This problem comes from a subfield of economics called decision theory, where we evaluate a decision from the distribution of outcomes it may entail according to a group of $N$ experts. According to expert $j$, there is a probability $p_{i,j}$ that the decision under consideration yield a utility of at least $\sigma_1 + \ldots + \sigma_i$. The expected utility of the decision is thus $u_j=\sum_{i=1}^{n}\sigma_{i}\cdot p_{i,j}$ according to expert $j$. To aggregate the $N$ experts' assessments, the decision-rule we consider aggregates the experts' decumulative density functions ($p_{1,j} \geq \ldots \geq p_{n,j}, \forall j$) using an aggregator function $I$, and evaluates the decision with the aggregated probabilities: $u=\sum_{i=1}^{n}\sigma_{i}I\left(p_{i,1},\ldots,p_{i,N}\right)$.

Here is my conjecture. Consider an aggregator $I$ such that when all experts agree on the utility ($u_j=x, \forall j$) the decision-rule also agrees with them ($u=x$); then $I$ must be an arithmetic mean. Proving this conjecture will help me show that two notions of being neutral to uncertainty coincide. Formally:

Let $N\in \mathbb{N}$. Let $I:[0;1]^N \rightarrow [0;1]$ be a continuous, symmetric, increasing function such that $I(p,\ldots,p)=p,\forall p\in [0;1]$. (By strictly increasing we mean that if $p_1 \geq q_1, \ldots, p_N \geq q_N$ then $I(\overrightarrow p) \geq I(\overrightarrow q)$ and the inequality is strict if $\overrightarrow p \neq \overrightarrow q$.)

Suppose the following property holds:

For all $n\in \mathbb{N}$, for all $(\sigma_{i})_{i\in \{1;\ldots;n\}}\geq 0$, for all $(p_{i,j})\in [0;1]^{nN}$, for all $x\in \mathbb{R} $ such that $p_{1,j} \geq \ldots \geq p_{n,j}, \forall j$ and $$\sum_{i=1}^{n}\sigma_{i}\cdot p_{i,j}=x,\forall j\in\left\{ 1;\ldots;N\right\}\quad (1)$$

then $$\sum_{i=1}^{n}\sigma_{i}I\left(p_{i,1},\ldots,p_{i,N}\right)=x\quad (2)$$

Show that $$\forall p_1,\ldots,p_N \in [0;1], \quad I\left(p_{1},\ldots,p_{N}\right)=\sum_{j=1}^{N}\frac{1}{N}p_{j}\quad (3)$$

To begin, notice that, using (1): $x=\sum_{j=1}^{N}\frac{1}{N}\sum_{i=1}^{n}\sigma_{i}\cdot p_{i,j}$, so that, using (2) $$\sum_{i=1}^{n}\sigma_{i}I\left(p_{i,1},\ldots,p_{i,N}\right)=\sum_{i=1}^{n}\sigma_{i}\sum_{j=1}^{N}\frac{1}{N}p_{i,j}\quad (4)$$

Thus, the problem of proving (3) amounts to "getting rid of the sum over $i$" in (4).

I also have a related conjecture, which is the same as above except that $(2)$ is replaced by $$\sum_{i=1}^{n}\sigma_{i}I\left(p_{i,1},\ldots,p_{i,N}\right)<x\quad (2^\prime)$$ and $(3)$ is replaced by $$\forall p_1,\ldots,p_N \in [0;1], \quad I\left(p_{1},\ldots,p_{N}\right)<\sum_{j=1}^{N}\frac{1}{N}p_{j}\quad (3^\prime)$$ and yet a similar conjecture where the strict inequalities are replaced by non-strict inequalities.

I believe that those conjectures are true but can't find a proof. Any idea?

