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I've asked the same question at the Quantitative Finance StackExchange.

Consider the following example:

"As a risk-averse consumer, you would want to choose a value of x so as to maximize expected utility, i.e.

Given actuarially fair insurance, where p = r, you would solve: max pu(w - px - L + x) + (1-p)u(w - px), since in case of an accident, you total wealth would be w, less the loss suffered due to the accident, less the premium paid, and adding the amount received from the insurance company.

Differentiating with respect to x, and setting the result equal to zero, we get the first-order necessary condition as: (1-p)pu'(w - px - L + x) - p(1-p)u'(w - px) = 0,

which gives us: u'(w - px - L + x) = u'(w - px)

Risk-aversion implies u" < 0, so that equality of the marginal utilities of wealth implies equality of the wealth levels, i.e.

w - px - L + x = w - px,

so we must have x = L.

So, given actuarially fair insurance, you would choose to fully insure your car. Since you're risk-averse, you'd aim to equalize your wealth across all circumstances - whether or not you have an accident.

However, if p and r are not equal, we will have x < L; you would under-insure. How much you'd underinsure would depend on the how much greater r was than p."

Now, how the condition u"<0 changes anything to reach the result expressed above?

John Doe
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    $u''<0$ means the agent is risk averse. Changing this $u''=0$ means the agent is risk neutral. Risk neutral agents buy insurance only if it is fair, never when the expected value from insuring is less. $u''>0$ are risk loving agents. They never buy insurance; they would pay to avoid insurance (or put differently, you can offer them unfair bets and they'd take it). Note that with $u''\ge 0$ you cannot use the first order condition; the solution to the maximization problem will be a corner solution. – Nameless Mar 28 '14 at 14:02
  • So u"<0 is just a condition to reach maximization by differentiating and setting this value to zero? I Think I get it. – John Doe Mar 28 '14 at 14:10
  • Put another way, you need $u''<0$ for the second order conditions for a maximum to hold. – Trurl Mar 28 '14 at 17:27
  • I think I understand it now. The thing is, by the example I've described above, I thought that u"<0 somehow implied that an equality of the derivatives of the same function would bring also the equality of the arguments -- which would be unnecessary, I think. But I get it as a condition for maximizing and having an interior solution. – John Doe Mar 28 '14 at 17:41
  • If you have $u^{\prime\prime}>0$ (risk-loving) you also get equality of the arguments but in this case the first-order condition gives you a minimum. Also if $u$ is neither risk-loving nor risk-averse, $u^\prime$ is not monotonic and you get not get that $\u^\prime(X)=u^\prime(Y)$ if and only if $X=Y$. – Sergio Parreiras Apr 15 '14 at 04:28

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You were right in thinking $u^{\prime\prime}<0$ is related to equating the arguments. It guarantees that $u^\prime$ is strictly decreasing so $u^\prime(X)=u^\prime(Y)$ if and only if $X=Y$.

If you have $u^{\prime\prime}>0$ (risk-loving) you also get equality of the arguments but in this case the first-order condition gives you a minimum.

But if $u$ is neither risk-loving nor risk-averse, $u^\prime$ is not strictly monotonic and you do not get that $u^\prime(X)=u^\prime(Y)$ if and only if $X=Y$.

Sergio Parreiras
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