1
$\begingroup$

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?

$\endgroup$
7
  • $\begingroup$ You are ignoring nonnegativity cobditions. Also, only relative prices matter; you can normalize them. $\endgroup$ May 10 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 $x^x$ $\endgroup$ May 10 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 $p_x = 1$. $\endgroup$ May 10 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. $\endgroup$ May 10 at 14:33
  • $\begingroup$ Using GeoGebra, I get that there are $2$ valid solutions: one near $p_y = 1.33$ and another one near $p_y = 3.597$ $\endgroup$ May 10 at 14:35

1 Answer 1

1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.