Decreasing absolute risk aversion, prudence and increased downside risk aversion

Abstract

Downside risk increases have previously been characterized as changes preferred by all decision makers u(x) with u′′′(x) > 0. For risk averse decision makers, u′′′(x) > 0 also defines prudence. This paper finds that downside risk increases can also be characterized as changes preferred by all decision makers displaying decreasing absolute risk aversion (DARA) since those changes involve random variables that have equal means. Building on these findings, the paper proposes using “more decreasingly absolute risk averse” or “more prudent” as alternative definitions of increased downside risk aversion. These alternative definitions generate a transitive ordering, while the existing definition based on a transformation function with a positive third derivative does not. Other properties of the new definitions of increased downside risk aversion are also presented.

Appendix

1.1 A1 Lemma 1

Lemma 1

F(x) is preferred or indifferent to G(x) for all decision makers v′(x) = δ′(x)·u′(x) with δ′′′(x) > 0 if and only if G(x) is an expected utility preserving increase in downside risk from F(x) for u(x).

Proof

The proof follows the similar steps in MGT′s proof of Theorem 1. Using v′(x) = δ′(x)·u′(x),

$$ {{\text{E}}_{\text{F}}}{\text{v}}\left( {\text{x}} \right) - {{\text{E}}_{\text{G}}}{\text{v}}\left( {\text{x}} \right) = \int {_{\text{a}}^{\text{b}}{\text{v}}\prime \left( {\text{x}} \right)\left( {{\text{G}}\left( {\text{x}} \right) - {\text{F}}\left( {\text{x}} \right)} \right){\text{dx}} = \int {_{\text{a}}^{\text{b}}\delta \prime \left( {\text{x}} \right){\text{u}}\prime \left( {\text{x}} \right)\left( {{\text{G}}\left( {\text{x}} \right) - {\text{F}}\left( {\text{x}} \right)} \right){\text{dx}}.} } $$

(The “only if” part)

Suppose E_Fv(x) ≥ E_Gv(x) for all δ′′′(x) > 0.

Let δ₁(x) = θx³/3 + x and δ₂(x) = θx³/3−x where θ > 0. Since δ₁′′′(x) = δ₂′′′(x) > 0, we have

$$ \begin{array}{*{20}{c}} {\int {_a^b\left( {\theta {x^2} + 1} \right)u\prime (x)\left( {G(x) - F(x)} \right)dx \geqslant 0} } \hfill \\ {\int {_a^b\left( {\theta {x^2} - 1} \right)u\prime (x)\left( {G(x) - F(x)} \right)dx \geqslant 0} } \hfill \\ \end{array} $$

which continues to hold as θ approaches zero. So $\int {_{\text{a}}^{\text{b}}{\text{u}}\prime \left( {\text{x}} \right)\left( {{\text{G}}\left( {\text{x}} \right) - {\text{F}}\left( {\text{x}} \right)} \right){\text{dx}} = 0}$, which is a) in Definition 4.

Note that when $\int {_{\text{a}}^{\text{b}}{\text{u}}\prime \left( {\text{x}} \right)\left( {{\text{G}}\left( {\text{x}} \right) - {\text{F}}\left( {\text{x}} \right)} \right){\text{dx}} = 0},$

$$ {{\text{E}}_{\text{F}}}{\text{v}}\left( {\text{x}} \right) - {{\text{E}}_{\text{G}}}{\text{v}}\left( {\text{x}} \right) = \int {_{\text{a}}^{\text{b}}\delta \prime \left( {\text{x}} \right){\text{u}}\prime \left( {\text{x}} \right)\left( {{\text{G}}\left( {\text{x}} \right) - {\text{F}}\left( {\text{x}} \right)} \right){\text{dx}}{.} = - \int {_{\text{a}}^{\text{b}}\delta \prime \prime \left( {\text{x}} \right)\int {_{\text{a}}^{\text{x}}{\text{u}}\prime \left( {\text{s}} \right)\left( {{\text{G}}\left( {\text{s}} \right) - {\text{F}}\left( {\text{s}} \right)} \right){\text{ds}} \cdot {\text{dx}}.} } } $$

