10-10 Rule 10% Selection Percentile
Freihaut states "the 10-10 rule says that a reinsurance contract exhibits risk transfer if there is at least a 10% chance of a 10% or greater loss for the reinsurer" - this would seem to imply that losses at the exact 90th percentile would be selected in the case of testing, if available.
I feel this goes against Spring 2016 #26, however. The examiner's report selects the first loss that is greater than a 10% chance of a loss, corresponding to 20,000. I would argue that, based on the wording of Freihaut, the loss that has at least a 10% chance of happening would be 30,000, as it is exactly at the 90th percentile.
Am I missing something here?
Comments
The 90th percentile is at 20000: chance of loss being <=20000 is 90%.
Sample 1 answer is a "lazy" demonstration. In the table, you see that prob of 20000 loss is 10%. 20000 is more than a 10% loss. So, he demonstrates by pointing out these facts.
Sample 2 answer is a full demonstration. He identifies the 4 loss outcomes that would fulfill the 10% loss condition. Then, he sums the probabilities of these outcomes to get 20%. Since this is >10%, the test is passed.