<p>The lottery you show is equivalent to having two lotteries, one where you get \$50 with certainty and one with the following payoffs:</p>

<ul>
<li>\$0 with probability 1/3</li>
<li>\$50 with probability 1/3</li>
<li>\$100 with probability 1/3</li>
</ul>

<p>The first lottery has a certainty equivalent of \$50. The second lottery has no negative payoffs and so shouldn't ever have a negative price. Therefore the certainty equivalent must be $\geq 50$.</p>

CC BY-SA 3.0
2015-03-15T00:06:30.220Z
2015-03-15T10:30:43.440Z
2015-03-15T10:30:43.440Z
BKay
<blockquote>
  <p><em>"Since agent is risk averse, we would expect that $U(E[g]) &lt; U(CE)$ ,
  where $CE$ is the certainty equivalent."</em></p>
</blockquote>

<p>This is wrong. I presume the Expected Utility Property holds here, so, if we denote the gamble by $G$, a discrete uniform random variable taking three values according to the setup, we have</p>

<p>$$U(CE) \equiv \sum_{i=1}^3p_iU(g_i) = E[U(G)] &lt; U[E(G)]$$</p>

<p>the inequality to the right due to Jensen's Inequality and the assumption that $U()$ is concave. This also gives us</p>

<p>$$ CE &lt; E(G)$$</p>

<p>which should be intuitive: a risk-neutral person would demand $E(G)$, the expected value of the gamble, in order not to take it. A risk-averse person would require <em>less</em>, to leave the gamble.</p>

<p>Having cleared this, the OP asks: <strong>Is it possible that $CE &lt; \min G$?</strong></p>

<p>The answer is : <strong>No.</strong> Assume that the gamble outcomes are ordered, so $\min G = g_1$.</p>

<p><strong>Ad absurdum</strong>, assume that $CE &lt; g_1$ holds. Then we will have</p>

<p>$$U(CE) &lt; U(g_1)$$</p>

<p>Using the definition of $CE$ we replace the left-hand side</p>

<p>$$\sum_{i=1}^3p_iU(g_i) &lt; U(g_1) \implies p_2U(g_2) + p_3U(g_3) &lt; (1-p_1) U(g_1)$$</p>

<p>$$\implies p_2U(g_2) + p_3U(g_3) &lt; p_2U(g_1) + p_3U(g_1)$$</p>

<p>$$\implies p_2[U(g_2)-U(g_1)] + p_3[U(g_3)-U(g_1)] &lt; 0$$</p>

<p>But this is impossible since $g_1 = \min\{g_1,g_2, g_3\}$ and so</p>

<p>$U(g_2)-U(g_1) &gt;0$ and $U(g_3)-U(g_1) &gt;0$.</p>

<p>So assuming $U(CE) &lt; U(g_1)$ led us to an impossible situation, and therefore  it cannot hold. </p>

<p>Intuitively, the <em>worst</em> outcome of being in the gamble is to receive the minimum payoff -so for a rational agent, even if it is risk-averse, it would be irrational to accept less than the worst outcome, since then it would <em>certainly</em> do worse than being in the gamble. Note that "aversion to risk" does <em>not</em> mean "take away <em>all</em> risk at <em>all</em> costs". </p>

CC BY-SA 3.0
2015-03-15T02:35:08.303Z
2015-03-15T16:00:25.423Z
2015-03-15T16:00:25.423Z
Alecos Papadopoulos
<p>Suppose there is an agent who faces the following gamble g: </p>

<ul>
<li>50\$ with probability 1/3</li>
<li>100\$ with probability 1/3</li>
<li>150\$ with probability 1/3</li>
</ul>

<p>Clearly, the E[g] = 100\$. Since agent is risk averse, we would expect that U(E[g]) &lt; U(CE) , where CE is the certainty equivalent. Now, my question is, is it hypothetically possible that the agent's Certainty Equivalent of this gamble will be below 50\$, or is it necessarily true that it will be somewhere between 50\$ and 100\$?</p>

<risk-aversion>
Can a risk-averse agent's Certainty Equivalent be lower than the lowest possible outcome of a gamble?
Paul
# Can a risk-averse agent's Certainty Equivalent be lower than the lowest possible outcome of a gamble?

(Asked by: username_0 on 2015-03-14)

Suppose there is an agent who faces the following gamble g:

* 50\$ with probability 1/3
* 100\$ with probability 1/3
* 150\$ with probability 1/3

Clearly, the E[g] = 100\$. Since agent is risk averse, we would expect that U(E[g]) < U(CE) , where CE is the certainty equivalent. Now, my question is, is it hypothetically possible that the agent's Certainty Equivalent of this gamble will be below 50\$, or is it necessarily true that it will be somewhere between 50\$ and 100\$?

**Tags:** risk-aversion


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## 2 answers

### Answer by username_2 on 2015-03-15. Score: 7 | Accepted answer

> *"Since agent is risk averse, we would expect that $U(E[g]) < U(CE)$ ,
> where $CE$ is the certainty equivalent."*

This is wrong. I presume the Expected Utility Property holds here, so, if we denote the gamble by $G$, a discrete uniform random variable taking three values according to the setup, we have

$$U(CE) \equiv \sum\_{i=1}^3p\_iU(g\_i) = E[U(G)] < U[E(G)]$$

the inequality to the right due to Jensen's Inequality and the assumption that $U()$ is concave. This also gives us

$$ CE < E(G)$$

which should be intuitive: a risk-neutral person would demand $E(G)$, the expected value of the gamble, in order not to take it. A risk-averse person would require *less*, to leave the gamble.

Having cleared this, the OP asks: **Is it possible that $CE < \min G$?**

The answer is : **No.** Assume that the gamble outcomes are ordered, so $\min G = g\_1$.

**Ad absurdum**, assume that $CE < g\_1$ holds. Then we will have

$$U(CE) < U(g\_1)$$

Using the definition of $CE$ we replace the left-hand side

$$\sum\_{i=1}^3p\_iU(g\_i) < U(g\_1) \implies p\_2U(g\_2) + p\_3U(g\_3) < (1-p\_1) U(g\_1)$$

$$\implies p\_2U(g\_2) + p\_3U(g\_3) < p\_2U(g\_1) + p\_3U(g\_1)$$

$$\implies p\_2[U(g\_2)-U(g\_1)] + p\_3[U(g\_3)-U(g\_1)] < 0$$

But this is impossible since $g\_1 = \min\{g\_1,g\_2, g\_3\}$ and so

$U(g\_2)-U(g\_1) >0$ and $U(g\_3)-U(g\_1) >0$.

So assuming $U(CE) < U(g\_1)$ led us to an impossible situation, and therefore it cannot hold.

Intuitively, the *worst* outcome of being in the gamble is to receive the minimum payoff -so for a rational agent, even if it is risk-averse, it would be irrational to accept less than the worst outcome, since then it would *certainly* do worse than being in the gamble. Note that "aversion to risk" does *not* mean "take away *all* risk at *all* costs".


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### Answer by username_1 on 2015-03-15. Score: 2

The lottery you show is equivalent to having two lotteries, one where you get \$50 with certainty and one with the following payoffs:

* \$0 with probability 1/3
* \$50 with probability 1/3
* \$100 with probability 1/3

The first lottery has a certainty equivalent of \$50. The second lottery has no negative payoffs and so shouldn't ever have a negative price. Therefore the certainty equivalent must be $\geq 50$.


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