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I am working on a model that assumes expected utility maximisation with a vNM function $u$ that exhibits some standard properties:

  1. $u'(w) > 0$ and $u''(w) < 0$ for all $w \geq 0$.
  2. $\lim_{w \rightarrow 0^+}u'(w) = \infty$.
  3. $-u''(w)/u'(w)$ is strictly decreasing in $w$ over $w \geq 0$.

As a leading example, I have chosen $u(w) = \ln(w)$. However, I now wonder if this satisfies (1) or (3) given that $\ln(w)$, and its derivative $1/w$, are both undefined when $w = 0$.

Relatedly, it is hard for me to argue that, in the optimal lottery, wealth is positive in every state: I want to say that $\ln(0) = -\infty$, which cannot be optimal, but this does not seem rigorous.

Assuming this is a problem, is there a simple solution?

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    $\begingroup$ How about $u(w)=\ln(1+w)$? It is continuous and well-defined at $0$. $\endgroup$
    – Amit
    Commented Jul 30 at 15:30
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    $\begingroup$ Why is it not rigorous to extend the function so that $u(0)=-\infty$? $\endgroup$ Commented Jul 30 at 20:00
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    $\begingroup$ It may be unrelated to your point, but one issue you may run in to is $\infty - \infty$, which is not well defined a priori. Consider the lottery that with $1/2$ probability gives $w=0$ and gives $w=\exp\{\frac{\pi^2}{3}n^2\}$ with probability $\frac{3}{\pi^2} \frac{1}{n^2}$ for $n\in \mathbb{N}$. The expected utility of such a lottery is $\frac{1}{2}u(0)+\sum_{n=1}^{\infty}\frac{3}{\pi^2} \frac{1}{n^2}u(\exp\{\frac{\pi^2}{3}n^2\})=\infty-\infty$. One solution is to define this $\infty-\infty$ to be something. Another is to restrict attention to lotteries with bounded wealth outcomes. $\endgroup$ Commented Aug 2 at 17:27
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    $\begingroup$ How about $u(w)=\sqrt{w}$? This function satisfy the desired properties 1 and 3 for all $w>0$, and also satisfy property 2. $\endgroup$
    – Amit
    Commented Aug 2 at 23:27
  • $\begingroup$ Thanks for the suggestions everyone; I'll need to think a bit more about this $\endgroup$
    – afreelunch
    Commented Aug 5 at 16:17

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