Economic consequences of Nth-degree risk increases and Nth-degree risk attitudes

Abstract

We study comparative statics of Nth-degree risk increases within a large class of problems that involve bidimensional payoffs and additive or multiplicative risks. We establish necessary and sufficient conditions for unambiguous impact of Nth-degree risk increases on optimal decision making. We develop a simple and intuitive approach to interpret these conditions : novel notions of directional Nth-degree risk aversion that are characterized via preferences over lotteries.

Appendices

Appendix A: Increases in Nth-degree risk

Proof of Lemma 1

The fact that 2. implies 1. results directly from Ekern (1980). For the sake of completeness we rederive it. Let $(\widetilde {\alpha }_{1},\widetilde { \alpha }_{2})$ be such that $\widetilde {\alpha }_{2}\succcurlyeq _{N} \widetilde {\alpha }_{1}.$ We have

$$\begin{array}{@{}rcl@{}} E[ q(\widetilde{\alpha }_{2})] -E\left[ q(\widetilde{\alpha }_{1}) \right] &=&\int_{0}^{B}q(x)dF_{\widetilde{\alpha }_{2}}(x)- \int_{0}^{B}q(x)dF_{\widetilde{\alpha }_{1}}(x) \\ &=&\sum\limits_{k=1}^{N}(-1)^{k-1}q^{(k-1)}(B)\left[ F_{\widetilde{\alpha }_{2}}^{[ k] }(B)-F_{\widetilde{\alpha }_{1}}^{[ k] }(B)\right] \\ &&+\int_{0}^{B}(-1)^{N}q^{(N)}(x)\left[ F_{\widetilde{\alpha }_{2}}^{\left[ N \right] }(x)-F_{\widetilde{\alpha }_{1}}^{[ N] }(x)\right] dx. \end{array} $$

Since q and its N first derivatives are continuous on $\mathbb {R}_{+}$, they are bounded on all compact subsets and all our integrals are well defined. □

By definition, all the terms in the sum are equal to 0 and $F_{\widetilde { \alpha }_{2}}^{[ N] }(x)-F_{\widetilde {\alpha }_{1}}^{\left [ N \right ] }(x)\geq 0$ on [0, B]. By 2., the integral is then nonnegative and $E[ q(\widetilde {\alpha }_{2})] \geq E\left [ q( \widetilde {\alpha }_{1})\right ]$.

Let us now prove that 1. implies 2. Let q be an N times continuously differentiable function on $\mathbb {R}_{+}$ such that $E\left [ q(\widetilde { \alpha }_{2})\right ] \geq E[ q(\widetilde {\alpha }_{1})]$ for all pair $(\widetilde {\alpha }_{1},\widetilde {\alpha }_{2})$ such that $\widetilde {\alpha }_{2}\succcurlyeq _{N}\widetilde {\alpha }_{1}.$ Let, if it exists, α be a nonnegative real number such that q ^(N)(α) ≠ 0 and let ε denote a positive real number such that q ^(N)(t) ≠ 0 for t ∈ [α, α + ε]. The sign of q ^(N)remains then constant on [α, α + ε]. Let $\widetilde { \beta }_{1}$ and $\widetilde {\beta }_{2}$ be two nonnegative bounded above random variables such that $\widetilde {\beta }_{2}\succcurlyeq _{N} \widetilde {\beta }_{1}.$ Let B be a common upper bound for $\widetilde { \beta }_{1}$ and $\widetilde {\beta }_{2}$ and let $\widetilde {\alpha }_{1}= \frac {\varepsilon }{B}\widetilde {\beta }_{1}+\alpha$ and $\widetilde {\alpha }_{2}=\frac {\varepsilon }{B}\widetilde {\beta }_{2}+\alpha .$ The random variables $\widetilde {\alpha }_{1}$ and $\widetilde {\alpha }_{2}$ take their values in [α, α + ε] and it is easy to check that $F_{\widetilde {\alpha }_{i}}^{[ k] }( t) =\left ( \frac {\varepsilon }{B}\right )^{k-1}F_{\widetilde {\beta }_{i}}^{[ k] }\left ( \frac {t-\alpha }{\varepsilon }B\right )$ for k = 1, 2. . . and i = 1, 2. Therefore, $\widetilde {\alpha }_{2}\succcurlyeq _{N} \widetilde {\alpha }_{1}$ which implies, by 1., that $E\left [ q(\widetilde { \alpha }_{2})\right ] \geq E[ q(\widetilde {\alpha }_{1})] .$ We have

$$\begin{array}{@{}rcl@{}} E[ q(\widetilde{\alpha }_{2})] -E\left[ q(\widetilde{\alpha }_{1}) \right] &=&\int_{\alpha }^{\alpha +\varepsilon }q(x)dF_{\widetilde{\alpha } _{2}}(x)-\int_{\alpha }^{\alpha +\varepsilon }q(x)dF_{\widetilde{\alpha } _{1}}(x) \\ &=&\sum\limits_{k=1}^{N}(-1)^{k-1}q^{(k-1)}(\alpha +\varepsilon )\left[ F_{ \widetilde{\alpha }_{2}}^{[ k] }(\alpha +\varepsilon )-F_{ \widetilde{\alpha }_{1}}^{[ k] }(\alpha +\varepsilon )\right] \\ &&+\int_{\alpha }^{\alpha +\varepsilon }(-1)^{N}q^{(N)}(t)\left[ F_{ \widetilde{\alpha }_{2}}^{[ N] }(t)-F_{\widetilde{\alpha }_{1}}^{[ N] }(t)\right] dt. \end{array} $$

By definition, we have $F_{\widetilde {\alpha }_{2}}^{[ k] }(\alpha +\varepsilon )=F_{\widetilde {\alpha }_{1}}^{[ k] }(\alpha +\varepsilon )$ for k = 1, . . . , N and $F_{\widetilde {\alpha }_{2}}^{[ N] }(t)-F_{\widetilde {\alpha }_{1}}^{[ N] }(t)\geq 0$ and the inequality is strict for some t in [α, α + ε], and even on a neighborhood of t by continuity of $F_{\widetilde {\alpha }_{1}}^{[ N] }$ and $F_{\widetilde {\alpha } _{2}}^{[ N] }.$ Since the sign of q ^(N)remains constant on [α, α + ε], this gives that ( − 1)^N q ^(N)(t) > 0 on α, α + ε and ( − 1)^N q ^(N)(x) > 0 for all x such that q ^(N)(x) ≠ 0 which completes the proof.

