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## #1 2011-04-18 17:31:45

getback0
Member
Registered: 2010-10-10
Posts: 25

### Expected Claim Count

An insurance company designates 10% of its customers as high risk and 90% as low risk.
The number of claims made by a customer in a calendar year is Poisson distributed with
mean θ and is independent of the number of claims made by a customer in the previous
calendar year. For high risk customers, θ = 0.6, while for low risk customers θ = 0.1.
Calculate the expected number of claims made in calendar year 1998 by a customer who
made one claim in calendar year 1997.

My solution:

Given that claims made by a customer in a year are independent of those made in the previous year, I disregarded the 1997 claim filing.

For any year, the probability that a customer makes one claim would be -

f(k) = [0.1 * (e ^ -0.6) * (0.6 ^ k) / k! ] + [0.9 * (e ^ -0.1) * (0.1 ^ k) / k! ]

E(x) = ∑ xf(x)

works out to 0.1*0.6 + 0.9*0.1 = 0.15

doesnt match the key

I am way off here, i guess!

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## #2 2011-04-18 17:46:53

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

### Re: Expected Claim Count

Hi getback();

I have this in my notes and .1 and .9 are not the probabilities of a high and low risk. Use Bayes theorem there.

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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## #3 2011-04-18 18:10:56

getback0
Member
Registered: 2010-10-10
Posts: 25

### Re: Expected Claim Count

Got it - so use baye's theorem to get the probability for high risk and low risk drivers in 1998 given that he had one claim and then use the logic I did

thanks

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## #4 2011-04-18 18:18:40

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

### Re: Expected Claim Count

Try that and see what you get.

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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