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}


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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