During a random process over a sample space $\Omega$, an agent incurs a cost (say, in money) given by a random variable $Z:\Omega \rightarrow R$, which is determined by the agent's action. The agent chooses the action which minimizes its "risk", which it calculates as some function $\rho(Z)$. Examples:
- A risk-neutral agent uses $\rho=E$, the expectation operator, to minimize expected cost.
- A minimax agent uses $\rho(Z)=\inf_\omega Z(\omega)$ to minimize the maximum possible cost.
- A maxipok agent uses $\rho(Z)=P(Z > z)$ to minimize the chance of a cost exceeding a threshold $z$.
The latter two agents are risk-averse: given two cost distributions with the same mean, they will associate more risk with whichever has higher variance. Risk-aversion is often quite sensible: would you rather certainly keep your house, or bet it against a second house on a fair coinflip? But expected utility theorists typically justify risk-aversion on the basis that an agent's true utility function is usually concave in their net worth–money has decreasing marginal utility. Equivalently, cost in utility is convex in monetary cost.
If this is the rationale for risk-aversion, then a risk-averse agent's risk metric should satisfy $\rho(Z)=E(f \circ Z)$, where $f:R \rightarrow R$ is a convex function that relates cost in money to cost in utility. It's not hard to show that the maxipok metric satisfies this relationship for $f(c) = I(c > z)$, but minimax does not for any $f$. Nor do many of the other most popular risk metrics in finance: none of the VaR$_\alpha$, CVaR$_\alpha$, and mean-variance metrics can be interpreted as minimizing the expectation of a latent disutility function. The only well-studied risk-averse metric besides maxipok with this property is entropic risk, which has $f(c) = e^{-ac}$.
Ok, here's the question: I'm curious whether there is a pleasant characterization of the set of risk metrics which compute the expectation of some disutility. In other words, when does $\rho:R^\Omega \rightarrow R \cup \{\pm\infty\}$ satisfy $\rho(Z) = E(f \circ Z)$ for some $f:R \rightarrow R$? Furthermore, if $f$ exists, when does it have nice properties, e.g. monotonicity, continuity, convexity?
The $|\Omega| < \infty$ case is probably the place to start, to make the measure theory less horrifying.
Best Answer
I now have a condition which I believe is necessary and sufficient in the special case of a finite sample space. My attempted proof is below.
Claim
Let $\Omega = [n]$, equipped with probability $Q=(q_1,\ldots,q_n)$ in the probability simplex in $R^n$. For a given risk metric $\rho:R^n \rightarrow R$, a latent disutility function $f:R \rightarrow R$ exists satisfying $\rho(Z)=E(f \circ Z)$ if and only if for every $\{x_1, \ldots, x_n\} \subseteq R$, $${\scr Q{\bf f}=R},$$ where $\scr Q$ is an $\underbrace{n \times \cdots \times n}_\text{$n+1$ times}$ tensor, ${\bf f} \in R^n$, and $\scr R$ is an $\underbrace{n \times \cdots \times n}_\text{$n$ times}$ tensor, with the components $${\scr q}_{\omega_1 \ldots \omega_n k}=\sum_j q_j I(\omega_j=k)$$ $${\bf f}_\omega = \rho(x_\omega, \ldots, x_\omega)$$ $${\scr r}_{\omega_1 \ldots \omega_n}=\rho(x_{\omega_1}, \ldots, x_{\omega_n}).$$.
Proof
$(\implies)$ Assume such an $f$ exists. A cost variable $Z=(z_1,\ldots,z_n)$ is an element of $R^n$, and $E(Z) =\sum_j q_j z_j$. If $Z$ is constant, then $z_j = c$ for each $j \in [n]$, and thus $$E(f \circ Z) =\sum_j q_jf(z_j)= \sum_j q_jf(c)=f(c).$$ Hence $f(c) = \rho(c,\ldots,c)$ for each $c \in R$.
For any $\{x_1, \ldots, x_n\} \subseteq R$, there are $n^n$ random variables supported on the $x_i$, each of which has the form $Z = (x_{\omega_1}, \ldots, x_{\omega_n})$. Each such $Z$ places a linear constraint on $\rho$: $$\rho(x_{\omega_1}, \ldots, x_{\omega_n})=\rho(Z)=E(f \circ Z)=\sum_j q_jf(x_{\omega_j})=\sum_j q_j\rho(x_{\omega_j}, \ldots, x_{\omega_j}).$$ The equation ${\scr Q{\bf f}=R}$ expresses all of these constraints in tensor form, so it must hold.
$(\impliedby)$ From above, $f:R \rightarrow R$ defined by $f(c) = \rho(c,\ldots,c)$ has the desired property. This completes the proof.
Discussion
This result allows one to easily reject risk functions which do not have associated latent disutilities, by evaluating $\rho$ for particular $Z$ and enforcing linear constraints. It also provides an explicit construction of the unique $f$ from $\rho$ in cases where such an $f$ exists. This makes it easy to check monotonicity/continuity/convexity.
Unfortunately, even though the condition I present is sufficient, it would be quite difficult to use this result to confirm risk functions which do have associated latent disutilities. I'd love to see if someone who knows more about crazy linear algebra could find a more easily verifiable sufficient condition. And of course, a generalization to infinite sample spaces would be incredible, although at this point I'm afraid I wouldn't understand it without a PhD in real analysis.