1.1 Proof of Proposition 1

Let us prove that 2. implies 1. Let $\left ( \widetilde {\alpha }_{1}, \widetilde {\alpha }_{2}\right )$ be such that $\widetilde {\alpha } _{2}\succcurlyeq _{N}\widetilde {\alpha }_{1}$ and let $q(\alpha )=g(x_{1}^{\ast },\alpha ,\mu ).$ We have $\frac {\partial ^{N}g}{\partial \alpha ^{N}}(x_{1}^{\ast },\alpha ,\mu )=-pU^{( 1,N) }\left ( K-x_{1}^{\ast }p,x_{1}^{\ast }\mu +\alpha \right ) +\mu U^{\left ( 0,N+1\right ) }( K-x_{1}^{\ast }p,x_{1}^{\ast }\mu +\alpha ) .$ By 2., we have $( -1)^{N}\frac {\partial ^{N}g}{\partial \alpha ^{N}} (x_{1}^{\ast },\alpha ,\mu )\geq 0$ and ( − 1)^N q ^(N)(α) ≥ 0. Since $\widetilde {\alpha }_{i}$ is bounded above ( i = 1, 2), $x_{1}^{\ast }\mu +\widetilde {\alpha }_{i}$ is bounded and bounded away from zero and U and its N + 1 first derivatives are bounded on the convex hull of the set of values taken by $\left ( K-x_{1}^{\ast }p,x_{1}^{\ast }\mu + \widetilde {\alpha }_{i}\right )$. The same applies to q on the convex hull of the set of values taken by $\widetilde {\alpha }_{i}$, i = 1, 2. By Lemma 1, this leads to $E[ q(\widetilde {\alpha }_{1})] \leq E [ q(\widetilde {\alpha }_{2})] .$ By definition, we have $E\left [ q(\widetilde {\alpha }_{1})\right ] =0,$ which gives $E\left [ q(\widetilde { \alpha }_{2})\right ] \geq 0$ or $E\left [ g(x_{1}^{\ast },\widetilde {\alpha } _{2},\mu )\right ] \geq 0.$ By concavity of U, it is easy to check that $E [ g(x,\widetilde {\alpha }_{2},\mu )]$ is a decreasing function of x. Since $x_{2}^{\ast }$ is characterized by $E\left [ g(x_{2}^{\ast }, \widetilde {\alpha }_{2},\mu )\right ] =0,$ we obtain that $x_{2}^{\ast }\geq x_{1}^{\ast }.$

Let us prove 1. implies 2. As in the proof of Lemma 1, we consider $\widetilde {\beta }_{1}$ and $\widetilde {\beta }_{2}$ two nonnegative random variables with a common upper bound B such that $\widetilde {\beta } _{2}\succcurlyeq _{N}\widetilde {\beta }_{1}.$ As above, we introduce the random variables $\widetilde {\alpha }_{1,\varepsilon }=\frac {\varepsilon }{B} \widetilde {\beta }_{1}+\alpha ^{\ast }$ and $\widetilde {\alpha } _{2,\varepsilon }=\frac {\varepsilon }{B}\widetilde {\beta }_{2}+\alpha ^{\ast }$ for some α ^∗ > 0 and some ε > 0. The random variables $\widetilde {\alpha }_{1,\varepsilon }$ and $\widetilde {\alpha } _{2,\varepsilon }$ take their values in [α ^∗, α ^∗ + ε] and $\widetilde {\alpha }_{2,\varepsilon }\succcurlyeq _{N}\widetilde {\alpha }_{1,\varepsilon }.$ Let us consider a given real number γ and let us define the random variables $\widetilde {\gamma }_{i,\varepsilon }$, i = 1, 2, as lotteries giving $\widetilde {\alpha }_{i,\varepsilon }$ with probability ε and γ with probability 1 − ε. The constant γ provides an additional degree of freedom that will prove useful in order to control the pair $( K-x_{1}^{\ast }p,x_{1}^{\ast }\mu +\alpha ) .$ We have $\widetilde {\gamma }_{2,\varepsilon }\succcurlyeq _{N}\widetilde {\gamma }_{1,\varepsilon }$ as an immediate consequence of the stability of this order relation under probability mixtures. Let $x_{1,\varepsilon }^{\ast }$ and $x_{2,\varepsilon }^{\ast }$ be respectively the solutions of $P_{U,K,p}( \widetilde {\gamma }_{1,\varepsilon },\mu )$ and $P_{U,K,p}(\widetilde {\gamma }_{2,\varepsilon },\mu ).$ By 1., we have $x_{2,\varepsilon }^{\ast }\geq x_{1,\varepsilon }^{\ast }.$ By definition, we have $E\left [ g(x_{1,\varepsilon }^{\ast },\widetilde {\gamma }_{1,\varepsilon },\mu ) \right ] =0$ and $E\left [ g(x_{2,\varepsilon }^{\ast },\widetilde {\gamma } _{2,\varepsilon },\mu )\right ] =0.$ By concavity of U, g is a decreasing function of x and we have $E\left [ g(x_{1,\varepsilon }^{\ast },\widetilde { \gamma }_{2,\varepsilon },\mu )\right ] \geq 0.$ Let q be defined by $q(\alpha )=g(x_{1,\varepsilon }^{\ast },\alpha ,\mu ).$ We have

$$\begin{array}{@{}rcl@{}} E[ q(\widetilde{\gamma }_{2,\varepsilon })] -E\left[ q(\widetilde{ \gamma }_{1,\varepsilon })\right] &=&\left( E\left[ q(\widetilde{\alpha } _{2,\varepsilon })\right] -E\left[ q(\widetilde{\alpha }_{1,\varepsilon }) \right] \right) \varepsilon \\ &=&\sum_{k=1}^{N}(-1)^{k-1}q^{(k-1)}(\alpha^{\ast }+\varepsilon ) \notag \\ &&\times\left[ F_{ \widetilde{\alpha }_{2},P}^{[ k] }(\alpha^{\ast }+\varepsilon )-F_{\widetilde{\alpha }_{1},P}^{[k] }(\alpha^{\ast }+\varepsilon )\right] \varepsilon \\ &&+\varepsilon \int_{\alpha^{\ast }}^{\alpha^{\ast }+\varepsilon }(-1)^{N}q^{(N)}(t)\left[ F_{\widetilde{\alpha }_{2},P}^{\left[ N\right] }(t)-F_{\widetilde{\alpha }_{1},P}^{[ N] }(t)\right] dt. \end{array} $$

