Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $u(x),v(x)$ be continuous bounded functions on $\mathbb{R}$ such that for any Borel probability measures $\mathbb{P}_{1},\mathbb{P}_2$ on $\mathbb{R}$ $$ \int u(x) \, \mathbb{P}_1(dx) \leqslant \int u(x) \, \mathbb{P}_{2}(dx) \;\;\; \left(\text{def: } \mathbb{P}_1 \preccurlyeq_{u} \mathbb{P}_2 \right) $$ holds if and only if $$ \int v(x) \, \mathbb{P}_{1}(dx) \leqslant \int v(x) \, \mathbb{P}_{2}(dx) \;\;\; (\mathbb{P}_1 \preccurlyeq_{v} \mathbb{P}_2) $$ How to show that in this case $u(x) = av(x) + b$, $a,b = \mathrm{const}$, $a>0$?

If we take $P_1 = \delta_{x_1}$, $P_2 = \delta_{x_2}$ we can obtain that $u(x_1) \leqslant u(x_2)$ holds if and only if $v(x_1) \leqslant v(x_2)$, so $u(x) = f(v(x))$, where $f(\cdot)$ is some increasing function. But what to do next?

share|cite|improve this question
up vote 1 down vote accepted

Assume first that $u(0)=v(0)=0$ and $u(1)=v(1)=1$. Choose $x$ not in $\{0,1\}$, and $\mathbb P_i=a_i\delta_1+b_i\delta_x+(1-a_i-b_i)\delta_0$ for some nonnegative $(a_i,b_i)$ such that $a_i+b_i\leqslant1$.

Then $\mathbb P_1\preccurlyeq_{u}\mathbb P_2$ if and only if $a_1+b_1u(x)\leqslant a_2+b_2u(x)$, likewise for $\mathbb P_1\preccurlyeq_v\mathbb P_2$. This means that, for every nonnegative $(a_i,b_i)$ such that $a_i+b_i\leqslant1$, $a_1+b_1u(x)\leqslant a_2+b_2u(x)$ if and only if $a_1+b_1v(x)\leqslant a_2+b_2v(x)$.

But, if $u(x)\ne v(x)$, there exists such $(a_i,b_i)$ such that $a_1+b_1u(x)\lt a_2+b_2u(x)$ and $a_1+b_1v(x)\gt a_2+b_2v(x)$. Hence $u(x)=v(x)$ for every $x$ not in $\{0,1\}$, that is, $u=v$.

The general case follows.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.