# A Characterization of Bounded Stochastic Dominance

Bar Light and Andres Perlroth Microsoft Research, NYC, USA. e-mail:
<EMAIL_ADDRESS>Google Research, CA, USA. e-mail<EMAIL_ADDRESS>
Abstract:

> Stochastic dominance orders are commonly used in the theory and practice of
> risk management and decision making under uncertainty. We provide a novel
> characterization for the $n$-th degree bounded stochastic dominance order
> which compares the $n$-th lower partial moment of the random variables under
> consideration on a bounded domain. This characterization establishes a
> connection between the risk tolerance of decision makers and bounded
> stochastic dominance orders, and hence, it provides a decision theoretic
> interpretation for these stochastic orders. As a by-product of our results,
> we provide non-trivial inequalities for $n$-convex functions and rankings
> over lotteries for decision makers with a globally bounded Arrow-Pratt
> measure of risk aversion.

Keywords: stochastic orders; lower partial moments; risk aversion.

##  1 Introduction

Stochastic orders are binary relations over random variables. They are used in
a broad array of applications including portfolio optimization, actuarial
sciences, and decision making under uncertainty. An important class of
stochastic orders that has received a significant attention from both
practitioners and researchers uses the lower partial moments (LPM) of the
random variables under consideration to determine the ordering between them
(Shaked and Shanthikumar, 2007). These stochastic orders enjoy well-known
desired theoretical properties (e.g., Bawa (1975), Fishburn (1976), Fishburn
(1980)) and computational properties (e.g., Dentcheva and Ruszczynski (2003),
Dentcheva et al. (2016), Chen et al. (2011), Post and Kopa (2017)).

The $n$-th LPM of a random variable with a distribution function $F$ at a
point $c$ is given by
$LPM_{n,c}(F)=\int_{-\infty}^{c}\max\\{c-x,0\\}^{n}dF(x)$. The LPM at a point
$c$ measures the random variable’s downside risk in the sense that only values
below the threshold $c$ are taken into consideration. Three different
stochastic orders based on the LPM were introduced and studied in the
literature. The most common between them is the $n$-th degree stochastic
dominance. For a positive integer $n$, we say that $F$ dominates $G$ in the
$n$-th degree stochastic dominance if $LPM_{n-1,c}(F)\geq LPM_{n-1,c}(G)$ for
all $c\in\mathbb{R}$ (Shaked and Shanthikumar, 2007). A second stochastic
order that is typically called the LPM risk measure considers the inequality
$LPM_{n-1,c}(F)\geq LPM_{n-1,c}(G)$ for a particular $c$ (Shapiro et al.,
2021). Finally, a third stochastic order that is called bounded stochastic
dominance considers the inequality $LPM_{n-1,c}(F)\geq LPM_{n-1,c}(G)$ for all
$c$ in some bounded interval $[a,b]$ (Fishburn, 1976). The main contribution
of this paper is to provide a novel decision theoretic characterization of the
bounded stochastic dominance orders.

A key advantage of the $n$-th degree stochastic dominance orders is that a
characterization of this stochastic order in terms of its generator is well-
known. Recall that a generator for a stochastic order $\succeq$ is a set of
utility functions $D$ such that $F\succeq G$ implies that every decision maker
that belongs to $D$ prefers $F$ to $G$, i.e., $\int u(x)dF(x)\geq\int
u(x)dG(x)$. The generator for the $n$-th degree stochastic dominance orders
consists of all the risk averse decision makers (i.e., decision makers with a
concave utility function) for $n=2$, all the risk averse and prudent decision
makers (i.e., decision makers with a concave utility function and a convex
marginal utility function) for $n=3$, and higher orders risk averse decision
makers for a larger $n$ (see Shaked and Shanthikumar (2007)). Because these
sets of decision makers are appealing from a decision theory point of view,
this characterization provides a decision theoretic motivation for using the
$n$-th degree stochastic dominance orders in applications. One limitation of
these stochastic orders is that they can be too strong in the sense that they
cannot compare many random variables of interest. On the other hand, the LPM
risk measure compares any two random variables. A limitation of this
stochastic order is that the choice of the threshold $c$ is typically somewhat
arbitrary. The bounded stochastic dominance orders can be seen as a middle
ground between the LPM risk measure and the $n$-th degree stochastic
dominance. The domain restriction when studying bounded stochastic dominance
orders has important theoretical consequences as it induces a new generator
for these stochastic orders that is not equal to the generator for the
standard $n$-th degree stochastic dominance orders (this fact was noted in
Fishburn (1976) and Fishburn (1980)). A generator for the bounded stochastic
dominance order that has a decision theoretic motivation is not known.

Our main result characterizes the bounded stochastic dominance orders by
providing a generator for these orders that has a decision theoretic
interpretation and is closely related to the Arrow-Pratt risk aversion measure
and the “$n$-convexity” of the utility functions that belong to the
generator.111A non-negative function $u$ is $n$-convex if $u^{1/n}$ is a
convex function We show that the generator for the $n+1$-th degree bounded
stochastic dominance consists of the utility functions $u$ satisfying (i) the
changing derivative property: the $k$-derivative satisfies
$(-1)^{k}u^{(k)}\leq 0$ for $k=1,\ldots,n+1$ and (ii) $u$ is $n,[a,b]$ convex,
i.e., $(u(b)-u(x))^{1/n}$ is a convex function on $[a,b]$. Alternatively, (ii)
can be replaced by the following condition (ii’) the Arrow-Pratt measure of
risk aversion $-u^{(2)}(x)/u^{(1)}(x)$ is bounded below by $(n-1)/(b-x)$ for
$x\in(a,b)$. The changing derivative property is a well-desired property from
a decision theoretic point of view and it characterizes the standard
$n$-degree stochastic dominance as we mentioned above. Condition (ii) and
(ii’) provide restrictions on the risk tolerance of the decision makers.
Intuitively, these restrictions imply that decision makers that belong to the
generator exhibit a minimal amount of risk aversion. We show that this
generator is essentially the maximal generator (Müller, 1997) of these
stochastic orders.

Our characterization has important consequences. First, it fills a theoretical
gap in the theory of stochastic dominance orders by providing decision
theoretic based generators for bounded stochastic dominance orders which
motivate the use of these stochastic orders in applications. Second, in
practice, because of computational considerations, to determine if $F$
dominates $G$ in the $n$-th degree stochastic dominance one restricts the
dominance relation to some bounded domain $[a,b]$ (see Shapiro et al. (2021)).
Our results shed light on how the set of decision makers under consideration
depends on this restriction as the generator we provide directly depends on
the interval $[a,b]$. Third, our result implies an equivalence between the
notion of $n$-convexity to functions with lower bounded Arrow-Pratt risk
aversion measure. We leverage this equivalence to derive novel inequalities
for $n$-convex functions such as Jensen-type inequalities. Finally, our
results can be used to show how decision makers with globally bounded from
below Arrow-Pratt risk aversion measure rank non-trivial lotteries.

##  2 Main Results

In this section we provide our main result. We first introduce some notations.

Let $C^{n}([a,b])$ be the set of all $n$ times continuously differentiable
functions $u:[a,b]\rightarrow\mathbb{R}$. For $k\geq 1$, we denote by
$u^{(k)}$ the $k$th derivative of a function $u$ and for $k=0$ we define
$u^{(0)}:=u$. As usual, the derivatives at the extreme points $u^{(k)}(a)$ and
$u^{(k)}(b)$ are defined by taking the left-side and right-side limits,
respectively. For a non-negative integer $n$ and real numbers $a<b$ we define
the following sets of utility functions that have the changing derivative
property:

${U}_{n,[a,b]}:=\\{u\in C^{n+1}([a,b]):\>(-1)^{k}u^{(k)}(x)\leq 0\text{
}\forall x\in[a,b]\text{ }\;\forall k=1,\ldots,n+1\\}.$

The sets of utility functions ${U}_{n,[a,b]}$ play an important rule in the
theory of stochastic orders (for example, see Hadar and Russell (1969),
Whitmore (1970), Menezes et al. (1980), and Ekern (1980)). Note that the set
$U_{1,[a,b]}$ consists of all risk averse decision makers and the set
$U_{2,[a,b]}$ consists of all risk averse decision makers that are also
prudent (Kimball, 1990). Stochastic orders that are generated by the sets
${U}_{n,[a,b]}$ have received a significant attention in the literature and
are typically called the $n$-th degree stochastic dominance orders (Müller and
Stoyan, 2002; Shaked and Shanthikumar, 2007). The utility functions that
belong to the sets $U_{n,[a,b]}$ have desired well-known theoretical
properties, and hence, stochastic orders that are generated by these sets are
well motivated from a decision theory point of view. In addition, there is a
known connection between these stochastic orders to the lower partial moments
(LPM) of the random variables under consideration (see Proposition 1 below).
This connection is fundamental for practical approaches that are used to
determine if a random variable dominates another random variable under these
stochastic orders (see Post and Kopa (2017) and Fang and Post (2022)). We now
define formally the $n$-th degree stochastic dominance.

###### Definition 1

Let $F$ and $G$ be two distribution functions on $[a,b]$ and let $n$ be a
positive integer. We say that $F$ dominates $G$ in $n$-th degree stochastic
dominance order and write $F\succeq_{n}G$, if and only if for all
$c\in\mathbb{R}$, we have

$\int_{-\infty}^{c}\max\\{c-x,0\\}^{n-1}dF(x)\geq\int_{-\infty}^{c}\max\\{c-x,0\\}^{n-1}dG(x)\;.$
(1)

For a proof of the following characterization of the $n$-th degree stochastic
dominance see, for example, Shaked and Shanthikumar (2007).

