2. Preliminaries and notation

In this paper, for any (cumulative) distribution function F, we let $F^{-1}(t) = \inf\{x \colon F(x) \geq t\}$ , $t \in (0, 1]$ , denote the quantile function of F. Let $\mathcal X$ denote the set of integrable random variables and let $\mathcal F$ be the set of distributions with finite mean. The terms ‘distributions’ and ‘distribution functions’ are treated as identical in this paper. For $F \in \mathcal F$ , ${Supp}(F)$ is the essential support of the distribution measure induced by F, which will be referred to as the support of F. For any distributions F, $G \in \mathcal F$ , we let $F\oplus G$ denote the distribution with quantile function $F^{-1}(t) + G^{-1}(t)$ , $t\in [0,1]$ . For a distribution F, $\mu(F)$ denotes the expectation of F. Throughout, ${\lceil x \rceil}$ and ${\lfloor x \rfloor}$ represent the ceiling and the floor of $x\in {\mathbb{R}}$ , respectively. A density function f is unimodal if there exists $a\in {\mathbb{R}}$ such that f is increasing on ${(-\infty, a] }$ and decreasing on $[a,\infty)$ .

The set $\mathcal D_n$ is related to the notion of convex order. A distribution $F\in \mathcal F$ is smaller than $G\in\mathcal F$ in convex order, denoted by $F \preceq_{cx} G$ , if

$$\int_{\mathbb{R}} \phi(x){\,\textrm{d}} F(x) \leq\int_{\mathbb{R}}\phi(x) {\,\textrm{d}} G(x) \quad \text{for all convex } \phi \colon {\mathbb{R}}\to {\mathbb{R}}{,}$$

provided that both expectations exist (finite or infinite). Standard references for convex order are [Reference Müller and Stoyan6] and [Reference Shaked and Shanthikumar10]. For a given distribution $F \in \mathcal F$ , let $\mathcal {C}(F)$ denote the set of all distributions dominated by F in convex order, that is,

$$ \mathcal C(F) = \{G \in \mathcal F \colon G\preceq_{cx} F\}.$$

Checking whether a distribution G is in $\mathcal C(F)$ can be conveniently done using an equivalent condition (e.g. [10, Theorem 3.A.1]) is

(1) $$\begin{align} \inf_{k\in {\mathbb{R}}} \bigg(\int_{\mathbb{R}} (x-k)_+ {\,\textrm{d}} F(x) -\int_{\mathbb{R}} (x-k)_+{\,\textrm{d}} G(x)\bigg)\geq 0. \label{eqn1} \end{align}$$

As the focus of this paper is the distribution of the sum of uniform random variables, we use the following simplified notation. For $n\in {\mathbb{N}}$ and $x\in {\mathbb{R}}_+$ , write

$$ \mathcal D^{\textrm{U}}_n = \mathcal D_n({\textrm{U}}[0,1],\ldots,{\textrm{U}}[0, 1]) \quad\text{and}\quad \mathcal C^{\textrm{U}}_x = \mathcal C({\textrm{U}}[0, x]),$$

where ${\textrm{U}}[a,b]$ stands for the uniform distribution over an interval $[a,b]\subset {\mathbb{R}}$ .

Below we list some basic properties of the sets $\mathcal D_n(\cdot)$ and $\mathcal C(\cdot)$ ; they also hold for $\mathcal D^{\textrm{U}}_n$ and $\mathcal C^{\textrm{U}}_x$ . First, note that if all distributions $F_1,\ldots,F_n$ are shifted by some constants or scaled by the same positive constant, then the elements in $\mathcal D_n(F_1,\ldots,F_n)$ are also simply shifted or scaled. Moreover, $\mathcal D_n(F_1,\ldots,F_n)$ is symmetric in the distributions $F_1,\ldots,F_n$ . These facts allow us to conveniently exchange the order of the distributions $F_1,\ldots,F_n$ and normalize these distributions by shifts and a common scale. For given distributions $F_1,\ldots,F_n \in\mathcal F$ , the distribution $F_1\oplus \dots \oplus F_n$ is the maximum in convex order in the set $\mathcal D_n(F_1,\ldots,F_n)$ . This fact is summarized in the following lemma.

3. Sums of two standard uniform random variables

In this section we look at the sum of two ${\textrm{U}}[0, 1]$ random variables. Unfortunately, as explained above, a full characterization of ${\mathcal D}_2^{\textrm{U}}$ appears difficult to obtain. We provide characterization results for four different types of distributions to be in ${\mathcal D}_2^{\textrm{U}}$ . We first present the main results on ${\mathcal D}_2^{\textrm{U}}$ in Section 3.1. Their proofs will be given in Section 3.2. Before presenting our main findings, we summarize some existing results on ${\mathcal D}_2^{\textrm{U}}$ . These facts can be derived from existing results on joint mixability in [Reference Wang and Wang13].

3.1. Main results

As a first new result in this paper, we show that the class of distributions with a unimodal density with the correct mean is contained in ${\mathcal D}_2^{\textrm{U}}$ .

Theorem 1. Let F be a distribution with a unimodal density on [0, 2] and mean 1. Then $F\in {\mathcal D}_2^{\textrm{U}}$ .

Both the two previous results in Proposition 1(ii,iii) are special cases of Theorem 1; thus Theorem 1 generalizes the existing results derived from [Reference Wang and Wang13].

The second result of the paper concerns the class of distributions which dominate a proportion of a uniform distribution.

Theorem 2. Let F be a distribution supported in $[a,a+b]$ with mean 1 and density function f. If there exists $h>0$ such that $f>3b/(4h)$ on $[1-h,1+h]$ , then $F\in { {\mathcal D}}_2^{\textrm{U}} $ .

In Theorem 2, the condition that the density of F dominates $3b/2$ times that of ${\textrm{U}}[1-h, 1+h]$ immediately implies the following admissible ranges of a, b, and h: $a\geq 1/3$ , $b\leq 2/3$ , and $h\leq 1/3$ . Hence, $\textrm{Supp}(F)\subset [0,2]$ , which is obviously necessary for $F\in {\mathcal D}_2^{\textrm{U}}(1, 1)$ . Theorem 2 also immediately implies the following fact: for a distribution F with mean 0 and bounded support, if F has a positive density $f > \varepsilon$ in a neighbourhood of 0 for some $ \varepsilon> 0$ , then $F \in {\mathcal D}_2(\textrm U[-m,m],\textrm U[-m,m])$ for sufficiently large $m > 0$ .

In addition to the two results on continuous distributions, we analyse discrete distributions. We shall obtain two results, one characterizing bi-atomic distributions in ${\mathcal D}_2^{\textrm{U}} $ , and one characterizing equidistant tri-atomic distributions in ${\mathcal D}_2^{\textrm{U}} $ .

Theorem 3. Let F be a bi-atomic distribution with mean 1 supported on $\{a,a + b\}$ with $b>0$ . Then $F \in {\mathcal D}^{\textrm{U}}_2 $ if and only if $1/b \in {\mathbb{N}}$ .

For a given $a\in [0,1)$ and $a+b\in (1,2]$ , there is a unique distribution on $\{a,a+b\}$ with mean 1. Hence, all the bi-atomic distributions that belong to ${\mathcal D}^{\textrm{U}}_2 $ have the corresponding distribution measures

$$ \{\nu \colon 1-\nu(\{a\})=\nu(\{a+1/k\}) = (1-a)k,~k\in{\mathbb{N}},~a\in [0,1)\}.$$

Note that many bi-atomic distributions supported on $\{a,a+b\}$ are in ${\mathcal C}^{\textrm{U}}_2$ but not in ${\mathcal D}_2^{\textrm{U}} $ , as long as $1/b\not\in {\mathbb{N}}$ . For example, one can choose a bi-atomic distribution F with equal probability on $\{1-1/\pi,1+ 1/\pi\}$ , and easily see that $F\in{\mathcal C}^{\textrm{U}}_2$ , whereas from Theorem 3 we find that F is not in ${\mathcal D}_2^{\textrm{U}} $ . Thus, Theorem 3 implies ${\mathcal D}_2^{U} \subsetneq {\mathcal C}^{U}_2$ , a fact noted in [Reference Mao and Wang4].

For a tri-atomic distribution F, write $F = (f_1, f_2, f_3)$ , where $f_1, f_2, f_3$ are the probability masses of F. Note that, given three points, the set of tri-atomic distributions with mean 1 has one degree of freedom. For tractability, we study the case of F having an equidistant support in the form of $\{a-b,\,a,\,a+b\}$ for some $b>0$ . We only consider the case $b\leq a \leq 1$ since the case $a > 1$ is symmetric.