By construction, the left side of the equality is nonnegative, all the terms in the sum are equal to zero and $F_{\widetilde {\alpha }_{2},P}^{\left [ N \right ] }(t)-F_{\widetilde {\alpha }_{1},P}^{[ N] }(t)$ is nonnegative and nonzero. Note that $\widetilde {\gamma }_{i,\varepsilon }$ is bounded and bounded away from zero for i = 1, 2. Therefore q and its first N derivatives are bounded on the convex hull of the set of values taken by $\widetilde {\gamma }_{i,\varepsilon }$, i = 1, 2. Therefore, ( − 1)^N q ^(N)(t) is nonnegative at least on a given subinterval of α ^∗, α ^∗ + ε. Let $\alpha _{\varepsilon }^{\ast }\$be in α ^∗, α ^∗ + ε such that $(-1)^{N}q^{(N)}(\alpha _{\varepsilon }^{\ast })\geq 0.$ We have then

$$ (-1)^{N+1}pU^{( 1,N) }\left( K-x_{1,\varepsilon }^{\ast }p,x_{1,\varepsilon }^{\ast }\mu +\alpha_{\varepsilon }^{\ast }\right) +(-1)^{N}\mu U^{( 0,N+1) }\left( K-x_{1,\varepsilon }^{\ast }p,x_{1,\varepsilon }^{\ast }\mu +\alpha_{\varepsilon }^{\ast }\right) \geq 0 $$

(6)

where $x_{1,\varepsilon }^{\ast }$ satisfies

$$E\left[ -pU^{( 1,0) }\left( K-x_{1,\varepsilon }^{\ast }p,x_{1,\varepsilon }^{\ast }\mu +\widetilde{\gamma }_{1,\varepsilon }\right) +\mu U^{( 0,1) }\left( K-x_{1,\varepsilon }^{\ast }p,x_{1,\varepsilon }^{\ast }\mu +\widetilde{\gamma }_{1,\varepsilon }\right) \right] =0 $$

or

$$ \varepsilon E\left[ g(x_{1,\varepsilon }^{\ast },\frac{\varepsilon }{B} \widetilde{\beta }_{1}+\alpha^{\ast },\mu )\right] +(1-\varepsilon )g(x_{1,\varepsilon }^{\ast },\gamma ,\mu )=0. $$

(7)

Remark that, until now, α ^∗, K and γ have been arbitrarily chosen. Let us now choose them carefully in order to derive our result. Let (Y, Z) be arbitrary in $( \mathbb {R}_{+}^{\ast })^{2}$. By our Inada condition, $\lim _{z\rightarrow 0}\frac {U^{\left ( 0,1\right ) }( Y,z) }{U^{( 1,0) }( Y,z) } =\infty$ and $\lim _{z\rightarrow \infty }\frac {U^{( 0,1) }\left ( Y,z\right ) }{U^{( 1,0) }( Y,z) }=0$ which gives that there exists some z ^∗ > 0 such that $\frac {U^{( 0,1) }( Y,z^{\ast }) }{U^{( 1,0) }( Y,z^{\ast }) }=\frac {p}{\mu }$ or − p U ^{(1, 0)}(Y, z ^∗) + μ U ^{(0, 1)}(Y, z ^∗) = 0. We choose x ^∗ > 0 such that μ x ^∗ < inf(Z, z ^∗) and α ^∗ and K are taken such that α ^∗ = Z − μ x ^∗ > 0 and K = Y + p x ^∗. Let γ be given by γ = z ^∗ − μ x ^∗ > 0. We have

$$\begin{array}{@{}rcl@{}} g(x^{\ast },\widetilde{\gamma },\mu ) &=&-pU^{( 1,0) }\left( K-x^{\ast }p,x^{\ast }\mu +\gamma \right) +\mu U^{( 0,1) }\left( K-x^{\ast }p,x^{\ast }\mu +\gamma \right) \\ &=&-pU^{( 1,0) }( Y,z^{\ast }) +\mu U^{\left( 0,1\right) }( Y,z^{\ast }) =0. \end{array} $$

The solution of Eq. 7 for ε = 0 is then given by x ^∗. In a well chosen neighborhood of x ^∗, x ↦ K − x p and $x\longmapsto x\mu +\frac {\varepsilon }{B}\widetilde {\beta }_{1}+\alpha ^{\ast }$ are bounded and bounded away from 0. The functions U ^{1, 0} and U ^{(0, 1)} being continuously differentiable, the function $(x,\varepsilon )\longmapsto \varepsilon E\left [ g(x,\frac {\varepsilon }{B}\widetilde {\beta }_{1}+\alpha ^{\ast },\mu )\right ] +(1-\varepsilon )g(x,\widetilde {\gamma },\mu )$ is then differentiable with respect to x at (x ^∗, 0). Furthermore, the derivative of this last function with respect to x at (x ^∗, 0) is nonzero (concavity of U ). The solution $x_{1,\varepsilon }^{\ast }$ of Eq. 7 is then continuous with respect to ε in a neighborhood of 0 which gives $\lim _{\varepsilon \rightarrow 0}x_{1,\varepsilon }^{\ast }=x^{\ast }.$ Furthermore, we clearly have $\lim _{\varepsilon \rightarrow 0}\alpha _{\varepsilon }^{\ast }=\alpha ^{\ast }.$ Taking the limit in Eq. 6 when ε tends to 0, we obtain

$$(-1)^{N+1}pU^{( 1,N) }\left(K\,-\,x^{\ast }p,x^{\ast }\mu \,+\,\alpha^{\ast }\right) +(-1)^{N}\mu U^{( 0,N+1) }\left( K\,-\,x^{\ast}p,x^{\ast }\mu +\alpha^{\ast }\right) \geq 0 $$

or, by construction

$$(-1)^{N+1}pU^{( 1,N) }( Y,Z) +(-1)^{N}\mu U^{\left( 0,N+1\right) }( Y,Z) \geq 0 $$

1.2 Proof of Proposition 2

Let us prove that 2. implies 1. Let $\left (\widetilde {\mu }_{1},\widetilde { \mu }_{2}\right )$ be such that $\widetilde {\mu }_{2}\succcurlyeq _{N} \widetilde {\mu }_{1}$ and let $q(\mu )=g(x_{1}^{\ast },\alpha ,\mu ).$ We have $q^{(N)}(\mu )=-p(x_{1}^{\ast })^{N}U^{\left ( 1,N\right )}(K-x_{1}^{\ast }p,x_{1}^{\ast }\mu +\alpha )+\mu (x_{1}^{\ast })^{N}U^{(0,N+1)}\left (K-x_{1}^{\ast }p,x_{1}^{\ast }\mu +\alpha \right )+N(x_{1}^{\ast })^{N-1}U^{(0,N)}\left (K-x_{1}^{\ast }p,x_{1}^{\ast }\mu +\alpha \right )$. Since we assumed that $x_{1}^{\ast }\geq 0$, by 2, we have

