simultaneous probability elicitation from multiple agents without an exogenous banker A proper scoring rule is a function $f:[0,1]\times\{0,1\}\to \Bbb R$ such that, if a subject will receive a reward of $f(x,0)$ for reporting his estimate of the likelihood of an event as $x$ if the event does not occur, and a reward of $f(x,1)$ for doing the same if the event does occur, then the subject maximizes his expected reward by reporting his true estimate of the likelihood of an event. Now suppose we have $n$ subjects. Can we find a rule $f:[0,1]^n\times \{0,1\}\to \Bbb R^n$ such that, for fixed $a_1,\ldots,a_{j-1},a_{j+1},\ldots a_n$, the function that takes $(x,y)$ to the $j$th coordinate of $f(a_1,\ldots,a_{j-1},x,a_{j+1},\ldots,a_n,y)$ is a proper scoring rule, and moreover the range of $f$ is confined to tuples that sum to zero? 
Edit: if it is hard to think directly about the general case, does anyone know the answer for the case $n=2$? I guess large groups of subjects can be decomposed into smaller groups, so solving the problem for $n=2$ would give a solution of all even $n$, and solving it for $n=2$ and $n=3$ would give a solution for all $n$. If possible, I would like a solution that treats all agents identically though.
 A: For $n\ge 2$, let $a_j\in[0,1]$ be the stated probability estimate of $j$ that $\theta=1$ (which means the event occurs). Moreover, $x_j$ is the net transfer $j$ receives (based on $\theta$ and $a_j$ of all $j$). 
Now consider a Brier scoring rule $BS_j=(a_j-\theta)^2$. Whenever the estimate $a_j$ deviates from the realization $\theta$, that score is positive. Now every $j$ must pay $BS_j$ into an account. Overall, we will collect $\sum_j BS_j$ in that account. This is positive iff at least one $j$ did not predict the outcome perfectly. The transfer everybody receives back is the average contribution, i.e.,
$$n^{-1}\sum_j BS_j.$$
The net transfer to $j$ is therefore 
$$x_j=-(a_j-\theta)^2 +n^{-1}\sum_j (a_j-\theta)^2.$$
Clearly, $\sum_j x_j=0$. It remains to be shown this makes truth telling the optimal strategy. I will assume agents are risk neutral, so every $j$  wants to maximize the expectation of his transfer,
$$\max_{a_j} E[x_j]=\max_{a_j} E[-(a_j-\theta)^2 +n^{-1}\sum_j (a_j-\theta)^2].$$
This is concave, and the first order condition is
$$-2a_j+2E[\theta]+1/n(2a_j-2E[\theta])=\frac{n-1}{n}\left(-2a_j+2E[\theta]\right)=0$$
so that
$$a_j=E[\theta] ~\forall j.$$
The expectation here refers to $j$'s subjective estimate of $\theta$, i.e., what you want to elicit. Hence, this scoring and transfer scheme induces risk neutral agents to reveal the truth for $n\ge 2$, because those who have smaller Brier score  than the average (=estimate closer to realization $\theta$) will earn money, while those who are off will lose money. The stakes can be raised or lowered by changing the constant factor of the scoring rule.
