Why insurance works
Tonight's idea
Suppose (invented numbers) each house has a 2% chance per year of a fire costing 20,000. The loss \(X\) is a random variable, as in The average outcome and how far it strays. Its expected value is \(\mu = 0.02 \times 20{,}000 = 400\). Its standard deviation is \(\sigma = 20{,}000\sqrt{0.02 \times 0.98} = 2800\).
An insurer charges a premium of, say, 500 a year. On average you lose 100 by buying. You are still glad to: a certain 500 is bearable and a sudden 20,000 is not.
The insurer can bear it because it holds many houses. For independent risks (one fire tells you nothing about another) variances add. So the total loss of \(n\) houses has variance \(n\sigma^2\) and standard deviation \(\sigma\sqrt{n}\). Divide by \(n\) for the average:
The average of \(n\) independent risks still has mean \(\mu\), but its standard deviation is \(\dfrac{\sigma}{\sqrt{n}}\).
With 10,000 houses the average loss is 400, give or take 28. The extra 100 pays for the little risk that is left.
Watch one: a pool of two
Two neighbours agree to split whatever either loses. Each pays the average of the two losses.
| Fires | Probability | Each pays |
|---|---|---|
| 0 | \(0.98^2 = 0.9604\) | 0 |
| 1 | \(2 \times 0.02 \times 0.98 = 0.0392\) | 10,000 |
| 2 | \(0.02^2 = 0.0004\) | 20,000 |
Mean: \(10{,}000 \times 0.0392 + 20{,}000 \times 0.0004 = 400\). Variance: \(10{,}000^2 \times 0.0392 + 20{,}000^2 \times 0.0004 - 400^2 = 3{,}920{,}000\), so the standard deviation is 1980. The rule predicts \(2800/\sqrt{2} = 1980\). Same average, less spread, with only two in the pool.
Watch many independent chances add up to a predictable pile: mathdetours.com/demos/galton.html.
Your turn
- A phone has a 5% chance per year of needing a 600 repair (invented numbers). Find the expected loss and its standard deviation. A policy costs 45. How much more than the expected loss is that?
- For the houses above, \(\sigma = 2800\). Find the standard deviation of the average loss in a pool of 400 houses. How big must the pool be for it to fall to 10?
- An insurer charges 500 per house. With 10,000 houses, find the premiums collected, the expected total claims, and the standard deviation of the total claims, \(\sigma\sqrt{n}\). How many standard deviations separate premiums from expected claims? Repeat for 100 houses. Which insurer is safe?
- A flood is different: either every house in a town is hit or none is. Suppose all \(n\) houses share a single 2% chance of losing 20,000 each. What is the standard deviation of the average loss now? Which word in the rule has failed?
Stretch: how big is big enough?
An insurer sets its premium at the expected loss plus two standard deviations of the average loss. Write the premium as a formula in \(n\) for the houses above. How many houses does it need before it can charge 450 or less?
Answers
- Expected loss \(0.05 \times 600 = 30\). Standard deviation \(600\sqrt{0.05 \times 0.95} \approx 131\). The policy costs 15 more than the expected loss.
- \(2800/\sqrt{400} = 140\). For 10 we need \(\sqrt{n} = 280\), so \(n = 78{,}400\).
- 10,000 houses: premiums 5,000,000; expected claims \(10{,}000 \times 400 = 4{,}000{,}000\); standard deviation \(2800 \times 100 = 280{,}000\). The gap of 1,000,000 is about 3.6 standard deviations. 100 houses: premiums 50,000; expected claims 40,000; standard deviation \(2800 \times 10 = 28{,}000\). The gap of 10,000 is only 0.36 of a standard deviation. The large insurer is safe; the small one will often pay out more than it collects.
- The average loss is 20,000 with probability 0.02 and 0 otherwise, exactly like one house. Its standard deviation is 2800 whatever \(n\) is. The risks are not independent, so variances do not simply add and pooling within the town does nothing. Insurers handle this by spreading over many regions.
Stretch. Premium \(= 400 + \dfrac{2 \times 2800}{\sqrt{n}} = 400 + \dfrac{5600}{\sqrt{n}}\). For this to be at most 450 we need \(\sqrt{n} \ge 112\), so \(n \ge 12{,}544\) houses.