To state our characterization of tri-atomic distribution in $\mathcal D_2^{\textrm U}$ , we introduce the following notation. For $x > 0$ , define a measure of non-integrity:

$$ \lceil x \rfloor = \min \bigg\{\frac{\lceil x \rceil}{x} - 1, 1 - \frac{\lfloor x \rfloor}{x}\bigg\}\in [0, 1].$$

Obviously $\lceil x \rfloor=0 \Leftrightarrow x \in {\mathbb{N}}$ .

Theorem 4. Suppose that $F = (f_1, f_2, f_3)$ is a tri-atomic distribution with mean 1 supported in $\{a - b, a, a + b\}$ and $0<b\leq a \leq1$ . Then $F \in \mathcal D_2^{\textrm U} $ if and only if one of the following three cases holds.

$a=1$ and $f_2\geq \lceil \frac{1}{(2b)}\rfloor$ .

$a<1$ and $\frac{1}{(2b)}\in {\mathbb{N}}$ .

$a<1$ , $\frac{1}{(2b)}-\frac{1}{2}\in {\mathbb{N}}$ and $f_2\geq a+b-\frac{1}{2}$ .

The corresponding distributions in Theorem 4 are summarized below. Write $c = a/b +1- 1/(2b)$ . Here $\operatorname{cx} (x, y)$ stands for the convex set generalized by some vectors x,y.

${(f_1,f_2,f_3)\in \operatorname{cx} \{(0,1,0), \frac{1}{2}(1-\lceil \frac{1}{2b} \rfloor, 2\lceil \frac{1}{2b} \rfloor, 1- \lceil \frac{1}{2b} \rfloor)\}}$ .

${(f_1,f_2,f_3)\in {cx} \{(0,c,1-c), \frac{1}{2}(c, 0, 2-c)\}}$ .

${(f_1,f_2,f_3)\in {cx} \{(0,c,1-c), \frac{1}{2}(c(1-b), 2bc, 2 - c(1-b))\}}$ .

3.2. Proofs of the main results

To prove Theorem 1, we need the following lemma. To state it, we introduce the notion of special simple unimodal functions. A function h is called a special simple unimodal (SSU) function on $[a,a+n)$ , $a\in{\mathbb{R}}$ , $n\in{\mathbb{N}}$ , if it is unimodal, and h(x) is a constant on $[a+k-1, a+k)$ for each $k=1,\ldots,n$ .

Lemma 3. Let F be a distribution function with density function f and support $[a,a+n)$ , $a\in{\mathbb{R}}_-$ , $n\in{\mathbb{N}}$ . Suppose that F has mean 0 and f is a SSU function on $[a,a+n)$ . Then, for $c\geq \max\{a+n,-a\}/2$ , we have $F\in {\mathcal D}_2({\textrm{U}}[-c,c],{\textrm{U}}[-c,c])$ .

Proof. We show the result by induction. For $n=1$ , then F is the uniform distribution on $[a,a+1)$ with mean 0. This means $a=-0.5$ , that is, F is the distribution of ${\textrm{U}}[-0.5,0.5]$ , and $c\geq 0.5$ . By [13, Theorem 3.1], we know that ${\textrm{U}}[-0.5,0.5]$ , ${\textrm{U}}[-c,c]$ , and ${\textrm{U}}[-c,c]$ are jointly mixable as the mean inequality $-0.5-2c+2c\leq 0 \leq 0.5+2c-2c$ is satisfied. This means $F\in {\mathcal D}_2({\textrm{U}}[-c,c],{\textrm{U}}[-c,c])$ .

Next, we assume that the result holds for $n\leq k$ and show that it holds for $n=k+1$ . Without loss of generality, we assume $a+n>-a$ . Otherwise consider the distribution of $X^*=-X$ with $X\sim F$ . Define a distribution H with density function $h\colon [a,a+n)\to{\mathbb{R}}_+$ defined as

$$ h(x)= \frac{n+1+2a}{n}\,{=\!:}\,\alpha,\quad x\in [a,a+1),\quad h(x) = \frac{-1-2a}{n(n-1)}\,{=\!:}\,\beta,\quad x\in[a+1,a+n).$$

It can be easily verified that it has mean 0 and $\alpha\geq 1/n\geq \beta>0$ as $2a+n>0$ . That is, h is a decreasing density function on $[a,a+n)$ with mean 0. Then, by [13, Theorem 3.2], H, ${\textrm{U}}[-c,c]$ and ${\textrm{U}}[-c,c])$ are jointly mixable. Hence, we have $H\in {\mathcal D}_2({\textrm{U}}[-c,c],{\textrm{U}}[-c,c])$ .

Denote $a_i = f(a+i)$ , $i=0,\ldots, n-1$ . Since f is unimodal, without loss of generality, assume that $a_1\leq \dots\leq a_k\geq \dots \geq a_n$ for some $k\in \{0,\ldots, n-1\}$ . Let

$$ \lambda \,{:\!=}\, \min\bigg\{ \frac{a_1}{\alpha},\frac{a_2}{\beta},\ldots, \frac{a_n}{\beta}\bigg\}.$$

By contradiction, it immediately follows from $\sum_{i=1}^na_i=1$ and $\alpha+(n-1)\beta=1$ that $\lambda \leq 1$ . If $\lambda=1$ , then $a_1=\alpha$ and $\alpha_i=\beta$ , $i=2,\ldots,n$ , that is, $F=H\in {\mathcal D}_2({\textrm{U}}[-c,c],{\textrm{U}}[-c,c])$ , which shows the statement in the lemma. Next, we consider the case $\lambda<1$ . We first assert that the sequence

$$ b_1 =a_1-\lambda\alpha,\quad b_i = a_i-\lambda\beta,\quad i=2,\ldots,n$$

is unimodal such that either $b_1$ or $b_n$ is 0. To see it, we only need to show it is unimodal by the definition of $\lambda$ . We consider the following two cases.

If $k\geq 2$ , then $a_1\leq a_2$ and hence $b_1=a_1-\lambda\alpha \leq a_2 -\lambda\beta =b_2$ as $\alpha\geq \beta$ . Also, note that $b_2\leq \dots\leq b_k\geq \dots \geq b_n$ . Hence, the sequence $\{b_i,\,i=1,\ldots,n\}$ is unimodal.

If $k=1$ , then $a_1\geq a_2\geq \dots\geq a_n$ , and hence $b_2\geq \dots\geq b_n$ . No matter whether $b_1\geq b_2$ or $b_1\leq b_2$ , we have that $\{b_i,\,i=1,\ldots,n\}$ is unimodal.

Define

$$F_0 = \frac{F-\lambda H}{ 1-\lambda}.$$

It is easy to check that $F_0$ is a distribution function with a density function $f_0$ taking value $b_i/(1-\lambda)$ on the interval $[a,a+i)$ , $i=1,\ldots,n$ . By the above observations on the sequence $\{b_i,\,i=1,\ldots,n\}$ , we know that $F_0$ is a distribution with mean 0 and with support on a subset of $[a,a+n-1)$ or $[a+1, a+n)$ . Then, by induction, we have $F_0\in {\mathcal D}_2({\textrm{U}}[-c,c], {\textrm{U}}[-c,c])$ . Since $F = (1 -\lambda)F_0 + \lambda H$ , and ${\mathcal D}_2$ is a convex set, we have $F\in {\mathcal D}_2({\textrm{U}}[-c,c], {\textrm{U}}[-c,c])$ .

Now we are ready to prove Theorem.1.

Proof of Theorem 1. Let f denote the density function of F, which is unimodal. Then there exists $x_0\in [0,2]$ such that $f(x_0)=\max\{ f(x),x\in[0,2]\}$ . Let X and Y be two independent random variables such that $X\sim F$ and $Y\sim {\textrm{U}}[0,1]$ , and define $X_m=\lfloor m(X-1)\rfloor +Y$ , $m\in{\mathbb{N}}$ . Then the density function of $X_m$ , denoted by $h_m$ , is given by

$$h_m(x)=\int_{j/m}^{(j+1)/m} f(x) {\,\textrm{d}} x\,{=\!:}\,p_j, \quad x\in [j-m,j-m+1),\ j=0,\ldots,2m-1.$$

Let $k=\lfloor mx_0\rfloor$ . Then we have

$$ p_0\leq p_1\leq \dots\leq p_{k-1},\quad p_{k+1}\geq p_{k+2}\geq \dots\geq p_{2m},$$

and, since f(x) is unimodal on $[k/m,(k+1)/m]$ ,

$$ p_k=\int_{k/m}^{(k+1)/m} f(x) {\,\textrm{d}} x \geq \frac1m\min \{f(k/m),\,f((k+1)/m) \} \geq \min\{p_{k-1},\,p_{k+1}\}.$$

That is, the sequence $\{p_k,\,k=1,\ldots,2m\}$ is unimodal. It can be verified that

$${\mathbb{E}}[X_m]={\mathbb{E}}[\lfloor m(X-1)\rfloor-m(X-1)]+ \frac 12 ,$$

which lies in $[-0.5,0.5]$ . Then the random variable $Y_m\,{:\!=}\,X_m-\mu(X_m)$ has mean 0 and takes values in a subset of $[-m-0.5,m+0.5]$ . By Lemma 3, the distribution of $Y_m$ belongs to ${\mathcal D}_2({\textrm{U}}[-c,c],{\textrm{U}}[-c,c])$ with $c=m/2+0.25$ . Then obviously, the distribution of $Y_m/(2m\,{+}\,0.5)+1$ belongs to ${\mathcal D}_2^{{\textrm{U}}} $ . Note that $Y_m/(2m+0.5)+1$ converges to X in the $L^\infty$ -norm as $m\to\infty$ . Hence, by the closure of ${\mathcal D}_2^{{\textrm{U}}} $ with respect to the $L^\infty$ -norm, we have $F\in{\mathcal D}_2^{{\textrm{U}}} $ . This completes the proof.