  • 1
    $\begingroup$ What ordering are you putting on $[0,1]^N$ --- lex/colex/reverse lex/componentwise/...? $\endgroup$ Apr 20, 2021 at 10:45
  • $\begingroup$ I've edited the question to specify which (partial) ordering we consider. $\endgroup$
    – bixiou
    Apr 20, 2021 at 13:39
  • $\begingroup$ If anyone is interested, I have written down a detailed proof for the main conjecture. It is tedious and technical so I don't reproduce here. And I don't give the idea of the proof either as I let the person who gave me the idea of the proof answer the question and get the credit. $\endgroup$
    – bixiou
    May 19, 2021 at 23:40
  • $\begingroup$ For the record, the last two conjectures are false. $\endgroup$
    – bixiou
    Sep 2, 2021 at 22:45

1 Answer 1


Yes, the conjecture is true. First, let me simplify slightly the problem by considering that $I(\bullet, \ldots, \bullet)$ is only defined on $\mathopen]0, 1\mathclose[^N$: this is not a true restriction, because of the continuity assumption. I will also assume that $I(\bullet, \ldots, \bullet)$ has $\mathrm{C}^1$ regularity (that is not a true restriction either, as it will turn out, but let us not discuss that point for now).

For $\vec{p}, \vec{q} \in \mathopen]0, 1\mathclose[^N$, I denote $\vec{p} \succ \vec{q}$ to mean that one has $p_j > q_j$ for all $j$. Let $\vec{p}_1, \vec{p}_2 \in \mathopen]0, 1\mathclose[^N$ with $\vec{p}_1 \succ \vec{p}_2$; my goal in the next lines will be to prove that one has $\partial_{p_1} I(\vec{p}_1) = \partial_{p_1}{I}(\vec{p}_2)$, where “$\partial_{p_1}{I}(\vec{p})$” refers to the derivative of $I(\bullet, \ldots, \bullet)$ along its first argument, assessed at point $\vec{p}$.

To prove that goal, the first observation is that it is possible to find $\vec{p}_3 \prec \vec{p}_2$ and $\sigma_3 > 0$ such that $$\vec{p}_1 + \vec{p}_2 + \sigma_3 \vec{p}_3 = (x, \ldots, x)$$ for some $x \in \mathbf{R}$: indeed, just pick arbitrarily $y > 0$ such that $(y, \ldots, y) \prec \vec{p}_2$; let $\epsilon > 0$ be small enough so that $(y, \ldots, y) - \epsilon(\vec{p}_1 + \vec{p}_2) \succ (0, \ldots, 0)$; and then take $\vec{p}_3 = (y, \ldots, y) - \epsilon(\vec{p}_1 + \vec{p}_2)$, $\sigma_3 = \epsilon^{-1}$ and $x = (2 + \sigma_3) y$. Such values being fixed, for $h \in \mathbf{R}$, denote $\vec{p}_1(h) = (p_{1, 1} + h, p_{1, 2}, \ldots, p_{1, N})$, resp. $\vec{p}_2(h) = (p_{2, 1} + h, p_{2, 2}, \ldots, p_{2, N})$: then, for all $h$ close enough to $0$, one has $\vec{p}_1(h) \succ \vec{p}_2(-h) \succ \vec{p}_3$ and $\vec{p}_1(h) + \vec{p}_2(-h) + \sigma_3 \vec{p}_3 = (x, \ldots, x)$; consequently, according to your assumptions, one must have $$I(\vec{p}_1(h)) + I(\vec{p}_2(-h)) + \sigma_3 I(\vec{p}_3) = x \text,$$ regardless of the value of $h$. Differentiating w.r.t. $h$ at $h = 0$, we then get that $\partial_{p_1}{I}(\vec{p}_1) - \partial_{p_1}{I}(\vec{p}_2) = 0$, which proves the goal that I had been setting in the previous paragraph.

That being established, one readily deduces that $\partial_{p_1}{I}(\bullet, \ldots, \bullet)$ has to be constant on the whole $\mathopen]0, 1\mathclose[^N$: to see that, in the first time show that the partial derivative has to be constant on the diagonal $\{(x, \ldots, x) \mathrel{|} x \in \mathopen]0, 1\mathclose[\}$; and then that for any $\vec{p} \in \mathopen]0, 1\mathclose[^N$ there is some $x \in \mathopen]0, 1\mathclose[$ such that $\vec{p} \succ (x, \ldots, x)$. Of course, on could similarly deduce that, for all $j \in \{2, \ldots, N\}$, $\partial_{p_j}{I}$ is constant too. Therefore $I(\bullet, \ldots, \bullet)$ is an affine application; and the only affine application being compatible with the imposed values on the diagonal and the symmetry assumption is the arithmetic mean. So, modulo that $\mathrm{C}^1$ regularity assumption, we have proved the conjecture indeed.