Now let δ₃(x) = θx³/3 + x ²/2 and δ₄(x) = θx³/3−x ²/2 where θ > 0. Since δ₃′′′(x) = δ₄′′′(x) > 0, we have

$$ \begin{array}{*{20}{c}} { - \int {_a^b\left( {2\theta x + 1} \right)\int {_a^xu\prime (s)\left( {G(s) - F(s)} \right)dsdx \geqslant 0} } } \hfill \\ { - \int {_a^b\left( {2\theta x - 1} \right)\int {_a^xu\prime (s)\left( {G(s) - F(s)} \right)dsdx \geqslant 0} } } \hfill \\ \end{array} $$

which continue to hold as θ approaches zero. So $\int {_{\text{a}}^{\text{b}}\int {_{\text{a}}^{\text{x}}{\text{u}}\prime \left( {\text{s}} \right)\left( {{\text{G}}\left( {\text{s}} \right) - {\text{F}}\left( {\text{s}} \right)} \right){\text{ds}} \cdot {\text{dx}} = 0} }$, which is the equality part of b) in Definition 4.

With $\int {_{\text{a}}^{\text{b}}{\text{u}}\prime \left( {\text{x}} \right)\left( {{\text{G}}\left( {\text{x}} \right) - {\text{F}}\left( {\text{x}} \right)} \right){\text{dx}} = 0\;{\text{and}}\;\int {_{\text{a}}^{\text{b}}\int {_{\text{a}}^{\text{x}}{\text{u}}\prime \left( {\text{s}} \right)\left( {{\text{G}}\left( {\text{s}} \right) - {\text{F}}\left( {\text{s}} \right)} \right){\text{ds}} \cdot {\text{dx}} = 0} } },$

$$ {{\text{E}}_{\text{F}}}{\text{v}}\left( {\text{x}} \right) - {{\text{E}}_{\text{G}}}{\text{v}}\left( {\text{x}} \right) = \int {_{\text{a}}^{\text{b}}\delta \prime \prime \prime \left( {\text{y}} \right)\int {_{\text{a}}^{\text{y}}\int {_{\text{a}}^{\text{x}}{\text{u}}\prime \left( {\text{s}} \right)\left( {{\text{G}}\left( {\text{s}} \right) - {\text{F}}\left( {\text{s}} \right)} \right){\text{ds}} \cdot {\text{dx}} \cdot {\text{dy}}.} } } $$

We prove the weak inequality of b) by contradiction. Assume $\int {_{\text{a}}^{\text{y}}\int {_{\text{a}}^{\text{x}}{\text{u}}\prime \left( {\text{s}} \right)\left( {{\text{G}}\left( {\text{s}} \right) - {\text{F}}\left( {\text{s}} \right)} \right){\text{ds}} \cdot {\text{dx}} < 0} }$ for some y₀ in [a,b]. Then there exists an interval in [a, b], [α, β], such that $\int {_{\text{a}}^{\text{y}}\int {_{\text{a}}^{\text{x}}{\text{u}}\prime \left( {\text{s}} \right)\left( {{\text{G}}\left( {\text{s}} \right) - {\text{F}}\left( {\text{s}} \right)} \right){\text{ds}} \cdot {\text{dx}} < 0} }$ for all y in [α, β]. Therefore, for

$$ {\delta_5}(x) = \left\{ {\begin{array}{*{20}{c}} {{x^3}/6,} \hfill & {x \in \left[ {\alpha, \beta } \right]} \hfill \\ {\theta {x^3}/6,} \hfill & {otherwise} \hfill \\ \end{array} } \right., $$

E_Fv(x) - E_Gv(x) < 0 for sufficiently small θ. Because δ₅′′′(x) > 0, that contradicts the original assumption that E_Fv(x) ≥ E_Gv(x) for all δ′′′(x) > 0.

Moreover, $\int {_{\text{a}}^{\text{y}}\int {_{\text{a}}^{\text{x}}{\text{u}}\prime \left( {\text{s}} \right)\left( {{\text{G}}\left( {\text{s}} \right) - {\text{F}}\left( {\text{s}} \right)} \right){\text{ds}} \cdot {\text{dx}} > 0} }$ for some y in (a, b) as long as G(x) and F(x) are not identical.