$(-1)^{N}p(x_{1}^{\ast })^{N}U^{(1,N) }(K-x_{1}^{\ast }p,x_{1}^{\ast }\mu +\alpha )\leq 0$ and

$$\begin{array}{@{}rcl@{}} &&(-1)^{N}\left(x_{1}^{\ast}\mu U^{(0,N+1)}(K-x_{1}^{\ast}p,x_{1}^{\ast}\mu+\alpha)+NU^{\left( 0,N\right)}(K-x_{1}^{\ast}p,x_{1}^{\ast}\mu+\alpha)\right)\\ &=&(-1)^{N}\frac{x_{1}^{\ast}\mu}{x_{1}^{\ast}\mu+\alpha} \left(\left( x_{1}^{\ast }\mu +\alpha\right)U^{(0,N+1)}\left(K-x_{1}^{\ast }p,x_{1}^{\ast}\mu +\alpha\right)\right.\\ &&\left.\qquad\;\;\;\,\,\qquad\qquad+NU^{\left( 0,N\right)}\left( K-x_{1}^{\ast}p,x_{1}^{\ast}\mu +\alpha\right)\right)\\ &&+(-1)^{N}\frac{\alpha N}{x_{1}^{\ast}\mu +\alpha}U^{\left( 0,N\right)}(K-x_{1}^{\ast }p,x_{1}^{\ast}\mu +\alpha) \\ &\geq &0 \end{array} $$

which gives ( − 1)^N q ^(N)(μ) ≥ 0. Since $\widetilde {\mu }_{i}$ is bounded above ( i = 1, 2), $x_{1}^{\ast }\widetilde { \mu }_{i}+\alpha$ is bounded and bounded away from zero and U and its N + 1 first derivatives are bounded on the convex hull of the set of values taken by $\left ( K-x_{1}^{\ast }p,x_{1}^{\ast }\widetilde {\mu }_{i}+\alpha \right )$. The same applies to q on the convex hull of the set of values taken by $\widetilde {\mu }_{i}$, i = 1, 2. By Lemma 1, this leads to $E[ q(\widetilde {\mu }_{1})] \leq E\left [ q(\widetilde {\mu } _{2})\right ] .$ By definition, we have $E\left [ q(\widetilde {\mu }_{1}) \right ] =0$ which gives $E[ q(\widetilde {\mu }_{2})] =E\left [ g(x_{1}^{\ast },\alpha ,\widetilde {\mu }_{2})\right ] \geq 0.$ By concavity of U, it is easy to check that g(x, α, μ) is a decreasing function of x. From there we derive that $x_{2}^{\ast }\geq x_{1}^{\ast }.$

Let us prove that 1. implies 2. As in the proof of Lemma 1 and Proposition 1, we consider $\widetilde {\beta }_{1}$ and $\widetilde {\beta }_{2}$ two nonnegative random variables with a common upper bound B such that $\widetilde {\beta }_{2}\succcurlyeq _{N}\widetilde { \beta }_{1}.$ As above, we introduce the random variables $\widetilde {\mu } _{1,\varepsilon }=\frac {\varepsilon }{B}\widetilde {\beta }_{1}+\mu ^{\ast }$ and $\widetilde {\mu }_{2,\varepsilon }=\frac {\varepsilon }{B}\widetilde { \beta }_{2}+\mu ^{\ast }$ for some μ ^∗ > 0 and some ε > 0. The random variables $\widetilde {\mu }_{1,\varepsilon }$ and $\widetilde {\mu }_{2,\varepsilon }$ take their values in [μ ^∗, μ ^∗ + ε] and $\widetilde {\mu }_{2,\varepsilon }\succcurlyeq _{N}\widetilde {\mu }_{1,\varepsilon }.$ Let us consider a given γ and let us define the random variables $\widetilde {\gamma } _{i,\varepsilon }$, i = 1, 2, as a lottery that takes the value $\widetilde { \mu }_{i,\varepsilon }(\omega )$ with probability ε and the value γ with probability 1 − ε. As previously, we have $\widetilde {\gamma }_{2,\varepsilon }\succcurlyeq _{N}\widetilde {\gamma } _{1,\varepsilon }.$ Let $x_{1,\varepsilon }^{\ast }$ and $x_{2,\varepsilon }^{\ast }$ be respectively the solutions of $P_{U,K,p}(\alpha ,\widetilde { \gamma }_{1,\varepsilon })$ and $P_{U,K,p}(\alpha ,\widetilde {\gamma } _{2,\varepsilon }).$ If $x_{1,\varepsilon }^{\ast }$ is nonnegative then, by 1., we have $x_{2,\varepsilon }^{\ast }\geq x_{1,\varepsilon }^{\ast }.$ By definition, we have $E\left [ g(x_{1,\varepsilon }^{\ast },\alpha ,\widetilde { \gamma }_{1,\varepsilon })\right ] =0$ and $E\left [ g(x_{2,\varepsilon }^{\ast },\alpha ,\widetilde {\gamma }_{2,\varepsilon })\right ] =0.$ By concavity of U, g is a decreasing function of x and we have $E\left [ g(x_{1,\varepsilon }^{\ast },\alpha ,\widetilde {\gamma }_{2,\varepsilon }) \right ] \geq 0.$ Let q be defined by $q(\mu )=g(x_{1,\varepsilon }^{\ast },\alpha ,\mu ).$ We have

$$\begin{array}{@{}rcl@{}} E[ q(\widetilde{\gamma }_{2,\varepsilon })] -E\left[ q(\widetilde{ \gamma }_{1,\varepsilon })\right] &=&\left( E\left[ q(\widetilde{\mu } _{2,\varepsilon })\right] -E\left[ q(\widetilde{\mu }_{1,\varepsilon }) \right] \right) \varepsilon \\ &=&\sum_{k=1}^{N}(-1)^{k-1}q^{(k-1)}(\mu^{\ast }+\varepsilon )\\ &&\times\left[ F_{ \widetilde{\mu }_{2},P}^{[ k] }(\mu^{\ast }+\varepsilon )-F_{ \widetilde{\mu }_{1},P}^{[ k] }(\mu^{\ast }+\varepsilon )\right] \varepsilon \\ &&+\varepsilon \int_{\mu^{\ast }}^{\mu^{\ast }+\varepsilon }(-1)^{N}q^{(N)}(t)\left[ F_{\widetilde{\mu }_{2},P}^{\left[ N\right] }(t)-F_{\widetilde{\mu }_{1},P}^{[ N] }(t)\right] dt. \end{array} $$