To prove Theorems 2–4, we need the following lemma.

Lemma 4. Let Z be a random variable with distribution F supported in $\{b-k,\, b-k+1,\ldots,b\}$ , $k\in{\mathbb{N}}$ , $b\in{\mathbb{R}}$ , satisfying

$${\mathbb{P}}(Z=b-i)=p_i\geq 0,\quad i=0,\ldots,k,\quad \text{and}\quad \sum_{i=0}^k p_i=1.$$

Then the following statements hold.

If $F\in {\mathcal D}_2({\textrm{U}}[0,T],{\textrm{U}}[-T,0])$ , then at least one of b and T is an integer.

If $k=1$ , then $F\in {\mathcal D}_2({\textrm{U}}[0,T],{\textrm{U}}[-T,0])$ if and only if $b\in (0,1)$ and $T\in{\mathbb{N}}$ .

If $k=2$ , then $F\in {\mathcal D}_2({\textrm{U}}[0,T],{\textrm{U}}[-T,0])$ if and only if one of the following three cases holds:

$b=1$ , $p_2=p_0=(1-p_1)/2$ , $p_1\geq r/T$ with $T=2m\pm r$ with $m\in{\mathbb{N}}$ and $r\in (0,1)$ , $b\in (0,1)\cup (1,2)$ , T is even, $p_1\in (0,\min\{b,2-b\})$ , $p_2=(b-p_1)/2$ , and $p_0=1-(b+p_1)/2$ ,

$b\in (0,1)\cup (1,2)$ , T is odd, $p_1\in (b^*/T,b^*)$ with $b^*=\min\{b,2-b\}$ , $p_2=(b-p_1)/2$ , and $p_0=1-(b+p_1)/2$ .

Proof. First we introduce the notation. For any random variable X and any set $L\subset {\mathbb{R}}$ , define random events

$$ A_X(L) \,{:\!=}\,\{X-n\in L~{\rm for~some}~n\in{\mathbb{N}}\}=\{X\ \text{mod}\, 1\in L\},$$

and a function

$$g_X(x) = \lim_{\delta\downarrow 0} \frac{{\mathbb{P}}(A_X([x-\delta,x+\delta]))}{2\delta} ,\quad x\in{\mathbb{R}},\quad \text{if the limit exists}.$$

Note that $Z \equiv b \ \text{mod}\, 1$ , ${\mathbb{E}}[Z]=0$ and hence $b-k\leq 0\leq b$ . If $F\in {\mathcal D}_2({\textrm{U}}[0,T],{\textrm{U}}[-T,0])$ , then there exist two random variables $X\sim {\textrm{U}}[0,T]$ and $Y\sim {\textrm{U}}[-T,0]$ such that $Z=X+Y$ a.s. Since $Z=X+Y \equiv b \ \text{mod}\, 1$ a.s., we have $A_X(L)=A_Y(b-L)$ a.s. for any $L\subset {\mathbb{R}}$ , where $b-L=\{b-x\colon x\in L\}$ . We first show that at least one of b and T is an integer. To this end, assume that T is not an integer. Then there exists $\ell \in {\mathbb{N}}$ such that $T=\ell+t$ with $t\in(0,1)$ . Note that $X\sim {\textrm{U}}[0,T]$ . We have

$$\begin{align*} g_X(x) & =\lim_{\delta\downarrow 0} \frac{{\mathbb{P}}(X\ \text{mod}\, 1\in [x-\delta,x+\delta])}{2\delta}\\* & = \frac1{2T} (\#\{n\in{\mathbb{N}}\colon n+x\in (0,T)\} +\#\{n\in{\mathbb{N}}\colon n+x\in [0,T]\} )\\* & = \begin{cases} \frac{\ell+1}{T} & \text{if $x\ \text{mod}\, 1\in (0,t)$},\\[9pt] \frac{\ell+1/2}{T} & \text{if $x\ \text{mod}\, 1= 0$ or \textit{t}},\\[9pt] \frac \ell {T} & \text{if $x\ \text{mod}\, 1\in (t,1)$}. \end{cases} \end{align*}$$

By the definition of $A_X(L)$ and $Y\stackrel{\rm d}= -X$ , we have $A_Y([x-\delta,x+\delta]) =A_{-X}([x-\delta,x+\delta])= A_{X}([-x-\delta,-x+\delta])$ a.s., and thus $g_Y(x)=g_X(-x)$ for $x\in{\mathbb{R}}$ . Also, note that $A_X(L)=A_Y(b-L)$ a.s., which implies $g_X(x)=g_Y(b-x)$ . Therefore, we have

$$g_X(x)=g_Y(b-x)=g_X(x-b)\quad \text{for any } x\in{\mathbb{R}}.$$

Note that there exists $x\in{\mathbb{R}}$ such that $x\ \text{mod}\, 1\in (0,t)$ and $x-b\ \text{mod}\, 1\in (t,1)$ , which contradicts the formula of $g_X$ . Hence, b must be an integer.

Next we consider the two cases $k=1$ and $k=2$ . For $k=1$ , note that for any $b\in{\mathbb{N}}$ , $\mu(F)$ could not be 0. Hence, we only need to show $F\in {\mathcal D}_2({\textrm{U}}[0,T],{\textrm{U}}[-T,0])$ when $b\in (0,1)$ and $T\in {\mathbb{N}}$ . By $\mu(F)=0$ , we have $p_1=b$ and $p_0=1-b$ . Let $X\sim {\textrm{U}}[0,T]$ and define a random variable Y such that

$$\begin{alignat*}{3} [Y|X\in[k,k+b]] &= b-1-X,&\quad \text{a.s.\quad for }& k=0,\ldots,T-1{,}\\ [Y|X\in(k+b,k+1]] &= b-X,&\quad \text{a.s.\quad for }& k=0,\ldots,T-1. \end{alignat*}$$

Then it is easy to see that $Y\sim {\textrm{U}}[-T,0]$ and $X+Y$ has the distribution F. Thus, we have $F\in\mathcal D_2({\textrm{U}}[0,T], {\textrm{U}}[-T,0])$ .

If $k=2$ , by the necessity condition that at least one of b and T is an integer, we consider the following three cases. Without loss of generality, assume $p_i={\mathbb{P}}(Z=b-i)>0$ , $i=0,1,2.$

(a) If b is an integer, for the mean-constraint to be satisfied, we have $b=1$ and

$$ {\mathbb{P}}(Z=-1) ={\mathbb{P}}(Z=1)=p_0>0 \quad \text{and} \quad {\mathbb{P}}(Z=0)=p_1 =1-2p_0.$$

Let $T= 2m\pm r$ with $r\in [0,1]$ and $m\in{\mathbb{N}}$ . We only need to find the smallest value of $p_1$ such that $F\in {\mathcal D}_2({\textrm{U}}[0,T],{\textrm{U}}[-T,0])$ as $\delta_0\in {\mathcal D}_2({\textrm{U}}[0,T],{\textrm{U}}[-T,0])$ and ${\mathcal D}_2({\textrm{U}}[0,T],{\textrm{U}}[-T,0])$ is closed under mixture, where $\delta_0$ is the point-mass at 0. Assume that there exist $X\sim {\textrm{U}}[0,T]$ and $Y\sim {\textrm{U}}[-T,0]$ such that $Z=X+Y$ a.s. If $T=2m+r$ , then since b is an integer, we have

$$ A_X((r,1)) = \{X\in \cup_{k=1}^{2m} (r+k-1,k) \} = A_Y(\{({-}1,-r)\}) = \{Y\in \cup_{k=1}^{2m} (-k, -r-k+1)\}\quad \text{a.s.}$$

We let

$$ [Y|X\in A_1] = -X-1\quad \text{and}\quad [Y|X\in A_2] = -X+1 \quad\text{a.s.}$$

where $A_1=\cup_{k=1}^{m} (r+2k-3,2k-1)$ and $A_2= \cup_{k=1}^{m} (r+2k-1,2k)$ . Then $[X+Y|X\in A_1\cup A_2]=[X+Y|A_X((r,1))]$ takes values on $\{-1,1\}$ . On the other hand, note that

$$\begin{align*} A_X((0,r)) &= \{X\in \cup_{k=0}^{2m} (k,k+r)\} = A_Y((-r,0)) = \{Y\in \cup_{k=0}^{2m} (-k-r,-k)\}. \end{align*}$$

