Steve.Newton
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JD, this is quite an open question. What type of insurance and I can give you some insight, but you will never be able to work it out yourself as the matrices used by each insurer is different with different rating factors which update the premium. The approach is broadly standard though.
I wasn't referring to a particular type of insurance per say, what I was trying to gauge was how many premiums would need to be charged to cover X% risk. I'll try and give you an example to explain myself better.
Say there was a 20% average chance that £1000 on average would have to be paid out.
1: How would the premium for that % amount of risk be worked out?
2: How many premiums would have to be collected to still be in profit after paying out the £1000 average?
I'm writing this at 5 in the morning after a 12 hour shift so I hope it makes sense.
Thanks for any help you can provide.
mb325
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Well if there's a 20% chance of a £1000 payout that's akin to paying everyone £200. Therefore premiums must be at least £200 + costs + profit.
In answer to your second question, theoretically a company should be able to insure one person, collect just one premium, and still make a profit. The more people they insure though, the more chance that this profit will tend to the expected level of profit, and that they won't make a loss - this is the law of large numbers, i.e. toss a coin 10 times and you might get 80% heads; toss a coin a million times and the number of heads will tend to 50%.
In answer to your second question, theoretically a company should be able to insure one person, collect just one premium, and still make a profit. The more people they insure though, the more chance that this profit will tend to the expected level of profit, and that they won't make a loss - this is the law of large numbers, i.e. toss a coin 10 times and you might get 80% heads; toss a coin a million times and the number of heads will tend to 50%.
For some types of claims or fraud they generate loss-frequency distributions using Monte Carlo simulation. One approach is to assume that the level of loss is independent of the frequency of that loss, thereby allowing pairs of events to be drawn randomly.
For an individual, Bayesian statistics are used, which are probability distributions conditioned on known factors of a particular individual, e.g. the age of a car driver.