By construction, the left side of the equality is nonnegative, all the terms in the sum are equal to zero and $F_{\widetilde {\mu }_{2},P}^{[ N] }(t)-F_{\widetilde {\mu }_{1},P}^{[ N] }(t)$ is nonnegative and nonzero. Note that $\widetilde {\gamma }_{i,\varepsilon }$ is bounded and bounded away from zero for i = 1, 2. Therefore q and its first N derivatives are bounded on the convex hull of the set of values taken by $\widetilde {\gamma }_{i,\varepsilon }$, i = 1, 2. Therefore, ( − 1)^N q ^(N) is nonnegative at least on a given subinterval of μ ^∗, μ ^∗ + ε. Let $\mu _{\varepsilon }^{\ast }\$be in μ ^∗, μ ^∗ + ε such that $(-1)^{N}q^{(N)}(\mu _{\varepsilon }^{\ast })\geq 0.$ We have then

$$\begin{array}{@{}rcl@{}} &&(-1)^{N+1}p( x_{1,\varepsilon }^{\ast })^{N}U^{\left( 1,N\right) }\left( K-x_{1,\varepsilon }^{\ast }p,x_{1,\varepsilon }^{\ast }\mu_{\varepsilon }^{\ast }+\alpha \right) \\ &&+(-1)^{N}\mu_{\varepsilon }^{\ast }\left( x_{1,\varepsilon }^{\ast }\right)^{N}U^{( 0,N+1) }\left( K-x_{1,\varepsilon }^{\ast }p,x_{1,\varepsilon }^{\ast }\mu_{\varepsilon }^{\ast }+\alpha \right) \\ &&+(-1)^{N}N( x_{1,\varepsilon }^{\ast })^{N-1}U^{\left( 0,N\right) }\left( K-x_{1,\varepsilon }^{\ast }p,x_{1,\varepsilon }^{\ast }\mu_{\varepsilon }^{\ast }+\alpha \right) \\ &\geq &0 \end{array} $$

(8)

where $x_{1,\varepsilon }^{\ast }$ satisfies

$$E\left[-pU^{( 1,0) }\left( K-x_{1,\varepsilon }^{\ast }p,x_{1,\varepsilon }^{\ast }\widetilde{\gamma }_{1,\varepsilon }+\alpha \right) +\widetilde{\gamma }_{1,\varepsilon }U^{( 0,1) }\left(K-x_{1,\varepsilon }^{\ast }p,x_{1,\varepsilon }^{\ast}\widetilde{\gamma} _{1,\varepsilon }+\alpha \right)\right]\,=\,0 $$

or

$$ \varepsilon E\left[ g\left(x_{1,\varepsilon }^{\ast },\alpha ,\frac{\varepsilon }{ B}\widetilde{\beta }_{1}+\mu^{\ast }\right)\right] +(1-\varepsilon )g(x_{1,\varepsilon }^{\ast },\alpha ,\gamma )=0. $$

(9)

Remark that, until now p, μ ^∗, K, α and γ have been arbitrarily chosen. Let us now choose them carefully in order to derive our result. We assume first that γ is equal to 1. Let x ^∗ > 0 be given and let (M, Y, Z) in $( \mathbb {R}_{+})^{3}$ such that M < Z. We take $p=\frac {U^{( 0,1) }(Y,x^{\ast }+Z-M)}{U^{( 1,0) }(Y,x^{\ast }+Z-M)}$, K = Y + p x ^∗, α = Z − M and $\mu ^{\ast }=\frac {M}{x^{\ast }}.$ By construction, we have α > 0 and we have

$$\begin{array}{@{}rcl@{}} g(x^{\ast },\alpha ,\gamma ) &=&-pU^{( 1,0) }\left( K-x^{\ast }p,x^{\ast }\gamma +\alpha \right) +\widetilde{\gamma }U^{( 0,1) }( K-x^{\ast }p,x^{\ast }\gamma +\alpha ) \\ &=&-pU^{( 1,0) }( K-x^{\ast }p,x^{\ast }+\alpha ) +U^{( 0,1) }( K-x^{\ast }p,x^{\ast }+\alpha ) \\ &=&-pU^{( 1,0) }( Y,x^{\ast }+Z-M) +U^{\left( 0,1\right) }( Y,x^{\ast }+Z-M) \\ &=&0. \end{array} $$

The solution of Eq. 9 for ε = 0 is then given by x ^∗. In a well chosen neighborhood of x ^∗, x ↦ K − x p and $x\longmapsto x\left (\mu ^{\ast }+\frac {\varepsilon }{B}\widetilde { \beta }_{1}\right ) +\alpha$ are bounded and bounded away from 0. The functions U ^{(1, 0)} and U ^{(0, 1)} being continuously differentiable, the function $(x,\varepsilon )\longmapsto \varepsilon E\left [ g(x,\alpha ,\frac {\varepsilon }{B}\widetilde {\beta }_{1}+\mu ^{\ast })\right ] +(1-\varepsilon )g(x,\alpha ,\gamma )$ is then differentiable with respect to x at (x ^∗, 0). Furthermore, the derivative with respect to x at (x ^∗, 0) is nonzero. The solution $x_{1,\varepsilon }^{\ast }$ of Eq. 9 is then continuous with respect to ε in a neighborhood of 0 which gives $\lim _{\varepsilon \rightarrow 0}x_{1,\varepsilon }^{\ast }=x^{\ast }$ and guarantees that $x_{1,\varepsilon }^{\ast }>0$ for ε small enough . Since we clearly have $\lim _{\varepsilon \rightarrow 0}\mu _{\varepsilon }^{\ast }=\mu ^{\ast },$ taking the limit in Eq. 8 when ε tends to 0, we obtain

$$(-1)^{N}(x^{\ast})^{N-1}\left(-px^{\ast}U^{(1,N)}(Y,Z)+\mu^{\ast}x^{\ast}U^{(0,N+1)}(Y,Z)+NU^{(0,N)}(Y,Z)\right)\geq 0 $$

or

$$ (-1)^{N}\left(-px^{\ast}U^{(1,N)}(Y,Z) +MU^{(0,N+1)}(Y,Z)+NU^{(0,N)}(Y,Z)\right) \geq 0 $$

(10)

