Can purchase of insurance be justified mathematically? [closed]
When I ask people to explain why they buy insurance, I often hear vaguely of "spreading the risk", but I am not actually sure what that means nor if insurance does this. How is an insurance company any different than a casino?
In a thought experiment where some large number of people who purchase insurance are compared against an equal number of people who do not, it seems to me when one takes into account the cost of insurance, the people who do not purchase it end up better financially than those who do not. It is argued that insurance is needed to protect against catastrophic events but isn't poverty in old age a catastrophic event also? I realize that these are not strictly mathematical questions but at its base, insurance must be either a good or bad choice based on statistics and probability.
EDIT: More succinctly: Buying insurance is making a bet with a negative expectation. If there is some way to justify this mathematically then are there other bets with negative expectation, like buying lottery tickets or roulette that can be justified and how?
EDIT: People are saying, this is not a mathematical question but the question: Is a person likely to be better off financially if the buy insurance is a pretty mathematical question. If you took 100 people and half bought insurance and the other did not, which group would have more money at the end of some period, is mathematical. I can answer this question about any negative-expectation betting, so why is insurance any different?
Solution 1:
One reason for certain types of insurance, particularly businesses taking out insurance against certain potential liabilities, but also personal car insurance, is that it is a legal requirement. In that case, of course, it doesn't have to be a bet worth taking in any mathematical sense.
However, other types of insurance can often be justified by a difference in utility functions. Say I risk some event which will cost me £10,000 if it happens, but only has a 1% chance of occurring. A company offers me insurance at £200. This is a good bet for them - they make many of these bets, they are essentially independent, and so it is very unlikely that they will fail to make a profit; even if they do so one year they can make up for it the next. Essentially they have a lot of money (much more than £10,000) and the utility of the various outcomes roughly corresponds essentially to the monetary value, so a positive monetary expectation is what they are looking for.
It could also be a good bet for me, because losing £10,000 is a very bad event, causing me more than 100 times the distress of paying £200. Because £10,000 is a significant proportion of what I have, my utility function is markedly non-linear over amounts of that magnitude (and concave, so losing £10,000 is worse than you might extrapolate from how I feel about losing £200). By "utility" I am talking about a function which translates losses of various amounts of money into how much my well-being would suffer; even though taking out insurance might decrease my expected wealth, it could still increase my expected well-being.
If the bad events are not really that bad - in that you could in extremis afford to pay for them without too much distress - it may well be that insurance isn't worth it to you.
Solution 2:
The risk of an accident doesn't change at all. What changes is the risk of bankruptcy, which is indeed spread.
If a major accident occurs, you are very likely to be bust (with a probability depending on the distribution of the accident severities), because you are charged the full amount.
But if you took part to risk mutualization (I mean if you contracted an insurance), the probability of company bankruptcy is insignificant and you remain safe as you are only charged the expectation of the damage plus the company fee.
The laws of probabilities do not apply to you alone, as you are dealing with rare events. They do for companies, for which they are daily, and the Central Limit Theorem comes into play.
The larger the spread of the accident severity, the more it is beneficial to take an insurance.
Solution 3:
I don't think this is strictly speaking a mathematical question - if you doubt this, try to come up with a mathematical definition of "justified" in the sense you want it to be understood. You can mathematically calculate the expected value of buying insurance - which will typically indeed be negative - but mathematics is silent on the question whether one should enter transactions with negative (or positive) expected value.
The question should most naturally be answered from the perspective of economics, which of course uses mathematics.
Most economists would use (expected) utility theory to approach the question whether a specific insurance policy should be purchased. Underlying is the assumption that the increase in "well-being" from wealth is not actually linear, but for most individuals a concave function of wealth, i.e. the more one already has, the less one gains from additional units (this is what economists call declining marginal utility).
Short answer: Expected value might be a bad guide for decisions. Typically, economists make the assumption that there is some utility function for each individual, mapping wealth to something like "well-being"; supposing this function is known, one could precisely answer the question whether entering such a contract is beneficial.
Suppose $u(\, )$ is said function for the agent in question; he currently posseses total wealth $w$. He is at risk of losing an amount $R$ with probability $p$ (say by a house fire). The agent could (completely) insure against this risk with a policy costing the fixed amount $c$. Then, accepting the policy is beneficial iff
$u(w-c) > p u(w-R) + (1-p) u(w)$
Typically, we observe $c> pR$. (That's how insurers earn money). If $u()$ is linear (or even convex), the agent is better of bearing the risk himself. However, if the curvature of u() around point $w$ is sufficiently concave, purchase of insurance is advised - i.e. it is better to take the sure loss here. (See wikipedia on risk aversion for graphics and more details).
If that's implausible to you, maybe thinking about it reversely can make it more intuitive: would you bet everything you have for a $ 1\%$ chance to win $101 \times$ of what you have? Simple expected value would advise you to, most people would sensibly decline.$^1$
Note that utility theory can of course also be used to think other cases in which accepting negative expected value transactions are advisable. For example, if the pleasure of betting $m$ on red outweighs the costs of losing $\frac 1 {37} m$ in expectation, then the individual might be well advised to play roulette.
Or, one could think about a situation where there's a certain amount of cash one needs desperately. If you have $1500\$$, but would need $3000 \$ $ for a life-saving operation, putting it all up for a biased coinflip might actually be a good course of action, assuming you die just the same whether you have $0\$ $ or $ 1500 \$ $.
Footnote 1: It might be interesting to you that expected utility theory in some sense was first used precisely to tackle the problem of a lottery which has unlimited expected value, and should therefore be "worth" more than any finite amount of money - even though it does not seem all that attractive to most: St. Peterburg paradox
Solution 4:
Note: This answer is similar to the one of Especially Lime but I try to focus on another aspect.
The point of insurance just is not to optimize the expected valued but instead it is to decrease variance.
Suppose every day you leave your home with about 30 dollars in cash for lunch, coffee and other smaller expenses during the day and assume further you have no credit card and cannot withdraw cash during the day (or borrow some etc.).
What do you prefer:
- losing 10 cents each and every day, or
- losing all your cash on a random day once a year?
I'd prefer the former by a lot even though the expected value is worse namely about 36,5 dollars loss per year versus about 30 dollars loss per year.
Why? Because I could care less about 10 cents a day and even over the span of my entire life the less than 7 dollars loss per year is not really relevant. But on that one random day each year it could be quite annoying not to have any cash at all during the day.
To return to the staring sentence in the first case I have expected value $-0.1$ dollar per day in the former scenario but about $-0.082$ in the latter (which is better).
However there is zero variance in the first case while there is significant variance in the second case, both informally as well as in the mathematical sense.
Summary: Insurance can allow to get a distribution with (much) smaller variance at the expense of a slightly worse expected value. And, this can be desirable.
Solution 5:
Short answer:
The key word is variance reduction.
When you are alone, the distribution of your loss has a huge spread.
When you are many, the expectation remains the same, but the variance is much narrower by the Central Limit Theorem.
This variance reduction has a cost: the insurance fee.