PDA

View Full Version : Another CAS 3 question

modelthry
10-31-2003, 06:53 PM
Just want to make sure I'm doing this correctly.

#13. The Allerton Insurance Company insures 3 indistinguishable populations. The claims frequency of each insured follows a Poisson process.

For population I, the expected time between claims is 12 months (15 months and 18 months for populations II and III). The probability of being in population I is 1/3 (same for pops II and III). The claims severity for each class is 1000.

Calculate the expected loss in year 2 for an insured that had no claims in year 1.

My solution: Thin the process into three populations. For pop I, E(loss)=1*1000, pop II E(Loss)=12/15*1000, pop III E(loss)=12/18*1000. (The expected number of claims in year 2 is independent of the number in year 1, by definition of the Poisson process).

However, the probability of being in the three classes has changed. P(In class 1 | no claims in year 1)=P(In class 1 and no claims in year 1)/P(No claims in year 1)=[exp(-1)/3)]/[exp(-1)/3+exp(-12/15)/3+exp(-12/18 )/3]=.276. For classes 2 and 3, the probabilities are .338 and .386.

DukeCrow
10-31-2003, 06:57 PM

ramanujan
10-31-2003, 07:12 PM
That is what I had as answer, before I showed over-intelligence and changed it to 822 (answer choice B), assuming that claims in period 1 are independent of those in period 0.

I guess that means a negetive 1.25 to my score. :shake:

Loki
10-31-2003, 08:22 PM
The probability of being in a class does not change based on the probability of a loss in the prior year. Because this is a Poisson process, the probability of a loss in year 1 is independent of having a loss in year 0. So P(loss year 1/loss year 0) = P(loss year 1). So there is no need to change the probability of being in a class.

Also, it is obvious that there is not a direct correlation between the probability of being in a closs and claims costs .... otherwise there would be only 1 class.

Loki

Loki
10-31-2003, 08:28 PM
lambdaI = 1.0
lambdaII = 0.8
lambdaIII = .6667

E(x) = 1.0*1/3*1000 + 0.8*1/3*1000 + .66667*1/3*1000 = 822.22

modelthry
10-31-2003, 08:28 PM
The probability of being in a class does not change based on the probability of a loss in the prior year. Because this is a Poisson process, the probability of a loss in year 1 is independent of having a loss in year 0. So P(loss year 1/loss year 0) = P(loss year 1). So there is no need to change the probability of being in a class.

Are you sure about this? If you give me an insured and tell me he had no claims in the past year, I think it's less likely that he's from the class with mean .34 than the one with mean .28.

Loki
10-31-2003, 08:31 PM
process. This was another question on the exam about what the criteria are for being a "counting process". One of the criteria is independent increments. In other words, the probability of a loss in any time period, no matter how big or small, does not depend on losses in prior periods. It only depends on the length of the period.

I would site the book, but I don't have it with me.

Loki

modelthry
10-31-2003, 08:35 PM
In other words, the probability of a loss in any time period, no matter how big or small, does not depend on losses in prior periods. It only depends on the length of the period.
Loki

I understand. But I think the probability of a given insured being in class I changes when you're given information about his losses in year one.

Loki
10-31-2003, 08:47 PM
Given that this is a Poisson process, the # of losses in a prior year have nothing to do with the probability of loss in the current year. The probability of loss in the current year is totally independent of loss in a prior year. So, if you do make the assumption that class is determined by probability of loss (in essence this is what you are doing since size of loss is consistant among classes ... differences in expected loss are totally defined by freqency, not severity), and probability of loss is independent of having had losses in the prior year, then having loss in the prior year won't change your probability of loss in the current year and therefore won't change class.

modelthry
10-31-2003, 08:58 PM
I wrote this in my solution: P(In class 1 | no claims in year 1)=
P(In class 1 and no claims in year 1)/P(No claims in year 1)=
[exp(-1)/3)]/[exp(-1)/3+exp(-12/15)/3+exp(-12/18 )/3]=.276.

Which part of this, specifically, do you disagree with?

Loki
10-31-2003, 09:32 PM
For Poisson, P(loss in year 1/loss in year 0) = P(loss in year 1)*P(loss in year 0)/P(loss in year 0) = P(loss in year 1)

Because the years are independent, P(loss in year 1 AND loss in year 0) = P(loss in year 1)*P(loss in year 0).

modelthry
10-31-2003, 09:34 PM
For Poisson, P(loss in year 1/loss in year 0) = P(loss in year 1)*P(loss in year 0)/P(loss in year 0) = P(loss in year 1)

Because the years are independent, P(loss in year 1 AND loss in year 0) = P(loss in year 1)*P(loss in year 0).

There aren't my equations. What's wrong with mine?

Loki
10-31-2003, 09:42 PM
I'm saying that P(being in class 1) is independent of loss in year 1.

So, P(class 1/loss year 1) = P(class 1 AND loss year 1)/P(loss year 1) =
P(class 1)*P(loss year 1)/P(loss year 1) = P(class 1). So P(class 1) doesn't change based on prior loss experience.

modelthry
10-31-2003, 09:48 PM
I'm saying that P(being in class 1) is independent of loss in year 1.

So, P(class 1/loss year 1) = P(class 1 AND loss year 1)/P(loss year 1) =
P(class 1)*P(loss year 1)/P(loss year 1) = P(class 1). So P(class 1) doesn't change based on prior loss experience.

But P(In class 1 and no Losses in year 1) is clearly P(In class 1)*P(no losses|class 1)=(1/3)*(exp(-1)). It's a different answer than the denominator P(No loss in year 1), because you know what class you're in.

ZamboniGuy
10-31-2003, 09:57 PM
I'm saying that P(being in class 1) is independent of loss in year 1.

So, P(class 1/loss year 1) = P(class 1 AND loss year 1)/P(loss year 1) =
P(class 1)*P(loss year 1)/P(loss year 1) = P(class 1). So P(class 1) doesn't change based on prior loss experience.

We're not looking at the chances of a claim but the chances of being in a population. Thus the independence of a Poisson process is irrelevant.

I didn't answer this, but am I the only one who thinks that the answer to #12 should be E, as the question clearly states that the probability that the driver is a good or bad driver is equal? I understand what they are going for and answered A, but I felt this one was written particularly badly.

Loki
10-31-2003, 09:59 PM
I have to answer you tomorrow ... it's almost 11 here and my hubby is mad that I am obsessing over this. I'll respond tomorrow. But I am 100% sure that you have to treat what happens during each time period as an independent event so what happens in one time period has nothing to do with what happens in another time period.

I'll repsond specifically to your formula tomorrow.
Loki

modelthry
10-31-2003, 10:03 PM
I have to answer you tomorrow ... it's almost 11 here and my hubby is mad that I am obsessing over this.

So you really are in Bermuda? I didn't know they had stress there. :P