<p>Let us assume all these functions exist and are differentiable. Furthermore let us denote the inverse of the function <span class="math-container">$\overline u^1(x_1)$</span> by <span class="math-container">$x_1\left(\overline u^1\right)$</span>.</p>

<p>Note that for any invertible function <span class="math-container">$f$</span> we have
<span class="math-container">$$
\left.\frac{\text{d} f(x)}{\text{d} x}\right|_{x=x_0} =
\frac{1}{\left.\frac{\text{d} f^{-1}(y)}{\text{d} y}\right|_{y=f(x_0)}}. 
$$</span>
In our case this means (with slightly simplified notation)
<span class="math-container">$$
\frac{\text{d} x_1(\overline u^1)}{\text{d} \overline u^1} =
\frac{1}{\frac{\text{d} \overline u^1(x_1)}{\text{d} x_1}}.
$$</span>
We will further simplify notation by writing
<span class="math-container">$$
\overline u{^i}' =  \frac{\text{d} \overline u^i(x_1)}{\text{d} x_1}.
$$</span>
We will denote second derivatives similarly by affixing <span class="math-container">$''$</span>.</p>

<hr>

<p>The question is under what conditions
<span class="math-container">$$
\overline u^2\left(x_1\left(\overline u^1\right)\right)
$$</span> 
is concave in <span class="math-container">$\overline u^1$</span>. This of course, depends on the sign of
<span class="math-container">$$
\frac{\text{d}^2 \overline u^2\left(x_1\left(\overline u^1\right)\right)}{\text{d} \overline {u}{^1}^2}.
$$</span>
Differentiating
<span class="math-container">\begin{align*}
\frac{\text{d} \overline u^2\left(x_1\left(\overline u^1\right)\right)}{\text{d} \overline {u}^1} &amp; =
\frac{\text{d} x_1(\overline u^1)}{\text{d} \overline u^1}
\frac{\text{d} \overline u^2(x_1)}{\text{d} x_1}
=
\frac{1}{\overline u{^1}' }\overline u{^2}' 
=
\frac{\overline u{^2}'}{\overline u{^1}' }.
\end{align*}</span>
Going one step further
<span class="math-container">\begin{align*}
\frac{\text{d}^2 \overline u^2\left(x_1\left(\overline u^1\right)\right)}{\text{d} \overline {u}{^1}^2} &amp; =
\frac{
\frac{1}{\overline u{^1}' }\overline u{^2}^{''} \overline u{^1}'
-
\overline u{^2}'\frac{1}{\overline u{^1}' }\overline u{^1}^{''} 
}
{\left(\overline u{^1}'\right)^2}.
\end{align*}</span>
This implies that the Pareto-frontier is concave when
<span class="math-container">\begin{align*}
\frac{\text{d}^2 \overline u^2\left(x_1\left(\overline u^1\right)\right)}{\text{d} \overline {u}{^1}^2} &amp; \leq 0 \\
\\
\frac{
\frac{1}{\overline u{^1}' }\overline u{^2}^{''} \overline u{^1}'
-
\overline u{^2}'\frac{1}{\overline u{^1}' }\overline u{^1}^{''} 
}
{\left(\overline u{^1}'\right)^2} &amp; \leq 0 \\
\\
\overline u{^2}^{''}
 &amp; \leq \frac{\overline u{^2}'}{\overline u{^1}' }\overline u{^1}^{''}
\end{align*}</span>
for all <span class="math-container">$x_1$</span>.</p>