All the above was stated under the regularity assumption. Without the regularity assumption, the basic idea remains the same, but everything gets quite more involved technically… Fix $h > 0$ very small and define the discretized partial derivative of $I$ at $\vec{p}$ as $$\tilde{\partial}{I}_{p_1}(\vec{p}) = \frac{I(p_1 + h, p_2, \ldots, p_N) - I(\vec{p})}{h} \text.$$

With this definition, the reasoning that we have been carrying above actually proves that $\tilde{\partial}{I}_{p_1}(\vec{p}_1) = \tilde{\partial}{I}_{p_1}(p_{2, 1} - h, p_{2, 2}, \ldots, p_{2, N})$ as long as $\vec{p}_1 \succ \vec{p}_2$, $p_{1, 1} < 1 - h$ and $p_{2, 1} > h$. (Note that, for everything to work correctly in this case, you have to impose that $y < p_{2, 1} - h$ at the point when you pick $y$).

From this, you get that $\tilde{\partial}{I}_{p_1}$ has the same value at $(1/4, \ldots, 1/4)$ and at $(3/4, \ldots, 3/4)$ (assuming that $h$ was chosen $< 1/2$); then that it is constant on $\{(x, \ldots, x) \mathrel| x \in \mathopen]0, 1 - h\mathclose[\}$ (indeed, assuming that $h$ was chosen $< 1/4$, you can equalize $\tilde{\partial}{I}_{p_1}(x, \ldots, x)$ with $\tilde{\partial}{I}_{p_1}(1/4, \ldots, 1/4)$ for $x \geq 1/2$, resp. with $\tilde{\partial}{I}_{p_1}(3/4, \ldots, 3/4)$ for $x < 1/2$); and finally you get that $\tilde{\partial}{I}_{p_1}(\bullet, \ldots, \bullet)$ is constant on (at least) $\mathopen]h, 1 - 2h\mathclose[^N$. By symmetry, all the other discretized partial derivatives also have to be constant on that set, and equal to the same value as for $\tilde{\partial}{I}_{p_1}$; and this value is necessarily $1/N$, because $$h = (1/2 + h) - 1/2 = I(1/2 + h, \ldots, 1/2 + h) - I(1/2, \ldots, 1/2) = h \tilde{\partial}{I}_{p_1}(1/2, \ldots, 1/2) + h \tilde{\partial}{I}_{p_2}(1/2, 1/2 + h, 1/2, \ldots, 1/2) + \cdots + h \tilde{\partial}{I}_{p_N}(1/2, \ldots, 1/2, \ldots, 1/2 + h) \text.$$

From that you deduce that for all the $\vec{p} \in \mathopen]h, 1 - 2 h\mathclose[^N$ such that $p_j \in 1/2 + \mathbf{Z} h\ \forall j$, $I(\vec{p})$ coincides with the arithmetic mean: indeed, start from the fact that it coincides at $(1/2, \ldots, 1/2)$, and then observe that, due to the values of the partial derivatives, coinciding at $\vec{p}$ implies coinciding at $\vec{q}$, where $\vec{q}$ differs from $\vec{p}$ by adding or substracting $h$ to any coordinate. Hence, using the increasing nature of $I(\bullet, \ldots, \bullet)$, it follows that $I(\vec{p})$ is at distance at most $h$ from the arithmetic mean for all $\vec{p} \in \mathopen]2 h, 1 - 3 h\mathclose[^N$. But since $h$ could be taken arbitrarily small, this implies that $I(\bullet, \ldots, \bullet)$ is actually the arithmetic mean on the whole $\mathopen]0, 1\mathclose[^N$, which concludes the proof in the general case.

  • $\begingroup$ I completely changed the terminal argument for the non-$\mathrm{C}^1$ case. Indeed, the convolution argument was actually not correct, because convoluting would not ensure that $\tilde{I}$ has the required value on the diagonal… :-| $\endgroup$ May 22, 2021 at 14:21

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