<p>This is just a consequence of the (here tacit) assumption that <span class="math-container">$u'&gt;0$</span>.</p>

CC BY-SA 4.0
2023-03-18T19:16:17.890Z
Apparently not, see my answer below
Philip Hartfield
CC BY-SA 4.0
2023-03-19T09:32:19.740Z
@PhilipHartfield, well, in your third equation below you take the $log$ of $u'$, which of course presupposes that $u'>0$.
VARulle
CC BY-SA 4.0
2023-03-19T11:44:53.207Z
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.
Philip Hartfield
CC BY-SA 4.0
2023-03-20T09:39:45.563Z
@PhilipHartfield, I'd rather say that $u'>0$ is presumed, but this should not be too surprising in the context of utility functions...
VARulle
CC BY-SA 4.0
2023-03-20T15:00:25.210Z
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.
Philip Hartfield
CC BY-SA 4.0
2023-03-03T11:06:31.950Z
2023-03-03T11:06:31.950Z
VARulle
<p>Here is the solution from class,</p>
<p><span class="math-container">\begin{align}
    -w\frac{u''(w)}{u'(w)} &amp;= \rho\\
    \frac{u''(w)}{u'(w)} &amp;= -\frac{\rho}{w}\\
    \frac{d}{dw}\big(\log u'(w)\big) &amp;= \frac{d}{dw}\big(\log w^{-\rho}\big)\\
     \int\frac{d}{dw}\big(\log u'(w)\big)dw &amp;= \int \frac{d}{dw}\big(\log w^{-\rho}\big)dw\\
    \log u'(w) &amp;= \log w^{-\rho}+C\\
    e^{\log u'(w)} &amp;= e^{\log w^{-\rho}+C}\\
    u'(w) &amp;= e^{C}w^{-\rho}\\
    u'(w) &amp;= \eta w^{-\rho}, \text{ }\text{ }\text{ }\eta\equiv e^C&gt;0
\end{align}</span>
As VARulle points out, in the 3rd line it is assumed that <span class="math-container">$u'(w)&gt;0$</span>.</p>

CC BY-SA 4.0
2023-03-18T19:05:23.073Z
2023-03-19T11:50:00.190Z
2023-03-19T11:50:00.190Z
Philip Hartfield
<p>Here is the question I am trying to tackle:</p>
<p>Suppose that we are given a utility function <span class="math-container">$u$</span> with relative risk aversion <span class="math-container">$R_u$</span>. Show that <span class="math-container">$R_u$</span> is constant and equal to <span class="math-container">$\rho$</span> iff there exist <span class="math-container">$\zeta\in\mathbb{R}$</span> and <span class="math-container">$\eta&gt;0$</span> such that
<span class="math-container">$$ u = \zeta + \eta D_\rho $$</span>
where
<span class="math-container">$$ D_\rho(w)=\begin{cases}\frac{w^{1-\rho}-1}{1-\rho},&amp;\text{if }\rho\neq 1\\ \log(w),&amp;\text{if } \rho=1\end{cases}$$</span>
In short, I do not understand the requirement for <span class="math-container">$\eta&gt;0$</span>. Here is my solution thus far:</p>
<p>Relative risk aversion is defined as
<span class="math-container">$$RRA=-w\frac{u''(w)}{u'(w)}$$</span>
and we require
<span class="math-container">\begin{align}
RRA&amp;=\rho\tag{1} \\
\frac{d}{dw}(RRA)&amp;=0\tag{2}
\end{align}</span>
We have,
<span class="math-container">\begin{align}
\frac{d}{dw}(RRA) &amp;= \frac{d}{dw}\Big(-wu''(u')^{-1}\Big)\\
&amp;= -u''(u')^{-1} -w\Big(u'''(u')^{-1}-(u'')^2(u')^{-1}\Big)
\end{align}</span>
From condition (2),
<span class="math-container">$$ u''+wu'''-w(u'')^2(u')^{-1}=0$$</span>
Substituting in condition (1), <span class="math-container">$-wu''(u')^{-1} = \rho$</span>,
<span class="math-container">\begin{align}
u'' + wu''' + \rho u'' &amp;=0\\
(-\rho-1)u'' &amp;= wu'''\rightarrow u'' = \kappa w^{-\rho-1}
\end{align}</span>
for some arbitrary constant <span class="math-container">$\kappa$</span>. Then
<span class="math-container">\begin{align}
u'&amp;=\int \kappa w^{-\rho-1} dw\\
&amp;= \eta w^{-\rho} + c
\end{align}</span>
for some arbitrary constants <span class="math-container">$\eta$</span> and <span class="math-container">$c$</span>. Then
<span class="math-container">\begin{align}
u = \int \eta w^{-\rho}+c\text{ }dw = \begin{cases}\eta\frac{w^{1-\rho}}{1-\rho}+cw+C,&amp;\text{if }\rho\neq 1\\ \eta\log(w)+cw+C,&amp;\text{if } \rho=1\end{cases}
\end{align}</span>
for some arbitrary constant <span class="math-container">$C$</span>. Clearly, condition (1) requires that <span class="math-container">$c=0$</span>, i.e.
<span class="math-container">\begin{align}
u &amp;= \begin{cases}\eta\frac{w^{1-\rho}}{1-\rho}+C,&amp;\text{if }\rho\neq 1\\ \eta\log(w)+C,&amp;\text{if } \rho=1\end{cases}
\end{align}</span>
Since <span class="math-container">$C$</span> is arbitrary (i.e. our two conditions are satisfied by any <span class="math-container">$C$</span>), we can rewrite this in the representation given by the question. Showing the reverse, that that the representation in the question is <em>sufficient</em> for the two conditions to be met, is trivial.</p>
<p>My question is, why do we <em>require</em> <span class="math-container">$\eta&gt;0$</span>? Is it a mistake in the question or am I missing something?</p>

<microeconomics><asset-pricing><risk-aversion>
Deriving the constant relative risk aversion utility function
Philip Hartfield
# Deriving the constant relative risk aversion utility function

(Asked by: username_0 on 2023-03-02)

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?

**Tags:** microeconomics, asset-pricing, risk-aversion

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## 2 answers

### Answer by username_1 on 2023-03-03. Score: 3

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

> **Comments:**
> **username_0**: Apparently not, see my answer below  
> **username_1**: @PhilipHartfield, well, in your third equation below you take the $log$ of $u'$, which of course presupposes that $u'>0$.  
> **username_0**: 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.  
> **username_1**: @PhilipHartfield, I'd rather say that $u'>0$ is presumed, but this should not be too surprising in the context of utility functions...  
> **username_0**: 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.  

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### Answer by username_0 on 2023-03-18. Score: 1

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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