Decision-making under risk: when is utility-maximization equivalent to risk-minimization?

Abstract

Motivated by the analysis of a general optimal portfolio selection problem, which encompasses as special cases an optimal consumption and an optimal debt-arrangement problem, we are concerned with the questions of how a personality trait like risk-perception can be formalized and whether the two objectives of utility-maximization and risk-minimization can be both achieved simultaneously. We address these questions by developing an axiomatic foundation of preferences for which utility-maximization is equivalent to minimizing a utility-based shortfall risk measure. Our axiomatization hinges on a novel axiom in decision theory, namely the risk-perception axiom.

Appendix

Proof of Theorem 1

$(2)\Rightarrow (1)$:

Let $U\left( {\textbf{x}}\right) =\min \left\{ {\mathbb {E}}_{\pi } \left[ {\textbf{x}}\right] :\pi \in D\right\}$ for all ${\textbf{x}}\in \ell _{\infty }$, and define the binary (preference) relation $\succsim$ on $\ell _{\infty }$ by ${\textbf{x}}\succsim {\textbf{y}}$ if and only if $U({\textbf{x}})\ge U({\textbf{y}})$ for all ${\textbf{x}},{\textbf{y}}\in \ell _{\infty }$. Since U is real-valued and D is closed, we have that

$$\begin{aligned} \min \left\{ {\mathbb {E}}_{\pi }\left[ {\textbf{x}}\right] :\pi \in D\right\} =\inf \left\{ {\mathbb {E}}_{\pi }\left[ {\textbf{x}}\right] :\pi \in D\right\} , \text { for all }{\textbf{x}}\in \ell _{\infty \text {.}}\text {.} \end{aligned}$$

(5)

Using (5), some straightforward algebra shows the following:

$$\begin{aligned} -U\left( {\textbf{x}}\right) =-\inf \left\{ {\mathbb {E}}_{\pi }\left[ {\textbf{x}} \right] :\pi \in D\right\} =\inf \left\{ t\in {\mathbb {R}}:{\textbf{x}}+t\textbf{ 1}\in P\right\} \text {,} \end{aligned}$$

where $P=\left\{ {\textbf{x}}\in \ell _{\infty }:{\mathbb {E}}_{\pi } \left[ {\textbf{x}}\right] \ge 0\text { for all }\pi \in D\right\} .$ Since P is a closed cone containing $\ell _{\infty }^{+}$, and $\mathbf {1\in }P$ is an order unit,¹⁶ by Lemma 1 and the above equality we have that $-U:\ell _{\infty }\rightarrow {\mathbb {R}}$ is a coherent risk measure, with respect to $\ell _{\infty }^{+}$ and the vector ${\textbf{1}}$, and ${\mathcal {A}}_{-U}=P$. Therefore, it follows from the four defining properties of coherent risk measures (in section 3) and the definition of acceptance set that:

$$\begin{aligned}{} & {} {\textbf{x}}-\mathbf {y\in }\ell _{\infty }^{+}\hbox { implies }U\left( {\textbf{x}} \right) \ge U\left( {\textbf{y}}\right) . \ (i)\\{} & {} U\left( \mathbf {x+}t{\textbf{1}}\right) =U\left( {\textbf{x}}\right) +t\hbox {, for all }{\textbf{x}}\in \ell _{\infty }\hbox { and }t\in {\mathbb {R}}. \ (ii)\\{} & {} U\left( \mathbf {x+y}\right) \ge U\left( {\textbf{x}}\right) +U\left( \textbf{ y}\right) \hbox {, for all }{\textbf{x}},{\textbf{y}}\in \ell _{\infty }. \ (iii) \\{} & {} U\left( \lambda {\textbf{x}}\right) =\lambda U\left( {\textbf{x}}\right) \hbox {, for all }{\textbf{x}}\in \ell _{\infty }\hbox { and }\lambda \ge 0. \ (iv) \\{} & {} \left\{ {\textbf{x}}\in \ell _{\infty }:U\left( {\textbf{x}}\right) \ge 0\right\} =A_{-U}=P \ (v). \end{aligned}$$

