In this case each loss's (including those small ones) expected contribution
is:
[ E(x)-E(x;100) + 100*[1-F(x100)] ] = $991.70
And multiply this by 10 losses.
You could go the John's route of calculating the expected non-zero payment,
but then you'd have to multiply it by the expected number of non-zero
payments 10 * [1-F(x)], so the [1-F(x)] would cancel.
Jason Israel, A.C.A.S., M.A.A.A.
Increased Limits and Rating Plans(16-1)
E-mail: JIsrael@ISO.COM
Phone:(212)-898-5829, Fax -5565
Visit our website: http://www.iso.com
> ----------
> From: Keith.Rogers@ercgroup.com[SMTP:Keith.Rogers@ercgroup.com]
> Sent: Wednesday, October 27, 1999 7:27 AM
> To: jbkelly@genre.com; studygroup4B@lists.casact.org
> Subject: 4b 5/1995 Q 22 - insurer's expected annual pmt
>
> I also have a problem determining which type of payment the questions is
> asking for. F97 #22 stumped me too. My assumption will be to include
> zero payments if the question does not mention them explicitly.
>
> Can any one else shed any light on this?
>
> Good Luck to all!
>
> -----Original Message-----
> From: jbkelly@genre.com [mailto:jbkelly@genre.com]
> Sent: Mittwoch, 27. Oktober 1999 01:08
> To: studygroup4B@lists.casact.org
> Subject: May 1995 Q 22 - insurer's expected annual pmt
>
> Losses follow a pareto (a=2, l=1000)
> 10 losses are expected each year
> insurer pays the whole amount of those claims above 100
>
> My question is why don't we divide by the probability of being greater
> than 100 [s(100)]?
> My thought (WRONG) was that the insurer's average payment
> is (the total amount of all the payments ) divided by (the number of
> payments.)
>
> ****
> When is it required to divide by s(100)? Will the problem specifically
> say NONZERO payments? Are there any other phrases that would indicate
> nonzero payments only?
> ****
>
> wait to scroll down if you want to work it out be for seeing the answer
>
>
>
>
> here's the answer: E[x] - E[X,100] + 100s(100)
>
I did this (which is wrong) : E[x] - E[X,100] + 100s(100)/ s(100).
> I interpeted the problem wrong and found the average of
> nonzero payments ( I think).
>
> -john
>
>