Nash Bargaining Equilibrium with exponential utilities

Question

I'm trying to derive the answer to the following question:

Two players play the classic divide-the-dollar game, which is an imperfect information version of the ultimatum class of games.

Utility functions of the players are given by:

$u_1(z)=z^{\gamma_1}$, $u_2(z)=z^{\gamma_2}$

The disagreement outcome is $d=(0,0)$

Now I want to find the Nash Bargaining solution:

$f^N(S,d)=\arg \max_{d\leq s \in S} (s_1-d_1)(s_2-d_2) $

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I tried to solve this using the FOC (Lagrange with only one variable):

$ \max_{\ x}\ \ x^{\gamma_1}(1-x)^{\gamma_2} $

$ \text{s. t.: } \gamma_1 < 1, \gamma_2 <1 \quad \text{(players are risk-averse)} $

$ f(x) = x^{\gamma_1}(1-x)^{\gamma_2} $

$ f'(x) = \gamma_1 x^{\gamma_1-1} (1-x)^{\gamma_2} - \gamma_2 x^{\gamma_1} (1-x)^{\gamma_2-1} = 0 $

$ \Rightarrow \frac{x}{1-x} = \frac{\gamma_1}{\gamma_2} $

But I don't get how this would lead me to the answer.

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According to the solution manual, the answer should be:

$ x = \frac{\gamma_1}{\gamma_1+\gamma_2}, \quad \quad 1-x=\frac{\gamma_2}{\gamma_1+\gamma_2} $

Does anybody see how to get there?

Answer 1

Let´s say $x=x_1$ and $1-x=x_2$. Your intermediate result becomes

$\frac{x_1}{x_2}=\frac{\gamma_1}{\gamma_2}$

Solving the equation for $x_1$: $x_1=x_2\frac{\gamma_1}{\gamma_2}$

The condition is, that $1=x_1+x_2$.

Inserting the expression for $x_1$: $1=x_2\frac{\gamma_1}{\gamma_2}+x_2$

Now you can solve the equation for $x_2$. At the end you should expand the fraction by $\gamma_2$ to get the given solution (same representation). And additionally you can substitute back $x_2$ to $1-x$.

Answer 2

Found it:

$ \frac{x}{1-x} = \frac{\gamma_1}{\gamma_2} $

$ \gamma_2 x = \gamma_1 - \gamma_1 x $

$ (\gamma_1 + \gamma_2) x = \gamma_1 $

$ x = \frac{\gamma_1}{\gamma_1 + \gamma_2} $

$ 1-x = 1-\frac{\gamma_1}{\gamma_1 + \gamma_2} = \frac{\gamma_1 + \gamma_2}{\gamma_1 + \gamma_2}-\frac{\gamma_1}{\gamma_1 + \gamma_2} = \frac{\gamma_2}{\gamma_1 + \gamma_2} $