This result being true for all (Y, Z) in $\left (\mathbb {R}_{+}^{\ast }\right )^{2},$ all M ∈ (0, Z) all x ^∗ > 0 and for $p=\frac {U^{(0,1)}(Y,x^{\ast }+Z-M)}{U^{(1,0)}(Y,x^{\ast }+Z-M)}.$ When x ^∗goes to 0, Y, M and Z being fixed, $x^{\ast }\frac {U^{(0,1)}\left (Y,x^{\ast }+Z-M\right )}{U^{( 1,0) }\left (Y,x^{\ast }+Z-M\right )}$ goes to 0 and we obtain that

$$(-1)^{N}\left(MU^{(0,N+1)}(Y,Z)+NU^{(0,N)}(Y,Z)\right) \geq 0 $$

for all (Y, Z) in $( \mathbb {R}_{+}^{\ast })^{2}$ and all M ∈ (0, Z) or equivalently

$$\begin{array}{@{}rcl@{}} (-1)^{N}&&\left(ZU^{(0,N+1)}(Y,Z)+NU^{(0,N)}(Y,Z)\right)\geq 0\\ &&\text{and}\,\,(-1)^{N}U^{(0,N)}(Y,Z)\geq 0\,\,\text{for all}\,\,(Y,Z)\,\,\text{in}\,\,\left(\mathbb{R}_{+}^{\ast}\right)^{2}. \end{array} $$

We assumed that $z\frac {U^{( 0,1) }(Y,z)}{U^{( 1,0)}(Y,z)}$ is unbounded. We then have either $\lim _{z\rightarrow 0}z\frac { U^{( 0,1) }(Y,z)}{U^{( 1,0) }(Y,z)}=\infty$ or $\lim _{z\rightarrow \infty }z\frac {U^{( 0,1) }(Y,z)}{U^{\left ( 1,0\right ) }(Y,z)}=\infty$. If $\lim _{z\rightarrow 0}z\frac {U^{\left ( 0,1\right ) }(Y,z)}{U^{( 1,0) }(Y,z)}=\infty$, it suffices to take $x^{\ast }=\frac {\zeta }{2}$ and $M=Z-\frac {\zeta }{2}$ for ζ arbitrarily small to make the quantity $px^{\ast }=\frac {U^{\left ( 0,1\right ) }(Y,x^{\ast }+Z-M)}{U^{( 1,0) }(Y,x^{\ast }+Z-M)} x^{\ast }=\frac {1}{2}\zeta \frac {U^{( 0,1) }(Y,\zeta )}{U^{\left ( 1,0\right ) }(Y,\zeta )}$ arbitrarily large. Since M, Y and Z are bounded, Eq. 10 gives then ( − 1)^N U ^{(1, N)}(Y, Z) ≤ 0. If $\lim _{z\rightarrow \infty }z\frac {U^{\left ( 0,1\right ) }(Y,z)}{U^{( 1,0) }(Y,z)}=\infty ,$ it suffices to take x ^∗sufficiently large to make the quantity x ^∗ + Z − M sufficiently large and $\frac {U^{( 0,1) }(Y,x^{\ast }+Z-M)}{ U^{( 1,0) }(Y,x^{\ast }+Z-M)}( x^{\ast }+Z-M)$ arbitrarily large. Since Z is kept fixed, $\frac {x^{\ast }}{x^{\ast }+Z-M}$ is arbitrarily close to 1 and p x ^∗arbitrarily large. Since M, Y and Z are bounded, Eq. 10 gives then ( − 1)^N U ^{(1, N)}(Y, Z) ≤ 0.The Proof of Corollary 1 follows directly by reversing the inequalities above.

1.3 Proof of Proposition 3

In an expected utility framework, Nth-degree risk aversion in the direction of (ρ _y , ρ _z ) is equivalent to

$$\begin{array}{@{}rcl@{}} &&\frac{1}{2}E\left[ U\left( y+x_{2}\rho_{y},z+x_{2}\rho_{z}+\widetilde{ \alpha }_{2}\right) \right] -\frac{1}{2}E\left[ U\left( y+x_{2}\rho_{y},z+x_{2}\rho_{z}+\widetilde{\alpha }_{1}\right) \right] \\ &>&\frac{1}{2}E\left[ U\left( y+x_{1}\rho_{y},z+x_{1}\rho_{z}+\widetilde{ \alpha }_{2}\right) \right] -\frac{1}{2}E\left[ U\left( y+x_{1}\rho_{y},z+x_{1}\rho_{z}+\widetilde{\alpha }_{1}\right) \right] . \end{array} $$

for all $( y,z,x_{1},x_{2}) \in \mathbb {R}_{+}^{4}$ with x ₂ > x ₁. The previous inequation is satisfied for all x ₂ > x ₁ if and only if $E\left [ U\left ( y+t\rho _{y},z+t\rho _{z}+\widetilde {\alpha } _{2}\right ) \right ] -E\left [ U\left ( y+t\rho _{y},z+t\rho _{z}+\widetilde { \alpha }_{1}\right ) \right ]$ is increasing in t or, equivalently, if and only if $E\left [ f\left ( y+t\rho _{y},z+t\rho _{z}+\widetilde {\alpha } _{2}\right ) \right ]$ is larger than $E\left [ f\left ( y+t\rho _{y},z+t\rho _{z}+\widetilde {\alpha }_{1}\right ) \right ]$ for all $(y,z,t)\in \mathbb {R} _{+}^{3}$. This is satisfied if and only if $E\left [ f\left ( y,z+\widetilde { \alpha }_{2}\right ) \right ]$ is larger than $E\left [ f\left ( y,z+\widetilde { \alpha }_{1}\right ) \right ]$ for all $(y,z)\in \mathbb {R}_{+}^{2}.$ Since we want this inequality to be true for all $(\widetilde {\alpha }_{1}, \widetilde {\alpha }_{2})$ such that $\widetilde {\alpha }_{2}$ is an increase in Nth-degree risk over $\widetilde {\alpha }_{1},$ this inequality is equivalent, by Lemma 1, to ( − 1)^N f ^{(0, N)} ≥ 0.

