Re: #23,#26

Damon Lay ( (no email) )
Fri, 07 May 1999 16:28:52 -0700

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I agree w/ Matt.
Mr. Klugman's error was in evaluating the limit as b goes to infinity of the
expression (a-1)/b (mode of the Gamma Distribution). He failed to account
for the fact that a = 0.1*b (because the mean a/b = 0.1). The limit should
be found by substituting, and then
lim b to inf of (0.1b-1)/b is (by our ol' buddy L'Hopital) 0.1.
This agrees with Matt.
- Damon
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From: matthew.fienman@milliman.com
To: studygroup4B@lists.casact.org
Subject: #23,#26
Date: May 07, 1999, 14:06
First and foremost I'd like to thank Stuart Klugman for the effort he put
into posting his solutions, as I'm sure we're all a bit antsy and on edge.
I did though have questions on two of the problems he worked, and I'm
looking to see what everyone else's thoughts are (incl. Stuart's).
Problem #23 - First, I didn't really understand Klugman's answer since f(0)
appears to equal zero for every combination of alpha and beta.
Here's how I approached it though:
H=theta, A=alpha, B=beta
For an exponential distribution: k=B, so to make credibility "very close to
zero", we want to maximize k, thus we want B to be very large
A=.1*B, so substitute .1*B for A and f(H)=e^(-BH)*B^A*H^(.1B-1)/Gamma(.1B).
Since B^(.1B)/Gamma(.1B) will be the same for all five answers, let's
eliminate it.
Thus we get our solution by maximizing e^(-BH)*H^(.1B-1).
Taking the natural log, and then the derivative, we solve for H=(.1B-1)/B =
..1 - 1/B which as B goes to infinity = .1 or Answer D.
As a check (today of course) I tried plugging in A=30,B=300 into f(H)
(excluding Gamma(A) of course) and got the following results:
A: 0
B: 1.1025*10^15
C: 1.1731*10^30
D: 1.9267*10^32
E: Calculator gives error when calcing e^(-300)
Also, f(1/15) = 3.32*10^31 and f(1/5)=9.68*10^27
Answer D with credibility equal to .0033223....
The second question is question #26
I ventured into this one trying to solve two things P(B given d) > P(A given
d) and P(B given d) > P(C given d).
Klugman's answer shows him solving Pr(A given d) > P(B given d). Is this
just reading the problem wrong. Reversing his analysis to get P(B)>P(A)
leads to a reverse of sign or d-10>2*(2ln2/3)^.5
That also means you can't eliminate looking at Type C, and if you do, you
get answer C.
-matt

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Re: #23,#26I agree w/ Matt.
Mr. Klugman's error was in evaluating the limit as b goes to infinity of th=e expression (a-1)/b (mode of the Gamma Distribution). He failed to account= for the fact that a =3D 0.1*b (because the mean a/b =3D 0.1). The limit should= be found by substituting, and then
lim b to inf of (0.1b-1)/b is (by our ol' buddy L'Hopital) 0.1.
This agrees with Matt.
- Damon
----------
From: matthew.fienman@milliman.com
To: studygroup4B@lists.casact.org
Subject: #23,#26
Date: May 07, 1999, 14:06
First and foremost I'd like to thank Stuart Klugman for the effort he put i=nto posting his solutions, as I'm sure we're all a bit antsy and on edge. I= did though have questions on two of the problems he worked, and I'm looking= to see what everyone else's thoughts are (incl. Stuart's).
Problem #23 - First, I didn't really understand Klugman's answer since f(0)= appears to equal zero for every combination of alpha and beta.
Here's how I approached it though:
H=3Dtheta, A=3Dalpha, B=3Dbeta
For an exponential distribution: k=3DB, so to make credibility "very clo=se to zero", we want to maximize k, thus we want B to be very large
A=3D.1*B, so substitute .1*B for A and f(H)=3De^(-BH)*B^A*H^(.1B-1)/Gamma(.1B).=
Since B^(.1B)/Gamma(.1B) will be the same for all five answers, let's elimi=nate it.
Thus we get our solution by maximizing e^(-BH)*H^(.1B-1).
Taking the natural log, and then the derivative, we solve for H=3D(.1B-1)/B =3D= .1 - 1/B which as B goes to infinity =3D .1 or Answer D.
As a check (today of course) I tried plugging in A=3D30,B=3D300 into f(H) (excl=uding Gamma(A) of course) and got the following results:
A: 0
B: 1.1025*10^15
C: 1.1731*10^30
D: 1.9267*10^32
E: Calculator gives error when calcing e^(-300)
Also, f(1/15) =3D 3.32*10^31 and f(1/5)=3D9.68*10^27
Answer D with credibility equal to .0033223....
The second question is question #26
I ventured into this one trying to solve two things P(B given d) > P(A g=iven d) and P(B given d) > P(C given d).
Klugman's answer shows him solving Pr(A given d) > P(B given d). Is thi=s just reading the problem wrong. Reversing his analysis to get P(B)>P(A=) leads to a reverse of sign or d-10>2*(2ln2/3)^.5
That also means you can't eliminate looking at Type C, and if you do, you g=et answer C.
-matt

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