Of course, $\succsim$ satisfies axiom A.1. Using the above properties (ii) and (iv) it is very easy to see that $\succsim$ satisfies axiom A.2. By the above properties (iii) and (iv), $-U$ is a sublinear function and one can readily see that it is bounded on every neighborhood of zero. Hence, U is continuous in the supremum norm-topology.¹⁷ In turn, continuity of U easily implies that $\succsim$ satisfies axiom A.3. That $\succsim$ satisfies axiom A.4. part (a) is an immediate consequence of property (i). As to part (b) of axiom A.4., suppose that there is some $\epsilon >0$, such that $x_{n}-y_{n}\ge \epsilon$ for every $n\in {\mathbb {N}}$. Then, ${\textbf{x}}-{\textbf{y}}$ lies in the norm-interior of $\ell _{\infty }^{+}$. Since $P\supseteq \ell _{\infty }^{+}$, it also belongs to the norm-interior of P. Therefore, since $-U\left( {\textbf{x}}\right) =\inf \left\{ t\in {\mathbb {R}}:{\textbf{x}}+t{\textbf{1}}\in P\right\}$ (see above), it follows from Lemma 2 that $U\left( {\textbf{x}}-{\textbf{y}}\right) >0$. On the other hand, as an easy consequence of property (iii), we get $U\left( {\textbf{x}}\right) -U\left( {\textbf{y}}\right) \ge U\left( {\textbf{x}}-{\textbf{y}}\right)$, which implies $U\left( {\textbf{x}}\right) >U\left( {\textbf{y}}\right)$, and hence, ${\textbf{x}}\succ {\textbf{y}}$. This establishes that $\succsim$ satisfies part (b) of axiom A.4. Next, a very straightforward application of properties (iii) and (iv) shows that $\succsim$ satisfies axiom A.5. Finally, property (iv) clearly yields $U\left( {\textbf{0}}\right) =0$. Thus, $\succsim$ satisfies axiom A.6 as an obvious implication of property (v).

$(1)\Rightarrow (2)$: To improve readability, the proof will be divided into three steps.

Step 1 (Existence of a utility function representing preferences):

Let $\succsim$ be a preference relation on $\ell _{\infty }$ satisfying axioms A.1-A.6. Pick any arbitrary $\mathbf {x\in }\ell _{\infty }$, and define the following sets:

$$\begin{aligned} A= & {} \left\{ t\in {\mathbb {R}}:t{\textbf{1}}\succsim {\textbf{x}}\right\} \text { } \\ B= & {} \left\{ t\in {\mathbb {R}}:{\textbf{x}}\succsim t{\textbf{1}}\right\} \text {.} \end{aligned}$$

We claim that A and B are both non-empty. To see this, note that, by definition of $\ell _{\infty }$, there exists some $M>0$ such that $-M\le x_{n}\le M$ for every $n\in {\mathbb {N}}$. Thus, by A.4 part (a) we get $M\mathbf {1\succsim x}\succsim -M{\textbf{1}}$. Therefore, $M\in A$ and $-M\in B$. Moreover, using axiom A.3 it is easy to check that A and B are both closed. Furthermore, completeness of preferences (see axiom A.1) implies $\mathbb {R=}A\cup B$. Thus, since ${\mathbb {R}}$ is connected, we must have that $A\cap B\ne \varnothing$. This readily implies (see the above definitions of sets A and B) that there exists a real number $t_{\textbf{ x}}$ such that $t_{{\textbf{x}}}\mathbf {1\sim x}$. Such a number is unique. For, suppose, by way of obtaining a contradiction, that there is some $s_{ {\textbf{x}}}\ne t_{{\textbf{x}}}$ such that $s_{{\textbf{x}}}\mathbf {1\sim x}$, and assume, without any loss of generality, that $s_{{\textbf{x}}}>t_{\textbf{x }}$. Then, on the one hand A.1 implies $s_{{\textbf{x}}}\mathbf {1\sim }t_{ {\textbf{x}}}{\textbf{1}}$, and on the other hand it follows from A.4 part (b) that $s_{{\textbf{x}}}{\textbf{1}}\succ t_{{\textbf{x}}}{\textbf{1}}$, which yields the desired contradiction. Therefore, we have established the existence of a function $U:\ell _{\infty }\rightarrow {\mathbb {R}}$ defined by

$$\begin{aligned} \ell _{\infty }\ni \mathbf {x\mapsto }U\left( {\textbf{x}}\right) :=t_{\textbf{x }}\text { with }t_{{\textbf{x}}}\mathbf {1\sim x}\text {.} \end{aligned}$$

