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I have a two-agent exchange economy where utility functions are given by: $u_{1}=x_{12}-x_{21}$ and $u_2=x_{21}x_{22}$. I am looking to find the set of pareto efficient allocations.

Since utility functions are differentiable, then at any interior pareto efficient allocation, we know that the MRS of either consumer must be equal. So we need that: $\frac{0}{1}=\frac{x_{22}}{x_{21}}$. Thus, in any interior pareto efficient allocation, $x_{22}=0$. But of course any such allocation is along the boundary of the edgeworth box (and so is not interior). Does this mean that there are no interior pareto efficient allocations?

Thank you.

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It seems like any allocation where $X_{11}=X_{12}$ is Pareto efficient.

Recall from Edgeworth box, $1 = X_{11} + X_{21}$ and $1 = X_{12} + X_{22}$ Start with the utility function: $$U_1 = X_{12} - X_{21} $$ Use the first Edgeworth box identity to substitute out $X_{21}$. $$\Rightarrow U_1 = X_{12} + (X_{11} - 1) $$ $$\Rightarrow MU_{11} = \frac{\partial U_1}{X_{11}} = 1 = \frac{\partial U_1}{X_{12}} = MU_{12} = 1 $$ At interior bundle utility maximization: $$\Rightarrow \frac{MU_{11}}{P_1} = \frac{1}{P_1} = \frac{MU_{12}}{P_2} = \frac{1}{P_2} \rightarrow P_1 = P_2$$ So the first guy is happy as long as the two goods have the same price. $$U_2 = X_{21} \cdot X_{22} $$ $$\Rightarrow MU_{21} = \frac{\partial U_2}{X_{21}} = X_{22} $$ $$\Rightarrow MU_{22} = \frac{\partial U_2}{X_{22}} = X_{21} $$ $$\Rightarrow \frac{MU_{21}}{P_1} = \frac{X_{22}}{P_1} = \frac{MU_{22}}{P_2} = \frac{X_{21}}{P_2} \rightarrow X_{21} = X_{22}$$ But, again because of the Edgeworth both, this implies $$ \Rightarrow 1- X_{11} = X_{21} = X_{22} = 1- X_{12} $$ $$ \Rightarrow X_{11} = X_{12} $$

Which, when I check the result, manually seems to work. Consider $$\Rightarrow X_{11} = X_{12} = 0.25 \Rightarrow X_{21} = X_{22} = 0.75$$ How many units of $X_{12}$ does person 1 want to give up 0.01 units of $X_{11}$? He wants .02 units: $$U_{1,old} = X_{12} + (X_{11} - 1) = 0.25 + 0.25 - 1 = -0.5$$ $$U_{1,new} = X_{12} + (X_{11} - 1) = 0.24 + 0.26 - 1 = -0.5$$ But that trade leaves the second person worse off: $$U_{2,old} = X_{21} \cdot X_{22} = 0.75^2 = 9 / 16 = 0.5625 $$ $$U_{2,new} = X_{21} \cdot X_{22} = 0.76 \cdot 0.74 = 0.562400365 < U_{2,old}$$ So there is no trade to do.

We should also check that the corner solutions aren't better.

  1. {1,0} is worse for person two
  2. {0,1} is worse for person two
  3. {1,1} is worse for person two
  4. {0,0} is worse for person one

More generally:

If $X_{11}=0 \rightarrow X_{21}=1$ and then $U_2 \geq 0.5625 \rightarrow X_{22} \geq 0.5625$ this in turn implies that $X_{12} \leq 0.4375 \rightarrow U_1 \leq 0.4375 - 1 = -0.5625 < -0.5$ which is not a pareto improvement.

If $X_{11}=1 \rightarrow X_{21}=0$ and then $U_2 \geq 0.5625$ has no valid solutions.

Similarly, if $X_{12}=1 \rightarrow X_{22}=0$ and then $U_2 \geq 0.5625$ has no valid solutions.

If $X_{12}=0 \rightarrow X_{22}=1$ and then $U_2 \geq 0.5625 \rightarrow X_{21} \geq 0.5625$ this in turn implies that $X_{11} \leq 0.4375 \rightarrow U_1 \leq 0.4375 - 1 = -0.5625 < -0.5$ which is not a pareto improvement.

So none of the corner solutions are pareto improvements. We could generalize this corner solution checking for arbitrary $X_{11}=X_{12}$, but I'll leave that as an exercise to the reader.

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In the Edgeworth box we have $x_{21}=1-x_{11}$, so, ignoring the constant, the first agent's utility function can simply be rewritten as $u_1=x_{11}+x_{12}$. Thus, for agent 1 the two goods are perfect substitutes and her indifference curves are straight lines with slope $-1$. Agent 2 has Cobb-Douglas indifference curves that are also symmetric between goods and therefore have slope $-1$ along the diagonal of the Edgeworth box. This diagonal is then also the set of Pareto efficient allocations.

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The pareto set is given by: boundary points $[(1,1), (0,0)]$ and $[(1,0), (0,1)]$, and then all points on the far left, bottom and right boundaries of the edgeworth box.

I can provide working if anyone would like.

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    $\begingroup$ Showing your work would greatly improve this answer. $\endgroup$
    – BKay
    Jun 17 '21 at 19:18
  • $\begingroup$ I haven't the time. Shall do so shortly. $\endgroup$
    – Charles
    Jun 17 '21 at 19:27

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