(The “if” part)

Suppose G(x) is an expected utility preserving increase in downside risk from F(x) for u(x). Then,

$$ \begin{array}{*{20}{c}} {{{\text{E}}_{\text{F}}}{\text{v}}\left( {\text{x}} \right) - {{\text{E}}_{\text{G}}}{\text{v}}\left( {\text{x}} \right) = \int {_{\text{a}}^{\text{b}}\delta \prime \prime \prime \left( {\text{y}} \right)\int {_{\text{a}}^{\text{y}}\int {_{\text{a}}^{\text{x}}{\text{u}}\prime \left( {\text{s}} \right)\left( {{\text{G}}\left( {\text{s}} \right) - {\text{F}}\left( {\text{s}} \right)} \right){\text{ds}} \cdot {\text{dx}} \cdot {\text{dy}} \geqslant 0\;{\text{for}}\;{\text{all}}\;\delta \prime \prime \prime \left( {\text{y}} \right) > 0.} } } } \hfill & {\text{QED}} \hfill \\ \end{array} $$

1.2 A2 Risk adjusted cumulative distribution functions

For continuously differentiable CDFs, note that the derivative of u′(x)(G(x)−F(x)) is u′(g–f) + u′′(G–F), where f and g are the densities associated with CDFs F and G. First assume u′′ ≥ 0. Then u′·g + u′′·G ≥ 0 and u′·f + u′′·F ≥ 0. These two expressions can be density functions if the probabilities sum to one. Integrate each over [a, b] to find the total value for each is u′(b) and thus the normalizing factor is t = 1/u′(b). Next assume u′′ < 0 and consider u′·g + u′′·F ≥ 0 and u′·f + u′′·G ≥ 0. Again these can be density functions if the probabilities add to one. Because $\int {_{\text{a}}^{\text{b}}\left( {{\text{u}}\prime \cdot {\text{g}} + {\text{u}}\prime \prime {\text{G}} - {\text{u}}\prime \cdot {\text{f}} - {\text{u}}\prime \prime {\text{F}}} \right){\text{dx}} = 0{,}\int {_{\text{a}}^{\text{b}}\left( {{\text{u}}\prime \cdot {\text{g}} + {\text{u}}\prime \prime \cdot {\text{F}}} \right){\text{dx}} = \int {_{\text{a}}^{\text{b}}\left( {{\text{u}}\prime \cdot {\text{f}} + {\text{u}}\prime \prime {\text{G}}} \right){\text{dx}} = 1/{\text{t}}} } }$ is the normalizing factor.

The probability distributions for these random variables have been “adjusted” by the reference decision maker with marginal utility u′(x) so that the means of $\widehat{G}\left( {\text{x}} \right)$ and $\widehat{F}\left( {\text{x}} \right)$ are the same. That is, they are risk adjusted so that for this person the choice between them depends only on their mean values. It is also the case that the conditions in Definition 4 imply that $\widehat{G}\left( {\text{x}} \right)$ is a downside increase in risk from $\widehat{F}\left( {\text{x}} \right)$ by the original MGT definition. Thus, the function u′(x)(G(x)−F(x)), properly normalized, adjusts the random alternatives associated with G(x) and F(x) so that he or she can act as a risk neutral person and choose between them based only on the mean values. This leads to an interpretation of Theorem 5. This theorem indicates that for those who are more downside risk averse than the person doing this risk adjustment, that is, those who are more decreasingly absolute risk averse than u′(x), this risk adjustment is not sufficient and $\widehat{F}\left( {\text{x}} \right)$ is preferred or indifferent to $\widehat{G}\left( {\text{x}} \right)$.

Consider the following example for F(x) and G(x) defined on [0, 1].

$$ {\text{Let}}\;{\text{G}}\left( {\text{x}} \right) = \begin{array}{*{20}{c}} 0 \hfill & {{\text{x}} < 0} \hfill \\ {.4{\text{x}}} \hfill & {0 \leqslant {\text{x}} < .75} \hfill \\ {.4{\text{x}} + .6} \hfill & {.75 \leqslant {\text{x}} < 1} \hfill \\ 1 \hfill & {{\text{x}} \geqslant 1} \hfill \\ \end{array} $$

and

$$ {\text{F}}\left( {\text{x}} \right) = \begin{array}{*{20}{c}} 0 \hfill & {{\text{x}} < .25} \hfill \\ {.8{\text{x}}} \hfill & {25 \leqslant {\text{x}} < .75} \hfill \\ {.6} \hfill & {.75 \leqslant {\text{x}} < 1} \hfill \\ 1 \hfill & {{\text{x}} \geqslant 1} \hfill \\ \end{array} $$

