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Old 04-21-2005, 03:27 PM
MortalityMan MortalityMan is offline
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Default sampel exam #141 E[Y^2] weird whoa

question...

A claim severity distribution is exponential with mean 1000. An insurance company will
pay the amount of each claim in excess of a deductible of 100.
Calculate the variance of the amount paid by the insurance company for one claim,
including the possibility that the amount paid is 0.


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Ok I get that E[y] = 1000*pr(x>100).

I dont get why E[y^2] = 1000^2 * 2 * pr(x>100)
where's the 2 coming from? something i should know i'm sure....
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Old 04-21-2005, 03:37 PM
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JJ_Franch JJ_Franch is offline
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Quote:
Originally Posted by wesimel
question...

A claim severity distribution is exponential with mean 1000. An insurance company will
pay the amount of each claim in excess of a deductible of 100.
Calculate the variance of the amount paid by the insurance company for one claim,
including the possibility that the amount paid is 0.


________________
Ok I get that E[y] = 1000*pr(x>100).

I dont get why E[y^2] = 1000^2 * 2 * pr(x>100)
where's the 2 coming from? something i should know i'm sure....
I think I comes from integrating by parts. The more important question is can you remember it?
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Old 04-21-2005, 04:05 PM
Jski Jski is offline
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Default sampel exam #141 E[Y^2] weird whoa

wesimel -- This problem is best attempted using the double expectation theorem for two variables. There should be no integrating, much less integration by parts. Let X be exp(theta = 1000) and I be Bernoulli with p = .9048 (this is the prob that X>100). Now, from Course 1

var = E[var(X|I)] + var[E(X|I)], where for the Bernoulli, n=1, p=.9048, q=.0952. Off you go.
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Old 04-21-2005, 04:41 PM
Brutè Brutè is offline
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It's a compound distribution with the chance of a claim over 100 being Bernoulli as described above by the previous poster. Due to the memoryless property of the Exponential, claims over 100 are also Exponential with theta = 1000. So Bernoulli is the primary distribution and Exponential with theta=1000 is the secondary distribution.

Variance = Mean(primary)*Var(secondary) + (Mean(secondary))^2 * Var(primary)

Last edited by Brutè; 04-21-2005 at 04:44 PM..
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Old 04-21-2005, 06:39 PM
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fishchaser fishchaser is offline
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Quote:
Originally Posted by wesimel
question...

A claim severity distribution is exponential with mean 1000. An insurance company will
pay the amount of each claim in excess of a deductible of 100.
Calculate the variance of the amount paid by the insurance company for one claim,
including the possibility that the amount paid is 0.


________________
Ok I get that E[y] = 1000*pr(x>100).

I dont get why E[y^2] = 1000^2 * 2 * pr(x>100)
where's the 2 coming from? something i should know i'm sure....
I think you are confused by the 2. Why not look up the inventory of continuous distribution? For exp distribution, E(x^k)=(theta^k)*K!, here k is 2. So of course, there is a 2.
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