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I'm having difficulties solving for multiple equilibria for competitive exchange economies.

Considering a quasi linear preference as such:

$U_{A}(x,y)=x+100(1-e^{-y/10})$

$U_{B}(x,y)=y+110(1-e^{-x/10})$

with initial endowments: $e_{A}=(40,0), e_{B}=(0,50)$

when I attempted to solve this I found:

$MRS_{A}= {\frac{1}{10e^{-y/10}}} = {\frac{p_1}{p_2}}$

$MRS_{B}= 11e^{-x/10} = {\frac{p_1}{p_2}}$

and then I plugged this into the endowment constraint but I'm stuck. Is my approach to this problem wrong?

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  • $\begingroup$ You are ignoring nonnegativity cobditions. Also, only relative prices matter; you can normalize them. $\endgroup$ Commented May 10, 2023 at 5:58
  • $\begingroup$ No matter what I try, I always get transcendental equations, such as a product of a linear term and a log, or in terms of the function $x^x$ $\endgroup$ Commented May 10, 2023 at 14:12
  • $\begingroup$ I used the non negativity constraints as Michael Greinecker pointed out, it only gave me that $p_y \in [1,10]$. Here I took the numeraire as $p_x = 1$. $\endgroup$ Commented May 10, 2023 at 14:30
  • $\begingroup$ I get the equation $\ln(11 p_y) = p_y \ln(\frac{10}{p_y})$, which can’t be solved algebraically. $\endgroup$ Commented May 10, 2023 at 14:33
  • $\begingroup$ Using GeoGebra, I get that there are $2$ valid solutions: one near $p_y = 1.33$ and another one near $p_y = 3.597$ $\endgroup$ Commented May 10, 2023 at 14:35

1 Answer 1

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Since you commented you only needed the agents’ demands as a function of the relative prices, I’m giving my answer.

I take as numeraire $p_x = 1$.

  • Agent $A$

The optimization program is

$\max x + 100 (1-e^{-\frac{y}{10}})$

s.t. $x + y p_y = 40$

The optimality condition is $MRS_A = \frac{p_x}{p_y}$

$\frac{1}{10 e^{-\frac{y}{10}}} = \frac{1}{p_y} \implies \frac{p_y}{10} = e^{-\frac{y}{10}} \implies -10 \ln(\frac{p_y}{10}) = y \implies {y_A}^\star = 10 \ln(\frac{10}{p_y})$

Isolating $x$ from the budget constraint we get

$x = 40 - p_y y$

Substituting Agent $A$’s $y$-demand we get

${x_A}^\star = 40 - 10 p_y \ln(\frac{10}{p_y})$

Notice our expression for ${y_A}^\star$ is negative exactly when $p_y >10$.

So when $p_y \leq 10$, Agent $A$’s demands are the expressions we got.

On the other hand, when $p_y > 10$, the actual demands are

${y_A}^\star = 0$

${x_A}^\star = 40$

We can show with calculus that the expression for ${x_A}^\star$ attains its minimum value at $p_y = \frac{10}{e}$.

Evaluating ${x_A}^\star$ at this argument, we get its minimum value is ${x_A}^\star = 4 - \frac{10}{e} > 4 - \frac{10}{2.5} = 0$.

Since the minimum value for the expression for ${x_A}^\star$ is positive, there is no need to worry about the non-negativity constraint on this one.

  • Agent $B$

The optimization program is

$\max y + 110 (1-e^{-\frac{x}{10}})$

s.t. $x + p_y y = 50 p_y$

The optimality condition is $MRS_B = \frac{p_x}{p_y}$

$11 e^{-\frac{x}{10}} = \frac{1}{p_y} \implies x = -10 \ln(\frac{1}{11 p_y}) \implies {x_B}^\star = 10 \ln(11 p_y)$

Isolating $y$ from the budget constraint we get

$y = 50 - \frac{x}{p_y}$

Substituting Agent $B$’s $x$-demand we get

${y_B}^\star = 50 - \frac{10}{p_y} \ln(11 p_y)$

Notice our expression for ${x_B}^\star$ is negative exactly when $p_y < 1$.

So when $p_y \leq 1$, Agent $B$’s demands are the expressions we got.

On the other hand, when $p_y <1$, the actual demands are

${x_B}^\star = 0$

${y_B}^\star = 50$

We can show with calculus that the expression for ${y_B}^\star$ attains its minimum value at $p_y = \frac{e}{11}$.

Evaluating ${y_B}^\star$ at this argument, we get its minimum value is ${y_B}^\star = 5 - \frac{11}{e} > 5 - \frac{11}{2.2} = 0$.

Since the minimum value for the expression for ${y_B}^\star$ is positive, there is no need to worry about the non-negativity constraint on this one.

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