Then, on the set $A_X((0,r))$ , $X+Y$ could not only take values on $\{-1,1\}$ . There is at least one $k\in\{0,\ldots,2m\}$ such that $X+Y=0$ on $\{X\in (k,k+r)\}$ . Hence, $p_1\geq (2m+1)r/(2m+1)T=r/T$ . We next show that $r/T$ can be attained by $p_1$ for $F\in {\mathcal D}_2({\textrm{U}}[0,T],{\textrm{U}}[-T,0])$ . It suffices to let $[Y|X\in (2m,2m+r)]=-X$ a.s., and for $k=0,\ldots,m-1$ ,

$$ [Y|X\in(2k,2k+r)] =-X-1,\quad [Y|X\in(2k+1,2k+1+r)] =-X+1\quad\text{a.s.}$$

In this case, we have $p_1=r/T.$ By symmetry, we can also get the same result if $T=2m-r$ . To see this, note that almost surely

$$ A_X((0,1-r)) = \{X\in \cup_{k=0}^{2m-1} (k,k+1-r)\} = A_Y((r-1,0)) = \{Y\in \cup_{k=0}^{2m-1} (-k-1+r,-k)\}.$$

We let

$$ [Y|X\in A_1] = -X-1\quad \text{and} \quad [Y|X\in A_2] = -X+1 \quad \text{a.s.}$$

where $A_1=\cup_{k=0}^{m-1} (2k,2k+1-r) $ and $A_2= \cup_{k=0}^{m-1} (2k+1,2k+2-r)$ . Then $[X+Y|X\in A_1\cup A_2]=[X+Y|A_X((0,1-r))]$ takes values on $\{-1,1\}$ . On the other hand, note that almost surely

$$ A_X((1-r,1)) = \{X\in \cup_{k=1}^{2m-1} (k-r,k)\} = A_Y(\{(-1,r-1)\}) = \{Y\in \cup_{k=1}^{2m-1} (-k, r-k)\}.$$

Then, on the set $A_X((1-r,1))$ , $X+Y$ could not only take values on $\{-1,1\}$ . There is at least one $k\in\{1,\ldots,2m-1\}$ such that $X+Y=0$ on $\{X\in (k-r,k)\}$ . Hence, $p_1\geq (2m\,{-}\,1)r/(2m-1)T=r/T$ . We next show that $r/T$ can be attained by $p_1$ for $F\in {\mathcal D}_2({\textrm{U}}[0,T], {\textrm{U}}[-T,0])$ . It suffices to let $[Y|X\in (2m-1-r,2m-1)]=-X$ a.s., and for $k=1,\ldots, m-1$ ,

$$ [Y|X\in(2k-r,2k)] =-X-1,\quad [Y|X\in(2k+1-r,2k+1)] =-X+1 \quad \text{a.s.}$$

In this case, we have $p_1=r/T.$ Hence, when b is an integer we have $Z\in {\mathcal D}_2({\textrm{U}}[0,T],{\textrm{U}}[-T,0])$ if and only if $b=1$ and ${\mathbb{P}}(Z=-1)={\mathbb{P}}(Z=1)\leq 1/2-r/(2T)$ .

(b) If b is not an integer and T is an even integer, without loss of generality, assume $b\in(0,1)$ as the case $b\in (1,2)$ can be discussed similarly by considering the symmetric distribution on $\{-b,1-b,2-b\}$ and noting that ${\mathcal D}_2({\textrm{U}}[0,T],{\textrm{U}}[-T,0])$ is closed under symmetric transform. To make sure ${\mathbb{E}}[Z]=0$ , we have $p_0-p_2=1-b$ . By part (i), we know that the distribution, denoted by $F_1$ , on $\{b-1,b\}$ with mean 0 belongs to ${\mathcal D}_2({\textrm{U}}[0,T],{\textrm{U}}[-T,0])$ , which is also closed under mixture. Hence, $F\in {\mathcal D}_2({\textrm{U}}[0,T],{\textrm{U}}[-T,0])$ implies

$$\begin{align*} \lambda F + (1-\lambda) F_0 & =\lambda p_2 \delta_{b-2} + (\lambda p_1 + (1-\lambda) b)\delta_{b-1} + (\lambda p_0 + (1-\lambda) (1-b) )\delta_{b} \\* & =\lambda p_2 \delta_{b-2} + (\lambda p_1 + (1-\lambda) b)\delta_{b-1} + (\lambda p_2 + 1-b )\delta_{b} \\* & \in {\mathcal D}_2({\textrm{U}}[0,T],{\textrm{U}}[-T,0])\quad \text{for any } \lambda\in [0,1], \end{align*}$$

where the second equality follows from $p_0-p_2=1-b$ . Therefore, we only need to find the smallest and largest values of $p_1$ such that $p_0+p_1+p_2=1$ , $p_0-p_2=1-b$ and $F\in {\mathcal D}_2({\textrm{U}}[0,T],{\textrm{U}}[-T,0])$ .

Note that $F\in {\mathcal D}_2({\textrm{U}}[0,T],{\textrm{U}}[-T,0])$ is equivalent to the existence of $X\sim {\textrm{U}}[0,T] $ and $Y\sim{\textrm{U}}[-T,0]$ such that $Z=X+Y$ a.s. Then, by $A_X(L)=A_Y(b-L)$ a.s., for any $L\subset {\mathbb{R}}$ , we have

$$ A_X((b,1)) = \{X\in \cup_{k=1}^{T} (b+k-1,k)\} = A_Y(\{(b-1,0)\}) = \{Y\in \cup_{k=1}^T (b-k,1-k) \}.$$

It is easy to verify that ${\mathbb{E}}[X+Y| A_X((b,1)) ]=b$ . Note that the $X+Y$ takes values on $\{b-2,b-1,b\}$ . Then $[X+Y|A_X((b,1))]=b$ a.s. This implies that, on $A_X((b,1))$ , $X+Y$ is not equal to $b-1$ a.s., which in turn implies that b is an upper bound of $p_1$ . On the other hand, note that

$$ A_X((0,b)) = \{X\in \cup_{k=1}^T (k-1,b+k-1) \} = A_Y(\{(0,b)\}) = \{Y\in \cup_{k=1}^T (-k,b-k) \}.$$

Letting $[Y| A_X((0,b))] = b-X-1$ a.s. and $[Y| A_X((b,1))] = b-X$ a.s. yields that b is the largest value of $p_1$ .

To find the smallest value of $p_1$ , we consider the two cases that T is even and odd. If T is even, then $T=2m$ for some $m\in{\mathbb{N}}$ . On the set $A_X(0,b)$ , for $k=0,1,\ldots,m-1$ , we let

$$ [Y|X\in (2k,b+2k)] = b-X-2 \quad \text{and}\quad [Y|X\in (2k+1,b+2k+1)] = b-X\quad \text{a.s.}$$

In this case $p_1$ is zero, which is the smallest possible value of $p_1$ .

(c) If b is not an integer and T is an odd integer, then, as for part (b), we only need to find the largest and smallest values of $p_1$ for the case $b\in (0,1)$ . The largest value of $p_1$ is b. To find the smallest value, note that $T=2m+1$ for some $m\in {\mathbb{N}}$ . On the set $A_X(0,b)$ , we let $[Y|X\in (2m,b+2m)] = b-X$ a.s., and for $k=0,1,\ldots,m-1$ ,

$$ [Y|X\in (2k,b+2k)] = b-X-2,\quad [Y|X\in (2k+1,b+2k+1)] = b-X\quad \text{a.s.}$$

In this case $p_1=b/T$ , which is the smallest possible value of $p_1$ . This is due to $ [X+Y|X\in (2m,b+2m)] >-1>b-2$ as $\{X\in (2m,b+2m)\}\subset A_Y(0,b)$ and $[Y|A_Y((0,b))]\geq -2m-1$ a.s.

Combining the above three cases, we complete the proof for the case of $k=2$ .

