September 2003 - Question 2

[MindFull]

In the past paper solution for this question, it seemed like the examiners used N as a negative binomial distribution, and the probability of an event as a binomial distribution. In order to get the expectation and variance of the total number of claims, I think they used the E[E[T|N]] and the (E[X])^2*V[N] + V[X]E[N]. But it seems that the mean of the neg. binomial distribution was squared instead of the mean of the binomial. Does anyone understand the examiners' answer to this question?

Thanks.

 

[John Lee]

In the past paper solution for this question, it seemed like the examiners used N as a negative binomial distribution, and the probability of an event as a binomial distribution. In order to get the expectation and variance of the total number of claims, I think they used the E[E[T|N]] and the (E[X])^2*V[N] + V[X]E[N]. But it seems that the mean of the neg. binomial distribution was squared instead of the mean of the binomial. Does anyone understand the examiners' answer to this question?

Thanks.

It's a type 1 negative binomial (rather than the usual type 2 negative binomial)

 

[MindFull]

It's a type 1 negative binomial (rather than the usual type 2 negative binomial)

I figured out that it's a Type 1 Neg. Binomial, but I'm still a bit lost on how the final answer, especially for the variance of the total number of claims was arrived at. For the total number of claims. Do we assume that the distribution of an event occurring is Binomial, with p = .02 and then multiply the variances to get the variance of the total number of claims?

Thanks much.

P.S. I'm also still unsure as to how to interpret the question. The neg. binomial type 2 is the # of failures until the kth success... So how does this definition make the distribution of the number of events given an event occurred distributed by a neg. binomial?

Thanks Mr. Lee

 

[John Lee]

I figured out that it's a Type 1 Neg. Binomial, but I'm still a bit lost on how the final answer, especially for the variance of the total number of claims was arrived at. For the total number of claims. Do we assume that the distribution of an event occurring is Binomial, with p = .02 and then multiply the variances to get the variance of the total number of claims?

There are three ways of answering this question.

This is describing an individual risk model - as there is a maximum of one event per risk. So you could just use the standard individual risk model formulae:

E(S) = sum (mq)
var(S) = sum (qs² + q(1-q)m²)

where q= probability of a claim (or in this case an event)
m = mean of each claim, s² = variance of each claim

Secondly, since each event for the risks is identical we could use the collective risk model formulae given on page 16 of the Tables with N~Bin(120,0.02).

Thirdly, we could torture ourselves with the horrendous conditional formulae.

P.S. I'm also still unsure as to how to interpret the question. The neg. binomial type 2 is the # of failures until the kth success... So how does this definition make the distribution of the number of events given an event occurred distributed by a neg. binomial?

It doesn't fit here it's simply used because it starts at 1.

 

[MindFull]

There are three ways of answering this question.

This is describing an individual risk model - as there is a maximum of one event per risk. So you could just use the standard individual risk model formulae:

E(S) = sum (mq)
var(S) = sum (qs² + q(1-q)m²)

where q= probability of a claim (or in this case an event)
m = mean of each claim, s² = variance of each claim

Secondly, since each event for the risks is identical we could use the collective risk model formulae given on page 16 of the Tables with N~Bin(120,0.02).

Thirdly, we could torture ourselves we the horrendous conditional formulae.



It doesn't fit here it's simply used because it starts at 1.

Thanks so much Mr. Lee.