So in one of my earlier posts on insurance fundamentals, I included an “exercise for the reader.” The question was to figure out how much money it would cost for the hypothetical insurance scheme in the example to buy re-insurance for those cases where they had not collected enough in premiums to cover the losses that actually occurred. I was thrilled that three people actually took the time to work through this. Andrew Parker answered in a comment, Gabriel Weinberg and Eddie Wharton sent emails.

All three rightly pointed out that my question about what it would take to “cover the conditional expectation of more than 2 losses” was a bit ambiguous. Was I asking about > 2 losses or >= 2 losses? It turns out that doesn’t really matter because either is a possible scenario for re-insurance but of course it means that for >=2 the reinsurance premium is higher and then the insurance company itself would cover only the case of no loss or exactly 1 loss.

Put differently it just represents a different split of the overall fair premium between the insurance company and the reinsurance company as the following table shows:

This has to be the case mathematically and also explains why no really fancy math was needed to figure this out – the numbers were already in the blog post. How so?

Well, with a 1 in 100 chance of a loss and 100 insurance participants, the expected number of losses is exactly 1! And so the following must be true: Expected number of losses X = 1.In the post I had written that the probabilities were as follows0 losses: 36%
1 loss: 36%
2 losses: 18%
So what is the expected number of losses 0.36 * 0 + 0.36 * 1 + 0.18 * 2 = 0.72And hence the expected number of losses > 2 is simply
1 - 0.72 = 0.28(The numbers in the table above are more precise from the binomial distribution but same in principle).Since the dollar losses are linear in the number of losses (just multiply by $50 in the example) we can easily calculate what the re-insurance premium should be. For the case of re-insuring > 2 losses it is thus $50 * 0.28 = $14 with the rounded numbers or $13.03 with the precise numbers.Now it is super important though to keep in mind that the reinsurance carrier must be well capitalized (or have access to capital on demand). Why? Because as you can see from the complete table of probabilities, there are cases in which it will be paying out a fair bit. For instance, the case of 4 losses ($200 total) still has a probability of 1.5%!

So in one of my earlier posts on insurance fundamentals, I included an “exercise for the reader.” The question was to figure out how much money it would cost for the hypothetical insurance scheme in the example to buy re-insurance for those cases where they had not collected enough in premiums to cover the losses that actually occurred. I was thrilled that three people actually took the time to work through this. Andrew Parker answered in a comment, Gabriel Weinberg and Eddie Wharton sent emails.

All three rightly pointed out that my question about what it would take to “cover the conditional expectation of more than 2 losses” was a bit ambiguous. Was I asking about > 2 losses or >= 2 losses? It turns out that doesn’t really matter because either is a possible scenario for re-insurance but of course it means that for >=2 the reinsurance premium is higher and then the insurance company itself would cover only the case of no loss or exactly 1 loss.

Put differently it just represents a different split of the overall fair premium between the insurance company and the reinsurance company as the following table shows:

This has to be the case mathematically and also explains why no really fancy math was needed to figure this out – the numbers were already in the blog post. How so?

Well, with a 1 in 100 chance of a loss and 100 insurance participants, the expected number of losses is exactly 1! And so the following must be true: Expected number of losses X = 1.In the post I had written that the probabilities were as follows0 losses: 36%
1 loss: 36%
2 losses: 18%
So what is the expected number of losses 0.36 * 0 + 0.36 * 1 + 0.18 * 2 = 0.72And hence the expected number of losses > 2 is simply
1 - 0.72 = 0.28(The numbers in the table above are more precise from the binomial distribution but same in principle).Since the dollar losses are linear in the number of losses (just multiply by $50 in the example) we can easily calculate what the re-insurance premium should be. For the case of re-insuring > 2 losses it is thus $50 * 0.28 = $14 with the rounded numbers or $13.03 with the precise numbers.Now it is super important though to keep in mind that the reinsurance carrier must be well capitalized (or have access to capital on demand). Why? Because as you can see from the complete table of probabilities, there are cases in which it will be paying out a fair bit. For instance, the case of 4 losses ($200 total) still has a probability of 1.5%!