Proof of Theorem 2. Note that $3b/4h \times 2h =3b/2 \leq1$ , hence $b\leq 2/3$ . Therefore, $[a,a\,{+}\,b]\subset [0,2]$ . Denote $\lambda=3b/2\geq 3h$ and let H be the distribution function of ${\textrm{U}}[1-h,1+h]$ . Define $G = (F - \lambda H)/(1-\lambda)$ . By the assumption on F, G is also a distribution with positive density and mean 1. There exists a sequence of distribution functions $\{G_n\}_{n\geq 2}$ with finite support on $[a,a+b]$ and mean 1 which converges to G in distribution as $n\to\infty$ . Define $F_n = \lambda H+(1-\lambda)G_n$ , $n\geq 2$ . Then $F_n$ converges to F in distribution as $n\to\infty$ . Note that ${ {\mathcal D}}_2^{\textrm{U}} $ is closed with respect to weak convergence by Lemma 2. We only need to show $F_n\in { {\mathcal D}}_2^{\textrm{U}} $ for $n\geq 2$ . Without loss of generality, assume

$$ G_n(\{x_i\}) = p_i,\quad i=1,\ldots,n \quad \text{with } p_1+\dots+p_n=1,\ a\leq x_1<\dots<x_n\leq a+b.$$

We show it by induction on n. Note that $a\leq 1-h\leq 1+h\leq b$ . For $n=2$ , without loss of generality, let $c,d>0$ be such that $x_1=1-c$ and $x_2=1+d>0$ , $3(c+d)/2\leq\lambda$ , and then

(2) $$\begin{equation} G_2(\{1-c\}) = \frac d{c+d}\quad \text{and}\quad G_2(\{1+d\}) = \frac c{c+d}. \label{eqn2} \end{equation}$$

Since $ 1 \geq \lambda \geq c + d + h $ , we have

$$\frac{ 1- h }{c+d} - \frac{1 - \lambda } {c+d} \geq 1 \quad \text{and}\quad \frac{ 1- h }{c+d} \geq 1.$$

Then there exists some integer $k \geq 1 $ such that

$$ \frac{ 1- h }{c+d} \geq k \geq \frac{1 - \lambda } {c+d} .$$

That is, $ 1- k (c+d) \geq h $ and $ k(c+d) \geq 1-\lambda .$ Denote

$$\theta \,{:\!=}\, \frac { k(c+d) + \lambda -1 }{ k(c+d) } \in [0, \lambda]\quad \text{as }\lambda - \theta = \frac { (1-\lambda)(1-k(c+d)) }{ k(c+d) } \geq 0 .$$

Note that the length of the support of $ {{\textrm{U}}} [0,k(c+d) ] $ ( ${{\textrm{U}}}[ 1-k(c+d), 1 ] $ ) is a multiple of ${(1+d)-(1-c)}$ . By Lemma 4(i), we have $G_2\in \mathcal D_2 ( {{\textrm{U}}} [0,k(c+d) ] , {{\textrm{U}}}[ 1-k(c+d), 1 ] ) $ . Similarly, we have $ H \in \mathcal D_2 ( {{\textrm{U}}}[k(c+d) ,1], {{\textrm{U}}}[ 0 , 1-k(c+d)]) $ . Also, by Theorem 3.1, we know $H \in \mathcal D_2^{\textrm{U}} $ . It follows that

$$ F_2=( 1- \lambda ) G_2 + ( \lambda - \theta ) H + \theta H \in \mathcal D_2^{\textrm{U}}.$$

Suppose that $F_n=( 1- \lambda ) G_n + \lambda H \in \mathcal D_2^{\textrm{U}}$ for $n\leq k$ , and we aim to show that $F_{k+1}= (1- \lambda) G_{k+1} + \lambda H \in \mathcal D_2^{\textrm{U}} $ . Let $G_{k+1,1}$ be defined by (2) with $c=1-x_1>0$ and $d=x_{k+1}-1>0$ . That is,

$$ G_{k+1,1}(\{x_1\}) = \frac {x_{k+1}-1}{x_{k+1}-x_1}\quad \text{and}\quad G_{k+1,1}(\{x_{k+1}\}) = \frac {1-x_1}{x_{k+1}-x_1}.$$

Let $\alpha \,{:\!=}\, \min\{p_1(c+d)/d,p_n(c+d)/c\}$ . Then $G_{2\colon k+1} \,{:\!=}\, (G_{k+1} - \alpha G_{k+1,1})/(1-\alpha)$ is a distribution function with support on $\{x_1,\ldots,x_{k}\}$ or $\{x_2,\ldots,x_{k+1}\}$ and mean 1. Then we have

$$\begin{align*} F_{k+1} & = ( 1- \lambda ) (\alpha G_{k+1,1} + (1-\alpha) G_{2\colon k+1}) + \lambda H \\* & = \alpha (( 1- \lambda ) G_{k+1,1} + \lambda H)+(1-\alpha)(( 1- \lambda ) G_{2\colon k+1}+ \lambda H)). \end{align*}$$

By induction, we have ${( 1- \lambda ) G_{k+1,1} + \lambda H \in \mathcal D_2^{\textrm{U}}}$ and ${( 1- \lambda ) G_{2\colon k+1}+ \lambda H\in \mathcal D_2^{\textrm{U}}}$ . Then, by the convexity of $\mathcal D_2^{\textrm{U}}$ from Lemma 2, we have $F_{k+1} \in \mathcal D_2^{\textrm{U}}$ . Thus, we complete the proof.

Proof of Theorems 3 and 4. We only give the proof of Theorem 4 as Theorem 3 can be proved similarly based on Lemma 4(ii). Note that the cumulative distribution function of a random variable Z belongs to $\mathcal D_2^{\textrm U}$ if and only if the cumulative distribution function of $T(Z-1)$ belongs to $\mathcal D_2({\textrm U}[0,T], {\textrm U}[-T,0])$ for $T\geq 0$ . Let Z be a random variable with distribution F and let $T\,{:\!=}\,1/b$ and $c \,{:\!=}\, (a+b-1)/b$ , that is, $b=1/T$ and $a=c/T+1-1/T$ . Then $Z_T\,{:\!=}\,T(Z-1)$ is a random variable satisfying

$${\mathbb{P}}(Z_T=c-i)=f_{i+1},\quad i=0,1,2.$$

By Lemma 4(iii), we know that the cumulative distribution function of $Z_T$ belongs to $\mathcal D_2^{\textrm U}$ if and only if one of the three cases of Lemma 4(iii) holds by replacing b with c and T with $1/b$ . That is,

$a+b-1/b=1$ , $f_2\geq rb$ with $1/b=2m\pm r$ with $m\in{\mathbb{N}}$ and $r\in (-1,1)$ ,

${(a+b-1)/b<1}$ , $1/b$ is even, $f_2\in (0, (a+b-1)/b)$ , $f_3=(c-f_2)/2$ and $f_1=1-((a+b-1)/b+f_2)/2$ ,

${(a+b-1)/b<1}$ , $1/b$ is odd, $f_2\in (a+b-1,(a+b-1)/b)$ , $f_3=(c-f_2)/2$ and $f_1=1-(c+f_2)/2$ .

Note that under the constraint $0<b\leq a\leq 1$ , we have $c\leq 1$ ; $c=1$ is equivalent to $a=1$ ; $rb=\lceil 1/(2b) \rfloor$ ; $1/b$ is even if and only if $1/({2b})\in {\mathbb{N}}$ ; $1/b$ is odd if and only if $1/({2b}) -1 /2 \in {\mathbb{N}}$ . Also, by ${\mathbb{E}}[Z]=1$ , we have $f_3=f_1+(1-a)/b$ and thus $f_2=1-2f_1-(1-a)/b$ , which implies that $f_2\in (0, (a+b-1)/b)$ always holds. Hence, the statement in Theorem 3.1 holds.

4. Sums of three or more standard uniform random variables

In this section we aim to show that for $n \geq 3$ the two sets $\mathcal D_n^{\textrm{U}}$ and $\mathcal C^{\textrm{U}}_n$ are identical, in sharp contrast to the case of $n=2$ analysed in Section 3. We start with the bi-atomic distribution. Let F be the distribution function of a random variable X such that ${\mathbb{P}}(X=a)=p$ and ${\mathbb{P}}(X=b)=1-p$ with ${\mathbb{E}}[X]=1/2$ , $a<b$ and $0<p<1$ , and let $T_{n} (F)$ be the distribution function of nX. That is,

$$ F = p \delta _{a} + (1- p) \delta _{ b }\quad \text{and}\quad T_{n} (F) = p\delta _{na} + (1- p) \delta _{n b},$$

where $\delta_x$ denotes the point-mass at $x\in {\mathbb{R}}$ .

Proof. It is easy to verify (iii) $\Rightarrow$ (i). It suffices to show (i) $\Rightarrow$ (ii) $\Rightarrow$ (iii). Let X be a random variable having distribution F and ${\mathbb{P}}(X=a)=p=1-{\mathbb{P}}(X=b)$ . Note that under the constraint of (i), we must have $a,b\in (0,1)$ and $\mu(F)=1/2$ , which implies $0<a<1/2<b<1$ ; under the constraint of (ii), by $\mu(F)=1/2$ , we have $a<1/2<b$ . Then, by $b-a\leq 1/2$ , we also have $a>0$ and $b<1$ . Thus, in the following proof, we always assume $0<a<1/2<b<1$ .

