Preference for safety under the Choquet model: in search of a characterization

Abstract

Victor prefers safety more than Ursula if whenever Ursula prefers a constant to an uncertain act, so does Victor. This paradigm, whose expected utility (EU) version is Arrow and Pratt’s more risk aversion concept, will be studied in the Choquet expected utility (CEU) model. Necessary condition Pointwise inequality between a function of the utility functions and another of the capacities is necessary and sufficient for the preference by Victor of safety over a dichotomous act whenever such is the preference of Ursula. However, increased preference for safety versus dichotomous acts does not imply preference by Victor of safety over a general act whenever such is the preference of Ursula. A counterexample will be provided, via the casino theory of Dubins and Savage. Sufficient condition Separation of the two functions by some convex function is sufficient for Victor to prefer safety more than Ursula, over general acts. Furthermore, a condition on the capacities will be presented for simplicity seeking, the preference by Victor over any act for some dichotomous act that leaves Ursula indifferent. This condition is met in particular if Victor’s capacity is a convex function of Ursula’s capacity. For these cases, the pointwise inequality (necessary) condition is a characterization of greater preference for safety, extending the Arrow–Pratt notion from EU to CEU and rank-dependent utility (RDU). These inequalities preserve the flavor of the “more pessimism than greediness” characterization of monotone risk aversion by Chateauneuf, Cohen and Meilijson in the RDU model and its extension by Grant and Quiggin to CEU. Preferences between safety and dichotomous acts are at the core of the biseparable preferences model of Ghirardato and Marinacci.

Appendices

Appendix 1: Proofs and further analysis

Proof of Proposition 1

The lower envelope function $\mathrm{{LE}}_w$ is the identity function $\mathrm{{LE}}_w(q)=q$ if and only if $w$ is concave. Indeed, if $w$ is not concave, then there is a triple $a < c=a+q(b-a) < b$ such that $(\mathrm{{LE}}_w(q) \le ) {{w(c)-w(a)} \over {w(b)-w(a)}}<q$. The other direction holds because ${{w(c)-w(a)} \over {w(b)-w(a)}} \ge {{c-a} \over {b-a}}=q$ can be made arbitrarily close to $q$ by letting $c$ be a point of continuity of $w$ in the interior of a small enough interval $(a,b)$. This argument also shows that $\mathrm{{LE}}_w$ is bounded above by the identity function. $\mathrm{{LE}}_w$ can certainly be zero everywhere except for $\mathrm{{LE}}_w(1)=1$. As will now be shown, except for these two extreme cases, $\mathrm{{LE}}_w$ is continuous, positive and below the identity function except at the endpoints. It will also be shown under the further assumption that $w$ is a differentiable function defined on some bounded closed interval, that (i) except for the two extreme cases above, $\mathrm{{LE}}_w$ is strictly increasing, and (ii) $\mathrm{{LE}}_w$ is self-zooming. These two properties may hold more generally, but we don’t have a proof.

If $w$ is continuous and strictly increasing but not concave, then there exist $a<c<b$ such that $w(c)<{{b-c} \over {b-a}} w(a) + {{c-a} \over {b-a}} w(b)$. Furthermore, there exist a maximal $a^{\prime } \in [a,c)$ and minimal $b^{\prime } \in (c,b]$ such that $w(a^{\prime })={{b-a^{\prime }} \over {b-a}} w(a) + {{a^{\prime }-a} \over {b-a}} w(b)$ and $w(c^{\prime })={{b-c^{\prime }} \over {b-a}} w(a) + {{c^{\prime }-a} \over {b-a}} w(b)$, respectively. But then the zoom $w_{[a^{\prime },c^{\prime }]}$ is strictly below the diagonal, whence so is $\mathrm{{LE}}_w$. Hence, lower envelope functions are either linear (if and only if $w$ is concave) or strictly below the diagonal except at the endpoints. The other properties stated above follow from Lemma 1.