###### Proposition 1

Let $F$ and $G$ be two distributions over $[a,b]$ and let $n$ be a positive
integer. Then $F\succeq_{n}G$ if and only if

$\int_{a}^{b}u(x)dG(x)\geq\int_{a}^{b}u(x)dF(x)$

for all $u\in U_{n-1,[a,b]}$.

As we discussed in the introduction, the $n$-th degree stochastic dominance
can be too strong in the sense that many random variables cannot be compared
by these stochastic orders. An alternative approach is to use a“restricted”
$n$-th degree stochastic dominance where inequality (1) is required to hold
only for $c\in[a,b]$ for some bounded interval $[a,b]$. These stochastic
orders were introduced by Fishburn (1976) and are called bounded stochastic
dominance orders. We now define formally bounded stochastic dominance (BSD).

###### Definition 2

Let $F$ and $G$ be two distribution functions on $[a,b]$ and let $n$ be a
positive integer. We say that $F$ dominates $G$ in the $n$-th degree bounded
stochastic dominance order and write $F\succeq_{n,[a,b]-BSD}G$, if and only if
for all $c\in[a,b]$, we have

$\int_{a}^{b}\max\\{c-x,0\\}^{n}dF(x)\geq\int_{a}^{b}\max\\{c-x,0\\}^{n}dG(x)\;.$
(2)

Clearly stochastic dominance implies bounded stochastic dominance. Notice
that, given Proposition 1, a generator of the $n$-th degree bounded stochastic
dominance can be given by a set of decisions makers $U_{n,[a,b]}\cap
W_{n,[a,b]}$ for some set $W_{n,[a,b]}$. Our main result provides the set
$W_{n,[a,b]}$ for each $n$ (see Theorem 1). We show that these sets contain
“very” risk averse decision makers in the sense that their utility functions
satisfy curvature conditions that are stronger than concavity. These curvature
conditions are related to the Arrow-Pratt risk aversion measure and the
$n$-convexity of the utility functions. We now introduce these sets.

We first discuss the set of $n,[a,b]$-concave functions that was studied in
Light and Perlroth (2021). We say that a function is $n,[a,b]$-concave if it
belongs to the set $LP_{n,[a,b]}$ where

${LP}_{n,[a,b]}:=\\{u:[a,b]\rightarrow\mathbb{R}:\left(u(b)-u(x)\right)^{1/n}\text{
is convex and and decreasing on }[a,b]\\},$

i.e., the function $u(b)-u(x)$ is decreasing and $n$-convex on $[a,b]$. For a
twice continuously differentiable concave and strictly increasing function $u$
with the normalization $u(b)=0$, we have $u\in LP_{n,[a,b]}$ if and only if

$\frac{\partial\ln(u^{(1)}(x))}{\partial\ln(u(x))}=\frac{u(x)u^{(2)}(x)}{\left(u^{(1)}(x)\right)^{2}}\geq\frac{n-1}{n}$

for all $x\in(a,b)$, i.e., the elasticity of the marginal utility function
with respect to the utility function is bounded below by $(n-1)/n$. The
elasticity of $u^{(1)}$ with respect to $u$ is a natural measure of the
concavity of $u$ and has an economic interpretation in terms of the trade-off
between risk and reward. Light and Perlroth (2021) study the properties of the
functions that belong to the set $LP_{n,[a,b]}$ and the stochastic orders that
are generated by these functions.

The second set we introduce is a set of decision makers that have a bounded
from below Arrow-Pratt measure of risk aversion. Recall that the Arrow-Pratt
measure of risk aversion is given by $-u^{(2)}/u^{(1)}=-\partial\ln(u^{(1)})$.

${AP}_{n,[a,b]}:=\\{u\in C^{2}([a,b]):u^{(1)}(x)\geq 0,\>u^{(2)}(x)\leq
0,\>(n-1)u^{(1)}(x)+u^{(2)}(x)(b-x)\leq 0\text{ }\;\forall x\in[a,b]\\}.$

For a twice continuously differentiable concave and strictly increasing
function $u$ we have $u\in AP_{n,[a,b]}$ if and only if

$-\partial\ln(u^{(1)}(x))=-\frac{u^{(2)}(x)}{u^{(1)}(x)}\geq\frac{n-1}{b-x}$

for all $x\in(a,b)$, i.e., the Arrow-Pratt measure of risk aversion is bounded
below by $(n-1)/(b-x)$. The Arrow-Pratt measure of risk aversion (Pratt, 1964)
has many well-known desired theoretical properties and is widely used in the
decision theory literature to measure the risk tolerance of decision makers.
Meyer (1977a) and Meyer (1977b) study the stochastic orders that are generated
by sets of decision makers that have bounded Arrow-Pratt measure of risk
aversion.

One disadvantage of the sets $AP_{n,[a,b]}$ and $LP_{n,[a,b]}$ is that they
are not a subset of $U_{n,[a,b]}$. For example, for $n=2$, the sets consist of
risk averse decision makers that might not be averse to downside risk
(prudent) which is a desired property in many applications of interest
(Kimball, 1990). Another disadvantage is that the stochastic orders that are
generated by these sets are not easy to characterize (see Meyer (1977a) and
Light and Perlroth (2021)). On the other hand, the sets $U_{n,[a,b]}\cap
LP_{n,[a,b]}$ or $U_{n,[a,b]}\cap AP_{n,[a,b]}$ include risk averse decision
makers that exhibit higher orders of risk aversion (e.g., prudence). For
example, for $n=2$ the set $U_{n,[a,b]}\cap AP_{n,[a,b]}$ consists of all risk
averse and prudent decision makers such that their Arrow-Prat measure of risk
aversion at a point $x$ is at least $1/(b-x)$. Our main result shows that the
stochastic orders generated by these sets have a simple characterization in
terms of lower partial moments. Theorem 1 shows that for every positive
integer $n$ these two sets are actually equal, i.e., $U_{n,[a,b]}\cap
LP_{n,[a,b]}=U_{n,[a,b]}\cap AP_{n,[a,b]}$ and that these sets generate the
$n$-th degree bounded stochastic dominance order.

The proof of Theorem 1 is provided in Section 4.

###### Theorem 1

Let $F$ and $G$ be two distributions over $[a,b]$ for some $a<b<\infty$ and
let $n$ be an integer. Then $F\succeq_{n,[a,b]-BSD}G$ if and only if

$\int_{a}^{b}u(x)dG(x)\geq\int_{a}^{b}u(x)dF(x)$ (3)

for all $u\in U_{n,[a,b]}\cap LP_{n,[a,b]}$ if and only if inequality (3)
holds for all $u\in U_{n,[a,b]}\cap AP_{n,[a,b]}$. In particular, we have
$U_{n,[a,b]}\cap AP_{n,[a,b]}=U_{n,[a,b]}\cap LP_{n,[a,b]}$.

Under suitable integrability conditions, Theorem 1 also holds for the case
that $a=-\infty$. On the other hand, when $b=\infty$ the bounded stochastic
order reduces to the standard unbounded stochastic dominance order (see
Proposition 1 above).

Theorem 1 shows that $U_{n,[a,b]}\cap AP_{n,[a,b]}=U_{n,[a,b]}\cap
LP_{n,[a,b]}$. One may wonder whether $AP_{n,[a,b]}=LP_{n,[a,b]}$. The next
proposition shows that this is not the case. The proof is deferred to the
Appendix.

###### Proposition 2

Let $a<b$ and $n\geq 2$. We have $AP_{n,[a,b]}\nsubseteq LP_{n,[a,b]}$.

###  2.1 The Maximal Generator of Bounded Stochastic Dominance

Theorem 1 provides a generator for the bounded stochastic dominance orders. In
this section we discuss the maximal generator of these stochastic orders.
Formally, define $F\succeq_{\mathfrak{F}}G$ if

$\int_{a}^{b}u(x)dF(x)\geq\int_{a}^{b}u(x)dG(x)$ (4)

for all $u\in\mathfrak{F}$ where $\mathfrak{F}$ is a subset of the set of all
bounded utility functions on $[a,b]$. The maximal generator $R_{\mathfrak{F}}$
of the stochastic order $\succeq_{\mathfrak{F}}$ is the set of all functions
$u$ with the property that $F\succeq_{\mathfrak{F}}G$ implies inequality (4).

When using a stochastic order in decision theory contexts, it is useful to
characterize the maximal generator of the stochastic order. If the maximal
generator is not known, it is not clear which decision makers are under
consideration when deciding if a random variable dominates another random
variable. Müller (1997) characterizes the properties of maximal generators. We
use the results in Müller (1997) to identify the maximal generator of the
bounded stochastic dominance orders. The following result follows from
Corollary 3.8 in Müller (1997).

###### Proposition 3

Suppose that $\mathfrak{F}$ is a convex cone containing the constant functions
and closed under pointwise convergence. Then $R_{\mathfrak{F}}=\mathfrak{F}$.

It is easy to see that $U_{n,[a,b]}$ is a convex cone that contains the
constant functions. The closure of $U_{n,[a,b]}$ denoted by $cl(U_{n,[a,b]})$
can be written by divided differences (Denuit et al., 1999). Light and
Perlroth (2021) show that $LP_{n,[a,b]}$ is a convex cone containing the
constant functions and closed under pointwise convergence. Hence, the maximal
generator of the bounded stochastic dominance order $\succeq_{n,[a,b]-BSD}$ is
$cl(U_{n,[a,b]})\cap LP_{n,[a,b]}$.