Without loss of generality, we also assume $ 0 < a \leq 1-b \leq 1/2$ by symmetry. Otherwise, consider another random variable $X^*=1-X$ with distribution $F^*$ , and it suffices to note that $\mu(F^*)=1/2$ , $b^*-a^*=b-a$ , and $F{\preceq_{cx}} {{U}}[0,1]$ is equivalent to $F^*{\preceq_{cx}} {{U}}[0,1]$ .

(i) $\Rightarrow$ (ii) By $F{\preceq_{cx}} {{U}}[0,1]$ , we have ${\mathbb{E}}[X]=1/2$ and $ {\mathbb{E}}[(X-t)_+] \leq {(1-t)^2}/{2}, $ for $t\in [0,1]$ , that is,

(3) $$\begin{equation} (b-t)(1-p)\leq \frac{(1-t)^2}{2},\quad a\leq t\leq b. \label{eqn3} \end{equation}$$

To show (ii), we only need to show $b-a\leq 1/2$ , that is, the largest possible value of $b-a$ is $1/2$ . To do this, we fix the value of p and denote $\ell_1(t)\,{:\!=}\, (b-t)(1-p)$ , $t\in [a,b]$ and $\ell_2(t)\,{:\!=}\, {(1-t)^2}/{2}$ , $t\in [0,1]$ . Note that $\ell_1(t)$ is a linear function with fixed and constant derivative equal to $p-1$ and lies below the quadratic curve $\ell_2(t)$ for $t\in [a,b]$ . Also note that as a decreases or b increases, $\ell_1$ moves upwards. Hence, when $b-a$ attains its largest possible value, the line $\ell_1$ and the quadratic curve $\ell_2$ are tangent, and the tangent point satisfies the equation $\ell_2'(t)=t-1=p-1$ , that is, the tangent point is $t=p$ . Then, by $\ell_1(p)=\ell_2(p)$ , we get $b=1+p/2$ , which is the largest possible value of b. Then, by ${\mathbb{E}}[X]=1/2$ , we can get $a=p/2$ , which is the smallest possible value of a. Therefore, the largest possible value of $b-a$ is $1/2$ .

(ii) $\Rightarrow$ (iii) Let $G_{x,y}$ be the distribution of $ {{\textrm{U}}}[x,y]$ . It suffices to show that $G_{0,1}= (1\,{-}\,p) H_1 + p H_2$ such that both $H_1$ and $H_2$ are n-CM, $\mu(H_1)=a$ and $\mu(H_2)=b$ . We first define two distributions $H_1^*$ and $H_2^*$ for the following three cases, with $1-b\geq a$ in mind.

(a) If $a \geq 1/n$ , define $H_1^* = H_2^* = G_{0,1}$ , that is, the distribution function of ${{\textrm{U}}}[0,1]$ .

(b) If $1- b \geq 1/n \geq a $ , define

$$ H_1^* = G_{0,na}\quad \text{and}\quad H_2^* = \frac{na -p}{1-p} G_{0,na}+ \frac{1-na}{1-p} G_{na,1}.$$

By ${\mathbb{E}}[X]=1/2$ , we have

$$p = \frac{b-1/2 }{b-a} = 1- \frac{1/2-a}{b-a} \leq 1- 2\bigg(\frac12-a\bigg) = 2a < na,$$

where the inequality follows from $b-a\leq 1/2$ . Hence, $H_2^*$ is a distribution with positive density function. It is obvious that $ \mu(H_1^*) = {na}/{2} \geq a $ and thus $\mu (H_2^*) \leq b$ as $p\, \mu (H_1^*) + (1\,{-}\,p) \mu(H_2^*) = 1/2$ and $pa+(1-p)b=1/2$ .

(c) If $1-b< 1/n$ , then we have $n=3$ as $b-a\leq 1/2$ and $a\leq 1-b$ . Define

$$ H_1^* = \frac{3b-2}{p } G_{0,3b-2} + \frac{p +2-3b}{p } G_{3b-2, 3a}\quad \text{and}\quad H_2^* = \frac{ 3a -p }{1-p} G_{3b-2, 3a} + \frac{1-3a}{1-p} G_{3a,1}.$$

Note that

$$p = \frac{b-1/2 }{b-a} \geq 2 \bigg(b-\frac{1}{2} \bigg) > 3b-2 \quad\text{and}\quad p < 3a.$$

Hence, both $H_1^*$ and $H_2^*$ are distribution functions with positive densities. It is easy to calculate that

$$ \mu(H_1^*)= \frac{3a + 3b}{2} -1 +\frac{3a }{p} -\frac{ 9 ab }{2p}. $$

Note that

$$\begin{align*} 2p(\mu(H_1^*) - a ) & = p(3b + a - 2) + 3a(2 - 3b)\\* & \geq (3b- 2) (3b + a - 2) - 2(b-a)3a(3b-2)\\ & = (3b-2)( 3b+a -2 +2(b-a)3a )\\ & \stackrel{\rm sgn}=3b - 2 + a + 2(b-a)3a\\* &\geq 0, \end{align*}$$

where the first inequality follows from $p>3b-2$ and $2(b-a)\leq 1$ , $A\stackrel{\rm sgn}=B$ represents that A and B have the same sign, and $\stackrel{\rm sgn}=$ is due to $1-b< 1/n$ and $n=3$ by the observations at the beginning of (c). Hence, we have $\mu(H_1^*)\geq a$ and similarly $\mu(H_2^*)\leq b$ .

In each of the above three cases, we have $G_{0,1}= p H_1^* + (1-p)H_2^*$ , $H_1^*$ has a decreasing density on [0,na] with $\mu(H_1^*) \geq a$ , and $H_2^*$ has an increasing density on $[1-n(1-b),1]$ with $\mu(H_2^*) \leq b$ .

On the other hand, let $H_1^0= G_{0,p}$ and $H_2^0 = G_{p,1}$ . Then it is obvious that $G_{0,1}=pH_1^0 + (1\,{-}\,p)H_2^0$ , $\mu(H_1^0) \leq a $ and $\mu(H_2^0)\geq b$ . Hence, we can find some $\alpha \in [0,1]$ such that $ \alpha \mu(H_1^*) + (1\,{-}\,\alpha) \mu(H_1^0) = a$ . Then the distribution $H_1\,{:\!=}\, \alpha H_1^* + (1-\alpha)H_1^0 $ is supported in [0,na] with decreasing density and $\mu(H_1)=a$ . By [13, Theorem 3.2], G is n-CM, that is, $\delta _{na} \in \mathcal D_n( G,\ldots,G )$ . Similarly, $H_2 = \alpha H_2^* + (1-\alpha)H_2^0 $ is supported in $[1-n(1-b),1]$ with an increasing density and $\mu(H_2)=b$ , which implies that $H_2$ is also n-CM. That is, $ \delta_ {nb} \in \mathcal D_n( H_2,\ldots,H_2)$ . By Lemma 2, we have

$$\begin{align*} T_{n} (F) & = p \delta _{n a} + (1- p) \delta _{n b } \in \mathcal D_n( p H_1 + (1-p )H_2,\ldots, p H_1 + (1-p )H_2) \\* &= \mathcal D_n( {{\textrm{U}}}[0,1] ,\ldots, {{\textrm{U}}}[0,1]). \end{align*}$$

Thus, we complete the proof.

Now we are ready to show our main result in dimension $n\geq 3$ . It turns out that for standard uniform distributions, the two sets $\mathcal D_n(F_1,\ldots, F_n) $ and $ \mathcal C(F_1\oplus\dots\oplus F_n)$ in Lemma 2 coincide. The following theorem is, to the best of our knowledge, the first analytical characterization of $\mathcal D_n$ for continuous marginal distributions.

Proof. As $\mathcal D_n^{\textrm{U}} \subseteq \mathcal C^{\textrm{U}}_n$ , it suffices to show $ \mathcal C^{\textrm{U}}_n\subseteq \mathcal D_n^{\textrm{U}},$ that is, for any distribution function F, $F{\preceq_{cx}} {{U}}[0,1]$ implies $\mathcal D_n^{\textrm{U}}(1/n)$ . Here and throughout the proof, we use the notation $ \mathcal D_n^{\textrm{U}}(x) = \mathcal D_n({\textrm{U}}[0,x],\ldots,{\textrm{U}}[0,x])$ for $x\in{\mathbb{R}}.$ For a distribution F, denote

$$ W_F(t) = {\mathbb{E}} [(X-t)_+] - \frac{(1-t)^2}{2},\quad t \in [0,1] ,$$

where X is a random variable having the distribution function F. We first consider the special case that F is a distribution function of a discrete random variable $(a_1,p_1;\ldots;a_m,p_m)$ with $0\leq a_1<\dots<a_m\leq 1$ and $G_{x,y}$ is the distribution function of ${{\textrm{U}}}[x,y]$ , $x<y$ . By Lemma 4, we know the result holds for $m=2$ . Next, we show that it holds for general $m\geq 2$ by induction.

