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Here is the question I am trying to tackle:

Suppose that we are given a utility function $u$ with relative risk aversion $R_u$. Show that $R_u$ is constant and equal to $\rho$ iff there exist $\zeta\in\mathbb{R}$ and $\eta>0$ such that $$ u = \zeta + \eta D_\rho $$ where $$ D_\rho(w)=\begin{cases}\frac{w^{1-\rho}-1}{1-\rho},&\text{if }\rho\neq 1\\ \log(w),&\text{if } \rho=1\end{cases}$$ In short, I do not understand the requirement for $\eta>0$. Here is my solution thus far:

Relative risk aversion is defined as $$RRA=-w\frac{u''(w)}{u'(w)}$$ and we require \begin{align} RRA&=\rho\tag{1} \\ \frac{d}{dw}(RRA)&=0\tag{2} \end{align} We have, \begin{align} \frac{d}{dw}(RRA) &= \frac{d}{dw}\Big(-wu''(u')^{-1}\Big)\\ &= -u''(u')^{-1} -w\Big(u'''(u')^{-1}-(u'')^2(u')^{-1}\Big) \end{align} From condition (2), $$ u''+wu'''-w(u'')^2(u')^{-1}=0$$ Substituting in condition (1), $-wu''(u')^{-1} = \rho$, \begin{align} u'' + wu''' + \rho u'' &=0\\ (-\rho-1)u'' &= wu'''\rightarrow u'' = \kappa w^{-\rho-1} \end{align} for some arbitrary constant $\kappa$. Then \begin{align} u'&=\int \kappa w^{-\rho-1} dw\\ &= \eta w^{-\rho} + c \end{align} for some arbitrary constants $\eta$ and $c$. Then \begin{align} u = \int \eta w^{-\rho}+c\text{ }dw = \begin{cases}\eta\frac{w^{1-\rho}}{1-\rho}+cw+C,&\text{if }\rho\neq 1\\ \eta\log(w)+cw+C,&\text{if } \rho=1\end{cases} \end{align} for some arbitrary constant $C$. Clearly, condition (1) requires that $c=0$, i.e. \begin{align} u &= \begin{cases}\eta\frac{w^{1-\rho}}{1-\rho}+C,&\text{if }\rho\neq 1\\ \eta\log(w)+C,&\text{if } \rho=1\end{cases} \end{align} Since $C$ is arbitrary (i.e. our two conditions are satisfied by any $C$), we can rewrite this in the representation given by the question. Showing the reverse, that that the representation in the question is sufficient for the two conditions to be met, is trivial.

My question is, why do we require $\eta>0$? Is it a mistake in the question or am I missing something?

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2 Answers 2

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This is just a consequence of the (here tacit) assumption that $u'>0$.

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  • $\begingroup$ Apparently not, see my answer below $\endgroup$ Commented Mar 18, 2023 at 19:16
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    $\begingroup$ @PhilipHartfield, well, in your third equation below you take the $log$ of $u'$, which of course presupposes that $u'>0$. $\endgroup$
    – VARulle
    Commented Mar 19, 2023 at 9:32
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    $\begingroup$ You are completely correct! Then as far as I can see $\eta > 0$ is just a consequence of their solution and is presumed; will have to raise this with the TA. $\endgroup$ Commented Mar 19, 2023 at 11:44
  • $\begingroup$ @PhilipHartfield, I'd rather say that $u'>0$ is presumed, but this should not be too surprising in the context of utility functions... $\endgroup$
    – VARulle
    Commented Mar 20, 2023 at 9:39
  • $\begingroup$ True, but still seems backwards for a question designed to be rigiorous to ask you to show $\eta$ is necessarily greater than zero when it isn't without making an assumption not given by the question. $\endgroup$ Commented Mar 20, 2023 at 15:00
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Here is the solution from class,

\begin{align} -w\frac{u''(w)}{u'(w)} &= \rho\\ \frac{u''(w)}{u'(w)} &= -\frac{\rho}{w}\\ \frac{d}{dw}\big(\log u'(w)\big) &= \frac{d}{dw}\big(\log w^{-\rho}\big)\\ \int\frac{d}{dw}\big(\log u'(w)\big)dw &= \int \frac{d}{dw}\big(\log w^{-\rho}\big)dw\\ \log u'(w) &= \log w^{-\rho}+C\\ e^{\log u'(w)} &= e^{\log w^{-\rho}+C}\\ u'(w) &= e^{C}w^{-\rho}\\ u'(w) &= \eta w^{-\rho}, \text{ }\text{ }\text{ }\eta\equiv e^C>0 \end{align} As VARulle points out, in the 3rd line it is assumed that $u'(w)>0$.

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