##  3 Examples and applications

###  3.1 Portfolio optimization

As an example for an implication of our characterization, consider the
following portfolio optimization problem with stochastic dominance constraints
(see Chapter 4 in Shapiro et al. (2021)). There are $n$ random variables
$R_{1},\ldots,R_{n}$ that represent future returns from different investments.
A typical problem in portfolio management is to choose weights
$x_{1},\ldots,x_{k}$ such that $x_{i}\geq 0$ and $\sum_{i=1}^{k}x_{i}=1$ to
maximize the portfolio’s expected return under some constraints that limit the
portfolio’s risk. A popular risk constraint is based on the lower partial
moments where the risk constraint implies that the portfolio
$\sum_{i=1}^{k}R_{i}x_{i}$ dominates a benchmark return $Y$ (e.g., an index
fund) in the sense that $LPM_{n,c}(\sum_{i=1}^{k}R_{i}x_{i})\geq LPM_{n,c}(Y)$
for all $c$ in some set $S$ that could also be a singleton ($LPM_{n,c}$ is the
lower partial moment of order $n$ at a point $c$, see the introduction for the
precise definition). In this case, the portfolio management problem reduces to
the following optimization problem:

$\max_{x\in\\{x\in\mathbb{R}^{k}:\sum_{i}x_{i}=1,\ x_{i}\geq 0\ \forall i\\}}\
\sum_{i=1}^{k}\mathbb{E}(R_{i})x_{i}\ \text{ s.t. }\
LPM_{n,c}\left(\sum_{i=1}^{k}R_{i}x_{i}\right)\geq LPM_{n,c}(Y)\ \forall c\in
S.$ (5)

Two popular choices are $S=\mathbb{R}$ so the constraints correspond to
stochastic dominance constraints and $S=\\{c^{\prime}\\}$ so the constraint
corresponds to the LPM risk measure constraint at a point $c^{\prime}$. Now
assume that the portfolio manager chooses the constraints
$LPM_{n,c}(\sum_{i=1}^{k}R_{i}x_{i})\geq LPM_{n,c}(Y)$ for all $c$ on a chosen
restricted domain $S=[a,b]$. What $LPM_{n,c}(\sum_{i=1}^{k}R_{i}x_{i})\geq
LPM_{n,c}(Y)$ implies under the restricted set? In particular, which decision
makers prefer the portfolio over the benchmark and how this set of decision
makers depends on the restriction of the domain? The characterization of the
bounded stochastic dominance orders we provide in Theorem 1 answers these
questions. We show that $LPM_{n,c}(\sum_{i=1}^{k}R_{i}x_{i})\geq LPM_{n,c}(Y)$
for all $c\in[a,b]$ means that every decision maker that belongs to
$U_{n,[a,b]}\cap AP_{n,[a,b]}=U_{n,[a,b]}\cap LP_{n,[a,b]}$ prefers the
portfolio over the benchmark.

###  3.2 Decision making and the Arrow-Pratt risk aversion measure

For a decision maker with a utility function $u$, Pratt (1964) recognized that
$-u^{(2)}/u^{(1)}$ is a useful measure for the decision maker’s risk aversion
strength. The theory provided in Pratt (1964) focuses on small changes in the
lottery that the decision maker faces and the risk premium that is associated
with that lottery. Theorem 1 provides a connection between the global
boundness of $-u^{(2)}/u^{(1)}$ to the decision maker’s ranking over
lotteries. In particular, $-u^{(2)}(x)/u^{(1)}(x)\geq(n-1)/(b-x)$ for all
$x\in(a,b)$ and $u\in U_{n,[a,b]}$ implies $F\succeq_{n,[a,b]-BSD}G$ for any
two lotteries.

As an example, consider a simple case where $G$ yields $a$ dollars with
probability $\theta^{n}$ and $b$ dollars with probability $1-\theta^{n}$, and
$F$ yields $\theta a+(1-\theta)b$ dollars with probability $1$ for
$\theta\in(0,1)$, $b>a$. Then it easy to check that $F\succeq_{n,[a,b]-BSD}G$.
Hence, Theorem 1 shows that if a decision maker has the changing derivative
property, i.e., $u\in U_{n,[a,b]}$ and a globally bounded Arrow-Pratt risk
aversion measure $-u^{(2)}(x)/u^{(1)}(x)\geq(n-1)/(b-x)$ for all $x\in(a,b)$,
then the decision maker prefers $F$ to $G$. Note that $G$’s expected value is
higher than $F$’s expected value so not every risk averse decision maker
prefers $F$ to $G$. Hence, our results provide non-trivial rankings of random
variables for decision makers that have a globally bounded from below Arrow-
Pratt measure of risk aversion.

Interestingly, a closely related class of utility functions to the popular
class of constant relative risk aversion utility functions
$u_{\gamma}(x)=x^{1-\gamma}/(1-\gamma)$ belongs to $U_{2,[a,b]}\cap
AP_{2,[a,b]}$. It can be easily shown that utility functions of the form

$v_{\gamma}(x)=u_{\gamma}(x)-\frac{x}{b^{\gamma}}$

for $\gamma>0$, $\gamma\neq 1$ and $u_{1}(x)=\log(x)-x/b$ for $\gamma=1$
belong to $U_{2,[a,b]}\cap AP_{2,[a,b]}$. Decision makers with a utility
function $v$ balance risk and excepted value. If the expected values of $F$
and $G$ are the same, they behave exactly as CRRA decision makers, but if the
expected values are different, then both the difference between the expected
values and the behavior of the CRRA decision maker are taken into account when
deciding between $F$ and $G$.

###  3.3 Relation to $n$-convex functions and inequalities

Theorem 1 has non-trivial implications for $n$-convex functions.222The set of
$n$-convex functions has been studied in convex geometry (e.g., Lovász and
Simonovits (1993)) and in economics (e.g., Jensen (2017)). In many
applications in economics, operations research, and mathematics a subset of
convex functions is studied in order to obtain stronger theoretical results
than those that can be proven for convex functions (for example, log-convex
functions, strongly convex functions, and $n$-convex functions). Corollary 1
establishes a useful characterization for $n$-convex functions. We provide a
connection between the convexity of $f(x)^{1/n}$, i.e., the $n$-convexity of
$f$, to the convexity of $f(x^{1/n})$. This Corollary can be used to prove
interesting inequalities for $n,[a,b]$-concave functions. For example,
Corollary 2 establishes a Jensen-type inequality for $n,[a,b]$-concave
functions that is tighter than the standard Jensen’s inequality for concave
functions. The proofs are deferred to the Appendix.

###### Corollary 1

Let $a<b$ and let $n$ be a positive integer. Assume that $f\in U_{n,[a,b]}$.
Then $g_{n}(x):=(f(b)-f(x))^{1/n}$ is convex on $[a,b]$ if and only if
$k_{n}(x):=-f(b-x^{1/n})$ is convex on $[0,(b-a)^{n}]$.

###### Corollary 2

Let $a<b$ and let $n$ be a positive integer. Assume that $f\in U_{n,[a,b]}$
and $(f(b)-f(x))^{1/n}$ is convex on $[a,b]$. Then

$\mathbb{E}f(X)\leq f\left(b-(\mathbb{E}(b-X)^{n})^{1/n}\right)\leq
f(\mathbb{E}(X))$

for every random variable $X$ on $[a,b]$.

##  4 Proof of Theorem 1

Fix an integer $n\geq 1$ and $a<b$. We first define the following set of
functions:

$\mathcal{G}_{n,[a,b]}:=\\{u\in U_{n,[a,b]}:u^{(k)}(b)=0\text{ }\forall
k=1,\ldots,n-1\\}.$

The proof proceeds with the following Lemmas. In Lemma 1 we show that
$\mathcal{G}_{n,[a,b]}=U_{n,[a,b]}\cap AP_{n,[a,b]}$. In Lemma 2 we show that
$\mathcal{G}_{n,[a,b]}=U_{n,[a,b]}\cap LP_{n,[a,b]}$, and hence,
$U_{n,[a,b]}\cap LP_{n,[a,b]}=U_{n,[a,b]}\cap AP_{n,[a,b]}$. Lemma 3 is a
technical Lemma that is needed in order to prove Lemma 4 and Lemma 5. In
Lemmas 4 and 5 we show that $F\succeq_{n,[a,b]-BSD}G$ if and only if

$\int_{a}^{b}u(x)d(F-G)(x)\leq 0\text{ for all }u\in\mathcal{G}_{n,[a,b]}.$

Combining this with Lemma 1 and Lemma 2 completes the proof of Theorem 1.

###### Lemma 1

We have $\mathcal{G}_{n,[a,b]}=U_{n,[a,b]}\cap AP_{n,[a,b]}$.

Proof. The proof is immediate for $n=1$. Assume that $n\geq 2$. Let
$u\in\mathcal{G}_{n,[a,b]}$. We will show that

$(n-1)u^{(1)}(x)+u^{(2)}(x)(b-x)\leq 0.$ (6)

for all $x\in[a,b]$.

