<p>In the slide, we're given the marginal utility (or the derivative of the utility function) as <span class="math-container">$m(x) = x^{-b}$</span>. The utility function whose derivative is <span class="math-container">$m(x)$</span> is</p>

<p><span class="math-container">\begin{eqnarray*} u(x) =  \int m(x) dx = \int x^{-b} dx = \begin{cases} \frac{1}{1- b} x^{1-b} &amp; \text{when } b \neq 1 \\  \ln x &amp; \text{when } b = 1 \end{cases} \end{eqnarray*}</span></p>

<p>Verify that <span class="math-container">$u'(x) = m(x)$</span>.</p>

CC BY-SA 4.0
2019-11-01T01:58:59.917Z
2019-11-01T01:58:59.917Z
Amit
<p>In this <a href="http://web.math.ku.dk/~rolf/teaching/slides_070409" rel="nofollow noreferrer">slide</a>, it says that constant relative risk-Aversion utility function have this form.</p>

<p><span class="math-container">$u(x) = \frac{1}{1-b} x^{1-b}$</span> for <span class="math-container">$b≠1$</span></p>

<p><span class="math-container">$u(x) = In(x)$</span> for <span class="math-container">$b=1$</span></p>

<p>When I tried to derive the utility function from <span class="math-container">$b = -x \frac{u''(x)}{u'(x)}$</span> (the representation of CRRA), I got <span class="math-container">$u(x) = x^{1-b}$</span>.</p>

<p>While I understand that preference will not be changed by linear transformation of the utility function, I wonder why is it necessary to make the utility function be <span class="math-container">$u(x) = \frac{1}{1-b} x^{1-b}$</span> for <span class="math-container">$b≠1$</span>.</p>

<p>Also how is the "<span class="math-container">$u(x) = In(x)$</span> for <span class="math-container">$b=1$</span>" obtained?</p>

<microeconomics><risk><risk-aversion>
How is the utility function with constant relative risk-aversion obtained?
Aqqqq
# How is the utility function with constant relative risk-aversion obtained?

(Asked by: username_0 on 2019-10-31)

In this [slide](http://web.math.ku.dk/~rolf/teaching/slides_070409), it says that constant relative risk-Aversion utility function have this form.

$u(x) = \frac{1}{1-b} x^{1-b}$ for $b≠1$

$u(x) = In(x)$ for $b=1$

When I tried to derive the utility function from $b = -x \frac{u''(x)}{u'(x)}$ (the representation of CRRA), I got $u(x) = x^{1-b}$.

While I understand that preference will not be changed by linear transformation of the utility function, I wonder why is it necessary to make the utility function be $u(x) = \frac{1}{1-b} x^{1-b}$ for $b≠1$.

Also how is the "$u(x) = In(x)$ for $b=1$" obtained?

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

---

## 1 answers

### Answer by username_1 on 2019-11-01. Score: 2 | Accepted answer

In the slide, we're given the marginal utility (or the derivative of the utility function) as $m(x) = x^{-b}$. The utility function whose derivative is $m(x)$ is

\begin{eqnarray\*} u(x) = \int m(x) dx = \int x^{-b} dx = \begin{cases} \frac{1}{1- b} x^{1-b} & \text{when } b \neq 1 \\ \ln x & \text{when } b = 1 \end{cases} \end{eqnarray\*}

Verify that $u'(x) = m(x)$.


---