1.4 Proof of Proposition 4

In an expected utility model, Nth-degree multiplicative-risk attraction in the direction of (ρ _y , ρ _z ) is equivalent to

$$\begin{array}{@{}rcl@{}} &&\frac{1}{2}E\left[ U\left( y+x_{2}\rho_{y},z+x_{2}\rho_{z}\widetilde{ \alpha }_{2}\right) \right] -\frac{1}{2}E\left[ U\left( y+x_{2}\rho_{y},z+x_{2}\rho_{z}\widetilde{\alpha }_{1}\right) \right] \\ &>&\frac{1}{2}E\left[ U\left( y+x_{1}\rho_{y},z+x_{1}\rho_{z}\widetilde{ \alpha }_{2}\right) \right] -\frac{1}{2}E\left[ U\left( y+x_{1}\rho_{y},z+x_{1}\rho_{z}\widetilde{\alpha }_{1}\right) \right] \end{array} $$

(11)

for all $(\widetilde {\alpha }_{1},\widetilde {\alpha }_{2})$ such that $\widetilde {\alpha }_{1}\preccurlyeq _{N}\widetilde {\alpha }_{2}$ and all (x ₁, x ₂) such that 0 < x ₁ < x ₂ and y + x _i ρ _y ≥ 0, i = 1, 2. This inequality is satisfied for all x ₂ > x ₁ if and only if $E \left [ U( y+t\rho _{y},z+t\rho _{z}\widetilde {\alpha }_{2}) \right ] -E\left [ U\left ( y+t\rho _{y},z+t\rho _{z}\widetilde {\alpha } _{1}\right ) \right ]$ is increasing on $T(y)=\left \{t\in \mathbb {R} _{+}:y+t\rho _{y}\geq 0\right \}$ or, equivalently, if and only if $E\left [ h(\widetilde {\alpha }_{2},y,z,t) \right ]$ is larger than $E [ h( \tilde {\alpha }_{1},y,z,t) ]$ for all $(y,z)\in \mathbb {R}_{+}^{2}$ and for t ∈ T(y) where h(α, y, z, t) = ρ _y U ^{(1, 0)} (y + t ρ _y , z + t ρ _z α) + α ρ _z U ^{(0, 1)} (y + t ρ _y , z + t ρ _z α). By Lemma 1, this last inequality is satisfied for every pair $(\widetilde {\alpha }_{1},\widetilde {\alpha }_{2})$ such that $\widetilde {\alpha }_{2}$ is an increase in Nth-degree risk over $\widetilde {\alpha }_{1}$ if and only if $(-1)^{N}\frac {\partial ^{N}h}{\partial \alpha ^{N}}\geq 0.$

Simple calculus gives $\frac {\partial ^{N}h}{\partial \alpha ^{N}}=t^{N}\rho _{y}\rho _{z}^{N}U^{(1,N)}+\alpha t^{N}\rho _{z}^{N+1}U^{\left ( 0,N+1\right ) }+Nt^{N-1}\rho _{z}^{N}U^{( 0,N) }$ where derivatives of U are taken at (y + t ρ _y , z + α t ρ _z ). Since t ≥ 0 and ρ _z ≥ 0, our necessary and sufficient condition can then be rewritten as ( − 1)^N t ρ _y U ^{(1, N)} + α t ρ _z U ^{(0, N + 1)} + N U ^{(0, N)} ≥ 0 for all $(\alpha ,y,z)\in \mathbb {R}_{+}^{3}\$and for t ∈ T(y). Denoting y + t ρ _y by Y and z + α t ρ _z by Z, our necessary and sufficient condition is equivalent to ( − 1)^N(Y − y)U ^{(1, N)} + (Z − z)U ^{(0, N + 1)} + N U ^{(0, N)} ≥ 0 for all $(Y,Z)\in \mathbb {R}_{+}^{2}$ and all (y, z) such that y ≥ Y and 0 ≤ z ≤ Z and where the derivatives are taken at (Y, Z). Taking successively y = Y and z = Z, y = Y and z = 0 we obtain that (− 1)^N U ^{(0, N)} ≥ 0 and ( − 1)^N Z U ^{(0, N + 1)} + N U ^{(0, N)} ≥ 0. Letting y go to ∞, we obtain ( − 1)^N U ^{(1, N)} ≤ 0. Conversely, if these 3 conditions are satisfied and for a given y ≥ Y, we have ( − 1)^N(Y − y)U ^{(1, N)} ≥ 0 and

$$\begin{array}{@{}rcl@{}} &&(-1)^{N}\left((Z-z)U^{(0,N+1)}+NU^{(0,N)}\right)\\ &=&\frac{Z-z}{Z}(-1)^{N}\left(ZU^{(0,N+1)}+NU^{(0,N)}\right)+\frac{z}{Z}(-1)^{N}NU^{(0,N)}\\ &\geq &0 \end{array} $$

which gives that our necessary and sufficient condition is satisfied.

A far as Nth-degree multiplicative-risk aversion is concerned, it is characterized by the fact that $E\left [ U\left ( y+t\rho _{y},z+t\rho _{z} \widetilde {\alpha }_{2}\right ) \right ] -E\left [ U\left ( y+t\rho _{y},z+t\rho _{z}\widetilde {\alpha }_{1}\right ) \right ]$ is decreasing on $T(y)=\left \{ t\in \mathbb {R}_{+}:y+t\rho _{y}\geq 0\right \}$ or, equivalently, by the fact that $E[ -h( \widetilde {\alpha }_{2},y,z,t) ]$ is larger than $E[ -h( \tilde {\alpha }_{1},y,z,t) ]$ for all $(y,z)\in \mathbb {R}_{+}^{2}$ and for t ∈ T(y) or, finally, by $( -1)^{N+1}\frac {\partial ^{N}h}{\partial \alpha ^{N}}\geq 0.$ The rest of the proof is then directly adapted from the multiplicative-risk attraction setting.

Appendix B: Nth-degree stochastic dominance

In this appendix we generalize Lemma 1, Propositions 1 and 2, and Corollary 1. We begin by formalizing the concept of Nth-degree stochastic dominance.

Definition 4 (Jean)

Let $\widetilde {\alpha }_{1}$ and $\widetilde {\alpha }_{2}$ denote two random variables with values in [0, B]. We say that $\widetilde {\alpha }_{2}$ is dominated in the sense of Nth-degree stochastic dominance by $\widetilde {\alpha }_{1},$ and we denote it by $\widetilde { \alpha }_{2}\succcurlyeq _{NSD}\widetilde {\alpha }_{1},$ if $F_{\widetilde { \alpha }_{2}}^{[ N] }(x)\geq F_{\widetilde {\alpha }_{1}}^{\left [ N \right ] }(x)$ for all x ∈ [0, B] where the inequality is strict for some x and $F_{\widetilde {\alpha }_{2}}^{[ k] }(B)\geq F_{\widetilde {\alpha }_{1}}^{[ k] }(B)$ for k = 1, . . . , N − 1.

Jean (1980) characterizes Nth-degree stochastic dominance: He establishes that $\widetilde {\alpha }_{2}$ is dominated in the sense of Nth-degree stochastic dominance by $\widetilde {\alpha }_{1}$ if and only if $E\left [ q( \widetilde {\alpha }_{2})\right ] >E[ q(\widetilde {\alpha }_{1})]$ for all N times continuously differentiable real valued function q such that ( − 1)^k q ^(k) > 0 for k = 1, . . . , N where $q^{(k)}=\frac { d^{k}q}{d\alpha ^{k}}.$ The following Lemma characterizes the set of N times continuously differentiable functions for which $E\left [ q(\widetilde { \alpha }_{2})\right ] \geq E[ q(\widetilde {\alpha }_{1})]$ for all pair $(\widetilde {\alpha }_{1},\widetilde {\alpha }_{2})$ where $\widetilde {\alpha }_{2}\succcurlyeq _{NSD}\widetilde {\alpha }_{1}.$

Lemma 2

Let q be a given real valued function that is N times continuously differentiable on ℝ₊. The following are equivalent.

