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pirates
04-05-2002, 08:08 AM
I am having a hard time with some of the problems in lesson 61. Could someone help me out?

I don't understand the following problems:

61.9, 61.13, 61.14, 61.15, 61.18.

I realize that there are a lot and I apologize for this but you guys and gals are my only source!

thank you!

MathGuy
04-05-2002, 10:12 AM
I apologize in advance for this :-?

61.9 I always approach Bayesian problems in the same way. I set up the following columns:

- Class (in this case, A and B for the Dice)
- A priori (the prob of being in each class, in this case .5 for each)
- P(Obs) (the probability of making the given observation for that class)
- Prob Weight (This is just A priori * P(Obs))
- Posterior (Prob Weight/(Sum of all Prob Weights))
- Expected Value (For each class)

The answer will be the sumof the Expected value column. (This is the method used in Mahler's NEAS notes)

It'll be a little hard to draw the columns here, so I'll do my best:

The classes are A and B.
The a priori probabilities are .5 and .5.
The probability of making the observation of (n-1) claims of size 2 in a row are: [(Prob of 1 claim)(Prob of Claim = 2)]^(n-1). Assuming Prob(Spinner X) = Prob(Spinner Y) = .5, we can compute this

For Die A: [(1/3)(1/2)(1/3+1)]^(n-1) = (2/9)^(n-1) = 2^(n-1)/9^(n-1)
For Die B: [(2/3)(1/2)(1/3+1)]^(n-1) = (4/9)^(n-1) = 2^(2n-2)/9^(n-1)

Thus, the probability weights are
For A: .5*2^(n-1)/9^(n-1) = 2^-1*2^(n-1)/9^(n-1) = 2^(n-2)/9^(n-1)
For B: .5*2^(2n-2)/9^(n-1) = 2^-1*2^(2n-2)/9^(n-1) = 2^(2n-3)/9^(n-1)

The sum of the weights is 2^(n-2)/9^(n-1) + 2^(2n-3)/9^(n-1) = 9^(1-n)[2^(n-2) + 2^(2n-3)].

The Posterior Probabilities are:

For Class A: 2^(n-2)/[2^(n-2) + 2^(2n-3)] = 1/(1 + 2^(n-1))
For Class B: 2^(2n-3)/[2^(n-2) + 2^(2n-3)] = 1/(2^(1-n) + 1)

Note that the 9^(1-n) terms all cancelled out

Anyway, this is a good time to take that limit: as n--> infinity, the term 2^(n-1) --> infinity. So, the posterior probability for Class A is 1/(1+infinity) = 0. Sinilarly, as n--> infinity, 2^(1-n) --> 0. So the posterior probability for Class B is 1.

Therefore, as n--> infinity, we get more and more sure that we must be talking about class B if we keep getting claims of size 2. What is the expected claim size for Class B? Prob(1 Claim)*E(Claim Size) = (2/3)(.5*(1/3)*2+.5*(2/3)*8 + .5*1*2 + .5*0*8 ) = (2/3)(4) = 8/3.

And there was much rejoicing. I'm very, very sorry.[/b]

MathGuy
04-05-2002, 12:50 PM
61.18 This is similar to the previous problem, only easier!

There are two urns. One contains two dice marked 1,2,3,4,5,6 and the other urn has one die marked 1,2,3,4,5,6 and the other marked 1,1,1,2,2,2. An urn is picked, and the two dice are rolled, with a total of 3.

What is the bayesian expected value of the total on the same two dice.

First, there are two classes, corresponding to the two urns: A and B.

The a priori chance of chossing a given urn is .5.

For urn A, there are 36 possible combos, with 2 giving a roll of 3: (1,2) and (2,1). Thus P(Obs) = 2/36.

For urn B, there are 36 possible combos, with six giving a roll of 3: (1,2), (1,2), (1,2), (2,1), (2,1) and (2,1) [because the second die has three 1's and three 2's.] Thus, P(Obs) = 6/36.

The probability weights (A Priori * P(Obs)) are: Class A: .5*2/36 = 1/36, and Class B: .5*6/36 = 3/36.

The sum of the weights is 1/36 + 3/36 = 4/36. Thus, the Posterior Probs are (Weights/Sum of Weights) are: Class A: (1/36)/(4/36) = .25, and Class B: (3/36)/(4/36) = .75.

The expected sum for urn A is 7*, and for urn B is 5*. Thus the Bayesian Expected Value is .25(7) + .75(5) = 5.5.

* I cheated and used Excel to do this part.

WQN
04-05-2002, 12:58 PM
I also had trouble with this section. My advice: try to work each problem and then write out the solutions from the manual trying to understand the logic. Then go back and do them all again without looking at the solutions. Its worth it to put the extra effort in by doing them again.

MathGuy
04-05-2002, 01:29 PM
Just so you start to see the pattern (which is key to learning this stuff)...

61.15 There are three classes: A (\$5,000 limit), B (\$10,000 limit) and C (\$20,000 limit). The A priori probabilities are A = .25, B = .25, C = .5.

The Observation is that a censored (i.e., after limit is applied) claim is between \$9K and \$11K. The probability of this happening for Class A is zero, since the limit is \$5K. The probability of this for B is: 1-F(9000) = (5/14)^2 = .12755. For C, we find the prob that the claim is between 9K and 11K, which is F(11000) - F(9000) = 1 - (5/16)^2 - (1 - (5/14)^2) = (5/14)^2 - (5/16)^2 = .029895.

The probability weights are:

A: .25(0) = 0
B: .25(.12755) = .0318875
C: .5(.029895) = .0149475

The sum of the weights is 0 + .0318875 + .0149475 = .046835, so the posterior probabilities are:

A: 0/.046835 = 0
B: .0318875/.046835 = .68085
C: .0149475/.046835 = .31915

We are looking for the posterior probability of a limit of \$10K (aka, class B). So the answer is (C) At least .55, but less than .75.

pirates
04-08-2002, 07:51 PM
Thank you for your help. You are awesome!