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?
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
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