This implies that

$$ {\text{G}}\left( {\text{x}} \right) - {\text{F}}\left( {\text{x}} \right) = \begin{array}{*{20}{c}} 0 \hfill & {{\text{x}} < 0} \hfill \\ {.4{\text{x}}} \hfill & {0 \leqslant {\text{x}} < .25} \hfill \\ { - .4{\text{x}}} \hfill & {.25 \leqslant {\text{x}} < .75} \hfill \\ {.4{\text{x}}} \hfill & {.75 \leqslant {\text{x}} < 1} \hfill \\ 0 \hfill & {{\text{x}} \geqslant 1} \hfill \\ \end{array} $$

Consider utility function u(x) = ln x so that u′(x) = 1/x. This implies that

$$ {\text{u}}\prime \left( {\text{x}} \right)\left( {{\text{G}}\left( {\text{x}} \right) - {\text{F}}\left( {\text{x}} \right)} \right) = \begin{array}{*{20}{c}} 0 \hfill & {x < 0} \hfill \\ {.4} \hfill & {0 \leqslant {\text{x}} < .25} \hfill \\ { - .4} \hfill & {.25 \leqslant {\text{x}} < .75} \hfill \\ {.4} \hfill & {.75 \leqslant {\text{x}} < 1} \hfill \\ 0 \hfill & {{\text{x}} \geqslant 1} \hfill \\ \end{array} $$

This sum of the increases (or decreases) for u′(x)(G(x)−F(x)) is 1.2 which is greater than 1 so a scaling factor less than or equal to 1/1.2 is needed to allow u′(x)(G(x)−F(x)) to represent the difference between two CDFs. These rescaled CDFs are given below and can be viewed as the difference between two risk adjusted CDFs where person with utility u(x) = ln x has adjusted CDFs F(x) and G(x) so that they have equal mean values. For that person the ranking of the two CDFs is the same, but for those more decreasing absolute risk averse, the adjustment is not sufficient, and $\widehat{\text{F}}\left( {\text{x}} \right)$ is preferred or indifferent to $\widehat{\text{G}}\left( {\text{x}} \right)$.

$$ \widehat{\text{G}}\left( {\text{x}} \right) - \widehat{\text{F}}\left( {\text{x}} \right) = \begin{array}{*{20}{c}} 0 \hfill & {{\text{x}} < 0} \hfill \\ {{1}/{3}} \hfill & {0 \leqslant {\text{x}} < .{25}} \hfill \\ { - {1}/{3}} \hfill & {.{25} \leqslant {\text{x}} < .{75}} \hfill \\ {{1}/{3}} \hfill & {.{75} \leqslant {\text{x}} < {1}} \hfill \\ 0 \hfill & {{\text{x}} \geqslant 1} \hfill \\ \end{array} $$

1.3 A3 Proof of Theorem 5′

Proof

(The “if” part) Suppose that G(x) is an expected utility preserving increase in downside risk from F(x) for u(x). Then for any utility function v(x),

$$ \begin{array}{*{20}{c}} {{{\text{E}}_{\text{F}}}{\text{v}}\left( {\text{x}} \right) - {{\text{E}}_{\text{G}}}{\text{v}}\left( {\text{x}} \right) = - \int {_{\text{a}}^{\text{b}}{\text{v}}\prime \prime \left( {\text{x}} \right)\int {_{\text{a}}^{\text{x}}\left( {{\text{G}}\left( {\text{s}} \right) - {\text{F}}\left( {\text{s}} \right)} \right){\text{dsdx}}} } } \\ { = - \int {_{\text{a}}^{\text{b}}\phi \prime {\text{u}}\prime \prime \left( {\text{x}} \right)\int {_{\text{a}}^{\text{x}}\left( {{\text{G}}\left( {\text{s}} \right) - {\text{F}}\left( {\text{s}} \right)} \right){\text{dsdx}}} } } \\ { = \int {_{\text{a}}^{\text{b}}\phi \prime \prime {\text{u}}\prime \prime \left( {\text{y}} \right)\int {_{\text{a}}^{\text{y}}{\text{u}}\prime \prime \left( {\text{x}} \right)\int {_{\text{a}}^{\text{x}}\left( {{\text{G}}\left( {\text{s}} \right) - {\text{F}}\left( {\text{s}} \right)} \right){\text{ds}} \cdot {\text{dx}} \cdot {\text{dy}}{.}} } } } \\ \end{array} $$

where the first equality is due to a) and the third equality due to b) in Definition 4′. Also from b) in Definition 4′, $\int {_{\text{a}}^{\text{y}}{\text{u}}\prime \prime \left( {\text{x}} \right)\int {_{\text{a}}^{\text{x}}\left( {{\text{G}}\left( {\text{s}} \right) - {\text{F}}\left( {\text{s}} \right)} \right){\text{ds}} \cdot {\text{dx}} \cdot {\text{dy}} \leqslant 0\;{\text{for}}\;{\text{all}}\;{\text{x}}} }$. Therefore, v(x) being more downside risk averse than u(x)–∅′′u′′(y) < 0 according to Lemma 2–implies E_Fv(x)–E_Gv(x) ≥ 0.