Suppose that $n$ is an even number. Note that if $n$ is an even number then
the fact that $u\in\mathcal{G}_{n,[a,b]}$ implies that $u^{(n-1)}$ is convex.
Hence, the derivative of $u^{(n-1)}$ is increasing, i.e., $u^{(n)}$ is
increasing. Using the fact that $u^{(n-1)}(b)=0$ implies that for all
$x\in[a,b]$ we have

$u^{(n-1)}(x)=u^{(n-1)}(b)-\int_{x}^{b}u^{(n)}(t)dt=-\int_{x}^{b}u^{(n)}(t)dt\leq-\int_{x}^{b}u^{(n)}(x)dt=-(b-x)u^{(n)}(x).$
(7)

Similarly, if $n$ is an odd number for all $x\in[a,b]$ we have

$u^{(n-1)}(x)=u^{(n-1)}(b)-\int_{x}^{b}u^{(n)}(t)dt=-\int_{x}^{b}u^{(n)}(t)dt\geq-\int_{x}^{b}u^{(n)}(x)dt=-(b-x)u^{(n)}(x).$
(8)

Thus, if $n=2$ then inequality (6) holds.

To prove that inequality (6) holds for $n\geq 3$, define

$z(x)=(n-1)u^{(1)}(x)+u^{(2)}(x)(b-x)\text{ on }[a,b].$

Note that $z^{(1)}(x)=(n-2)u^{(2)}(x)+u^{(3)}(x)(b-x)$, and

$z^{(k)}(x)=(n-1-k)u^{(k+1)}(x)+u^{(k+2)}(x)(b-x)$

for all $k=0,\ldots,n-1$. Because $u^{(k)}(b)=0$ for all $k=1,\ldots,n-1$, we
have $z^{(k)}(b)=0$ for all $k=0,\ldots,n-2$.

Suppose first that $n$ is an even number. Inequality (7) yields
$z^{(n-2)}(x)\leq 0$ for all $x\in[a,b]$. Thus, $z^{(n-3)}$ is a decreasing
function. Combining this with the fact that $z^{(n-3)}(b)=0$ implies that
$z^{(n-3)}(x)\geq 0$ for all $x\in[a,b]$. Using the same argument, repeatedly,
it follows that $z^{(j)}(x)\leq 0$ for all $x\in[a,b]$ and for all even
numbers $0\leq j\leq n-2$ and $z^{(j)}(x)\geq 0$ for all $x\in[a,b]$ and for
all odd numbers $0\leq j\leq n-2$. In particular, $z^{(0)}(x):=z(x)\leq 0$ for
all $x\in[a,b]$. When $n$ is an odd number, using inequality (8), an analogous
argument to the argument above shows that $z(x)\leq 0$ for all $x\in[a,b]$. We
conclude that inequality (6) holds for every integer $n\geq 1$, i.e., $u\in
U_{n,[a,b]}\cap AP_{n,[a,b]}$.

Now assume that $u\in U_{n,[a,b]}\cap AP_{n,[a,b]}$. Then $z(x)\leq 0$ on
$[a,b]$ which implies that $z(b)=(n-1)u^{(1)}(b)\leq 0$. The fact that
$u^{(1)}\geq 0$ implies $u^{(1)}(b)=0$. Thus, we have $z(b)=0$.

Assume in contradiction that there exists some $j=2,\ldots,n-1$ such that
$u^{(j)}(b)\neq 0$. Then there exists an integer $2\leq m\leq n-1$ that
satisfies $u^{(m)}(b)\neq 0$ and $u^{(k)}(b)=0$ for all $k=1,\ldots,m-1$. We
have $z^{(k)}(b)=0$ for all $k=0,\ldots,m-2$ and
$z^{(m-1)}(b)=u^{(m)}(b)(n-m)<0$ if $m$ is an even number and $z^{(m-1)}(b)>0$
if $m$ is an odd number. Thus, if $m$ is an even number, then the continuity
of $z^{(m-1)}$ implies that $z^{(m-1)}(x)<0$ for all $x\in[b-\epsilon,b]$ for
some $\epsilon>0$. Combining this with the fact that $z^{(m-2)}(b)=0$ implies
that there exists a $\delta>0$ such that $z^{(m-2)}(x)>0$ for all
$x\in(b-\delta,b)$. Using the fact that $z^{(k)}(b)=0$ for all
$k=0,\ldots,m-2$ and the argument above repeatedly show that
$z^{(0)}(x):=z(x)>0$ for some $x$ close to $b$ which is a contradiction to
$z(x)\leq 0$ on $[a,b]$. Similarly, if $m$ is an odd number then
$z^{(m-1)}(b)>0$ and $z^{(m-2)}(b)=0$ imply that $z^{(m-2)}(x)<0$ for all
$x\in(b-\epsilon,b)$ for some $\epsilon>0$ which leads to a contradiction from
the argument above. We conclude that $u^{(k)}(b)=0$ for all $k=1,\ldots,n-1$.
Thus, $u\in\mathcal{G}_{n,[a,b]}$.

###### Lemma 2

We have $\mathcal{G}_{n,[a,b]}=U_{n,[a,b]}\cap LP_{n,[a,b]}$.

Proof. The proof is by induction on $n$. The base case, $n=1$, trivially
holds.

For the sake of exposition, we divide the proof to 9 steps. Step 1 shows that
we can restrict the analysis to a simpler class of functions. Steps 2-4 prove
that $\mathcal{G}_{n,[a,b]}\subseteq U_{n,[a,b]}\cap LP_{n,[a,b]}$. Steps 5-9
prove that $U_{n,[a,b]}\cap LP_{n,[a,b]}\subseteq\mathcal{G}_{n,[a,b]}$.

Step 1: If $\mathcal{G}_{n,[a,b]}=U_{n,[a,b]}\cap LP_{n,[a,b]}$ for functions
$u$ satisfying $|u^{(k)}(x)|>0$ for every $x<b$ and $k=1,\ldots,n$ then
$\mathcal{G}_{n,[a,b]}=U_{n,[a,b]}\cap LP_{n,[a,b]}$.

Suppose that $\mathcal{G}_{n,[a,b]}=U_{n,[a,b]}\cap LP_{n,[a,b]}$ for
functions $u$ satisfying $|u^{(k)}(x)|>0$ for every $x<b$ and $k=1,\ldots,n$.
Let $u\in\mathcal{G}_{n,[a,b]}$ and observe that
$-\epsilon(b-x)^{n}\in\mathcal{G}_{n,[a,b]}$ for every $\epsilon>0$. The set
$\mathcal{G}_{n,[a,b]}$ is convex so the function
$u_{\epsilon}(x)=u(x)-\epsilon(b-x)^{n}$ belongs to the set
$\mathcal{G}_{n,[a,b]}$. Moreover, $|u_{\epsilon}^{(k)}(x)|>0$ for $x<b$ and
$k=1,\ldots,n$. Thus, we conclude that $u_{\epsilon}\in U_{n,[a,b]}\cap
LP_{n,[a,b]}$. Because the set $LP_{n,[a,b]}$ are closed under the pointwise
topology, we have that when $\epsilon\to 0$, $u\in LP_{n,[a,b]}$, and hence,
$u\in U_{n,[a,b]}\cap LP_{n,[a,b]}$. Because $-\epsilon(b-x)^{n}\in
LP_{n,[a,b]}$, a similar argument implies the inclusion $U_{n,[a,b]}\cap
LP_{n,[a,b]}\subseteq\mathcal{G}_{n,[a,bg]}$. Hence,
$\mathcal{G}_{n,[a,b]}=U_{n,[a,b]}\cap LP_{n,[a,b]}$.

Let $u\in\mathcal{G}_{n,[a,b]}$. For the next steps we assume that
$|u^{(k)}(x)|>0$ for every $x<b$ and $k=1,\ldots,n$.

We define the function

$R(x)=\frac{(u(x)-u(b))u^{(2)}(x)}{u^{(1)}(x)^{2}}\text{ on }(a,b).$

Because $u^{(1)}(x)>0$ for $x\in(a,b)$, the function $R$ is well-defined and
differentiable on $(a,b)$. From the differentiability of $u$ it follows that
showing that $u\in LP_{n,[a,b]}$ is equivalent to showing that for all
$x\in(a,b)$ we have

$R(x)=\frac{(u(x)-u(b))u^{(2)}(x)}{u^{(1)}(x)^{2}}\geq\frac{n-1}{n}.$ (9)

Step 2: We claim that

$\displaystyle R^{(1)}(x)=\frac{1}{u^{(1)}(x)^{2}}\bigg{[}$
$\displaystyle-u^{(1)}(x)u^{(2)}(x)\left(\frac{n}{n-1}R(x)-1\right)$
$\displaystyle+(u(x)-u(b))\left(u^{(3)}(x)-\frac{n-2}{n-1}\frac{u^{(2)}(x)^{2}}{u^{(1)}(x)}\right)\bigg{]}$
(10)

for all $x\in(a,b)$.

