C4 Nov 2002, Q.14

[KGB Actuary]

Does anyone find this question to be defective? I perfectly understand why SOA came up with the answer choice B, but according to the official readings (and every study manual that I have seen), in case of Poisson claim count we don't replace 1 with the ratio of the variance to the mean. While it's not said explicitly, it can be interpreted that how lambda is distributed is immaterial. Looks like the "popular answer key" back up this assumption. Shouldn't the correct answer be A?

[Course 5 Zoombie]

i am putting A too. i have the same view as yours.

[casquetboy]

If I remember well, the frequency is not poisson, but rather a poisson/gamma process, which makes the frequency negative binomial distributed. This is why you can't plug in a 1.

[gymkam]

I agree with Casquetboy. You can only use 1 when claim frequency is Poisson Distributed with FIXED lambda, because then, Mean = Variance for the claim frequency.

In Q14, however, that is not the case due to lambda's prior distribution.

[bbe431]

Can anyone show how the result is B instead of E? I got exactly E when I did the calculation...

[thing]

(1.645/0.05)²(1.2+1²)=2381.302

Presumably everyone's okay with the (1.645/0.05)² part.

As previously noted, the Poisson-Gamma gives a negative binomial distribution with r=1.5, β=0.2 -- this gives σ²_f/μ_f = rβ(1+β)/rβ = 1.2

For the exponential distribution, Var(X) = θ², E[X] = θ, so CV_S = 1.

There you go. Glenn, why isn't my Greek working?

[Can't Take a Joke]

I agree with bbe431, not that it is correct. Actex has the formula as (y/k)^2 * Var(S)/E(S)^2. This is incorrect according to the Philbrick Study Guide. I answered E which is what you would get if you studied using Actex!!!

:swear:

[DukeBlue]

The formula that you're quoting Actex as having is correct if the issue is claim frequencies. But for this question the issue is aggregate losses, in which case Nf = (y/k)^2 * (VAR(f) / MEAN(f) + VAR(s)/MEAN(s)^2).

Obviously, if you're accurately quoting what Actex has printed for aggregate losses, then you have a beef.

Hope this sheds some light.

[casquetboy]

I believe the formula in the ACTEX is correct if you are determining number of exposures, not claims.

[KGB Actuary]

I don't recall anything stating that lambda has to be fixed. A typical example or problem would state that a claim count is Poisson with mean lambda. Since nothing is said anywhere about the distribution of lambda, it is open to interpretation. I don't disagree with those who chose to answer #14 based on the assumption of the marginal distribution (Neg. Binomial) for the claim count- I came up with that answer, too. But let's not forget what these exams are about- not how you interpret but rather what the official text. Not recalling anything to dismiss the Poisson case, I went with the answer consistent with the readings, which was A. That's why, IMHO, the question is defective and either A or B should be allowed as a possible answer.

Anyone appealing this question?

[DukeBlue]

Since nothing is said anywhere about the distribution of lambda,

"...where (lambda) has a gamma distribution with (alpha) = 1.5 . and (theta) = 0.2..."

[IOWA ACTUARY]

I wasn't sure about the answer, but I knew it wasn't A.

I figured that there had to be more required than A, since lamda varied so you are adding some uncertainty.

Ended up guessing right at the end.

[KGB Actuary]

Since nothing is said anywhere about the distribution of lambda,

"...where (lambda) has a gamma distribution with (alpha) = 1.5 . and (theta) = 0.2..."

I meant that there was nothing in the readings discussed about the distribution of lambda when applied to limited fluctuation credibility.

[Course 5 Zoombie]

IMHO this questions is defeative, and need appealing to SoA. either A or B will do.