(The “only if” part) Suppose u(x) is risk averse. The case in which u(x) is risk loving can be similarly dealt with. Now suppose that F(x) is preferred or indifferent to G(x) for all risk averse decision makers v(x) who are more downside risk averse than u(x). That is ${{\text{E}}_{\text{F}}}{\text{v}}\left( {\text{x}} \right) - {{\text{E}}_{\text{G}}}{\text{v}}\left( {\text{x}} \right) = \int {_{\text{a}}^{\text{b}}{\text{v}}\prime \left( {\text{x}} \right){\text{G}}\left( {\text{x}} \right) - {\text{F}}\left( {\text{x}} \right){\text{)dx}} = \int {_{\text{a}}^{\text{b}}\phi \left( {{\text{u}}\prime } \right)\left( {{\text{G}}\left( {\text{x}} \right) - {\text{F}}\left( {\text{x}} \right)} \right){\text{dx}} \geqslant 0\;{\text{for}}\;{\text{all}}\;\phi \;{\text{such}}\;{\text{that}}\;\phi \prime \prime > 0} }$.

Let ∅(u′) be θ(u′)² +1 and θ(u′)² -1, respectively, where θ > 0. That the above inequality holds for both ∅(u′) and for all positive θ implies $\int {_{\text{a}}^{\text{b}}\left( {{\text{G}}\left( {\text{x}} \right) - {\text{F}}\left( {\text{x}} \right)} \right){\text{dx}} = 0}$, which is a) in Definition 4′.

With $\int {_{\text{a}}^{\text{b}}\left( {{\text{G}}\left( {\text{x}} \right) - {\text{F}}\left( {\text{x}} \right)} \right){\text{dx}} = 0}$, we have ${{\text{E}}_{\text{F}}}{\text{v}}\left( {\text{x}} \right) - {{\text{E}}_{\text{G}}}{\text{v}}\left( {\text{x}} \right) = \int {_{\text{a}}^{\text{b}}\phi \left( {{\text{u}}\prime } \right)\left( {{\text{G}}\left( {\text{x}} \right) - {\text{F}}\left( {\text{x}} \right)} \right){\text{dx}} = - \int {_{\text{a}}^{\text{b}}\phi \prime {\text{u}}\prime \prime \left( {\text{x}} \right)\int {_{\text{a}}^{\text{x}}\left( {{\text{G}}\left( {\text{s}} \right) - {\text{F}}\left( {\text{s}} \right)} \right){\text{dsdx}} \geqslant 0} } }$. Let ∅(u′) be θ(u′)² + u′ and θ(u′)²–u′, respectively, where θ > 0. Again, that this inequality holds for these ∅(u′) and for all positive θ implies $\int {_{\text{a}}^{\text{b}}{\text{u}}\prime \prime \left( {\text{x}} \right)\int {_{\text{a}}^{\text{x}}\left( {{\text{G}}\left( {\text{s}} \right) - {\text{F}}\left( {\text{s}} \right)} \right){\text{dsdx}} = 0} }$, which is the equality in b) in Definition 4′.

With $\int {_{\text{a}}^{\text{b}}\left( {{\text{G}}\left( {\text{x}} \right) - {\text{F}}\left( {\text{x}} \right)} \right){\text{dx}} = 0\;{\text{and}}\;} \int {_{\text{a}}^{\text{b}}{\text{u}}\prime \prime \left( {\text{x}} \right)\int {_{\text{a}}^{\text{x}}\left( {{\text{G}}\left( {\text{s}} \right) - {\text{F}}\left( {\text{s}} \right)} \right){\text{ds}} \cdot {\text{dx}} = 0} }$, we have ${{\text{E}}_{\text{F}}}{\text{v}}\left( {\text{x}} \right) - {{\text{E}}_{\text{G}}}{\text{v}}\left( {\text{x}} \right) = \int {_{\text{a}}^{\text{b}}\phi \prime \prime {\text{u}}\prime \prime \left( {\text{y}} \right)\int {_{\text{a}}^{\text{y}}{\text{u}}\prime \prime \left( {\text{x}} \right)\int {_{\text{a}}^{\text{x}}\left( {{\text{G}}\left( {\text{s}} \right) - {\text{F}}\left( {\text{s}} \right)} \right){\text{ds}} \cdot {\text{dx}} \cdot {\text{dy}} \geqslant 0} } }$ for all ∅′′ > 0, which implies $\int {_{\text{a}}^{\text{y}}{\text{u}}\prime \prime \left( {\text{x}} \right)\int {_{\text{a}}^{\text{x}}\left( {{\text{G}}\left( {\text{s}} \right) - {\text{F}}\left( {\text{s}} \right)} \right){\text{ds}} \cdot {\text{dx}} \leqslant 0} }$, the weak inequality of b) in Definition 4′.