Indeed, we have

$\displaystyle R^{(1)}(x)$
$\displaystyle=\frac{1}{u^{(1)}(x)^{4}}\bigg{[}u^{(1)}(x)^{3}u^{(2)}(x)+(u(x)-u(b))u^{(1)}(x)^{2}u^{(3)}(x)-2(u(x)-u(b))u^{(1)}(x)u^{(2)}(x)^{2}\bigg{]}$
$\displaystyle=\frac{1}{u^{(1)}(x)^{2}}\bigg{[}u^{(1)}(x)u^{(2)}(x)+(u(x)-u(b))u^{(3)}(x)-2\frac{(u(x)-u(b))u^{(2)}(x)^{2}}{u^{(1)}(x)}\bigg{]}$
$\displaystyle=\frac{1}{u^{(1)}(x)^{2}}\bigg{[}u^{(1)}(x)u^{(2)}(x)\bigg{(}1-2\underbrace{\frac{(u(x)-u(b))u^{(2)}(x)}{u^{(1)}(x)^{2}}}_{R(x)}\bigg{)}+(u(x)-u(b))u^{(3)}(x)\bigg{]}$
$\displaystyle=\frac{1}{u^{(1)}(x)^{2}}\bigg{[}-u^{(1)}(x)u^{(2)}(x)(2R(x)-1)+(u(x)-u(b))u^{(3)}(x)\bigg{]}$
$\displaystyle=\frac{1}{u^{(1)}(x)^{2}}\bigg{[}-u^{(1)}(x)u^{(2)}(x)\left(\frac{n}{n-1}R(x)-1\right)-\frac{n-2}{n-1}\underbrace{u^{(1)}(x)u^{(2)}(x)R(x)}_{\frac{(u(x)-u(b))u^{(2)}(x)^{2}}{u^{(1)}(x)}}+(u(x)-u(b))u^{(3)}(x)\bigg{]}$
$\displaystyle=\frac{1}{u^{(1)}(x)^{2}}\bigg{[}-u^{(1)}(x)u^{(2)}(x)\left(\frac{n}{n-1}R(x)-1\right)+(u(x)-u(b))\left(u^{(3)}(x)-\frac{n-2}{n-1}\frac{u^{(2)}(x)^{2}}{u^{(1)}(x)}\right)\bigg{]}.$

Step 3: Let $R(b):=\liminf_{x\to b^{-}}R(x)$, then we assert that
$R(b)\geq\frac{n-1}{n}$.

First, observe that $\lim_{x\to b^{-}}{(u(x)-u(b))u^{(2)}(x)}=0$ and
$\lim_{x\to b^{-}}u^{(1)}(x)^{2}=u^{(1)}(b)^{2}=0$. Hence,

$\displaystyle R(b)$ $\displaystyle=\liminf_{x\to
b^{-}}\frac{(u(x)-u(b))u^{(2)}(x)}{u^{(1)}(x)^{2}}$
$\displaystyle\geq\liminf_{x\to
b^{-}}\frac{u^{(1)}(x)u^{(2)}(x)+(u(x)-u(b))u^{(3)}(x)}{2u^{(1)}(x)u^{(2)}(x)}$
$\displaystyle=\frac{1}{2}+\liminf_{x\to
b^{-}}\frac{(u(x)-u(b))u^{(3)}(x)}{2u^{(1)}(x)u^{(2)}(x)}$
$\displaystyle=\frac{1}{2}+\liminf_{x\to
b^{-}}\frac{1}{2}\frac{(u(x)-u(b))u^{(2)}(x)}{u^{(1)}(x)^{2}}\frac{u^{(1)}(x)u^{(3)}(x)}{u^{(2)}(x)^{2}}$
$\displaystyle\geq\frac{1}{2}+\frac{1}{2}\liminf_{x\to
b^{-}}\frac{1}{2}\frac{(u(x)-u(b))u^{(2)}(x)}{u^{(1)}(x)^{2}}\liminf_{x\to
b^{-}}\frac{u^{(1)}(x)u^{(3)}(x)}{u^{(2)}(x)^{2}}$
$\displaystyle=\frac{1}{2}+\frac{1}{2}\;R(b)\;\liminf_{x\to
b^{-}}\frac{u^{(1)}(x)u^{(3)}(x)}{u^{(2)}(x)^{2}}.$ (11)

The first inequality follows from the general version of L’Hopital rule. The
second inequality follows using that $\liminf_{x\to
b^{-}}f(x)g(x)\geq\liminf_{x\to b^{-}}f(x)\liminf_{x\to b^{-}}g(x)$ for any
functions $f$ and $g$.

Second, note that the function $v(x)=-u^{(1)}(x)$ is in the set
$\mathcal{G}_{n-1,[a,b]}$. The inductive hypothesis implies that $v\in
LP_{n-1,[a,b]}$. Then, we have that

$\frac{(v(x)-v(b))v^{(2)}(x)}{v^{(1)}(x)^{2}}\geq\frac{n-2}{n-1}.$

Using $v(x)=-u^{(1)}(x)$ and $u^{(1)}(b)=0$, we derive that

$\frac{u^{(1)}(x)u^{(3)}(x)}{u^{(2)}(x)^{2}}\geq\frac{n-2}{n-1}.$

Hence, $\liminf_{x\to
b^{-}}\frac{u^{(1)}(x)u^{(3)}(x)}{u^{(2)}(x)^{2}}\geq\frac{n-2}{n-1}.$

To conclude this step we combine the last inequality with Inequality (11)
obtaining

$R(b)\geq\frac{1}{2}+\frac{1}{2}\;R(b)\;\frac{n-2}{n-1}.$

This simplifies to $R(b)\geq\frac{n-1}{n}$.

Step 4: We claim that $u\in LP_{n,[a,b]}$. That is, for all $x\in(a,b)$ we
have that $R(x)\geq\frac{n-1}{n}$.

Suppose for the sake of contradiction that a $y\in(a,b)$ exists such that
$R(y)<\frac{n-1}{n}$. Let $z=\inf\\{w\in[y,b]|R(w)>R(y)\\}$. Step 3 shows that
$\liminf_{x\to b^{-}}R(x)\geq\frac{n-1}{n}>R(y)$, which in turn, implied that
$z$ is well-defined and $z<b$. Moreover, because $R$ is continuous we have
that $R(z)=R(y)<\frac{n-1}{n}$. From the definition of the infimum we have
that

$\displaystyle\forall\epsilon>0\;\exists w_{\epsilon}\in(z,z+\epsilon)\mbox{
such that }R(w_{\epsilon})>R(y)=R(z).$ (12)

As in Step 3, we take use $v(x)=-u^{(1)}(x)$. Note that $v$ is an increasing
function. From the inductive hypothesis to obtain that

$v^{(2)}(z)\leq\frac{n-2}{n-1}\frac{v^{(1)}(z)^{2}}{v(z)-v(b)}$

Because $v(b)=-u^{(1)}(b)=0$, we conclude that

$u^{(3)}(z)-\frac{n-2}{n-1}\frac{u^{(2)}(z)^{2}}{u^{(1)}(z)}\geq 0.$

Combining the above inequality and $R(z)<\frac{n-1}{n}$ and the facts that $u$
is strictly increasing and strictly concave, we get that

$-u^{(1)}(z)u^{(2)}(z)\left(\frac{n}{n-1}R(z)-1\right)+(u(z)-u(b))\left(u^{(3)}(z)-\frac{n-2}{n-1}\frac{u^{(2)}(z)^{2}}{u^{(1)}(z)}\right)<0$

From Step 2 we conclude that $R^{(1)}(z)<0$. Thus,

$\displaystyle\exists\epsilon>0\mbox{ such that }\forall
w\in(z,z+\epsilon)\text{ we have }R(w)<R(z).$ (13)

conditions (12) and (13) are mutually impossible. Therefore, we conclude that
for every $x\in(a,b)$, $R(x)\geq\frac{n-1}{n}$.

We now show that if $u\in U_{n,[a,b]}\cap LP_{n,[a,b]}$ then
$u\in\mathcal{G}_{n,[a,b]}$.

Step 5: For $k=1,\ldots,n-2$ we have that $u^{(k)}(b)=0$.

It can be shown that $LP_{n,[a,b]}\subseteq LP_{n-1,[a,b]}$ (see for example
Light and Perlroth (2021)). Hence, if $u\in U_{n,[a,b]}\cap LP_{n,[a,b]}$ then
$u\in U_{n,[a,b]}\cap LP_{n-1,[a,b]}\subseteq U_{n-1,[a,b]}\cap
LP_{n-1,[a,b]}$. By the inductive hypothesis we conclude that
$u\in\mathcal{G}_{n-1,[a,b]}$. Thus, $u^{(k)}(b)=0$ for $k=1,\ldots,n-2$.

The next three steps conclude the proof by showing that $u^{(n-1)}(b)=0$.

Step 6: For $k=0,\ldots,n-2$, let

$R_{u^{(k)}}(b)=\limsup_{x\to
b^{-}}\frac{(u^{(k)}(x)-u^{(k)}(b))u^{(k+2)}(x)}{u^{(k+1)}(x)^{2}}.$

We assert that $R_{u^{(k)}}(b)\geq 0$ for $k=0,\ldots,n-2$.

Because $u$ is increasing and concave we have $(u(x)-u(b))u^{(2)}\geq 0$.
Thus, Step 6 holds for $k=0$. For $1\leq k\leq n-2$, because $u^{(k)}(b)=0$
(Step 5), we get that

$(u^{(k)}(x)-u^{(k)}(b))u^{(k+2)}(x)=u^{(k)}(x)u^{(k+2)}(x)=(-1)^{k}u^{(k)}(x)\cdot(-1)^{k+2}u^{(k+2)}(x)\geq
0$

where the last inequality follows because $(-1)^{(k)}u^{k}(x)<0$ and
$(-1)^{(k+2)}u^{k+2}(x)<0$ for every $x\in(a,b)$. We conclude that
$R_{u^{(k)}}(b)\geq 0$.

