The general model

The model is as follows. There are $n$ agents (nations), they can trade some type of good, and they use the same currency. Every agent may produce or consume the good. The benefit function of the $j$th agent is $B_j$, and the cost function is $C_j$. The amount of export from the $j$th agent to the $k$th agent is $T_{j,k}$. The amount of trade cost by the $j$th agent is $S_j$. Now, we want to find the amount $Q_j$ that every agent produce and the amount $T_{j,k}$ that every agent import from other agents. Assume that $S$ is only related to $T$ and does not depend on $Q$. Also, assume that there is no externality (i.e. whenever $j\ne k$, $\partial_kB_j=0$ and $\partial_kC_j=0$). Also, assume that every agent is rational and with perfect information.

Now, consider the profit $\Pi_j$ of the $j$th agent. Subtract the cost from the benefit, and we have $$\textstyle \Pi_j=B_j\!\left(Q_j+\sum_kT_{j,k}\right)-C_j\!\left(Q_j\right)-S_j\!\left(T\right).$$ According to the fundamental theorem of welfare economics, $T$ and $Q$ is Pareto optimal under market equilibrium. We assume that this case happens at the stationary point of the social benefit, and the social benefit is the sum of the profit of every agent. We can then get the equations $$\begin{align*} &0=\frac{\partial}{\partial Q_l}\sum_j\Pi_j =B_l'\!\left(Q_l+\sum_kT_{l,k}\right)-C_l'\!\left(Q_l\right),\quad\forall l;\\ &0=\frac{\partial}{\partial T_{l,k}}\sum_j\Pi_j =B_l'\!\left(Q_l+\sum_kT_{l,k}\right)-B_m'\!\left(Q_m+\sum_kT_{m,k}\right) -\sum_j\frac{\partial S_j}{\partial T_{l,m}}\!\left(T\right),\quad\forall l<m. \end{align*}$$ Here are $n+\frac{n\left(n-1\right)}2$ equations, and exactly $Q$ and $T$ have $n+\frac{n\left(n-1\right)}2$ degrees of freedom in total (note that $T$ is anti-symmetric). In principle, we are able to solve $Q$ and $T$.

Zero trade cost

For the case where there is no trade cost, we can see that the domestic prices are all equal, and the price may be called the world price.

However, given $S=0$, the equations above are not independent. Actually, there are only $2n-1$ independent equations (all $2n$ components of $B'$ and $C'$ are equal). This means that, for $n>2$, the free trade with zero trade cost is an indeterminate system.

This phenomenon looks counter-intuitive, but it is actually understandable: under zero trade cost, every two agents may trade arbitrary amount of goods under the same world price, this provides extra degrees of freedom to the model. To be specific, if $(Q,T)$ is a solution to the model, then $(Q,T+\Delta T)$ is also a solution, where the anti-symmetric matrix $\Delta T$ satisfies $$\sum_k\Delta T_{j,k}=0,\quad\forall j,$$ where there are $n-1$ independent equations in the $n$ equations. Therefore, the total number of degrees of freedom in the solution of the model is $$n+\frac{n\left(n-1\right)}2-\left(\frac{n\left(n-1\right)}2-\left(n-1\right)\right)=2n-1.$$ Now, the useful quantities that we can solve is the production and the net-import $T_j\coloneqq\sum_kT_{j,k}$ of every agent. Note that the net-import actually has only $n-1$ degrees of freedom because of the restriction $\sum_jT_j=0$.

The middleman (re-exportation)

It is worth pointing out that the existence of the middleman or re-exportation is completely due to the presence of trade cost. Here we consider a simplified problem: there are three agents playing respectively as the producer, the retailer, and the customer. The producer does not consume (the benefit is $0$); the customer does not produce (the cost and the marginal cost is infinity); and the retailer does not produce or consume. Assume that the trade between any two of them does not bring cost to the third one. Then, the social benefit is $$\Pi=B\!\left(T_{\mathrm c,\mathrm r}+T_{\mathrm c,\mathrm p}\right) -C\!\left(T_{\mathrm c,\mathrm p}+T_{\mathrm r,\mathrm p}\right) -S_\mathrm c\!\left(T_{\mathrm c,\mathrm r},T_{\mathrm c,\mathrm p}\right) -S_\mathrm r\!\left(T_{\mathrm c,\mathrm r},T_{\mathrm r,\mathrm p}\right) -S_\mathrm p\!\left(T_{\mathrm c,\mathrm p},T_{\mathrm r,\mathrm p}\right).$$

A (binary) voting system is a tuple $(P,V,q)$, where $P$ is any set, called the set of proposals, and $V$ is a finite set of preference relations on $P$, called the set of voters, and $q$ is an integer between (inclusive) $0$ and $\left|V\right|$, called the quota.