Assume that $\mathrm{{LE}}_w(q)>0$ for some $q \in (0,1)$. It follows from the first inequality in (11) that for all $n$, $\mathrm{{LE}}_w(q^n) \ge \mathrm{{LE}}_w(q)^n$. Hence, $\mathrm{{LE}}_w$ is strictly positive throughout $(0,1)$. The second inequality in (11) implies that $1-\mathrm{{LE}}_w(q^n) \le (1-\mathrm{{LE}}_w(q))^n$. Send $q$ to 1 to obtain that for all $n$, $1-\mathrm{{LE}}_w(1-) \le (1-\mathrm{{LE}}_w(1-))^n$. Hence, $\mathrm{{LE}}_w(q-)=1$, or, $\mathrm{{LE}}_w$ is continuous at $1$. Now send $p$ to $1$ in the first inequality in (11) to see that $\mathrm{{LE}}_w$ is continuous everywhere.

Consider the zoom $w_{a,b}$ of $w$ on an arbitrary interval $[a,b]$. Then, (with $\mathrm{{LE}}=\mathrm{{LE}}_w$)

$$\begin{aligned} {{{\text{ LE }}(y)-{\text{ LE }}(x)} \over {{\text{ LE }}(z)-{\text{ LE }}(x)}}&\ge {{{\text{ LE }}(y)-{\text{ LE }}(x)} \over {w_{a,b}(z)-{\text{ LE }}(x)}} \ge {{LE(y)-w_{a,b}(x)} \over {w_{a,b}(z)-w_{a,b}(x)}} \nonumber \\&= {{w_{a,b}(y)-w_{a,b}(x)} \over {w_{a,b}(z)-w_{a,b}(x)}} - {{w_{a,b}(y)-{\text{ LE }}(y)} \over {w_{a,b}(z)-w_{a,b}(x)}} \nonumber \\&\ge {\text{ LE }}(q) - {{w_{a,b}(y)-{\text{ LE }}(y)} \over {w_{a,b}(z)-w_{a,b}(x)}} = {\text{ LE }}(q) - { \text{ RATIO } } \end{aligned}$$

(18)

where $q = {{y-x} \over {z-x}}$. To infer that LE is self-zooming, we must check that

$$\begin{aligned} {{{\text{ LE }}(y)-{\text{ LE }}(x)} \over {{\text{ LE }}(z)-{\text{ LE }}(x)}} \ge {\text{ LE }}(q) \end{aligned}$$

(19)

Since (18) holds for all $[a,b]$, suppose that the zoom on some $[a,b]$ achieves the envelope value at $y$. Then, the numerator in $\text{ RATIO }$ is zero, the denominator is positive, and we are done. If $\mathrm{{LE}}(y)$ is an infimum over zooms, under the assumption that $w$ is defined on a bounded closed interval, any sequence $[a_i,b_i]$ along which the infimum is achieved has a convergent subsequence. Clearly, the limit achieves the infimum as a minimum as long as $a = \lim a_i < \lim b_i = b$. In this case, consider any $z \in (0,y)$. Then ${\text{ LE }}_w(z) \le w_{a,b}(z)<w_{a,b}(y)=\mathrm{{LE}}_w(y)$, or, $\mathrm{{LE}}_w$ is strictly increasing on $[0,y]$.$\square$

Remark 1

There is a problem if $a=b$, that is, if $\mathrm{{LE}}(y)$ is achieved infinitesimally, because then the numerator of $\text{ RATIO }$ may not go to zero faster than the denominator, in which case $\text{ RATIO }$ would not go to zero. However, if $w$ is differentiable everywhere and $a=b$ then $\mathrm{{LE}}(y)$ can only be $y$ (the diagonal), a case we are not interested in, the concave case.$\square$