Step 7: Suppose $n>2$, then for $k=0,\ldots,n-3$ we assert that

$R_{u^{(k)}}(b)\leq\frac{1}{2}\left(1+R_{u^{(k)}}(b)R_{u^{(k+1)}}(b)\right).$
(14)

Step 5 shows that $\lim_{x\to b^{-}}u^{(k+1)}(x)=u^{(k+1)}(b)=0$ and
$\lim_{x\to b^{-}}(u^{(k)}(x)=u^{(k)}(b))$. Now, observe that

$\displaystyle R_{u^{(k)}}(b)$ $\displaystyle=\limsup_{x\to
b^{-}}\frac{(u^{(k)}(x)-u^{(k)}(b))u^{(k+2)}(x)}{u^{(k+1)}(x)^{2}}$
$\displaystyle\leq\limsup_{x\to
b^{-}}\frac{u^{(k+1)}(x)u^{(k+2)}(x)+(u^{(k)}(x)-u^{(k)}(b))u^{(k+3)}(x)}{2u^{(k+1)}(x)u^{(k+2)}(x)}$
$\displaystyle=\frac{1}{2}+\limsup_{x\to
b^{-}}\frac{(u^{(k)}(x)-u^{(k)}(b))u^{(k+3)}(x)}{2u^{(k+1)}(x)u^{(k+2)}(x)}$
$\displaystyle=\frac{1}{2}+\limsup_{x\to
b^{-}}\frac{1}{2}\frac{(u^{(k)}(x)-u^{(k)}(b))u^{(k+2)}(x)}{u^{(k+1)}(x)^{2}}\frac{u^{(k+1)}(x)u^{(k+3)}(x)}{u^{(k+2)}(x)^{2}}$
$\displaystyle\leq\frac{1}{2}+\frac{1}{2}\;R_{u^{(k)}}(b)\;\limsup_{x\to
b^{-}}\frac{u^{(k+1)}(x)u^{(k+3)}(x)}{u^{(k+2)}(x)^{2}}$
$\displaystyle=\frac{1}{2}\left(1+R_{u^{(k)}}(b)R_{u^{(k+1)}}(b)\right).$

The inequalities follow from the same arguments as in Step 3.

Step 8: Consider $n>2$. Then for $k=1,\ldots,n-2$ and a positive integer $m$
satisfying $R_{u^{(k)}}(b)\leq\frac{m-1}{m}$, we have that
$R_{u^{(k-1)}}(b)\leq\frac{m}{m+1}$.

Observe that

$\displaystyle
R_{u^{(k-1)}}(b)\leq\frac{1}{2}\left(1+R_{u^{(k-1)}}(b)R_{u^{(k)}}(b)\right)\leq\frac{1}{2}\left(1+R_{u^{(k-1)}}(b)\frac{m-1}{m}\right).$

The first inequality comes from Step 7. The second inequality holds due to
Step 6 and that $R_{u^{(k)}}(b)\leq\frac{m-1}{m}$.

Hence, a simple algebraic manipulation leads to

$\displaystyle 2mR_{u^{(k-1)}}(b)\leq m+R_{u^{(k-1)}}(b)(m-1)$
$\displaystyle\Longleftrightarrow R_{u^{(k-1)}}(b)\leq\frac{m}{m+1},$

which proves Step 8.

Step 9: We assert that $u\in\mathcal{G}_{n,[a,b]}$.

First, consider the case $n=2$ and suppose for the sake of a contradiction
that $u^{(1)}(b)>0$. The continuity of $u^{(1)}$ implies that a $\delta>0$
exists such that $\lim_{x\to b^{-}}u^{(1)}(x)^{2}>\delta$. Combining this
inequality with

$\lim_{x\to b^{-}}(u(x)-(b))u^{(2)}(x)=\underbrace{\lim_{x\to
b^{-}}(u(x)-u(b))}_{0}\underbrace{\lim_{x\to
b^{-}}u^{(2)}(x)}_{u^{(2)}(b)}=0\;,$

implies that

$R(b)=\lim_{x\to
b^{-}}\frac{(u(x)-u(b))u^{(2)}(x)}{u^{(1)}(x)^{2}}=\frac{\lim_{x\to
b^{-}}(u(x)-u(b))u^{(2)}(x)}{\lim_{x\to b^{-}}u^{(1)}(x)^{2}}=0\;.$

Thus, $u\notin LP_{2,[a,b]}$ which is a contradiction. We conclude that
$u\in\mathcal{G}_{2,[a,b]}$.

Second, consider that $n>2$. We assert that $R_{u^{(n-2)}}(b)>0$. Suppose that
not and $R_{u^{(n-2)}}(b)\leq 0$. Step 6 implies that $R_{u^{(n-2)}}(b)=0$.
Then using Step 8 for $k=n-2$ and $m=1$ we get that
$R_{u^{(n-3)}}(b)\leq\frac{1}{2}$. Reiterating Step 8 $k$ times, we obtain
that $R_{u^{(n-2-k)}}(b)\leq\frac{k}{1+k}$. In particular,
$R_{u^{(0)}}(b)\leq\frac{n-2}{n-1}<\frac{n-1}{n}$. Thus, $u\notin
LP_{2,[a,b]}$ which is a contradiction. Therefore, $R_{u^{(n-2)}}(b)>0$.

To conclude the proof, because $R_{u^{(n-2)}}(b)>0$ and $\lim_{x\to
b^{-}}u^{(n-2)}(x)-u^{(n-2)}(b)=0$, we obtain that $\lim_{x\to
b^{-}}u^{(n-1)}(x)=0$. Thus, $u^{(n-1)}(b)=0$. We conclude from Step 5 that
$u\in\mathcal{G}_{n,[a,b]}$.

###### Lemma 3

Consider a convex and decreasing function $u:[a,b]\to\mathbb{R}$ such that
$u^{(1)}(a)$ exists and is finite. Then there is a sequence of infinitely
differentiable, decreasing and convex functions $u_{n}:[a,b]\to\mathbb{R}$
such that $u_{n}$ converges uniformly to $u$.

Proof. We first extend $u$ to be defined on all $\mathbb{R}$. Define
$\hat{u}:\mathbb{R}\to\mathbb{R}$ by

$\hat{u}(x)=\begin{cases}u(a)+u^{(1)}(a)(x-a)&\mbox{ if }x<a\\\ u(x)&\mbox{ if
}x\in[a,b]\\\ u(b)&\mbox{ if }x>b\end{cases}.$

By construction the restriction of the function $\hat{u}$ to $[a,b]$,
$\hat{u}|_{[a,b]}$, is the function $u$. We claim that $\hat{u}$ is decreasing
and convex.

  * •

Monotonicity: Because $u^{(1)}(a)\leq 0$, we have that $\hat{u}$ is decreasing
for $x<a$. For $x\in[a,b]$, $\hat{u}$ is decreasing because $u$ is decreasing.
For $x>b$, the function is constant. Because $\lim_{x\to a+}\hat{u}(x)=u(a)$
and $\lim_{x\to b^{-}}\hat{u}(x)=u(b)$, we conclude that $\hat{u}$ is
decreasing on $\mathbb{R}$.

  * •

Convexity: Take $x_{1},x_{2}\in\mathbb{R}$, with $x_{1}<x_{2}$, and
$\lambda\in(0,1)$.

For $\lambda x_{1}+(1-\lambda)x_{2}\geq b$, by the monotonicity of $\hat{u}$,
we have that

$\hat{u}(\lambda
x_{1}+(1-\lambda)x_{2})=u(b)=\min_{x\in\mathbb{R}}\hat{u}(x).$

Hence, $\hat{u}(\lambda
x_{1}+(1-\lambda)x_{2})\leq\lambda\hat{u}(x_{1})+(1-\lambda)\hat{u}(x_{2})$.

For $\lambda x_{1}+(1-\lambda)x_{2}\in[a,b]$, we split to three subcases. If
$x_{1},x_{2}\in[a,b]$, we have $\hat{u}(\lambda
x_{1}+(1-\lambda)x_{2})\leq\lambda\hat{u}(x_{1})+(1-\lambda)\hat{u}(x_{2})$
because $u$ is convex.

If $x_{1}<a$ and $x_{2}\in[a,b]$, then $\lambda
x_{1}+(1-\lambda)x_{2}=\hat{\lambda}a+(1-\hat{\lambda})x_{2}$, with
$\hat{\lambda}=\lambda\frac{x_{2}-x_{1}}{x_{2}-a}$.333Because $\lambda
x_{1}+(1-\lambda)x_{2}\in[a,x_{2}]$, we have that $\hat{\lambda}\in[0,1]$. The
convexity of $u$ implies that444Because $u$ is convex and differentiable at
$a$, we have $u^{(1)}(a)(x-a)+u(a)\leq u(x)$ for every $x\in[a,b]$.

$\displaystyle\hat{u}(x_{2})-\hat{u}(a)\geq u^{(1)}(a)(x_{2}-a)$
$\displaystyle\Longleftrightarrow$
$\displaystyle\underbrace{(x_{1}-a)}_{(x_{1}-x_{2})+(x_{2}-a)}(\hat{u}(x_{2})-\hat{u}(a))\leq(x_{1}-a)\quad
u^{(1)}(a)(x_{2}-a)$ $\displaystyle\Longleftrightarrow$
$\displaystyle(x_{1}-x_{2})(\hat{u}(x_{2})-\hat{u}(a))\leq(x_{2}-a)(\underbrace{u^{(1)}(a)(x_{1}-a)+\hat{u}(a)}_{\hat{u}(x_{1})}-\hat{u}(x_{2}))$
$\displaystyle\Longleftrightarrow$
$\displaystyle\frac{x_{2}-x_{1}}{x_{2}-a}(\hat{u}(a)-\hat{u}(x_{2}))\leq\hat{u}(x_{1})-\hat{u}(x_{2})\qquad$
$\displaystyle\Longleftrightarrow$
$\displaystyle\underbrace{\lambda\;\frac{x_{2}-x_{1}}{x_{2}-a}}_{\hat{\lambda}}\;(\hat{u}(a)-\hat{u}(x_{2}))\leq\lambda\;(\hat{u}(x_{1})-\hat{u}(x_{2}))\;$
$\displaystyle\Longleftrightarrow$
$\displaystyle\hat{\lambda}\hat{u}(a)+(1-\hat{\lambda})\hat{u}(x_{2})\leq\lambda\hat{u}(x_{1})+(1-\lambda)\hat{u}(x_{2})\;.$

Using that $\hat{u}$ is convex over $[a,b]$, we obtain

$\displaystyle\hat{u}(\lambda x_{1}+(1-\lambda)x_{2})$
$\displaystyle=\hat{u}(\hat{\lambda}a+(1-\hat{\lambda})x_{2})\leq\hat{\lambda}\hat{u}(a)+(1-\hat{\lambda})\hat{u}(x_{2})$

which proves subcase (ii).