For general $m\geq 2$ , by $F\preceq_{cx} G_{0,1}$ , we have $W_F(t)\leq 0$ for $t\in[0,1]$ , that is, for $k=2,\ldots,m$ , we have

(4) $$\begin{equation} (a_k-t)p_k+\dots+(a_m-t)p_m\leq \frac{(1-t)^2}{2},\quad a_{k-1}\leq t\leq a_{k}. \label{eqn4} \end{equation}$$

Next, we consider two cases.

(a) If there exists $t\in [a_{k-1}, a_{k})$ such that the equality of (4) holds, then $p_1+\dots+p_{k-1}=t$ . To see this, denote $\ell_1(t)\,{:\!=}\, (a_k-t)p_k+\dots+(a_m-t)p_m$ , and $\ell_2(t)={(1-t)^2}/{2}$ , $t\in [a_{k-1},a_k]$ . Note that $\ell_1$ is a linear function which lies below the decreasing quadratic curve $\ell_2$ . Thus, $\ell_1$ and the quadratic curve $\ell_2$ are tangent at the point t and the tangent point t satisfies the equation $t-1= -p_k-\dots-p_m$ , that is, $p_1+\dots+p_{k-1}=t$ . Let $X_1$ and $X_2$ be two random variables satisfying ${\mathbb{P}}(X_1=a_i)=p_i/t$ , $i=1,\ldots,k-1$ , and ${\mathbb{P}}(X_2=a_i)=p_i/(1-t)$ , $i=k,\ldots,m$ . Let $F_1$ and $F_2$ denote the distributions of $X_1$ and $X_2$ , respectively. It is easy to verify that $F=tF_1+(1-t)F_2$ , $F_1\preceq_{cx} G_{0,t}$ and $F_2\preceq_{cx} G_{t,1}$ . Then, by induction, we have $F_1\in \mathcal D_n^{\textrm{U}}(t/n)$ and $F_2\in \mathcal D_n({\textrm{U}}[t/n,1/n],\ldots,{\textrm{U}}[t/n,1/n])$ . That is, there exist $X_{11},\ldots,X_{1n}\sim {\textrm{U}}[0,t/n]$ and $X_{21},\ldots,X_{2n}\sim {\textrm{U}}[t/n,1/n]$ such that

$$ X_{11}+\dots+X_{1n}\sim F_1 \quad \text{and}\quad X_{21}+\dots+X_{2n}\sim F_2.$$

Without loss of generality, assume that $X_{11},\ldots,X_{1n}$ are independent of $X_{21},\ldots,X_{2n}$ . Let A be a random event independent of $X_{11},\ldots,X_{1n}$ , $X_{21},\ldots,X_{2n}$ such that ${\mathbb{P}}(A)=p_1+\dots+p_{k-1}=t$ . Define

$$ Y_{i} = X_{1i}1_A + X_{2i} 1_{A^c},\quad i=1,\ldots,n.$$

It is obvious that $Y_i\sim {\textrm{U}}[0,1]$ , $i=1,\ldots,n$ , and $Y_1+\dots+Y_n\sim F$ . This means that $F\in \mathcal D_n^{\textrm{U}}(1/n,\ldots, 1/n)$ .

(b) If the inequality of (4) is strict for every $t\in (0,1)$ , define two functions G and H by

$$G = \lambda \delta_{a_1} +(1-\lambda)\delta_{a_m} \quad \text{and} \quad H = \frac{F - \theta G }{1-\theta},$$

where $\delta_{a}$ denotes the degenerate distribution at point a,

$$\lambda=\frac{a_m - 1/2 } {a_m-a_1} \quad \text{and} \quad \theta = \min \bigg\{ \frac {p_1} \lambda, \frac{p_m} {1-\lambda}\bigg\}< 1.$$

It is easy to verify that G and H are two distribution functions satisfying $ F = \theta G + (1-\theta) H $ . We also assert that $ H {\preceq_{cx}} F$ . To see it, let $X\sim F,$ $Y \sim G ,$ and $Z \sim H $ . Then, for any convex function $\phi$ , we have

$$\phi(X) \leq \frac{a_m - X }{ a_m - a_1 } \phi(a_1) + \frac{ X-a_1 }{ a_m - a_1 } \phi(a_m) \quad \text{a.s.}$$

It then follows that

$$ {\mathbb{E}}[\phi(X)] \leq \frac{a_m - {\mathbb{E}} [X] }{ a_m - a_1 } \phi(a_1) + \frac{ {\mathbb{E}}[X]-a_1 }{ a_m - a_1 } \phi(a_m) = \lambda \phi(a_1) + (1-\lambda) \phi(a_m) = {\mathbb{E}} [\phi(Y)]{.}$$

Also, note that $F = \theta G + (1-\theta) H$ which implies $ {\mathbb{E}}[\phi(X)] = \theta {\mathbb{E}}[\phi(Y)] + (1- \theta ) {\mathbb{E}}[\phi(Z)]$ . Combined with $ {\mathbb{E}}[\phi(X)] \leq {\mathbb{E}}[\phi(Y)]$ , we have ${\mathbb{E}}[\phi(Z)] \leq {\mathbb{E}}[\phi(X)],$ that is, $ H {\preceq_{cx}} F$ . This implies $H{\preceq_{cx}} {{U}}[0,1]$ , and note that the support of distribution H has at most $m-1$ points. By induction hypothesis, we have $H \in \mathcal D_n^{\textrm{U}}(1/n)$ .

On the other hand, if $ a_m - a_1 \leq \frac{1}{2} $ , we know $ G \preceq_{cx} {{U}}[0,1] $ , $G \in \mathcal D_n^{\textrm{U}}(1/n,\ldots, 1/n)$ . Then $F = \theta G + (1-\theta) H \in \mathcal D_n^{\textrm{U}}(1/n)$ . Otherwise, if $ a_m - a_1 > \frac{1}{2} $ , then $ G \preceq_{cx} {{U}}[0,1]$ , that is, $ W_G(t_0) > 0 $ for some $t_0\in(0,1)$ . Define a function

$$\alpha(t) = -\frac{ W_{G} (t) }{ W_{F} (t)} ,\quad t\in (0,1){,}$$

which is a continuous function satisfying $\alpha (0+) = \alpha (1- ) = - 1 $ as $ W_F(t) = W_G(t) = - {t^2}/{2}$ for $t\in [0,a_1] $ and $W_F(t) = W_G(t) = -{(1-t)^2}/{2}$ for $t\in [a_m,1] $ . Hence, $\alpha(t) $ takes its maximum value at some $t\in (0,1)$ and the maximum value is positive. Without loss of generality, assume $\alpha_0 = \alpha (t_1) > 0 $ is its maximum value. Then we have

(5) $$\begin{equation} W_G (t) + \alpha_0 W_F(t) \leq 0 \quad \text{for all } t\in [0,1]\ \text{and} \ W_G (t_1) + \alpha_0 W_F(t_1) = 0. \label{eqn5} \end{equation}$$

Define a distribution $F_0 \,{:\!=}\, (G + \alpha_0 F )/({1+ \alpha_0})$ . Then, by (5), we have $F_0 {\preceq_{cx}} {{U}}[0,1]$ and $W_{F_0} (t_1 ) = 0 $ . By case (a), we have $F_0 \in \mathcal D_n^{\textrm{U}}(1/n)$ .

Note that $ G = (1+\alpha_0) F_0 - \alpha_0 F $ and $F = \theta G + (1-\theta) H $ , which implies $ F = (1+\alpha_0) \theta F_0 - \alpha_0 \theta F + (1-\theta) H, $ that is,

$$F= \frac { ( \theta+ \alpha_0 \theta ) F_0 + (1-\theta) H }{ 1 + \alpha_0 \theta }.$$

Then, by Lemma 2, we have $F \in \mathcal D_n^{\textrm{U}}(1/n)$ .

If F is a general distribution function such that $F\preceq_{cx} {U}[0,1]$ , then ${Supp(F)\subset [0,1]}$ and it has no mass on 0 and 1. For any $n\in{\mathbb{N}}$ , define $F_n$ as the distribution function of $X_n$

$$ X_n = \sum_{k=1}^{n} {\mathbb{E}} \bigg[X \biggm| \frac{k-1}{n} \leq X < \frac{k}{n} \bigg] 1_{\{ \frac{k-1}{n} \leq X < \frac{k}{n} \}}$$

where X is a random variable having distribution function F. Then $F_n$ converges to F in weak convergence as $n\to\infty$ , and $F_n\preceq_{cx} {U}[0,1]$ . By the above proof for discrete distributions with finite support, we have $F_n\in \mathcal D_n^{\textrm{U}}(1/n)$ for each $n\in{\mathbb{N}}$ . Then, by Lemma 2, we have $F\in \mathcal D_n^{\textrm{U}}(1/n)$ . Thus, we complete the proof.