Proof of Lemma 2

To prove (i), observe that if $g$ is convex, $Z_{g,q,r}(\cdot )$ is non-increasing. To prove (ii), observe that by continuity of $g$, it is enough to show monotonicity at some arbitrary dense set. Suppose the extreme $\varDelta$ occurs at $q-\varDelta r=0$ and consider the dense set of points $q$ at which $g$ is differentiable. At any such point, send $\varDelta$ to zero to obtain via $Z_{g,q,r}(0+)=r$ that semi-convexity implies $r \ge {{g(q)} \over {g({q \over r})}}$, or ${{g(q)} \over q} \le {{g({q \over r})} \over {q \over r}}$. Similarly if the extreme $\varDelta$ occurs at $q+\varDelta (1-r)=1$. To prove one direction of (iii), the star-shaped property is satisfied by convexity and is closed under taking minima. To prove the other direction, every $T$-function that is star-shaped at $0$ and $1$ is the minimum over $q$ of special convex $T$-functions, maxima of two linear functions with breakpoint at $(q,g(q))$.$\square$

Proof of Theorem 2

Theorem 2 will be proved first in RDU language. The proof will then be extended to cover CEU. Consider a random variable $X$ supported by $x_1 < x_2 < \cdots < x_n$ with $n \ge 3$ and $P(X > x_i)=q_i \ ; \ 1 \le i < n$.

Semi-convexity will be shown to imply the following statement: There is another distribution, with the same mean and support as $X$, except for missing one of the three leftmost atoms, leaving all other atoms intact in value and probability, preferred by the DM.

This proof is an adaptation of the proof of Smith’s theorem (Smith 1967) by Dubins (1972).$\square$

Analysis will be split according to whether $E[X|X \le x_3]$ is to the left or to the right of $x_2$.

Case 1

($E[X|X \le x_3] \ge x_2$) The removal of the atom $x_1$ and the assignment $p_1=q_2-(1-q_1){{x_2-x_1} \over {x_3-x_2}}$ entail the comparison

$$\begin{aligned}&(1-f(q_1)) u(x_1)+(f(q_1)-f(q_2)) u(x_2) +(f(q_2)-f(q_3) u(x_3)) + \cdots \nonumber \\&\quad \le (1-f(p_1)) u(x_2) + (f(p_1)-f(q_3)) u(x_3) + \cdots \end{aligned}$$

(20)

or

$$\begin{aligned} {{u(x_2)-u(x_1)} \over {u(x_3)-u(x_1)}} \ge {{f(q_2)-f(p_1)} \over {1-f(q_1)+f(q_2)-f(p_1)}} \end{aligned}$$

(21)

between two (objective) distributions with the same mean. Similarly, the removal of the atom $x_2$ and the assignment $p_2=q_1 {{x_2-x_1} \over {x_3-x_1}} + q_2 (1-{{x_2-x_1} \over {x_3-x_1}})$ entail, as before, the comparison

$$\begin{aligned}&(1-f(q_1)) u(x_1)+(f(q_1)-f(q_2)) u(x_2) +(f(q_2)-f(q_3) u(x_3)) + \cdots \nonumber \\&\quad \le (1-f(p_2)) u(x_1) + (f(p_2)-f(q_3)) u(x_3) + \cdots \end{aligned}$$

(22)

or

$$\begin{aligned} {{u(x_2)-u(x_1)} \over {u(x_3)-u(x_1)}} \le {{f(p_2)-f(q_2)} \over {f(q_1)-f(q_2)}} \end{aligned}$$

(23)

between distributions with equal means.

At least one of (21) and (23) is satisfied if and only if

$$\begin{aligned} {{f(q_2)-f(p_1)} \over {1-f(q_1)+f(q_2)-f(p_1)}} \le {{f(p_2)-f(q_2)} \over {f(q_1)-f(q_2)}} \end{aligned}$$

(24)

In other words, if and only if

$$\begin{aligned} {{f(p_2)-f(p_1)} \over {1-f(p_1)}} \le {{f(p_2)-f(q_2)} \over {f(q_1)-f(q_2)}} \end{aligned}$$

(25)

Inequality (25) is clearly satisfied when $f$ is semi-convex: The LHS and RHS of (25) are zooms of $f$ around $p_2$ with objective rate

$$\begin{aligned} r={{p_2-p_1} \over {1-p_1}}={{p_2-q_2} \over {q_1-q_2}}={{x_2-x_1} \over {x_3-x_1}} \end{aligned}$$

(26)

As such, the LHS of (25) is a maximal zoom-out of the RHS: fix $p_2$, fix $r$ and let $q_1$ be defined by (26) as a function of $q_2$. Then, if $f$ is semi-convex, the LHS of (25) is the minimum of the RHS.