• 1. For all pair $(\widetilde {\alpha }_{1},\widetilde {\alpha }_{2})$ such that $\widetilde {\alpha }_{2}\succcurlyeq _{NSD}\widetilde {\alpha }_{1},$ we have $E[ q(\widetilde {\alpha }_{2})] \geq E\left [ q(\widetilde { \alpha }_{1})\right ] .$

• 2. For all x ≥ 0, we have ( − 1)^k q ^(k) ≥ 0 for k = 1, . . . , N.

Proof

The fact that 2. implies 1. results directly from Jean (1980). Let us now prove that 1. implies 2. Let q be an N times continuously differentiable function on $\mathbb {R}_{+}$ such that $E\left [ q(\widetilde {\alpha }_{2})\right ] \geq E\left [ q(\widetilde {\alpha }_{1}) \right ]$ for all pair $(\widetilde {\alpha }_{1},\widetilde {\alpha }_{2})$ such that $\widetilde {\alpha }_{2}\succcurlyeq _{NSD}\widetilde {\alpha } _{1}.$ Let k be in {1, . . . , N} and let $(\widetilde {\alpha }_{1},\widetilde {\alpha }_{2})$ such that $\widetilde {\alpha } _{2}\succcurlyeq _{k}\widetilde {\alpha }_{1}.$ By definition, we have $F_{ \widetilde {\alpha }_{2}}^{[ k] }(x)\geq F_{\widetilde {\alpha } _{1}}^{[ k] }(x)$ for all x ∈ [0, B] where the inequality is strict for some x and $F_{\widetilde {\alpha }_{2}}^{\left [ i \right ] }(B)=F_{\widetilde {\alpha }_{1}}^{[ i] }(B)$ for i = 1, . . . , k − 1. By integration, we obtain $F_{\widetilde {\alpha }_{2}}^{\left [ j\right ] }(x)\geq F_{\widetilde {\alpha }_{1}}^{[ j] }(x)$ for all j ≥ k and all x ∈ [0, B] and, in particular, $\widetilde {\alpha }_{2}\succcurlyeq _{NSD}\widetilde {\alpha }_{1}$ hence $E [ q(\widetilde {\alpha }_{2})] \geq E\left [ q(\widetilde {\alpha } _{1})\right ] .$ By Lemma 2 we have ( − 1)^k q ^(k) ≥ 0 on 0, B for k = 1, . . . , N. □

The following Proposition establishes the effect on optimal choice of an Nth-degree stochastically dominated shift on the initial endowment α.

Proposition 5

Let U be a given increasing, strictly concave and infinitely differentiable utility function satisfying Assumption A1. Let us consider p and μ as given. The following properties are equivalent:

• 1. For all initial endowment $(K,\widetilde {\alpha }_{1})$, a switch from $\widetilde {\alpha }_{1}$ to $\widetilde {\alpha }_{2}$ such that $\widetilde {\alpha }_{2}\succcurlyeq _{NSD}\widetilde {\alpha }_{1}$ increases the optimal level of the choice variable, i.e., $x_{2}^{\ast }\geq x_{1}^{\ast }$.

• 2. For all (y, z), we have ( − 1)^k (− p U ^{(1, k)}(y, z) + μ U ^{(0, k + 1)}(y, z)) ≥ 0 for k = 1, . . . , N.

Proof

Let us prove that 2. implies 1. Let $\left ( \widetilde { \alpha }_{1},\widetilde {\alpha }_{2}\right )$ be such that $\widetilde { \alpha }_{2}\succcurlyeq _{NSD}\widetilde {\alpha }_{1}$ and let $q(\alpha )=g(x_{1}^{\ast },\alpha ,\mu ).$ By 2., we have ( − 1)^k q ^(k)(α) ≥ 0 for k = 1, . . . , N. By Lemma 2, this leads to $E[ q(\widetilde {\alpha }_{1})] \leq E\left [ q( \widetilde {\alpha }_{2})\right ] .$ By definition, we have $E\left [ q( \widetilde {\alpha }_{1})\right ] =0,$ which gives $E\left [ q(\widetilde { \alpha }_{2})\right ] \geq 0$ or $E\left [ g(x_{1}^{\ast },\widetilde {\alpha } _{2},\mu )\right ] \geq 0.$ By concavity of U, it is easy to check that $E [ g(x,\widetilde {\alpha }_{2},\mu )]$ is a decreasing function of x. Since $x_{2}^{\ast }$ is characterized by $E\left [ g(x_{2}^{\ast }, \widetilde {\alpha }_{2},\mu )\right ] =0,$ we obtain that $x_{2}^{\ast }\geq x_{1}^{\ast }.$ □

Let us prove 1. implies 2. As in the proof of the Lemma 2, it suffices to remark that, for k = 1, . . . , N, $\widetilde {\alpha } _{2}\succcurlyeq _{k}\widetilde {\alpha }_{1}$ implies $\widetilde {\alpha } _{2}\succcurlyeq _{NSD}\widetilde {\alpha }_{1}.$

Finally, the following Proposition establishes the effect on optimal choice of an Nth-degree stochastically dominated shift on the multiplicative variable μ.

Proposition 6

Let U be a given increasing, strictly concave and infinitely differentiable utility function satisfying Assumption A2. The following properties are equivalent:

• 1. For all initial endowment (K, α) and all asset cost and payoff $(p,\widetilde {\mu }_{1})$ such that $x_{1}^{\ast }\geq 0$, and any switch from $\widetilde {\mu }_{1}$ to $\widetilde {\mu }_{2}$ with $\widetilde {\mu } _{2}\succeq _{NSD}\widetilde {\mu }_{1}$ increases (decreases) the optimal level of the choice variable, i.e., $x_{2}^{\ast }\geq x_{1}^{\ast }$.

• 2. For all (y, z), we have ( − 1)^k U ^{1, k}(y, z) ≤ 0(r e s p. ≥ 0)R ^k(y, z) ≤ N and( − 1)^k U ^{0, k}(y, z) ≥ 0(r e s p. ≤ 0) for k = 1, . . . , N.

Proof

Follows the proof of Proposition 5. □