For any random variable $X\sim F$ with mean 0, we have $F_a \preceq_{cx} F$ for any $a\in [0,1]$ , where $F_a$ is the distribution of aX. Hence, we immediately get the following corollary. Note that this corollary, although it looks simple, does not seem to allow for an elementary proof without using Theorem 5.

5. An application

In risk management, one often needs to optimize a statistical functional, mapping $\mathcal F$ to ${\mathbb{R}}$ (such as a risk measure), over the set of $\mathcal D_n(F_1,\ldots,F_n)$ , and this type of problem is called risk aggregation with dependence uncertainty (see e.g. [Reference Bernard, Jiang and Wang2] and [Reference Embrechts, Puccetti and Rüschendorf3]). These problems are typically quite difficult to solve in general, as the set $\mathcal D_n(F_1,\ldots,F_n)$ is a complicated object. For uniform marginal distributions, using results in this paper (in particular, Theorem 5), we are able to translate many optimization problems on $\mathcal D_n^{\textrm{U}}$ to $\mathcal C^{\textrm{U}}_n$ for $n\geq 3$ , which is a convenient object to work with.

We study an application of the problem of minimizing or maximizing, for a given interval A, the value of ${\mathbb{P}}(S\in A)$ , where S is the sum of n standard uniform random variables. A special case of the problem concerning bounds on ${\mathbb{P}}(S\leq x)$ for $x\in {\mathbb{R}}$ , i.e. bounds on F(x) for $F\in \mathcal D^{\textrm U} _n $ , is studied in [Reference Rüschendorf8]. Using Theorem 5, we are able to solve the problem of ${\mathbb{P}}(S\in A)$ completely.

Proposition 2. For $n\geq 3$ , and $0\leq a\leq a+b\leq n$ , we have

(6) $$\begin{align} \min_{F_S\in \mathcal D^{\textrm U}_n} {\mathbb{P}}(S\in (a,a+b)) = \bigg(\frac{2b}n-1\bigg)_+, \label{eqn6} \end{align}$$

and

(7) $$\begin{align} \max_{F_S\in \mathcal D^{\textrm U}_n }{\mathbb{P}}(S\in [a,a+b]) = \min\bigg\{\frac{2(a+b)}n, \frac{2(n-a)}n,1\bigg\}, \label{eqn7} \end{align}$$

where $F_S$ stands for the cdf of S.

Proof. As Theorem 5 gives $ \mathcal D^{\textrm U} _n=\mathcal C^{\textrm{U}}_n$ , it suffices to look at the optimization problems for $ \mathcal C^{\textrm{U}}_n$ . For $0\leq u\leq v\leq n$ , let $\mathcal A_{u,v}$ be the sigma field generated by $\{U\leq u\}$ and $\{U\leq v\}$ . $S={\mathbb{E}}[U|\mathcal A_{u,v}]$ is tri-atomically distributed with distribution measure

(8) $$\begin{align} \frac un \delta_{u/2} + \frac{v-u}{n}\delta_{(u+v)/2}+ \frac{n-v}{n}\delta_{(n+v)/2}.\label{eqn8} \end{align}$$

Note that $F_S\preceq_{cx} F_U$ because S is a conditional expectation of U, and thus $F_S \in \mathcal C^{\textrm{U}}_n= \mathcal D^{\textrm U} _n$ .

We first verify the following inequality (indeed, it is [8, Theorem 1]). For any S which is the sum of n standard uniform random variables and $x\in {\mathbb{R}}$ , we have

(9) $$\begin{equation} {\mathbb{P}}(S\leq x)\leq 2x/n \quad\text{and}\quad {\mathbb{P}}(S\geq x)\leq 2(n-x)/n.\label{eqn9} \end{equation}$$

Equation (9) can be shown using the equivalent condition of convex order (see [10, Theorem 3.A.5]). Using $F_S\preceq_{cx} F_U$ ,

$$\begin{align*} F^{-1}_S(\alpha)\alpha \geq \int_0^{\alpha} F^{-1}_S(t) {\,\textrm{d}} t \geq \int_0^{\alpha } F^{-1}_U(t){\,\textrm{d}} t = \frac{n\alpha^2}{2}, \quad \alpha \in (0,1) . \end{align*}$$

Therefore, $F^{-1}_S(\alpha)\geq \frac{n\alpha}{2}$ , $\alpha \in (0,1)$ , and equivalently $F_S(x)\leq 2x/n$ , $x\in {\mathbb{R}}$ . The other inequality in (9) is symmetric by noting that $n-S$ is still the sum of n standard uniform random variables and ${\mathbb{P}}(S\geq x)={\mathbb{P}}(n-S\leq n-x)$ .

We now analyse the problem of the minimum in (6).

Suppose $b\leq n/2$ . Since $(n+u)/2-u/2 =n/2>b$ , we can find $u\in [0,n]$ such that $u/2\leq a$ and ${(n+u)/2\geq b+a}$ . By letting $S={\mathbb{E}}[U|\mathcal A_{u,u}]$ and using (8), we have ${\mathbb{P}}(S=u/2) = u/n$ and ${\mathbb{P}}(S=(n+u)/2)=(n-u)/n$ . In this case, ${\mathbb{P}}(S\in (a,a+b))=0$ ; thus (6) holds.

Suppose $b>n/2$ , which implies $a<n/2$ and $a+b>n/2$ . Let S be given by ${\mathbb{E}}[U|\mathcal A_{u,v}]$ where $u=2a$ and $v=2(a+b)-n$ . Note that ${\mathbb{P}}(S\in (a,a+b))= \frac{v-u}{n} = \frac{2b}n-1.$ This shows the ‘ $\leq $ ’ direction of (6). On the other hand, by (9), for any S which is the sum of n standard uniform random variables, ${\mathbb{P}}(S\leq a)\leq 2a/n$ and ${\mathbb{P}}(S\geq a+b)\leq 2(n-a-b)/n.$ Thus

$${\mathbb{P}}(S\in (a,a+b))\geq 1-\frac{2a}n - \frac{2(n-a-b)}{n} = \frac{2b}n-1.$$

This shows the ‘ $\geq$ ’ direction of (6).

Next, we analyse the problem of the maximum in (7).

If $a+b> n/2$ and $a< n/2$ , then $n/2\in [a,a+b]$ . Taking $S={\mathbb{E}}[U]=n/2$ gives ${\mathbb{P}}(S\in [a,a+b])=1$ ; thus (7) holds.

Suppose $a+b\leq n/2$ . By (9), for any S which is the sum of n standard uniform random variables, ${\mathbb{P}}(S\in [a,a+b])\leq {\mathbb{P}}(S\leq a+b) \leq 2(a+b)/n$ . To see that such a bound is attainable, take $S={\mathbb{E}}[U|\mathcal A_{u,u}]$ where $u=2(a+b)$ . Then, by (8), we have

$${\mathbb{P}}(S\in [a,a+b])\geq {\mathbb{P}}(S=a+b) = {\mathbb{P}}\bigg(S=\frac u2\bigg) = \frac un = \frac{2(a+b)}n.$$

Therefore, (7) holds.

Suppose $a\geq n/2$ . Similarly to the above case, by (9), for any S which is the sum of n standard uniform random variables, ${\mathbb{P}}(S\in [a,a+b])\leq {\mathbb{P}}(S\geq a) \leq 2(n-a)/n$ . To see that such a bound is attainable, take $S={\mathbb{E}}[U|\mathcal A_{u,u}]$ where $u=2a-n$ . Then, by (8), we have

$$\pi(S\in [a,a+b])\geq {\mathbb{P}}(S=a) = {\mathbb{P}}\bigg(S=\frac {n+u}2\bigg) = \frac {n-u}n = \frac{2(n-a)}n.$$

Therefore, (7) holds.

Remark 2. Based on the proof of Proposition 5, we can identify some minimizing distributions for (6) and some maximizing distributions for (7). For $0\leq u\leq v\leq n$ , write the distribution

$${F_{u,v}} = \frac{u}{n}{\delta _{u/2}} + \frac{{v - u}}{n}{\delta _{(u + v)/2}} + \frac{{n - v}}{n}{\delta _{(n + v)/2}}.$$

There are a few cases. For the minimum in (6):

If $b\leq n/2$ , then $F_{u,u}$ attains (6) for $u \in [2a+2b-2n, 2a]$ .

If $b> n/2$ , then $F_{u,v} $ attains (6) where $u=2a$ and $v=2(a+b)-n$ .

For the maximum in (7):

If $a+b> n/2$ and $a< n/2$ , then $\delta_{n/2}$ attains (7).

If $a+b\leq n/2$ , then $F_{u,u}$ attains (7) where $u=2(a+b)$ .

If $a\geq n/2$ , then $F_{u,u}$ attains (7) where $u=2a-n$ .