= Solution
First, equality of two $U_0$ values forces equality of the corresponding $V_0$ values: neither strict comparison holds for $U_0$, and the same is then true for $V_0$ in both directions. Thus $V_0=g\circ U_0$ for a well-defined <strictly increasing function> $g$ on $J=U_0(\mathcal P)$. Since $\mathcal P$ is a <convex set> and $U_0,V_0$ are <affine functions>, $J$ is a <real interval> and
$$
g(pu+(1-p)v)=pg(u)+(1-p)g(v),\qquad u,v\in J,\quad0\leq p\leq1.
$$
If $J$ contains $u<v$, let $a=[g(v)-g(u)]/(v-u)>0$ and $b=g(u)-au$. The mixture identity gives $g(w)=aw+b$ for $w\in[u,v]$. For $w>v$, express $v$ as a <convex combination> of $u$ and $w$ and solve the same identity for $g(w)$; for $w<u$, express $u$ as a mixture of $w$ and $v$. Hence the formula holds on all of $J$, not merely between the chosen anchors. We obtain <positive affine uniqueness of affine preference representations>:
$$
\boxed{V_0(\lambda)=aU_0(\lambda)+b\quad\text{for every }\lambda\in\mathcal P,\qquad a>0.}
$$
If $U_0$ is constant, the same comparisons force $V_0$ to be constant; choose $a=1$ and the appropriate $b$. If the probability-measure set is empty, the assertion is vacuous. These degenerate cases do not require distinct anchors.
Back to article page