Case 2

($E[X|X \le x_3] \le x_2$) The removal of the atom $x_3$ and the assignment $p_3=q_1+q_2{{x_3-x_2} \over {x_2-x_1}}$ preserves means and entails the comparison

$$\begin{aligned}&(1-f(q_1)) u(x_1)+(f(q_1)-f(q_2)) u(x_2) +(f(q_2)-f(q_3) u(x_3)) + \cdots \nonumber \\&\quad \le (1-f(p_3)) u(x_1) + (f(p_3)-f(q_3)) u(x_2) + (f(q_3)-f(q_3)) u(x_3)+\cdots \nonumber \\ \end{aligned}$$

(27)

or

$$\begin{aligned} (f(p_3)-f(q_1)) (u(x_1)-u(x_2))+f(q_2)(u(x_3)-u(x_2)) \le f(q_3)(u(x_3)-u(x_2))\nonumber \\ \end{aligned}$$

(28)

Inequality (28) is satisfied for all $q_3 \ge 0$ if and only if it is satisfied for the most stringent value $q_3=0$, in which case the inequality becomes

$$\begin{aligned} {{u(x_2)-u(x_1)} \over {u(x_3)-u(x_1)}} \ge {f(q_2) \over {f(q_2)+f(p_3)-f(q_1)}} \end{aligned}$$

(29)

As in Case 1, at least one of (23) and (29) is satisfied if and only if

$$\begin{aligned} {{f(q_2)} \over {f(q_2)+f(p_3)-f(q_1)}} \le {{f(p_2)-f(q_2)} \over {f(q_1)-f(q_2)}} \end{aligned}$$

(30)

In other words, if and only if

$$\begin{aligned} {{f(p_2)} \over {f(p_3)}} \le {{f(p_2)-f(q_2)} \over {f(q_1)-f(q_2)}} \end{aligned}$$

(31)

This inequality is of the same nature as (25), satisfied by semi-convex $f$ for the same reason.

The proof of Theorem 2 under RDU is essentially finished. We have seen that DMs with semi-convex $f$ always prefer some equal-mean distribution missing one of the three leftmost atoms. The recursive application of this idea will end up with a preferred dichotomous equal-mean distribution.

In CEU, with $f_{\mu ,\nu }=f$, express the Choquet index for Victor in terms of Ursula’s utility on wealth $u_i = U(x_i)$ and capacity $q_i=\mu (A_i)$ as

$$\begin{aligned} {\text{ CEU }}_V[X]&= w(u_1)+\nu (A_1)(w(u_2)-w(u_1))+\nu (A_2)(w(u_3)-w(u_2)+\cdots \nonumber \\&\le w(u_1)+f(q_1)(w(u_2)-w(u_1))+f(q_2)(w(u_3)-w(u_2)+\cdots \nonumber \\ \end{aligned}$$

(32)

Now apply the RDU result verbatim, with Ursula’s Choquet index $\mathrm{{CEU}}_U[X]$ playing the role of expectation.

Appendix 2: The value iteration routine in dynamic programming (Shapley 1953; Bellman 1957)

To perform numerical evaluation of the $U$ function in a discretized version of market $M_3$, let wealth be an integer multiple of ${1 \over N}$. The algorithm may start from any lower bound $U$ on $U$ (such as $1$ from the goal onwards and zero elsewhere), to be improved in every iteration by trying at every wealth level (that is an integer multiple of ${1 \over N}$) the effect of investing (just on the first time period) on each of the two risky assets individually. This move leads to a (dichotomous) new wealth level, at which the player collects the $U$-value. The best such choice is iteratively recorded as the new lower bound on $U$. $U$ is obtained as the limit of these progressively increasing lower bounds.

The following straightforward MATLAB program will do the job. Let ITER be large enough, such as $\mathrm{{ITER}}=50$.