Moreover, $\int {_{\text{a}}^{\text{y}}{\text{u}}\prime \prime \left( {\text{x}} \right)\int {_{\text{a}}^{\text{x}}\left( {{\text{G}}\left( {\text{s}} \right) - {\text{F}}\left( {\text{s}} \right)} \right){\text{ds}} \cdot {\text{dx}} < 0} }$ for some y in (a, b) as long as G(x) and F(x) are not identical. QED

1.4 A4 Proof of Corollary 1′

Proof

Suppose v(x) is more downside risk averse than u(x). Then for any expected utility preserving increase in downside risk from F(x) to G(x) with u(x) as the reference decision maker,

$$ \begin{array}{*{20}{c}} {{{\text{E}}_{\text{F}}}{\text{v}}\left( {\text{x}} \right) - {{\text{E}}_{\text{G}}}{\text{v}}\left( {\text{x}} \right) = - \int {_{\text{a}}^{\text{b}}{\text{v}}\prime \prime \left( {\text{x}} \right)\int {_{\text{a}}^{\text{x}}\left( {{\text{G}}\left( {\text{s}} \right) - {\text{F}}\left( {\text{s}} \right)} \right){\text{dsdx}}} } } \\ { = - \int {_{\text{a}}^{\text{b}}\delta \prime \prime \left( {\text{x}} \right){\text{u}}\prime \prime \left( {\text{x}} \right)\int {_{\text{a}}^{\text{x}}\left( {{\text{G}}\left( {\text{s}} \right) - {\text{F}}\left( {\text{s}} \right)} \right){\text{dsdx}}} } } \\ { = \int {_{\text{a}}^{\text{b}}\delta \prime \prime \prime \left( {\text{y}} \right)\int {_{\text{a}}^{\text{y}}{\text{u}}\prime \prime \left( {\text{x}} \right)\int {_{\text{a}}^{\text{x}}\left( {{\text{G}}\left( {\text{s}} \right) - {\text{F}}\left( {\text{s}} \right)} \right){\text{ds}} \cdot {\text{dx}} \cdot {\text{dy}}{.}} } } } \\ \end{array} $$

where the first equality is due to a) and the third equality due to b) in Definition 4′.

If we further assume that v′′(x)·u′′(x) > 0 in [a,b], then v(x) is more downside risk averse than u(x) implies δ′′′(y) ≤ 0. Therefore E_Fv(x)–E_Gv(x) ≥ 0 since $\int {_{\text{a}}^{\text{y}}{\text{u}}\prime \prime \left( {\text{x}} \right)\int {_{\text{a}}^{\text{x}}\left( {{\text{G}}\left( {\text{s}} \right) - {\text{F}}\left( {\text{s}} \right)} \right){\text{ds}} \cdot {\text{dx}} \leqslant 0} }$ according to b) in Definition 4′.

Conversely, if E_Fv(x)–E_Gv(x) ≥ 0 for all expected utility preserving downside risk increases with u(x) as the reference decision maker, then it has to be the case that δ′′′(y) ≤ 0 and v(x) is more downside risk averse than u(x) (with the additional assumption that v′′(x)·u′′(x) > 0 in [a, b]). The proof of this is by contradiction with arguments parallel to those in Keenan and Snow (2009). Suppose δ′′′(y) > 0 at some point y₀ in [a, b]. Then there would exist a surrounding interval within [a, b] on which δ′′′(y) > 0, which implies by choosing G(x)–F(x) to be concentrated on the same interval, one would obtain E_Fv(x)–E_Gv(x) < 0, contradicting the initial assumption that E_Fv(x)–E_Gv(x) ≥ 0. QED