Finally, if $x_{2}>b$ we have

$\hat{u}(\lambda x_{1}+(1-\lambda)x_{2})\leq\hat{u}(\lambda
x_{1}+(1-\lambda)b)\leq\lambda\hat{u}(x_{1})+(1-\lambda)\hat{u}(b)=\lambda\hat{u}(x_{1})+(1-\lambda)\hat{u}(x_{2})$

which proves subcase (iii). The first inequality follows because $\hat{u}$ is
decreasing. The second inequality follows from subcases (i) and (ii).

For, $\lambda x_{1}+(1-\lambda)x_{2}<a$ we claim that

$\hat{u}(\lambda
x_{1}+(1-\lambda)x_{2})=\lambda\hat{u}(x_{1})+(1-\lambda)(u^{(1)}(a)(x_{2}-a)+u(a))\leq\lambda\hat{u}(x_{1})+(1-\lambda)\hat{u}(x_{2})\;.$

Thus, we need to show that $u^{(1)}(a)(x_{2}-a)+u(a)\leq\hat{u}(x_{2})$. If
$x_{2}<a$, the inequality holds from the definition of $\hat{u}$. If
$x_{2}\in[a,b]$ the inequality holds from the convexity of $u$. If $x_{2}>b$
we have that

$u^{(1)}(a)(x_{2}-a)+u(a)\leq
u^{(1)}(a)(b-a)+u(a)\leq\hat{u}(b)=\hat{u}(x_{2}).$

The first inequality follows because $u^{(1)}(a)\leq 0$. The second inequality
follows from the convexity of $u$.

We prove that $\hat{u}$ is decreasing and convex. Because a convex function is
continuous on the interior of the domain, we have that $\hat{u}$ is a
continuous function.

The next step is based on a mollification argument (see Appendix C in Evans
(2010)). Consider a positive mollifier $g$, with a compact support and define
$g_{n}(x)=ng(nx)$. Then,

$\hat{u}_{n}(x)=\hat{u}*g_{n}(x)=\int\hat{u}(x-y)g_{n}(y)dy$

is infinitely differentiable. Since $\hat{u}$ is continuous, we have that
$\hat{u}_{n}$ converges uniformly to $\hat{u}$ on compact subsets of
$\mathbb{R}$ (see Appendix C, Theorem 6, in Evans (2010)). In particular, we
have that $\hat{u}_{n}$ converges uniformly to $\hat{u}=u$ on $[a,b]$.

We assert that $\hat{u}_{n}$ is convex, decreasing, which implies that
$u_{n}:=\hat{u}_{n}|_{[a,b]}$ is convex decreasing, infinitely differentiable,
such that $u_{n}$ converges uniformly to $u$.

Indeed, take $x_{1}<x_{2}$ and $\lambda\in(0,1)$, then:

  * •

Monotonicity: Because $\hat{u}$ is decreasing and $g\geq 0$, we have that for
every $y$,

$\hat{u}(x_{1}-y)g_{n}(y)\geq\hat{u}(x_{2}-y)g_{n}(y)\;.$

Hence, integrating over $y$, we obtain that
$\hat{u}_{n}(x_{1})\geq\hat{u}_{n}(x_{2})$.

  * •

Convexity: Because $\hat{u}$ is convex and $g\geq 0$, we have that for every
$y$,

$\hat{u}(\lambda
x_{1}+(1-\lambda)x_{2}-y)g_{n}(y)=\hat{u}(\lambda(x_{1}-y)+(1-\lambda)(x_{2}-y))g_{n}(y)\leq\lambda\hat{u}(x_{1}-y)g_{n}(y)+(1-\lambda)\hat{u}(x_{2}-y)g_{n}(y)\;.$

Hence, integrating over $y$, we obtain that $\hat{u}_{n}(\lambda
x_{1}+(1-\lambda)x_{2})\leq\lambda\hat{u}_{n}(x_{1})+(1-\lambda)\hat{u}_{n}(x_{2})$.

###### Lemma 4

$F\succeq_{n,[a,b]-BSD}G$ implies

$\int_{a}^{b}u(x)d(F-G)(x)\leq 0\text{ for all }u\in\mathcal{G}_{n,[a,b]}.$

Proof. For every positive integer $j$ and a distribution function $W$ on
$[a,b]$ define recursively $W_{j+1}(x):=\int_{a}^{x}W_{j}(z)dz$ for all
$x\in[a,b]$ where $W_{1}:=W$. Integration by parts yields

$W_{j+1}(c)=\frac{1}{j!}\int_{a}^{c}(c-x)^{j}dW(x)=\frac{1}{j!}\int_{a}^{b}M_{j,c}(x)dW(x)$
(15)

for every positive integer $j$ where $M_{j,c}(x)=\max\\{c-x,0\\}^{j}$.

Assume that $F\succeq_{n,[a,b]-BSD}G$. Let $u\in\mathcal{G}_{n,[a,b]}$. Using
integration by parts for a Lebesgue-Stieltjes integral multiple times yield

$\displaystyle\int_{a}^{b}u(x)d(F-G)(x)$
$\displaystyle=u(x)\left(F(x)-G(x)\right)\Big{|}_{a^{-}}^{b^{+}}-\int_{a}^{b}u^{(1)}(x)\left(F(x)-G(x)\right)dx$
$\displaystyle=-u^{(1)}(x)\int_{a}^{x}\left(F(z)-G(z)\right)dz\Big{|}_{a}^{b}+\int_{a}^{b}u^{(2)}(x)\int_{a}^{x}\left(F(z)-G(z)\right)dzdx$
$\displaystyle=u^{(2)}(x)\int_{a}^{x}\int_{a}^{y}\left(F(z)-G(z)\right)dzdy\Big{|}_{a}^{b}-\int_{a}^{b}u^{(3)}(x)\int_{a}^{x}\int_{a}^{y}\left(F(z)-G(z)\right)dzdydx$
$\displaystyle=u^{(2)}(b)\int_{a}^{b}\int_{a}^{y}\left(F(z)-G(z)\right)dzdy-\int_{a}^{b}u^{(3)}(x)\int_{a}^{x}\int_{a}^{y}\left(F(z)-G(z)\right)dzdydx$
$\displaystyle=u^{(2)}(b)(F^{3}(b)-G^{3}(b))-\int_{a}^{b}u^{(3)}(x)(F^{3}(x)-G^{3}(x))dx.$

In the second equality we use the fact that
$F(b^{+})-G(b^{+})=F(a^{-})-G(a^{-})=0$. In the third equality we use the fact
that $u\in\mathcal{G}_{n,[a,b]}$ implies that $u^{(1)}(b)=0$. Continuing
integrating by parts and using the fact that $u\in\mathcal{G}_{n,[a,b]}$
implies that $u^{(k)}(b)=0$ for all $k=1,\ldots,n-1$ yield

$\displaystyle\int_{a}^{b}u(x)d(F-G)(x)$
$\displaystyle=(-1)^{n}u^{(n)}(b)(F^{n+1}(b)-G^{n+1}(b))+(-1)^{n+1}\int_{a}^{b}u^{(n+1)}(x)(F^{n+1}(x)-G^{n+1}(x))dx$
$\displaystyle=\frac{1}{n!}(-1)^{n}u^{(n)}(b)\left(\int_{a}^{b}M_{n,b}(x)d(F-G)(x)\right)$
$\displaystyle+\frac{1}{n!}\int_{a}^{b}(-1)^{n+1}u^{(n+1)}(x)\left(\int_{a}^{b}M_{n,x}(z)d(F-G)(z)\right)dx\leq
0.$

The equality follows from Equation (15). The inequality follows from the facts
that $(-1)^{k}u^{(k)}\leq 0$ for $k=n,n+1$ and $F\succeq_{n,[a,b]-BSD}G$.

###### Lemma 5

The inequality $\int_{a}^{b}u(x)d(F-G)(x)\leq 0\text{ for all
}u\in\mathcal{G}_{n,[a,b]}$ implies $F\succeq_{n,[a,b]-BSD}G$.

Now assume that $\int_{a}^{b}v(x)dG(x)\geq\int_{a}^{b}v(x)dF(x)$ for every
function $v\in\mathcal{G}_{n,[a,b]}$. In what follows, we show that for every
lower partial moment function $u(x)=\max\\{c-x,0\\}^{n}$, there exist a
sequence of functions $(u_{j})$ such that $u_{j}\in\mathcal{G}_{n,[a,b]}$ for
every integer $j$ and $u_{j}$ converges to $-u$ weakly, i.e.,
$\lim_{j\rightarrow\infty}\int_{a}^{b}u_{j}(x)dW(x)=-\int_{a}^{b}u(x)dW(x)$
for every distribution function $W$ on $[a,b]$. Therefore, proving that
$F\succeq_{n,[a,b]-BSD}G$.