For each voter $v\in V$ and two proposals $x,y\in P$, we denote “$v$ prefers $x$ to $y$” by $$x\succeq_vy.$$ A proposal $x\in P$ is a defeat of $y\in P$ if $$\left|\left\{v\in V\,\middle|\,x\succeq_vy\right\}\right|\geq q,$$ denoted as $x\succsim_{V,q}y$ (despite this notation, $\succsim_{V,q}$ is not necessarily a preference relation on $P$ because it is not transitive generally, which is actually a well-known example of irrationality).

The core $\mathcal C(P,V,q)$ of the voting system is the set of such element $x\in P$: $x$ does not have any defeat other than $x$ itself (non-trivial defeat).

Pareto sets are common concepts in economics. To clarify, I also give the mathematical definition of them here.

Let $P$ be a set and $Q$ be a family of preference relations on $P$. Then, $x\in P$ is called a (weak) $Q$-Pareto improvement of $y\in P$ if $\forall v\in V:x\succeq_vy$, denoted as $x\succsim_Qy$ (despite the notation, $\succsim_Q$ is not necessarily a preference relation on $P$).

The Pareto set $\mathcal P(P,Q)$ is the set of all such element $x\in P$: $x$ does not have any $Q$-Pareto improvement other than $x$ itself (non-trivial $Q$-Pareto improvement).

Here is the main result. For a voting system $(P,V,q)$, $$\mathcal C(P,V,q)=\bigcap_{Q\subseteq V,\left|Q\right|=q}\mathcal P(P,Q).$$

Proof. To prove this, we need to show that $x\in P$ does not have any non-trivial Pareto improvement for any $q$ voters iff $x$ does not have any non-trivial defeat.

To prove the forward direction, suppose that $x\in P$ does not have any non-trivial Pareto improvement for any $q$ voters. Let $y\in P$ such that $y\ne x$, and the goal is to prove that $y$ is not a defeat of $x$.

Let $$Y\coloneqq\left\{v\in V\,\middle|\,y\succeq_vx\right\}.$$ Then, $y$ is a $Y$-Pareto improvement of $x$, so we have $\left|Y\right|<q$ (because otherwise there is a subset of $Y$ with $q$ voters for which $y$ is a Pareto improvement of $x$). Therefore, $y$ is not a defeat of $x$.

To prove the backward direction, suppose that $x\in P$ has a non-trivial $Q$-Pareto improvement, where $Q\subseteq V$ and $\left|Q\right|=q$. Denote the improvement as $y$. Let $$Y\coloneqq\left\{v\in V\,\middle|\,y\succeq_vx\right\}.$$ because $y$ is a $Q$-Pareto improvement of $x$, we have $Q\subseteq Y$. Therefore, $\left|Y\right|\geq\left|Q\right|=q$. Therefore, $y$ is a defeat of $x$. $\square$

Specially, we have $$\mathcal C\!\left(P,V,\left|V\right|\right)=\mathcal P(P,V).$$

Here is an example. Suppose we have 5 voters, and the set of proposals is $\mathbb R^2$. Each voter has an ideal point and prefers points nearer to the ideal point. The 5 ideal points form a convex pentagon. Then we can find the core easily by the conclusion above:

First, define the Lorenz curve: it is the curve that consists of all points $(u,v)$ such that the poorest $u$ portion of population in the country owns $v$ portion of the total wealth.

The Gini coefficient $G/\mu$ is defined as the area between the Lorenz curve and the line $u=v$ divided by the area enclosed by the three lines $u=v$, $v=0$, and $u=1$.

Now, suppose the wealth distribution in the country is $p(X)$, where $p\!\left(x\right)\mathrm dx$ is the portion of population that has wealth in the range $[x,x+\mathrm dx]$.

Then, the Lorenz curve is the graph of the function $g$ defined as $$g(F(x))=\frac1\mu\int_{-\infty}^xtp\!\left(t\right)\mathrm dt,$$ where $$F\!\left(x\right)\coloneqq\int_{-\infty}^xp\!\left(t\right)\mathrm dt$$ is the cumulative distribution function of $p(X)$, and $$\mu\coloneqq\int_{-\infty}^{+\infty}tp\!\left(t\right)\mathrm dt$$ $$(1)$$ is the average wealth of the population, which is just $\mathrm E[\mathrm X]$ ($X$ is a random variable such that $X\sim p(X)$).

