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?

  • 1
    $\begingroup$ could you provide some more background? Some formulas and perhaps a general setup ? $\endgroup$ – Probilitator Mar 28 '14 at 7:16

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

  • $\begingroup$ But even if you had a convex function, why wouldn't the arguments be equal if the first derivatives (of the same utility function) are also equal? $\endgroup$ – John Doe Mar 28 '14 at 14:03
  • $\begingroup$ @JohnDoe i dont understand your question $\endgroup$ – Aksakal almost surely binary Mar 28 '14 at 14:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.