，若E(x)=.9*k，那么，E(x^2)就等于.9*2*k^2
欢迎拍砖^ ^

感谢LS提供的思路，但我感觉好像不太对。。。

不知还有高人提供见解吗？非常感谢！

lz能否斧正一下？

darth vender rule只能用来求期望吧...那个是有严格证明的...2阶距的证明麻烦给一下吧...

具体内容，你可以看一下guo的p 第十版，195-196页 6# wbwd1997

感谢ls的思路，非常感谢。

还有一个小问题，why E((X-d)^2| X>d) =E(X^2) ? because of  the memoryless property of the exponential distribution?

再补充一下上面我的问题，我了解E((X-d)| X>d) =E(X)，E((X| X>d) =E(X)+d，那么 E((X-d)^2| X>d) =E(X^2)？ E((X^2| X>d) =？