Course 4 November 2000 Question #14

[sokura]

Im wondering if anyone has looked at this problem while practicing problems like I am.

Without going into too much detail for the entire problem, is the information stated in parts (i) and (ii) not necessary? I usually expect problems too give you minimally sufficient information in order for you to solve the problem (for the most part that is). I was under the impression that you know the distribution behavior (poisson|gamma) then you should estimate the parameters by the method of moments which leads me to a variance calculation as if it were biased (dividing by n). I got the correct answer but not the same exact number. the solutions I think used an unbiased sample variance (dividing by n-1).

Is my assumption that I know the frequency distribution behavior false????

The link at least that works today is

[url]http://www.soa.org/files/pdf/course4_1100.pdf[/url]

[NoMoreExams]

Which parameters are you talking about? You find E(.) and V(.) of the compound dist. What are you doing?

[sokura]

so when you find E(.) and Var(.) of the compound distribution you use the formula

E[S] = E[N]E[X] and Var[S] = E[N]Var[X]+Var[N]E[X]^2

You know that
E[X]^2 = 1500^2 and Var[X] = 6750000 bc they give you that already (X is Pareto)

BUT
When you calculate E[N] and Var[N], what do you do?

so I felt like N|lambda is Poisson and Lambda is Gamma so N is Negative Binomial. Therefore E[N] = r*Beta and Var[N] = r*Beta*(1+Beta)

so now
E[S] = r*Beta*1500 and Var[S] = r*Beta*6750000+r*Beta*(1+Beta)*1500^2

then with the given data in part (iii), you can estimate the parameters for r and Beta.

So here is my dilemma. Ordinarily with this set up I was under the impression that you plug in r and Beta as estimates obtained from the method of moments estimates.

So really all this is just saying is that
750/1000 = Sample mean = r*Beta = E[N]
And
1494/1000 = Sample second moment = E[N^2] = r*Beta*(1+Beta) + r^2*Beta^2

So
E[N] = r*Beta = 750/1000 = 0.75
and
Var[N] = *Beta*(1+Beta) = 1494/1000 – (750/1000)^2 = 0.9315

E[S] = 0.75*1500 = 1125 and Var[S] = 0.75*6750000 + 0.9315*1500^2 =7158375

Now by the inequality for full credibility we have:

n_full >= (1.96/0.05)^2 * 7158375/1125^2 = 8691.23584

If you look at the solutions, they have a slightly different answer than mine. The difference I see is that their Var[N] is equal to 0.932432 which to me suggests they just computed it by finding the (unbiased) sample variance from the data in part (iii) and completely ignoring information from part (i) and (ii).

Am I making a false assumption somewhere??? I feel like if you know the distribution for the frequency and have a sample, you would compute their expectations and variances by first estimating their parameters. If I was only given sample data and did not know the distribution, then I would have been ok here and done exactly what the solutions did.