<p>Let <span class="math-container">$F$</span> be the set of feasible allocations i.e. the points in the Edgeworth-box. Consider any allocation <span class="math-container">$((x_1, y_1), (x_2, y_2)) \in F$</span>, to check for efficiency of this allocation - consider the two sets:</p>
<ul>
<li><span class="math-container">$B_1 = \{((x_1', y_1'), (x_2', y_2'))\in F | u_1(x_1', y_1') \geq u_1(x_1, y_1) \}$</span> (Upper level set for individual 1)</li>
<li><span class="math-container">$B_2 = \{((x_1', y_1'), (x_2', y_2'))\in F | u_2(x_2', y_2') \geq u_2(x_2, y_2) \}$</span> (Upper level set for individual 2)</li>
</ul>
<p>If at every allocation <span class="math-container">$((x_1', y_1'), (x_2', y_2')) \in B_1\cap B_2$</span>, we have</p>
<ul>
<li><span class="math-container">$u_1(x_1', y_1') = u_1(x_1, y_1)$</span> and</li>
<li><span class="math-container">$u_2(x_2', y_2') = u_2(x_2, y_2)$</span></li>
</ul>
<p>then the allocation <span class="math-container">$((x_1, y_1), (x_2, y_2))$</span> is Pareto efficient. If not, <span class="math-container">$((x_1, y_1), (x_2, y_2))$</span> is not Pareto efficient.</p>

CC BY-SA 4.0
2022-05-05T11:09:36.873Z
2022-05-05T11:09:36.873Z
Amit
<p>I have been reviewing calculations to find walrasian equilibrium and pareto efficient allocations.</p>

<p>Assume we are in an environment with two consumers, $A$ and $B$, and two goods, $x$ and $y$. From <a href="https://en.wikipedia.org/wiki/Pareto_efficiency" rel="nofollow">this article</a> in wikipedia, we have that a sufficient condition for pareto efficiency is that the marginal rates of substitution are equal for both consumers. </p>

<p>However for practical purposes this seems only useful when utility functions are differentiable. Assuming preferences are continuous and locally non-satiated, is there any "fast and loose" rule for pareto efficiency when utility functions are not differentiable? </p>

<p>For example with leontieff preferences $u(x,y) = \min\{ x,2y \}$ in which case there might be a corner solution.</p>

CC BY-SA 3.0
2015-12-17T16:39:35.880Z
Are we in a setting where the non-differentiable utility functions for both consumers are weakly monotonically increasing  in both goods and continuous?
BKay
CC BY-SA 3.0
2015-12-17T16:51:36.323Z
@Bkay Yes (otherwise I think the question becomes somewhat pathological). I will edit the question.
möbius
CC BY-SA 3.0
2015-12-17T17:22:28.537Z
If the non-differentiability is caused by kinks in the indifference curves (is in the extreme case of Leontieff) then one should be able to calculate a left-hand MRS and a right-hand MRS (being the MRS to the left and right of the kink respectively. Pareto optimality should then reduce to a comparison between these left- and right-hand MRSs. In fact, think about a standard 2-person Edgeworth box. When we say the MRSs should be equal in a standard differentiable, convex case, we are implicitly saying that each agent's left hand MRS should be greater than the other guy's right-hand MRS.
Ubiquitous
<pareto-efficiency>
Corner Solutions for Pareto Efficiency
möbius
# Corner Solutions for Pareto Efficiency

(Asked by: username_0 on 2015-12-17)

I have been reviewing calculations to find walrasian equilibrium and pareto efficient allocations.

Assume we are in an environment with two consumers, $A$ and $B$, and two goods, $x$ and $y$. From [this article](https://en.wikipedia.org/wiki/Pareto_efficiency) in wikipedia, we have that a sufficient condition for pareto efficiency is that the marginal rates of substitution are equal for both consumers.

However for practical purposes this seems only useful when utility functions are differentiable. Assuming preferences are continuous and locally non-satiated, is there any "fast and loose" rule for pareto efficiency when utility functions are not differentiable?

For example with leontieff preferences $u(x,y) = \min\{ x,2y \}$ in which case there might be a corner solution.

**Tags:** pareto-efficiency

> **Comments:**
> **username_1**: Are we in a setting where the non-differentiable utility functions for both consumers are weakly monotonically increasing  in both goods and continuous?  
> **username_0**: @Bkay Yes (otherwise I think the question becomes somewhat pathological). I will edit the question.  
> **username_2**: If the non-differentiability is caused by kinks in the indifference curves (is in the extreme case of Leontieff) then one should be able to calculate a left-hand MRS and a right-hand MRS (being the MRS to the left and right of the kink respectively. Pareto optimality should then reduce to a comparison between these left- and right-hand MRSs. In fact, think about a standard 2-person Edgeworth box. When we say the MRSs should be equal in a standard differentiable, convex case, we are implicitly saying that each agent's left hand MRS should be greater than the other guy's right-hand MRS.  

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## 1 answers

### Answer by username_3 on 2022-05-05. Score: 2

Let $F$ be the set of feasible allocations i.e. the points in the Edgeworth-box. Consider any allocation $((x\_1, y\_1), (x\_2, y\_2)) \in F$, to check for efficiency of this allocation - consider the two sets:

* $B\_1 = \{((x\_1', y\_1'), (x\_2', y\_2'))\in F | u\_1(x\_1', y\_1') \geq u\_1(x\_1, y\_1) \}$ (Upper level set for individual 1)
* $B\_2 = \{((x\_1', y\_1'), (x\_2', y\_2'))\in F | u\_2(x\_2', y\_2') \geq u\_2(x\_2, y\_2) \}$ (Upper level set for individual 2)

If at every allocation $((x\_1', y\_1'), (x\_2', y\_2')) \in B\_1\cap B\_2$, we have

* $u\_1(x\_1', y\_1') = u\_1(x\_1, y\_1)$ and
* $u\_2(x\_2', y\_2') = u\_2(x\_2, y\_2)$

then the allocation $((x\_1, y\_1), (x\_2, y\_2))$ is Pareto efficient. If not, $((x\_1, y\_1), (x\_2, y\_2))$ is not Pareto efficient.


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