CC BY-SA 4.0
2020-01-14T21:05:51.623Z
2020-01-14T21:05:51.623Z
Giskard
<p>Consider an exchange economy with two agents.
Each agent <span class="math-container">$i \in \{1,2\}$</span> derives utility <span class="math-container">$u^i(x_1,x_2) \in \mathbb R$</span> by consuming <span class="math-container">$(x_1,x_2) \in \mathbb R_+^2$</span>. 
Let <span class="math-container">$u_j^i(x_1,x_2) = \partial u^i(x_1,x_2) / \partial x_j$</span> for <span class="math-container">$(i,j) \in \{1,2\}^2$</span>.
The set of efficient consumption bundles is given by
<span class="math-container">\begin{align}
C = \left\{(x_1,x_2) \in \mathbb R_+^2 ~ \bigg| ~ \frac{u_1^1(x_1,x_2)}{u_2^1(x_1,x_2)} = \frac{u_1^2(x_1,x_2)}{u_2^2(x_1,x_2)}\right\}
\end{align}</span>
Suppose the function <span class="math-container">$x_2^c :\mathbb R_+ \to \mathbb R_+$</span> solves 
<span class="math-container">\begin{align}
\frac{u_1^1(x_1,x_2^c(x_1))}{u_2^1(x_1,x_2^c(x_1))} = \frac{u_1^2(x_1,x_2^c(x_1))}{u_2^2(x_1,x_2^c(x_1))}
\end{align}</span>
The contract curve is the graph <span class="math-container">$\mathcal G(x_2^c) = \{(x_1,x_2^c(x_1)) \mid x_1 \in \mathbb R_+\}$</span>.
Let <span class="math-container">$\overline u^i(x_1) = u^i(x_1,x_2^c(x_1))$</span>.
The Pareto frontier is now given by the following utility allocations
<span class="math-container">\begin{align}
P = \{(\overline u^1(x_1), \overline u^2(x_1)) \mid x_1 \in \mathbb R_+\} \subset \mathbb R^2.
\end{align}</span></p>

<p>Suppose that the Parento frontier is downward sloping w.r.t. <span class="math-container">$x_1$</span> such that
<span class="math-container">\begin{align}
\frac{\overline u_1^2(x_1)}{\overline u_1^1(x_1)} &lt; 0.
\end{align}</span></p>

<blockquote>
  <p><strong>Q</strong>: What conditions must be met such that the Pareto frontier is concave?</p>
</blockquote>

<p>I'm especially wondering if the following condition is sufficient: 
<span class="math-container">\begin{align}
\frac{\overline u_{11}^2(x_1)}{\overline u_{11}^1(x_1)} &lt; 0.
\end{align}</span></p>

<microeconomics><pareto-efficiency><pure-exchange-economy><edgeworth-box>
Contract curve and Pareto frontier
clueless
# Contract curve and Pareto frontier

(Asked by: username_0 on 2020-01-14)

Consider an exchange economy with two agents.
Each agent $i \in \{1,2\}$ derives utility $u^i(x\_1,x\_2) \in \mathbb R$ by consuming $(x\_1,x\_2) \in \mathbb R\_+^2$.
Let $u\_j^i(x\_1,x\_2) = \partial u^i(x\_1,x\_2) / \partial x\_j$ for $(i,j) \in \{1,2\}^2$.
The set of efficient consumption bundles is given by
\begin{align}
C = \left\{(x\_1,x\_2) \in \mathbb R\_+^2 ~ \bigg| ~ \frac{u\_1^1(x\_1,x\_2)}{u\_2^1(x\_1,x\_2)} = \frac{u\_1^2(x\_1,x\_2)}{u\_2^2(x\_1,x\_2)}\right\}
\end{align}
Suppose the function $x\_2^c :\mathbb R\_+ \to \mathbb R\_+$ solves
\begin{align}
\frac{u\_1^1(x\_1,x\_2^c(x\_1))}{u\_2^1(x\_1,x\_2^c(x\_1))} = \frac{u\_1^2(x\_1,x\_2^c(x\_1))}{u\_2^2(x\_1,x\_2^c(x\_1))}
\end{align}
The contract curve is the graph $\mathcal G(x\_2^c) = \{(x\_1,x\_2^c(x\_1)) \mid x\_1 \in \mathbb R\_+\}$.
Let $\overline u^i(x\_1) = u^i(x\_1,x\_2^c(x\_1))$.
The Pareto frontier is now given by the following utility allocations
\begin{align}
P = \{(\overline u^1(x\_1), \overline u^2(x\_1)) \mid x\_1 \in \mathbb R\_+\} \subset \mathbb R^2.
\end{align}

Suppose that the Parento frontier is downward sloping w.r.t. $x\_1$ such that
\begin{align}
\frac{\overline u\_1^2(x\_1)}{\overline u\_1^1(x\_1)} < 0.
\end{align}

> **Q**: What conditions must be met such that the Pareto frontier is concave?