Then, the Lorenz curve is $$v=g(u)\coloneqq\frac1\mu\int_{-\infty}^{F^{-1}(u)}tp\!\left(t\right)\mathrm dt.$$

According to the definition of the Gini coefficient, $$\begin{align*} G&\coloneqq2\mu\int_0^1\left(u-g(u)\right)\mathrm du\\ &=\mu-2\mu\int_0^1g\!\left(u\right)\mathrm du\\ &=\mu-2\int_{u=0}^1\int_{t=-\infty}^{F^{-1}(u)}tp\!\left(t\right)\mathrm dt\,\mathrm du. \end{align*}$$ Interchange the order of integration, and we have $$\begin{align*} G&=\mu-2\int_{t=-\infty}^{+\infty}\int_{u=F(t)}^1tp\!\left(t\right)\mathrm dt\,\mathrm du\\ &=\mu-2\int_{-\infty}^{+\infty}\left(1-F(t)\right)tp\!\left(t\right)\mathrm dt. \end{align*}$$ Substitute Equation 1 into the above equation, and we have $$\begin{align*} G&=\int_{-\infty}^{+\infty}2tF\!\left(t\right)p\!\left(t\right)\mathrm dt-\mu\\ &=\int_{-\infty}^{+\infty}\left(2tF\!\left(t\right)-1\right)tp\!\left(t\right)\mathrm dt\\ &=\int_0^1\left(2u-1\right)F^{-1}\!\left(u\right)\mathrm du. \end{align*}$$ Now here is the neat part. Separate it into two parts, and write them in double integrals: $$\begin{align*} G&=\int_0^1uF^{-1}\!\left(u\right)\mathrm du-\int_0^1\left(1-u\right)F^{-1}\!\left(u\right)\mathrm du\\ &=\int_{u_2=0}^1\int_{u_1=0}^{u_2}F^{-1}\!\left(u_2\right)\mathrm du_1\,\mathrm du_2 -\int_{u_1=0}^1\int_{u_2=u_1}^1F^{-1}\!\left(u_1\right)\mathrm du_1\,\mathrm du_2. \end{align*}$$ Interchange the order of integration of the second term, and we have $$\begin{align*} G&=\int_{u_2=0}^1\int_{u_1=0}^{u_2}\left(F^{-1}\!\left(u_2\right)-F^{-1}\!\left(u_1\right)\right)\mathrm du_1\,\mathrm du_2\\ &=\frac12\int_{u_2=0}^1\int_{u_1=0}^1\left|F^{-1}\!\left(u_2\right)-F^{-1}\!\left(u_1\right)\right|\mathrm du_1\,\mathrm du_2\\ &=\frac12\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}\left|x_2-x_1\right|p\!\left(x_1\right)p\!\left(x_2\right)\mathrm dx_1\,\mathrm dx_2\\ &=\frac12\mathrm E\!\left[\left|X_2-X_1\right|\right], \end{align*}$$ where $X_1$ and $X_2$ are two independent random variables with $p$ being their respective distribution functions: $\left(X_1,X_2\right)\sim p\!\left(X_1\right)p\!\left(X_2\right)$.

By this result, we can easily see how the Gini coefficient represents the statistical dispersion.

We can apply similar tricks to the variance $\sigma_X^2$. $$\begin{align*} \sigma_X^2&=\mathrm E\!\left[X^2\right]-\mathrm E\!\left[X\right]^2\\ &=\int_{-\infty}^{+\infty}t^2p\!\left(t\right)\mathrm dt -\left(\int_{-\infty}^{+\infty}tp\!\left(t\right)\mathrm dt\right)^2\\ &=\int_0^1F^{-1}\!\left(u\right)^2\,\mathrm du -\left(\int_0^1F^{-1}\!\left(u\right)\mathrm du\right)^2. \end{align*}$$ Separate the first into two halves, and write the altogether three terms in double integrals: $$\begin{align*} \sigma_X^2&=\frac12\int_0^1F^{-1}\!\left(u_2\right)^2\,\mathrm du_2\int_0^1\mathrm du_1\\ &\phantom{=~}{}-\int_0^1F^{-1}\!\left(u_1\right)\mathrm du_1\int_0^1F^{-1}\!\left(u_2\right)\mathrm du_2\\ &\phantom{=~}{}+\frac12\int_0^1F^{-1}\!\left(u_1\right)^2\,\mathrm du_1\int_0^1\mathrm du_2\\ &=\frac12\int_0^1\int_0^1 \left(F^{-1}\!\left(u_2\right)^2-2F^{-1}\!\left(u_1\right)F^{-1}\!\left(u_2\right)+F^{-1}\!\left(u_1\right)^2\right) \mathrm du_1\,\mathrm du_2\\ &=\frac12\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} \left(x_2-x_1\right)^2p\!\left(x_1\right)p\!\left(x_2\right)\mathrm dx_1\,\mathrm dx_2\\ &=\frac12\mathrm E\!\left[\left(X_2-X_1\right)^2\right]. \end{align*}$$ Then we can derive the relationship between the Gini coefficient and the variance: $$2\sigma_X^2-4G^2=\sigma_{\left|X_2-X_2\right|}^2.$$