Why Fair Insurance Leads a Risk-Averse Consumer to Full Coverage

# Why is that a risk averse consumer buys the optimum insurance when there is actuarially fair insurance?


# Why is that a risk averse consumer buys the optimum insurance when there is actuarially fair insurance?












I think I understand the fact that when marginal utilities of the same function are equal (a consequence of the actuarially fair insurance), the independent variables in it must be equal -- right? But what it is the reason in this for a consumer being risk averse? What a $u''<0$ changes in comparison to a $u">0$ condition?

Edit: Example found here

> "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: $\max \left[pu(w - px - L + x) + (1-p)u(w - px)\right]$, 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?

## Answer by ajc3 (score 2)

https://quant.stackexchange.com/a/11089

It is the second derivative test.

From your example: For $u'(w-px-L+x)-u'(w-px)=0$ to be at a maximum, we need

\begin{eqnarray} &\frac{d}{dx}&\left[u'(w-px-L+x)-u'(w-px)\right]\\ &=&(1-p)u''(w-px-L+x)+pu''(w-px)<0. \end{eqnarray} For a risk averse individual, $u''(x)<0$ because of Jensen's Inequality, hence the condition is met.

A more thorough walkthrough than your example can be found here

## Answer by Aksakal almost surely binary (score 1)

https://quant.stackexchange.com/a/10741

this is related to the concept of Jensen inequality. basically, $\frac{f(x-|\delta|)+f(x+|\delta|)}{2}\ne f(x)$, for convex functions it's $>f(x)$, and for concave ones $<f(x)$. risk averse guys have concave utilities, that's the relation you need to look at