I'm especially wondering if the following condition is sufficient:
\begin{align}
\frac{\overline u\_{11}^2(x\_1)}{\overline u\_{11}^1(x\_1)} < 0.
\end{align}

**Tags:** microeconomics, pareto-efficiency, pure-exchange-economy, edgeworth-box


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## 1 answers

### Answer by username_1 on 2020-01-14. Score: 0 | Accepted answer

Let us assume all these functions exist and are differentiable. Furthermore let us denote the inverse of the function $\overline u^1(x\_1)$ by $x\_1\left(\overline u^1\right)$.

Note that for any invertible function $f$ we have
$$
\left.\frac{\text{d} f(x)}{\text{d} x}\right|\_{x=x\_0} =
\frac{1}{\left.\frac{\text{d} f^{-1}(y)}{\text{d} y}\right|\_{y=f(x\_0)}}.
$$
In our case this means (with slightly simplified notation)
$$
\frac{\text{d} x\_1(\overline u^1)}{\text{d} \overline u^1} =
\frac{1}{\frac{\text{d} \overline u^1(x\_1)}{\text{d} x\_1}}.
$$
We will further simplify notation by writing
$$
\overline u{^i}' = \frac{\text{d} \overline u^i(x\_1)}{\text{d} x\_1}.
$$
We will denote second derivatives similarly by affixing $''$.

---

The question is under what conditions
$$
\overline u^2\left(x\_1\left(\overline u^1\right)\right)
$$
is concave in $\overline u^1$. This of course, depends on the sign of
$$
\frac{\text{d}^2 \overline u^2\left(x\_1\left(\overline u^1\right)\right)}{\text{d} \overline {u}{^1}^2}.
$$
Differentiating
\begin{align\*}
\frac{\text{d} \overline u^2\left(x\_1\left(\overline u^1\right)\right)}{\text{d} \overline {u}^1} & =
\frac{\text{d} x\_1(\overline u^1)}{\text{d} \overline u^1}
\frac{\text{d} \overline u^2(x\_1)}{\text{d} x\_1}
=
\frac{1}{\overline u{^1}' }\overline u{^2}'
=
\frac{\overline u{^2}'}{\overline u{^1}' }.
\end{align\*}
Going one step further
\begin{align\*}
\frac{\text{d}^2 \overline u^2\left(x\_1\left(\overline u^1\right)\right)}{\text{d} \overline {u}{^1}^2} & =
\frac{
\frac{1}{\overline u{^1}' }\overline u{^2}^{''} \overline u{^1}'
-
\overline u{^2}'\frac{1}{\overline u{^1}' }\overline u{^1}^{''}
}
{\left(\overline u{^1}'\right)^2}.
\end{align\*}
This implies that the Pareto-frontier is concave when
\begin{align\*}
\frac{\text{d}^2 \overline u^2\left(x\_1\left(\overline u^1\right)\right)}{\text{d} \overline {u}{^1}^2} & \leq 0 \\
\\
\frac{
\frac{1}{\overline u{^1}' }\overline u{^2}^{''} \overline u{^1}'
-
\overline u{^2}'\frac{1}{\overline u{^1}' }\overline u{^1}^{''}
}
{\left(\overline u{^1}'\right)^2} & \leq 0 \\
\\
\overline u{^2}^{''}
& \leq \frac{\overline u{^2}'}{\overline u{^1}' }\overline u{^1}^{''}
\end{align\*}
for all $x\_1$.


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