CAS 5A 11/98 Q.23 (question 4.85 in Mahler aggregate dist.

[shluffer]

The distribution of aggregate claim, S, is a compound poisson with mean 3. individual calim amounts are distributed as follows:

x p(x)
1 .4
2 .2
3 .4

Z is a standard normal variable. Which of the following is the closest to the normal aproximation of Pr[S>9)?

A. Pr[Z>.67]
B. Pr[Z>.71]
C. Pr[Z>.75]
D. Pr[Z>.79]
E. Pr[Z>.85]

Answer is D. I get D without the discrete corection (+-.5) and something else with it. How do I know if I should use the correction? I thought we allways use it when the distribution is discete but here it is discrete and don't use it.

[Been There Done That]

It appears that the solution is in error. The half-integer correction should be used here.

Here is a solution:

Let X1, X2, and X3 represent the sums of the claims of sizes 1, 2, and 3, respectively. Observe that each Xi is a Poison process and that they are independent.

Their means are 1.2, 1.2, and 3.6, respectively
and their variances are 1.2(1), 0.6(4), 1.2(9), respectively

Since they are independent, for the sum of the Xi we have:

Mean = 6
Variance = 14.4

so Standard deviation = 3.8

Prob( sum >9 ) = Prob ( sum > 9.5) = Prob ( (sum - mean)/st. dev > (9.5 - 6.0)/ 3.8 )

approximate by normal and get:

Prob (Z > 0.92)

without the half integer correction you get Prob (Z > 0.79).

[Avi]

Originally Posted by Been There Done That

Prob( sum >9 ) = Prob ( sum > 9.5)

In this case, shouldn't it be P (S ≥ 8.5) ==> P(Z>.6588)

I would think that we would move back left ½ to compensate for the discrete variable since we want a probability to the right.

[Gandalf]

Why are you using a half-integer correction? Wouldn't that be appropriate if the possible alternatives were 8 and 9?

Since here 8.8 is a possible outcome that you want to classify as less than 9, you do not want the probability that S > 8.5.

[Avi]

How is 8.8 a possible outcome?

All claims are either 1, 2, or 3 and claims are Possion, λ = 3. So the actual aggregate claim amount has to be integral.

[Been There Done That]

We were asked for the probability that the sum was greater than 9. Since there are no possible observations between 9 and 10, when using a continuous approximation to a discrete distribution, it makes sense to use the midpoint.

[Gandalf]

My mistake. I misread the table and thought the 0.4 and 0.2 column was outcomes. Maybe aggregate probability > 1 should have been a tip-off. :P

But I think the sign of your [edited to add: Avi's] correction is wrong. For P(S>9) aren't you distinguishing between discrete outcomes S <=9 and S >= 10.

[Been There Done That]

Yes.

In the given problem P(S>9) = P(S>=10)

When approximating the sum by a continuous distribution, I get:

P(S>9) = P(Z > 9.5) = P(Z >= 9.5) = P(S>=10)

[shluffer]

I was under the impression that there is a rule, for discreet distibutions you use the correction, for continuose you don't. If this is true, this wuestion should use it.

[Been There Done That]

Generally speaking, when you use a normal approximation to a distribution, it is assumed that you used the "best" one. If the distribution that you are approximating is continuous, then it often makes sense to use:

P(X>k) = P((X-mu)/sigma > (k-mu)/sigma) = P(Z > (k-mu)/sigma).

Suppose that Prob(X>k) = Prob(X+epsilon>k) for some positive epsilon, in this case any value between X and X+epsilon yields the same probability ... as a first approximation, evaluating the continuous approximating function (the normal distribution) at the midpoint will give the best estimate of the probability for this interval. That is why it is chosen.