# Multiple Equilibria

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?

• You are ignoring nonnegativity cobditions. Also, only relative prices matter; you can normalize them. May 10 at 5:58
• 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 $x^x$ May 10 at 14:12
• 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 $p_x = 1$. May 10 at 14:30
• I get the equation $\ln(11 p_y) = p_y \ln(\frac{10}{p_y})$, which can’t be solved algebraically. May 10 at 14:33
• Using GeoGebra, I get that there are $2$ valid solutions: one near $p_y = 1.33$ and another one near $p_y = 3.597$ May 10 at 14:35

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.