Why Insurance Diversification Depends on Standard Error

# Why is standard error used to show diversification effect for unsystematic risk?


# Why is standard error used to show diversification effect for unsystematic risk?












quite long text incoming, sorry for that: While reading a corporate finance textbook, i came across a section describing the effect of diversification as well as the systematic and unsystematic risk. The book provides an example in order to show the effect of diversification:

A city where there is a 1% chance of a house being robbed and a 1% chance of the city being hit by an earthquake. An insurance company takes out 100,000 policies of theft insurance and 100,000 policies of earthquake insurance. It is to be expected that 0.01⋅100.000=1.000 houses per year will be robbed.

The textbook quantifies the difference in diversifiability using the standard deviation of the percentage of losses (assuming a 1% probability of a house being robbed and a 1% probability of an earthquake occurring). At the end of the year the damage either occurred (100%) or did not occur (0%).

First, the standard deviation of a claim for a homeowner is calculated:

Standard Deviation_Homeowner (Theft Insurance) = √(0,99*(0-0,01)^2+0,01*(1-0.01)^2) =9,95%

Standard Deviation_Home Owner (Earthquake Insurance) = √(0,99*(0-0,01)^2+0,01*(1-0.01)^2) =9,95%

I find it difficult to interpret the standard deviation in this context I would go with: The probability that my house will be robbed or hit by an earthquake is 1%. So on average, out of 100 robberies/earthquakres my house be affected once, right? The SD tells me that, on average, the probability that my house will be robbed or hit by an earthquake fluctuates around 9.95%. But this interpretation makes no sense for me, since the probability 1%-9.95%=-8.95% would be negative.

In a next step, the risk for the insurance company is calculated.

The book states: In the case of earthquake insurance, because the risk is common, the percentage of claims is either 100% or 0%, just as it was for the homeowner.

However, to calculate the standard deviation, e.g. the risk for theft insurance, it is written that "When the risks are independent and identical, the standard deviation of the average is called the standard error". I don't understand why in the case of theft insurance from an insurance perspective, the standard error is suddenly calculated to quantify the risk of the insurance company.

TL;DR: Why is the standard error used to show the diversification effect for unsystematic risk?

Thanks a lot

## Answer by demully (score 1)

https://quant.stackexchange.com/a/58175

OK, short answer, think of it this way. For a more formal explanation, google the distinction between what statisticians call a "confidence interval" versus a "prediction interval".

So imagine that 1 in 10,000 people contract "Jumping Jack Flash Syndrome" every year. An insurer sells insurance against this with a 1 million payout. The fair price of that insurance would be 100. The insurer makes say a 5% combined ratio (ie profit margin) on their policy, so I pay 105. So I have a payout with a mean loss of 5 every year (their margin) and a standard deviation of 10,000. But I'm covered for the disaster, at no noticeable cost to my ability to thrive without getting it.

The risk to the insurer is a little different. They are indifferent to whether I versus any other of their customers gets JJFS. They just don't want me and them to get it together. They want to be sure it's 1-in-10,000, not 1-in-9000. Rather they care about the uncertainty around that 1-in-10000 estimate of the frequency. Which is exactly the "systematic" versus "idiosyncratic" error the textbook mentions, and you asked about.

If the 1-in-10,000 is correct, then the insurer would love to double the size of their insurance book; and double their profits (at least in the long-run). But if it was 1-in-5,000, then doubling the book would spell faster and greater losses, and potential insolvency.

And even if the 1-in-10,000 risk was correct, they might just get unlikely. Doubling the normal cases in any year might wipe out the aggregate profitability of that product for decades! So the risk to the insurer isn't the volatility of the payout to you; but the volatility of their payouts across you plus the other million customers. Which is the "standard error" around the 0.01% payout rate.

So you pay 105, expect to on average lose 5, to get 1,000,000 given JJFS, giving you an insurance risk (standard deviation) of ~10,000.

The insurer has a million customers like you. Their risk per contract is the same 10,000 per contract divided by the root of a million contracts, equals 10 per contract (for 5 expected gain). Or the same 10,000 per contract times the root of a million contracts equals 10 million in aggregate (for 5 million expected profit). Either way, they are making an expected profit of X by running 2X (normalised) risk, akin to a Sharpe Ratio of 0.5 (on that policy alone).

And then the insurer is hoping that its JJFS book is uncorrelated with its "Eleanor Rigby Syndrome","Twisted Firestarter", and "Buffalo Soldier" risk books. If all 4 were uncorrelated, then they're getting 1 return for 1 vol. Etc. Etc...

hope this clarifies.