View Full Version : ASM Manual P 37 Ex 2.8

mallkins

January 8th 2006, 10:28 PM

In the question for Ex 2.8 it says "Calculate the probability that there will be at least four months in which no accidents occur before the fourth month in which at least one accident occurs."

I can follow the working, but should it be "...probability that there will be at least three months in which no accidents occur..." ?

If not, what is the reason?

In the solution for Ex 2.9 the calculation for E(Y) contains

...2000e^-0.6 sigma (2=<k<infinity) 0.6^k/k!

This then changes to

...2000(1 - e^-0.6 - 0.6.e^-0.6)

What is the theory behind this change, and is there a reference in the manual?

Cheers,

Miles

krzysio

January 9th 2006, 10:40 PM

In the question for Ex 2.8 it says "Calculate the probability that there will be at least four months in which no accidents occur before the fourth month in which at least one accident occurs."

I can follow the working, but should it be "...probability that there will be at least three months in which no accidents occur..." ?

If not, what is the reason?

In the solution for Ex 2.9 the calculation for E(Y) contains

...2000e^-0.6 sigma (2=<k<infinity) 0.6^k/k!

This then changes to

...2000(1 - e^-0.6 - 0.6.e^-0.6)

What is the theory behind this change, and is there a reference in the manual?

Cheers,

Miles

For your first question: you know that the wording of the exercise comes directly from the actuarial exam. So you must be thinking of something else than what the question asks. The question asks to count the months without accident in relation to a series of months until the fourth month with an accident. There is absolutely no connection to number three.

For your second question:

Pr (Poisson => 2) = 1 - Pr (Poisson = 0) - Pr (Poisson = 1)

I am sorry this is not clear in the text, I thought it was. I will add an explanation.

Yours,

Krzys'

krzysio

January 9th 2006, 10:51 PM

In the question for Ex 2.8 it says "Calculate the probability that there will be at least four months in which no accidents occur before the fourth month in which at least one accident occurs."

I can follow the working, but should it be "...probability that there will be at least three months in which no accidents occur..." ?

Here is what the solution says:

"Consider a Bernoulli trial in which a month with no accident is a success, and a month with an accident is a failure. Then the probability of success is Now consider a negative binomial random variable, which counts the number of failures (months with accident) until 4 successes (months without accidents) ..."

Where does three come from?

I am perplexed. I am an actuary, and actuaries are bad at communication ... notoriously. I do not want to be ...

Could you give me your thoughts?

Yours,

Krzys'

P.S. This was one of the most twisted questions asked since 2000 ...

mallkins

January 9th 2006, 10:58 PM

That's great, thanks.

I think the first question issue is just getting used to the wording of questions on my part, and the second one is clear as a whistle now.

Hope you don't mind me posting these questions.

Miles

mallkins

January 9th 2006, 11:05 PM

Here is what the solution says:

"Consider a Bernoulli trial in which a month with no accident is a success, and a month with an accident is a failure. Then the probability of success is Now consider a negative binomial random variable, which counts the number of failures (months with accident) until 4 successes (months without accidents) ..."

Where does three come from?

I am perplexed. I am an actuary, and actuaries are bad at communication ... notoriously. I do not want to be ...

Could you give me your thoughts?

Yours,

Krzys'

P.S. This was one of the most twisted questions asked since 2000 ...

On rereading the question it seems that the key is in the "at least", which I now think must mean that it guarantees the three months prior to the accident.

In the sentence it says about the four months prior with no accident, but later in the same sentence says that an accident will occur in the fourth month, which seems to contradict the first part. The three comes from the failure in the fourth month meaning that there must be three months with success that we are concerned about.

Miles

mallkins

January 9th 2006, 11:09 PM

PS

Looking over the gamma distribution, do we need to rember with n and lambda or alpha and beta for the exam?

Miles

krzysio

January 11th 2006, 11:11 PM

On rereading the question it seems that the key is in the "at least", which I now think must mean that it guarantees the three months prior to the accident.

In the sentence it says about the four months prior with no accident, but later in the same sentence says that an accident will occur in the fourth month, which seems to contradict the first part. The three comes from the failure in the fourth month meaning that there must be three months with success that we are concerned about.

Miles

At least four ... so it can't be three. The bad thing in this is that the number of failures and the number of successes that we want in the resulting negative binomial are both four, so the wording is ackward. But just separate them, using the definition of a success and of a failure that I gave in the manual (and in a posting above). Good luck.

Yours,

Krzys'

mallkins

January 12th 2006, 07:36 AM

Thanks Krzys'

I think the key word is awkward. Having the same number made me think that they were tied in.

Miles

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