e(x) = integral x to infinity [(t) f(t) dt] / (1 -
F(x) - x
therefore : integral x to infinity [(t) f(t) dt] = (e(x) + x) (1 -
F(x))
What is in red is what we are looking for e(x) for Pareto: (lambda + x)
/ (alpha - 1) = (1000 + 100) / (2 - 1) = 1100
therefore : integral x to infinity [(t) f(t) dt] = (e(x) + x) (1 - F(x)) =
1200 (1000 / 1100)^2 = 991.74
I think the trick is the expected annual payments include the zero payment
claims.
----Original Message-----
> From: jbkelly@genre.com [SMTP:jbkelly@genre.com]
> Sent: Tuesday, October 26, 1999 7:08 PM
> 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
>