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Given a Leontief model with $n \times n$ input-output matrix $B$, whose diagonal elements are positive and off-diagonal elements are non-positive. There is a single unit of labor available to this economy. Let $x$ be the vector of labor allocated to each of the $n$ sectors. A vector $y$ is a net output of the system provided $y=Bx$ for some $x\ge 0,$ and it is feasible whenever $\sum_{j=1}^n x_j \le 1$. A vector $\bar{y}$ is efficient if $\bar{y}$ is feasible and $y\ge \bar{y}, y \ne \bar{y}$ implies $y$ is not feasible. The set of efficient net outputs is denoted $E(B)$. Finally, a Leontief model is productive if there exists $y^*>>0, x^*\ge0$ such that $y^*=Bx^*.$

I would like to prove the following claim: Suppose the Leontief model is productive. Then, there is a price vector $p>>0$ such that $$E(B)=\{y\in\mathbb{R}_+^n:p \cdot y=1\}.$$

Here, the labor wage rate is the numeraire and normalized to 1.

tvk
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  • Restrict to positive inputs for the moment. Consider the image of the standard simplex $\Delta_n$ under the linear map $B$. This is a convex set and your $E(B)$ is a face of it. Productivity means this convex set is full-dimensional, therefore the face in question is part of a hyperplane (cut down to the positive quadrant). That hyperplane is the price vector you're looking for. – Michael Dec 13 '13 at 04:24

1 Answers1

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The set of feasible outputs is the image $B(H)$ of the half-space $H = \{ x \in \mathbb{R}^n, \sum x_i \leq 1\} $ under the linear map $B$. $B(H)$ is a convex set and $E(B)$ is a face of it.

Productivity implies that $B(H)$ is full-dimensional and has compact intersection with $\mathbb{R}^n_+$, because an element $x^*$ in the standard simplex $\Delta_n$ is mapped to the interior of $\mathbb{R}^n_+$.

In other words, $B(H)$ is itself a half-space of the form $\{ y \in \mathbb{R}^n_+:\; p \cdot y \leq 1\}$. So you're done.

Michael
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  • Thank you for the answer. It seems that my mathematical background, mainly in real analysis, is insufficient to understand your proof. May I know what areas of mathematics I should study to understand this proof? Specifically, I don't quite understand what is defined to be a "face," why $B(H)$ is a half-space, etc. Thank you very much! – tvk Dec 14 '13 at 01:03
  • Try drawing some pictures in $\mathbb{R}^2$. $H$ is the set ${ x + y \leq 1}$. You want to show its image under $B$ is of the form ${ ax + by \leq c}$ where $a$, $b$, and $c$ are strictly positive. The image of $x + y = 1$ is determined by where $B$ takes the basis vectors. Once you get this line, you have to check it intersects the positive quadrant and that you get the right half. The productivity assumption gives you that. – Michael Dec 14 '13 at 03:52