Moreover, using axioms A.1 and A.4 it is a simple exercise to verify that ${\textbf{x}}\succsim {\textbf{y}}$ if and only if $U\left( \textbf{x }\right) \ge U\left( {\textbf{y}}\right)$. This proves that the above $U:\ell _{\infty }\rightarrow {\mathbb {R}}$ is a utility function representing the given preferences. Incidentally, observe that, by construction of the utility function, $U\left( c{\textbf{1}}\right) =c$ for every real number c.¹⁸

Step 2 (Showing that the utility function is a functional a la Gilboa and Schmeidler):

We claim that axioms (A.1)-(A.5) imply the existence of a non-empty, closed, and convex set C of finitely additive probability measures on $2^{{\mathbb {N}}}$ such that

$$\begin{aligned} U\left( {\textbf{x}}\right) =\min \left\{ {\mathbb {E}}_{\pi }\left[ {\textbf{x}} \right] :\pi \in C\right\} ,\text { for all }{\textbf{x}}\in \ell _{\infty }. \end{aligned}$$

To prove the claim, we begin by noting that A.4 part (a) immediately implies that the utility function is monotonic, i.e., satisfies property (i) listed above. Next, we will prove that the utility function U satisfies property (iv) listed above. That (iv) holds true for $\lambda =0$ and $\lambda =1$ is obvious. Therefore, assume, to begin with, that $0<\lambda <1$, and pick any $\mathbf {x\in }\ell _{\infty }$. Note that $U\left( \lambda {\textbf{x}}\right) =t_{\lambda {\textbf{x}}}$ where $t_{\lambda \textbf{ x}}\mathbf {1\sim }\lambda {\textbf{x}}$, and $U\left( {\textbf{x}}\right) =t_{ {\textbf{x}}}$, where $t_{{\textbf{x}}}\mathbf {1\sim x}$. Since $t_{{\textbf{x}}} \mathbf {1\sim x}$ and $0<\lambda <1$, it follows from A.2 that $\lambda t_{ {\textbf{x}}}\mathbf {1\sim }\lambda {\textbf{x}}$. On the other hand, we know that $t_{\lambda {\textbf{x}}}\mathbf {1\sim }\lambda {\textbf{x}}$; therefore, A.1 implies $t_{\lambda {\textbf{x}}}\mathbf {1\sim }\lambda t_{{\textbf{x}}} {\textbf{1}}$; hence, $U\left( t_{\lambda {\textbf{x}}}{\textbf{1}}\right) =t_{\lambda {\textbf{x}}}=U\left( \lambda t_{{\textbf{x}}}{\textbf{1}}\right) =\lambda t_{{\textbf{x}}}$, which finally implies $U\left( \lambda {\textbf{x}} \right) =\lambda U\left( {\textbf{x}}\right)$. We are left with showing that property (iv) holds true also for $\lambda >1$, as follows: we know that $t_{\lambda {\textbf{x}}}\mathbf {1\sim }\lambda {\textbf{x}}$, $t_{{\textbf{x}}} \mathbf {1\sim x}$, and that $0<\frac{1}{\lambda }<1$. Therefore, by A.2 we get $\frac{1}{\lambda }t_{\lambda {\textbf{x}}}\mathbf {1\sim x}$, so A.1 readily implies $\frac{1}{\lambda }t_{\lambda {\textbf{x}}}\mathbf {1\sim }t_{ {\textbf{x}}}{\textbf{1}}$. Thus

$$\begin{aligned} t_{{\textbf{x}}}=U\left( t_{{\textbf{x}}}{\textbf{1}}\right) =U\left( \frac{1}{ \lambda }t_{\lambda {\textbf{x}}}{\textbf{1}}\right) =\frac{1}{\lambda }U\left( t_{\lambda {\textbf{x}}}{\textbf{1}}\right) =\frac{1}{\lambda }t_{\lambda {\textbf{x}}}, \end{aligned}$$

which yields immediately $\lambda U\left( {\textbf{x}}\right) =U\left( \lambda {\textbf{x}}\right)$. Next, we show that the utility function U enjoys property (ii) listed above. Toward this end, pick any $\textbf{x }\in \ell _{\infty }$ and $t\in {\mathbb {R}}$. Define

$$\begin{aligned} \beta =U\left( 2{\textbf{x}}\right) =2U\left( {\textbf{x}}\right) \text {,} \end{aligned}$$

