P(Y>=1) = 2P(NY>=1) where Y - youthful driver, NY - nonyouth
Let P(NY>=1)=p
P(NY<1)=1-p
900(1-2p) = 800(1-p)
p=0.1
900(1-2p)=720
Hope it helps. Do you know how to do #12-14, what do you get for #3??
Thanks in advance!
On Mon, 19 Apr 1999, Scherr, Vickie wrote:
> From: Vickie_Scherr@ewb.com
> To: studygroup4b@lists.casact.org
> Subject: Old Exam Solutions
>
>
> I'm having trouble solving the following problems. Any help would be
> appreciated!
>
> Thanks in advance,
> Vickie
>
> USE THE FOLLOWING INFORMATION FOR QUESTION 8 AND 9.
>
> * A portfolio consists of 75 liability risks and 25 property risks.
>
> * The risks have identical claim count distributions.
>
> * Loss sizes for the liability risks follow a Pareto distribution with
> lamba=300 and alpha=4.
>
> * Loss sizes for the property risks follow a Pareto distribution with
> lamba=1,000 and alpha=3.
>
>
> 8. Determine the variance of the claim size distribution for this
portfolio
> of a single claim.
>
> ANSWER At least 225,000, but less than 300,000
>
> 9. A risk is randomly selected from the portfolio and a claim of size k
is
> observed.
>
> Determine the limit of the posterior probability that this risk is a
> liability risk as k goes to zero.
>
>
> ANSWER 40/43
>
>
>
> 16. You are given the following:
>
> * A portfolio of automobile risks consists of 900 youthful drivers and
800
> nonyouthful drivers.
>
> * A youthful driver is twice as likely as a nonyouthful driver to incur
at
> least one claim during the next year.
>
> * The expected number of youthful drivers (n) who will be claim-free
during
> the next year is equal to the expected number of nonyouthful drivers who
> will be claim-free during the next year.
>
> Determine n.
>
> ANSWER: At least 550, but less than 750
>
>
>