ASM Chapter 13, Problem #19

[sophie]

hi guys,

can anyone give me a hand on this one?

the problem goes like this:

Losses follow an inverse Pareto distribution with tao = 2 and theta = 300.
The number of nonzero payments N on a coverage with ordinary deductible of 200 has the following distribution:

n = 0; P(N=0) = 0.60
n = 1; P(N=1) = 0.30
n = 2; P(N=2) = 0.05
n = 3; P(N=3) = 0.05

The deductible is raised to 500.

Calculate the probability of exactly one nonzero payment on the coverage with the revised deductible.

The solution says:

The probability of one loss above 500 is the sum of the probability of:
1. 1 loss above 200 w/c is also above 500
2. 2 losses above 200, one below 500 and one above 500
3. 3 losses above 200, two below 500 and one above 500

this part is what i really did not get in the solution! how can i figure that out in an actual exam?

[joni308]

I don't know how the exact solution goes since I don't have the manual but there's not much to it really. All you are doing here is a conditional probability. Ideally, you would like to condition on the number of losses but you are not given [the distribution of] exactly that. You are given, instead, the distribution of the number of payments. So we'll go with that.

Denote:

# of payments with new deductible = EOP
# of payments with old deductible = EOP*

You are asked to find P(EOP=1). Then:

P(EOP=1) = P(EOP=1|EOP*=0)P(EOP*=0) + P(EOP=1|EOP*=1)P(EOP*=1) + P(EOP=1|EOP*=2)P(EOP*=2) + P(EOP=1|EOP*=3)P(EOP*=3)
=P(EOP=1 AND EOP*=0) + P(EOP=1 AND EOP*=1)+ P(EOP=1 AND EOP*=2) + P(EOP=1 AND EOP*=3)


The first term is 0. Now think about the other 3 terms in my equation above and you'll see these are the 3 probabilities you listed in words. Of course, in my opinion, the most interesting part if how you proceed from here. I'm not sure how the author does it but it'sa nice problem.

[sophie]

yey! thank u very much! that helped a lot!

the rest of the solution goes like this:

P(X>500)/P(X>200) = (1 - (200/500)^2) / (1 - (500/800)^2) = 0.725446

P(EOP* = 1 & EOP = 1) = 0.30*0.725446
P(EOP* = 2 & EOP = 1) = 0.05*2*0.725446*(1-0.725446)
P(EOP* = 3 & EOP = 1) = 0.05*3*0.725446*(1-0.725446)^2

the sum of the 3 probabilities is 0.24575.

many, many thanks!