Economics Stack Exchange is a question and answer site for professional and academic economists and analysts. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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\$?

share|improve this question
up vote 4 down vote accepted

"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".

share|improve this answer
Thank you so much, that is a great explanation! – Paul Mar 15 '15 at 3:20
@Paul You're welcome. – Alecos Papadopoulos Mar 15 '15 at 3:41

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 $\geq 50$.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.