Problem #23 - First, I didn't really understand Klugman's answer since f(0)=20=
appears to equal zero for every combination of alpha and beta=2E
Here's how I approached it though:
H=3Dtheta, A=3Dalpha, B=3Dbeta
For an exponential distribution: k=3DB, so to make credibility "very close to=20=
zero", we want to maximize k, thus we want B to be very large
A=3D=2E1*B, so substitute =2E1*B for A and f(H)=3De^(-BH)*B^A*H^(=2E1B-1)/Gamma(=2E=
1B)=2E
Since B^(=2E1B)/Gamma(=2E1B) will be the same for all five answers, let's elimina=
te=20=
it=2E
Thus we get our solution by maximizing e^(-BH)*H^(=2E1B-1)=2E
Taking the natural log, and then the derivative, we solve for H=3D(=2E1B-1)/B =3D=
=2E1=20=
- 1/B which as B goes to infinity =3D =2E1 or Answer D=2E
As a check (today of course) I tried plugging in A=3D30,B=3D300 into f(H)=20=
(excluding Gamma(A) of course) and got the following results:
A: 0
B: 1=2E1025*10^15
C: 1=2E1731*10^30
D: 1=2E9267*10^32
E: Calculator gives error when calcing e^(-300)
Also, f(1/15) =3D 3=2E32*10^31 and f(1/5)=3D9=2E68*10^27
Answer D with credibility equal to =2E0033223=2E=2E=2E=2E
The second question is question #26
I ventured into this one trying to solve two things P(B given d) > P(A given d)=20=
and P(B given d) > P(C given d)=2E
Klugman's answer shows him solving Pr(A given d) > P(B given d)=2E Is this just=20=
reading the problem wrong=2E Reversing his analysis to get P(B)>P(A) leads to a=20=
reverse of sign or d-10>2*(2ln2/3)^=2E5
That also means you can't eliminate looking at Type C, and if you do, you get=20=
answer C=2E
-matt