An easier way to think about it : average payment > 100 = e(100) + 100
To get the average over all payments: 0 F(100) + {e(100) + 100} [1 -
F(100)}.
> -----Original Message-----
> From: Timothy_Devine@CGUUSA.com [SMTP:Timothy_Devine@CGUUSA.com]
> Sent: Wednesday, October 27, 1999 10:35 AM
> To: jbkelly@genre.com; studygroup4B@lists.casact.org
> Subject: RE: May 1995 Q 22 - insurer's expected annual pmt
>
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> Mahler's Section 28 - e(x) = integral x to infinity (t - x) f(t) dt /
> (
> 1 - F(x) ) which also equals
>
> 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
> >
>