4
$\begingroup$

Let $$\mathbb R_{++}^2\equiv\{(x_1,x_2)\in\mathbb R^2\,|\,x_1>0\text{ and }x_2>0\}$$ be the positive orthant in the Cartesian plane. Suppose that $f:\mathbb R_{++}^2\to\mathbb R$ is a function satisfying the following properties:

(Continuity) The function $f$ is continuous.

(Translation invariance of value comparisons) For any $\mathbf x,\mathbf y,\mathbf z\in\mathbb R^2_{++}$, one has $$f(\mathbf x)\geq f(\mathbf y)\quad\text{if and only if}\quad f(\mathbf x+\mathbf z)\geq f(\mathbf y+\mathbf z).$$

(Monotonicity) The function $f$ is strictly increasing in the first argument and strictly decreasing in the second. That is, whenever $x>y>0$ and $z>0$, \begin{align} f(x,z)>&\;f(y,z)\text{ and}\\\\ f(z,x)<&\;f(z,y). \end{align}

Conjecture: There exists a transformation $T:f(\mathbb R_{++}^2)\to\mathbb R$, where $f(\mathbb R_{++}^2)\subseteq\mathbb R$ is the range of $f$, such that

  • $T$ is strictly increasing; and
  • The composite function $T\circ f$ is linear; that is, there exist $\alpha_1,\alpha_2\in\mathbb R$ such that $$T(f(x_1,x_2))=\alpha_1 x_1+\alpha_2 x_2$$ for any $(x_1,x_2)\in\mathbb R^2_{++}.$

Intuitively, translation invariance indicates that the level sets of $f$ in the plane are parallel straight lines (which are upward sloping and the direction of increase is toward southeast, given the monotonicity assumption), which suggests that $f$ is “linear up to relabeling,” if you will, according to some monotonic transformation. Yet, I have found this intuitive reasoning quite hard to formalize. Any hint would be appreciated.


Note #1 We are working in the positive orthant of $\mathbb R^2$, so unfortunately the scope of exploiting the full vector-space structure of $\mathbb R^2$ seems limited (in particular, $(0,0)\notin\mathbb R^2_{++}$).

Note #2 The case when $f$ is strictly $\textit{in}\text{creasing}$ in both arguments has an elegant solution (due to Bruno de Finetti, I think); in that case, one can define $$\varphi(x_1,x_2)\equiv\inf\big\{t\in[\min\{x_1,x_2\},\max\{x_1,x_2\}]\,\big|\;f(t,t)\geq f(x_1,x_2)\big\}$$ for each $(x_1,x_2)\in\mathbb R_{++}^2$. That is, a new function is constructed by considering the equivalent-value projection of each point onto the 45$^{\circ}$ line. One then can show (with some work) that $\varphi$ is linear and preserves the same order structure on $\mathbb R_{++}^2$ as does $f$, so that $f$ and $\varphi$ are order-theoretically the “same” modulo some strictly increasing transformation. That said, I find it surprisingly and frustratingly difficult to adapt de Finetti’s nice proof to the current case in which $f$ is strictly $\textit{de}\text{creasing}$ in its second argument.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.