[casquetboy]

I'd have to disagree with you guys. If lambda for the Poisson distribution is not fixed, then the Poisson distribution is conditional on lambda. Whenever you have distributions conditioned on a variable parameter, then you have a compound process, and you have to apply the conditional expectation and variance formulas. Notice the conditional variance formula: Var(N)=E(Var(N/lambda)) + Var(E(N/lambda). For Poisson, this is the same as Var(N)=E(lambda) + Var(lambda). If lambda is fixed, then Var(lambda)=0 and E(lambda)=lambda, so Var(N)=lambda. If lambda is not fixed, then it will have some variance associated with it, and the aforementioned would not be true [Var(lambda) will have a non-zero value, and Var(N) = lambda + Var(lambda)]. This is one of the main topics of this exam and is recurrent throughout the material. The question made it clear that lambda had a variance associated with it, and I don't see room for ambiguity.

[Howard Mahler]

As has been stated, if you are given a Gamma-Poisson situation, the marginal (prior mixed) distribution is Negative Binomial. It is the frequency distribution for the portfolio, in this case Negative Binomial, one needs to work with.

See for example, 4, 11/00, Q.14.

Howard Mahler

P.S. I am speaking for myself and not on behalf of the CAS/SOA.

I don't recall anything stating that lambda has to be fixed. A typical example or problem would state that a claim count is Poisson with mean lambda. Since nothing is said anywhere about the distribution of lambda, it is open to interpretation. I don't disagree with those who chose to answer #14 based on the assumption of the marginal distribution (Neg. Binomial) for the claim count- I came up with that answer, too. But let's not forget what these exams are about- not how you interpret but rather what the official text. Not recalling anything to dismiss the Poisson case, I went with the answer consistent with the readings, which was A. That's why, IMHO, the question is defective and either A or B should be allowed as a possible answer.

Anyone appealing this question?

[Flying Squirrel]

Let me state that I got this problem wrong (I answered E). I think we all agree now that the claim frequency (N) is N.B. with r = 1.5 and B = 0.2. Using our course 3 knowledge, I get E[S] = E[N]*E[X] = .3*5000 = 1500
VAR(S) = E[N]*VAR(X) + E[X]^2*VAR(N) = .3(25M) + .36(25M) = 16.5M

I used how-to-pass, and the formula Gordo uses is:

n = (1.645/.05)^2 * (Var(S)/E[S]^2)
= 1082.41*(16.5M/2.25M)
= 1082.41*7.3333 = 7937.67 - this is the total exposure credibility

I left off this step...
Claim credibility is exposure credibility * E[N] = 7937.67 * 0.3 = 2381.3

[Kenny]

There was nothing spectacular about this problem. They threw in the Poisson-Gamma combo for the frequency distribution to trick people into using the Poisson frequency shortcut for credibility.

The Poisson frequncy formulas are just that, they can only be used if frequency is Poisson and forthis particular problem it was not.

[Julietn528]

ASM addresses this issue:

"The problem occurs when the individual distribition is Poisson, but the parameter varies among the grouop. In this case, the group distribution is not Poisson, and in fact the variance must be greater than the mean by the conditional variance formula. Of the 4 published exams on classical credibility, 2 of them were of this nature! Multitudes of students fell for this trap, don't be one of them!"

[DrNO811]

The safe way to deal with these is to just calculate the variance of the frequency over the mean. To my knowledge, there are no shortcuts other than the Poisson shortcut. However, in response to your question, there are several conjugate prior pairs (Binomial/Beta, Normal/Normal, Inverse Exponential/Gamma, Inverse Gamma/Exponential, Poisson/Gamma, and more that are even less likely to show up on this exam)

[FSAwannabe]

The safe way to deal with these is to just calculate the variance of the frequency over the mean. To my knowledge, there are no shortcuts other than the Poisson shortcut. However, in response to your question, there are several conjugate prior pairs (Binomial/Beta, Normal/Normal, Inverse Exponential/Gamma, Inverse Gamma/Exponential, Poisson/Gamma, and more that are even less likely to show up on this exam)

I think Inverse Exponential/ Gamm is a two parameter pareto.

[DrNO811]

Yes, that's the predictive distribution for that combo.