(6)

and note that $\beta =t_{2{\textbf{x}}}$ with $\beta \mathbf {1\sim }2 {\textbf{x}}$. Observe that $2t\mathbf {1\in }\ell _{\infty }^{C}$. Therefore, letting $\alpha =\frac{1}{2}$ and using A.2 we get $\frac{1}{2}2\mathbf {x+} \frac{1}{2}2t\mathbf {1\sim }\frac{1}{2}\beta \mathbf {1+}\frac{1}{2}2t\textbf{ 1}$, i.e., $\mathbf {x+}t\mathbf {1\sim }(\frac{1}{2}\beta +t){\textbf{1}}$. The previous condition and (6) readily imply

$$\begin{aligned} U\left( \mathbf {x+}t{\textbf{1}}\right) =U\left( (\frac{1}{2}\beta +t)\textbf{1 }\right) =\frac{1}{2}\beta +t=U\left( {\textbf{x}}\right) +t\text {,} \end{aligned}$$

as was to be proven. We proceed to show that the utility function is superadditive (property (iii) listed above). To this end, a straightforward application of property (iv) reveals that it will suffice to prove that for all ${\textbf{x}}$ and ${\textbf{y}}$ in $\ell _{\infty }$

$$\begin{aligned} U\left( \frac{1}{2}\mathbf {x+}\frac{1}{2}{\textbf{y}}\right) \ge \frac{1}{2} U\left( {\textbf{x}}\right) +\frac{1}{2}U\left( {\textbf{y}}\right) \text {.} \end{aligned}$$

To establish the above inequality, we pick any ${\textbf{x}}$ and ${\textbf{y}}$ in $\ell _{\infty }$ and distinguish three exhaustive cases. Case (a): $U\left( {\textbf{x}}\right) =U\left( {\textbf{y}}\right)$; case (b): $U\left( {\textbf{x}}\right) >U\left( {\textbf{y}}\right)$; case (c): $U\left( {\textbf{x}}\right) <U\left( {\textbf{y}}\right)$. We first deal with case (a), as follows: $U\left( {\textbf{x}}\right) =U\left( {\textbf{y}} \right)$ implies $\mathbf {x\sim y}$. Therefore, it follows from axiom A.5 that $\frac{1}{2}\mathbf {x+}\frac{1}{2}\mathbf {y\succsim x}$. Hence, $U\left( \frac{1}{2}\mathbf {x+}\frac{1}{2}{\textbf{y}}\right) \ge U\left( {\textbf{x}}\right) =\frac{1}{2}U\left( {\textbf{x}}\right) +\frac{1}{2}U\left( {\textbf{y}}\right)$, as was to be proven. Regarding case (b), let

$$\begin{aligned} t=U\left( {\textbf{x}}\right) -U\left( {\textbf{y}}\right) \text {,} \end{aligned}$$

(7)

and set ${\textbf{c}}=\mathbf {y+}t{\textbf{1}}$. By property (ii) and (7) above, we get

$$\begin{aligned} U\left( {\textbf{c}}\right) =U\left( \mathbf {y+}t{\textbf{1}}\right) =U\left( {\textbf{y}}\right) +t=U\left( {\textbf{y}}\right) +U\left( {\textbf{x}}\right) -U\left( {\textbf{y}}\right) =U\left( {\textbf{x}}\right) \text {.} \end{aligned}$$

Now, using again property (ii), we see that

$$\begin{aligned} U\left( \frac{1}{2}\mathbf {x+}\frac{1}{2}{\textbf{c}}\right) =U\left( \frac{1}{ 2}\mathbf {x+}\frac{1}{2}\mathbf {y+}\frac{1}{2}t{\textbf{1}}\right) =U\left( \frac{1}{2}\mathbf {x+}\frac{1}{2}{\textbf{y}}\right) +\frac{1}{2}t\text {.} \end{aligned}$$

(8)

