<p>Assume that you have an income of <span class="math-container">$z$</span> and receive an additional <span class="math-container">$x_i$</span> with probability <span class="math-container">$\pi_i$</span>. Then the certainty equivalent <span class="math-container">$c_1$</span> is implicity defined by:
<span class="math-container">$$
\sum_i \pi_i u(z+x_i) = u(z+c)
$$</span>
If we use the CARA utility function this gives:
<span class="math-container">$$
\begin{align*}
&amp;\sum_i \pi_i (-\exp(-b(z + x_i))) = (-\exp(-b (z + c_1))\\
\leftrightarrow &amp;\sum_i \pi_i \exp(-b(z + x_i)) = \exp(-b (z + c_1) \tag{1}
\end{align*}
$$</span>
Now consider a lottery where at every state <span class="math-container">$i$</span> you receive an additional amount <span class="math-container">$g$</span> then the certainty equivalent <span class="math-container">$c_2$</span> is given by:
<span class="math-container">$$
\begin{align*}
&amp;\sum_i \pi_i (-\exp(-b(z + g+ x_i))) = (-\exp(-b (z + c_2)),\\
\leftrightarrow &amp;\sum_i \pi_i \exp(-b(z + g+ x_i)) = \exp(-b (z + c_2)
\end{align*}
$$</span>
Using the properties of the exponential and using <span class="math-container">$(1)$</span>, we can rewrite this as:
<span class="math-container">$$
\begin{align*}
&amp;\exp(-bg)  \sum_i \pi_i \exp(-b(z +x_i)) = \exp(-b (z + c_2),\\
\leftrightarrow &amp;\exp(-bg) \exp(-b(z + c_1) = \exp(-b(z + c_2)),\\
\leftrightarrow &amp; \exp(-b(g + z + c_1 - z - c_2) = 1,\\
\leftrightarrow &amp; \exp(-b(g + c_1 - c_2) = 1,\\
\to &amp; c_1 + g = c_2.
\end{align*}
$$</span></p>
<p>This shows that if the mean of a lottery increases by an amount <span class="math-container">$g$</span> (by adding to every state the same amount) then the certainty equivalence increases by the same amount.</p>

CC BY-SA 4.0
2021-11-08T15:38:40.820Z
2021-11-08T15:38:40.820Z
tdm
<p>suppose an agent has <span class="math-container">$u(z)=-e^{-bz}$</span> where <span class="math-container">$b&gt;0$</span> as her Bernoulli utility function and faces two gambles:<br />
G1: win 1000 dollars with probability <span class="math-container">$\frac{1}{2}$</span> and zero with probability <span class="math-container">$\frac{1}{2}$</span> and his certainty equivalent is 488 dollars<br />
G2: win 1500 dollars with probability <span class="math-container">$\frac{1}{2}$</span> and 500 with probability <span class="math-container">$\frac{1}{2}$</span><br />
Question is : what can we say about his certainty equivalent for gamble2?<br />
How can I approach this problem?</p>

CC BY-SA 4.0
2021-11-02T00:38:50.803Z
"I tried to derive his certainty equivalent by plugging numbers but I could not find it." -- How exactly did you try? What formula did you plug the numbers in?
Herr K.
CC BY-SA 4.0
2021-11-02T01:26:57.080Z
Thanks fpr response. After you said I tried again and found 988 as certainty equivalent. How can I relate this to absolute risk aversion and relative risk aversion?
Mrnobody
<microeconomics><utility><expected-utility><risk-aversion><uncertainty>
certainty equivalent and lotteries
Mrnobody
# certainty equivalent and lotteries

(Asked by: username_0 on 2021-11-01)

suppose an agent has $u(z)=-e^{-bz}$ where $b>0$ as her Bernoulli utility function and faces two gambles:  
G1: win 1000 dollars with probability $\frac{1}{2}$ and zero with probability $\frac{1}{2}$ and his certainty equivalent is 488 dollars  
G2: win 1500 dollars with probability $\frac{1}{2}$ and 500 with probability $\frac{1}{2}$  
Question is : what can we say about his certainty equivalent for gamble2?  
How can I approach this problem?

**Tags:** microeconomics, utility, expected-utility, risk-aversion, uncertainty
> **Comments:**
> **username_1**: "I tried to derive his certainty equivalent by plugging numbers but I could not find it." -- How exactly did you try? What formula did you plug the numbers in?  
> **username_0**: Thanks fpr response. After you said I tried again and found 988 as certainty equivalent. How can I relate this to absolute risk aversion and relative risk aversion?  

---

## 1 answers

### Answer by username_2 on 2021-11-08. Score: 0 | Accepted answer

Assume that you have an income of $z$ and receive an additional $x\_i$ with probability $\pi\_i$. Then the certainty equivalent $c\_1$ is implicity defined by:
$$
\sum\_i \pi\_i u(z+x\_i) = u(z+c)
$$
If we use the CARA utility function this gives:
$$
\begin{align\*}
&\sum\_i \pi\_i (-\exp(-b(z + x\_i))) = (-\exp(-b (z + c\_1))\\
\leftrightarrow &\sum\_i \pi\_i \exp(-b(z + x\_i)) = \exp(-b (z + c\_1) \tag{1}
\end{align\*}
$$
Now consider a lottery where at every state $i$ you receive an additional amount $g$ then the certainty equivalent $c\_2$ is given by:
$$
\begin{align\*}
&\sum\_i \pi\_i (-\exp(-b(z + g+ x\_i))) = (-\exp(-b (z + c\_2)),\\
\leftrightarrow &\sum\_i \pi\_i \exp(-b(z + g+ x\_i)) = \exp(-b (z + c\_2)
\end{align\*}
$$
Using the properties of the exponential and using $(1)$, we can rewrite this as:
$$
\begin{align\*}
&\exp(-bg) \sum\_i \pi\_i \exp(-b(z +x\_i)) = \exp(-b (z + c\_2),\\
\leftrightarrow &\exp(-bg) \exp(-b(z + c\_1) = \exp(-b(z + c\_2)),\\
\leftrightarrow & \exp(-b(g + z + c\_1 - z - c\_2) = 1,\\
\leftrightarrow & \exp(-b(g + c\_1 - c\_2) = 1,\\
\to & c\_1 + g = c\_2.
\end{align\*}
$$

This shows that if the mean of a lottery increases by an amount $g$ (by adding to every state the same amount) then the certainty equivalence increases by the same amount.


---