Let $u(x)=\max\\{c-x,0\\}^{n}$ be a lower partial moment function. First,
notice that the function $-u$ is $n-1$ times differentiable with
$-u^{(n-1)}(x)=n!(-1)^{n}\max\\{c-x,0\\}$. Hence, $-u^{(n-1)}$ is either a
concave and increasing function or a convex and decreasing function. For
simplicity of the proof, assume that $n$ is even, i.e., $-u^{(n-1)}$ is convex
and decreasing (the proof for the concave and increasing case is the same).
Because $-u^{(n-1)}(a)$ exists, Lemma 3 implies the existence of a sequence
$(g_{j})$ of decreasing, convex, and infinitely differentiable functions such
that $g_{j}$ converges uniformly to $-u^{(n-1)}$.555The Lemma states the
result for $f$ that is convex, decreasing, such that $f^{(1)}(a)$ exists.
Clearly, it also holds for the case that $f$ is a concave and increasing
function such that $f^{(1)}(a)$ exists.

Define recursively $G_{j,k}(x)=\int_{a}^{x}(G_{j,k-1}(z)-G_{j,k-1}(b))dz$ for
all $x\in[a,b]$ and every positive integer $k$ where
$G_{j,1}(x):=g_{j}(x)-g_{j}(b)$. Let $u_{j}(x):=G_{j,n}(x)$, in the next two
steps we show that (i) $u_{j}\in\mathcal{G}_{n,[a,b]}$ and (ii) $u_{j}$
converges to $-u$ weakly.

Step 1. Notice that because the function $g_{j}$ is infinitely differentiable
we have $u_{j}\in C^{n+1}([a,b])$. In addition, $u_{j}$ satisfies

$u_{j}^{(k)}(x)=\begin{cases}G_{j,n-k}(x)-G_{j,n-k}(b)&\mbox{ if
}k=1,\ldots,n-1\\\ g_{j}^{(k+1-n)}(x)&\mbox{ if }k=n,n+1.\end{cases}$

Hence, $u_{j}^{(k)}(b)=0$ for $k=1,\ldots,n-1$. It remains to show that
$u_{j}\in U_{n,[a,b]}$. Because $g_{j}$ is convex and decreasing and $n$ is an
even number, we have that $(-1)^{k}u_{j}^{(k)}(x)\leq 0$ for $k=n,n+1$. For
$k\leq n-1$, we show by induction that the function $G_{j,k}$ is decreasing if
$k$ is odd and increasing if $k$ is even. The base case $k=1$ is
straightforward since $G_{j,1}(x)=g_{j}(x)-g_{j}(b)$ and $g_{j}$ is a
decreasing function. For the induction step, notice that
$G^{(1)}_{j,k}(x)=G_{j,k-1}(x)-G_{j,k-1}(b)$. Therefore if $G_{j,k-1}$ is a
decreasing (increasing) function we have that $G^{(1)}_{j,k}(x)$ is positive
(negative) and, hence, $G_{j,k}$ is an increasing (decreasing) function. We
conclude that $(-1)^{k}u_{j}^{(k)}\leq 0$ for every $k=0,\ldots,n+1$. Thus,
$u_{j}\in U_{n,[a,b]}$ and, therefore, $u_{j}\in\mathcal{G}_{n,[a,b]}$.

Step 2. We claim that for any distribution function $F$ on $[a,b]$ we have
$\int_{a}^{b}u_{j}(x)dF(x)\to-\int_{a}^{b}u(x)dF(x)$. Because
$u_{j}^{(k)}(b)=0$ and $u^{(k)}(b)=0$ for $k=1,\ldots,n-2$, integration by
parts implies that

$\displaystyle\int_{a}^{b}u_{j}(x)dF(x)$
$\displaystyle=(-1)^{n-1}\int_{a}^{b}u_{j}^{(n-1)}(x)F^{n-1}(x)dx,\text{ and
}$ (16) $\displaystyle-\int_{a}^{b}u(x)dF(x)$
$\displaystyle=(-1)^{n-1}\int_{a}^{b}-u^{(n-1)}(x)F^{n-1}(x)dx.$ (17)

Because $u_{j}^{(n-1)}(x)=g_{j}(x)-g_{j}(b)$, $g_{j}$ converges uniformly to
$-u^{(n-1)}$ and $\lim g_{j}(b)=0$, we conclude using the dominated
convergence theorem that
$\lim_{j\to\infty}\int_{a}^{b}u_{j}(x)dF(x)=-\int_{a}^{b}u(x)dF(x)$.

##  5 Summary

In this paper we provide a generator for the bounded stochastic dominance
orders. We identify sets of risk averse decision makers that satisfy curvature
conditions such as bounded from below Arrow-Pratt measure of risk aversion or
$n$-convexity, that generate the bounded stochastic dominance orders. This
result fills a theoretical gap in the literature on stochastic orders as it
provides a decision theoretic motivation to use the bounded stochastic orders.
Our results also imply novel inequalities for $n$-convex functions such as
Jensen-type inequalities and non-trivial rankings over lotteries for decision
makers with bounded from below Arrow-pratt measure of risk aversion.

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##  6 Appendix

Proof of Proposition 2. Without loss of generality assume that $a=0$. Consider
the function

$g(x)=-(b-x)^{n+1}(1-\gamma(b-x))$

where $\gamma$ will be determined later in the proof. Note that

$\displaystyle g^{(1)}(x)$
$\displaystyle=(b-x)^{n}\left(n+1-\gamma(n+2)(b-x)\right)$ $\displaystyle
g^{(2)}(x)$ $\displaystyle=-(n+1)(b-x)^{n-1}\left(n-\gamma(n+2)(b-x)\right)$
$\displaystyle\psi_{g}(x)$
$\displaystyle:=(n-1)g^{(1)}(x)+g^{(2)}(x)(b-x)=-(b-x)^{n}\left(n+1-2\gamma(n+2)(b-x)\right).$

Simple inspection implies that $g^{(1)}(x)>0$, $g^{(2)}(x)<0$ and
$\psi_{g}(x)\leq 0$ for every $x\in[0,b]$ if and only if $g^{(1)}(0)>0$,
$g^{(2)}(0)<0$ and $\psi_{g}(0)\leq 0$. Thus, $\gamma=\frac{n+1}{2b(n+2)}$
makes the three inequalities to hold. Therefore, $g\in AP_{n,[0,b]}$ To
conclude, we show that $g\notin LP_{n,[0,b]}$. To this extent, we use the
characterization of $n,[0,b]$-concave differentiable functions and show that
for $x$ close to zero we have

$R(x)=\frac{(g(x)-g(b))g^{(2)}(x)}{g^{(1)}(x)^{2}}<\frac{n-1}{n}.$

Noting that $g(b)=0$, we have

$\displaystyle R(x)$
$\displaystyle=\frac{(n+1)(1-\gamma(b-x))\left(n-\gamma(n+2)(b-x)\right)}{\left(n+1-\gamma(n+2)(b-x)\right)^{2}}.$

Taking $x=0$ and using $\gamma=\frac{n+1}{2b(n+2)}$, we obtain
$R(0)=\frac{(n+3)(n-1)}{(n+2)(n+1)}.$ Because for every $n\geq 2$,
$\frac{n+3}{(n+2)(n+1)}<\frac{1}{n}$, we conclude that $R(0)<\frac{n-1}{n}$.
The continuity of $g$ further implies that for $x$ close to zero,
$R(x)<\frac{n-1}{n}$. Thus, $g\notin LP_{n,[0,b]}$.

Proof of Corollary 1. Note that $k_{n}(x):=-f(b-x^{1/n})$ is convex on
$[0,(b-a)^{n}]$ if and only if $k_{n}^{(2)}(x)\geq 0$ on $[0,(b-a)^{n}]$,
i.e., if and only if $x^{1/n}f^{(2)}(b-x^{1/n})+(n-1)f^{(1)}(b-x^{1/n})\leq 0$
for all $x\in[0,(b-a)^{n}]$. Letting $y=b-x^{1/n}$ the last inequality is
equivalent to $(b-y)f^{(2)}(y)+(n-1)f^{(1)}(y)\leq 0$ for all $y\in[a,b]$.
This implies that $f$ belongs to $AP_{n,[a,b]}$, and hence, Theorem 1 implies
that $f$ is $n,[a,b]$-concave, i.e., $(f(b)-f(x))^{1/n}$ is convex on $[a,b]$.
Similarly, if $(f(b)-f(x))^{1/n}$ is convex on $[a,b]$, then Theorem 1 implies
that $f$ belongs to $AP_{n,[a,b]}$ which implies that
$(b-y)f^{(2)}(y)+(n-1)f^{(1)}(y)\leq 0$ for all $y\in[a,b]$. Thus,
$-f(b-x^{1/n})$ is convex on $[0,(b-a)^{n}]$.

Proof of Corollary 2. Let $X$ be a random variable on $[a,b]$. From Corollary
1 the function $-f(b-x^{1/n})$ is convex on $[0,(b-a)^{n}]$. Hence, from
Jensen’s inequality we have

$\mathbb{E}f(b-Y^{1/n})\leq f\left(b-\mathbb{E}(Y)^{1/n}\right).$

for the random variable $Y=(b-X)^{n}$. Thus,

$\mathbb{E}f(X)\leq f\left(b-(\mathbb{E}(b-X)^{n})^{1/n}\right).$

For the second inequality note that $f$ is increasing since $f\in U_{n,[a,b]}$
and use the monotonicity of the $L^{p}$ norm to conclude that
$(\mathbb{E}(b-X)^{n})^{1/n}\geq\mathbb{E}(b-X)$.