On the other hand, since $U\left( {\textbf{c}}\right) =U\left( {\textbf{x}}\right)$, we can rely on the previous case (a) and invoke property (ii) together with (8) above to conclude

$$\begin{aligned}{} & {} U\left( \frac{1}{2}\mathbf {x+}\frac{1}{2}{\textbf{y}}\right) +\frac{1}{2} t=U\left( \frac{1}{2}\mathbf {x+}\frac{1}{2}{\textbf{c}}\right) \ge \frac{1}{2} U\left( {\textbf{x}}\right) +\frac{1}{2}U\left( {\textbf{c}}\right) \\{} & {} \quad =\frac{1}{2}U\left( {\textbf{x}}\right) +\frac{1}{2}U\left( \mathbf {y+}t {\textbf{1}}\right) =\frac{1}{2}U\left( {\textbf{x}}\right) +\frac{1}{2}\left[ U\left( {\textbf{y}}\right) +t\right] =\frac{1}{2}U\left( {\textbf{x}}\right) + \frac{1}{2}U\left( {\textbf{y}}\right) +\frac{1}{2}t; \end{aligned}$$

hence, $U\left( \frac{1}{2}\mathbf {x+}\frac{1}{2}{\textbf{y}}\right) \ge \frac{1}{2}U\left( {\textbf{x}}\right) +\frac{1}{2}U\left( {\textbf{y}} \right)$, as was to be proven. As for case (c), it turns out it can be handled very similarly to case (b).

So far, we have proven that axioms (A.1)-(A.5) give rise to a utility function U, representing preferences, which satisfies properties $(i)-(iv)$ and which is such that $U\left( {\textbf{1}}\right) =1$.¹⁹. Therefore, by Lemma 3.5 in Gilboa and Schmeidler (1989) there exists a non-empty, closed and convex set C of finitely additive probability measures on $2^{{\mathbb {N}}}$ such that

$$\begin{aligned} U\left( {\textbf{x}}\right) =\min \left\{ {\mathbb {E}}_{\pi }\left[ {\textbf{x}} \right] :\pi \in C\right\} ,\text { for all }{\textbf{x}}\in \ell _{\infty }. \end{aligned}$$

(9)

Step 3 (Pinpointing the ambiguity set C ):

In what follows, we show that the risk-perception axiom A.6 pins down the ambiguity set (i.e., $C=D$), hence the exact form of the utility function U.

We know from Step 2 that $-U$ is a coherent risk measure on $\ell _{\infty }$, when the latter is ordered by $\ell _{\infty }^{+}$ and ${\textbf{1}}$ is an order unit. Therefore, by (3) above we get $-U\left( {\textbf{x}}\right) =\inf \left\{ t\in {\mathbb {R}}:{\textbf{x}}+t{\textbf{1}}\in {\mathcal {A}}_{-U}\right\}$ for all $\mathbf {x\in }\ell _{\infty }$. Also, it readily follows from the definition of acceptance set and axiom A.6 that ${\mathcal {A}}_{-U}=P=\left\{ {\textbf{x}}\in \ell _{\infty }:{\mathbb {E}}_{\pi } \left[ {\textbf{x}}\right] \ge 0\text { for all }\pi \in D\right\}$. Therefore

$$\begin{aligned} -U\left( {\textbf{x}}\right) =\inf \left\{ t\in {\mathbb {R}}:{\textbf{x}}+t\textbf{ 1}\in P\right\} \text {.} \end{aligned}$$

(10)

Performing some straightforward algebra in (10), we get $U\left( {\textbf{x}}\right) =\inf \left\{ {\mathbb {E}}_{\pi }\left[ {\textbf{x}}\right] :\pi \in D\right\} ,$ for all ${\textbf{x}}\in \ell _{\infty }$. Since every $\pi \in D$ is a probability measure and any ${\textbf{x}}\in \ell _{\infty }$ is bounded, the $\inf$ in the previous equation is a real number. Furthermore, because D is closed, by assumption, such infimum is actually attained. Therefore, we come to

$$\begin{aligned} U\left( {\textbf{x}}\right) =\min \left\{ {\mathbb {E}}_{\pi }\left[ {\textbf{x}} \right] :\pi \in D\right\} ,\text {for all }{\textbf{x}}\in \ell _{\infty }. \end{aligned}$$

(11)

Finally, one can readily see that preferences satisfying axiom (A.6) are non-degenerate.²⁰ Thus, as in Gilboa and Schmeidler (1989), the ambiguity set C in (9) above must be unique. Therefore, since D is by assumption closed and convex, (9) and (11) imply $C=D$. The proof is now complete. $\square$