Chapter 11 and qs 3.10 of Q&A3

[Cheng]

Hi!

I have a few questions.

For question 3.10 in Q&A Bank Part 3, why is it that when I calculate P(Sr=s) where s=0,1,2 in the following way, I wouldn't get the same as the answer provided.

P(Sr=0)= P(S=0) + P(S=1) + P(S=2)
P(Sr=1)= P(S=3)
P(Sr=2)= P(S=4)

and then calculating P(S=s) using the formula in p17 of the tables.

Also, could you explain further how the low discrepancy points/ Latin Hypercube (p29 of ch 11) works?

thanks!

 

[Ian Senator]

Q3.10: Not sure exactly how you're calculating things here. If you post/email your full solution we can have a look to see what's going wrong.

Latin Hypercube sampling: Normally when sampling from a distribution, eg using Monte Carlo, the tails don't see much action (and it's often these that you're interested in). What stratified sampling does (of which LH is an example) is to 'force' the sampling to come from parts of the distribution. So any simulation model should be more representative of the true underlying distribution and hence converge more quickly to whatever it is you're modelling.

Hope this helps!

 

[Cheng]

Thanks for the reply!

What do you mean by 'forcing' the sampling to come from parts of the distribution?

For Qs 3.10, what I did was as follow, I don't get the same answer but I'm not sure what went wrong.

P(Sr=0)= P(S=0) + P(S=1) + P(S=2)

P(S=0) =P(N=0) = e^-4
P(S=1) =4P(X=1)P(S=0) = 4(0.35)e^-4 = 1.4e^-4
P(S=2) = 2[ P(X=1)P(S=1) + 2P(X=2)P(S=0)] = 2[0.35(1.4e^-4) + 2(0.3)e^-4] = 2.18e^-4

thus, P(Sr=0) = e^-4 + 1.4e^-4 + 2.18e^-4 = 4.58e^-4

and P(Sr=1) =P(S=3) = 4/3[P(X=1)P(S=2) + 2P(X=2)P(S=1) + 3P(X=3)P(S=0)] = 4/3[ 0.35(2.18e^-4) + 2(0.3)(1.4e^-4) + 3(0.2)e^-4] = 2.9373e^-4

and P(Sr=2) = P(S=4) = P(X=1)P(S=3) + 2P(X=2)P(S=2) + 3P(X=3)P(S=1) + 4P(X=4)P(S=0) = 0.35(2.9373e^-4) + 2(0.3)(2.18e^-4) + 3(0.2)(1.4e^-4) + 4(0.15)e^-4 = 3.7761e^-4

 

[Katherine Young]

What do you mean by 'forcing' the sampling to come from parts of the distribution?

When Ian used the word “force”, he didn’t mean anything very technical. He only meant that we tell the software to take its sample from specific areas of the distribution.

The course notes give a pretty good explanation of how the latin hypercube works, which is:

"Just taking points at random from the distribution, we might proceed as follows:
(1) Use the 10th, 20th, 30th, etc percentiles to subdivide the distribution into 10 ranges of values, each of which contains one tenth of the probability distribution.
(2) Create a simulated sample value from each of these ranges, using the appropriate model within the range.
This will ensure that we have a sample that reflects the shape of the distribution, but which will create sample results that converge more quickly to the desired distribution."

If you're still unsure, would you mind telling me which part of this explanation you don't understand, and we can take it from there.

For Qs 3.10, what I did was as follow, I don't get the same answer but I'm not sure what went wrong.

P(Sr=0)= P(S=0) + P(S=1) + P(S=2)

This question is talking about an individual excess of loss arrangement. However, your workings are based on an aggregate excess of loss arrangement. In other words, the reinsurance pays out for X>2, not for S>2.

So, for example, your first row states: P(Sr=0)= P(S=0) + P(S=1) + P(S=2)

We can see why this is wrong, by considering the possibility that S= 1 +1 +1 =3, in which case we would still have Sr=0.

Therefore, it's better to work out the distributions of the number and amount of claims involving the reinsurer, and then use Panjer's formula.