I think this has to do with the definition of concavity and the fact that a risk averse person has a concave utility function, but I'm not sure how that helps.


1 Answer 1


Suppose that the vector $W=\left(w_1,w_2,\dots,w_n\right)$ represents wealth in $n$ possible states. In addition, assume the probability of each state occurring is represented by the vector $\pi=\left(\pi_1,\pi_2,\dots,\pi_n\right)$. We can express this as the simple gamble:

$$g = \left(\pi_1\circ w_1,\pi_2\circ w_2, \dots, \pi_n\circ w_n\right)$$

The expected value of the simple gamble $g$ is:


Suppose a consumer has a sub-utility function $u(w)$. Further, suppose that the consumers preferences over gambles are such that they can be represented by a von Neumann Morgenstern utility function (has the expected utility property). The utility of the gamble can be expressed as the expected utility across all states:


Now, if we assume that the sub-utility function $u(w)$ is strictly concave (risk averse consumer), by the concavity of functions it must be the case that:

$$u\left(\mathbb{E}[g]\right) > u(g)$$

This implies that the consumer would prefer to receive the expected value of the gamble with certainty rather than the gamble itself. A question we could ask; how much would I need to give to the consumer such that they are indifferent between choosing the gamble and receiving some amount of wealth with certainty (Certainty Equivalent)? The certainty equivalent $CE$ must satisfy the following:

$$u\left(CE\right) = u(g) \implies u\left(\mathbb{E}[g]\right) > u\left(CE\right) \implies \mathbb{E}[g] > CE$$

Since $u(w)$ is increasing in wealth, there exists a $P>0$ such that the following will hold:

$$u\left(\mathbb{E}[g]-P\right)= u\left(CE\right) \implies \mathbb{E}[g] -P= CE$$

$P$ represents the risk premium. It is the amount you would be willing to pay to get rid of uncertainty (receive the certainty equivalent). The fact that the risk premium is positive does arise from the fact that we assume the sub-utility function is strictly concave.

  • $\begingroup$ Ahhh that makes a lot of sense. Thanks for your insight! $\endgroup$ Dec 2, 2019 at 19:58

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