Deriving the constant relative risk aversion utility function

Question

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?

Answer 1

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

Answer 2

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 $u'(w)>0$.