[Math] Finding best response function

calculuseconomicsgame theoryself-learning

Consider two players

Player 1 choose $x_1$

Player 2 choose $x_2$

$x_1, x_2 \in [0,b]$ and $b>0$

The payoffs for players 1 and 2 are identical and equal

$$Z=u(x_1+x_2)+v(2b-x_1-x_2)$$

$u$ and $v$ are strictly concave functions.

I want to find best response functions for both players and unique Nash equilibrium. And what can I say about the slope of the best response functions?

Solution

For player 1

$\partial Z /\partial x_1=0$

$$Z_1=u_1(x_1+x_2)*(1)+v_1(2b-x_1-x_2)(-1)=0$$

I obtain best response of player 1 to player 2. I.e. $R_1(x_2)$

For player 2

$\partial Z /\partial x_2=0$

$$Z_2=u_2(x_1+x_2)*(1)+v_2(2b-x_1-x_2)(-1)=0$$

I obtain best response of player 2 to player 1. I.e. $R_2(x_1)$

So far I obtain best response functions.

Now, if I solve these above two equations simultaneously,

That’s, $R_2(x_1)=x_2$ and $R_1(x_2)=x_1$

Then I will get results $x_1^*$ and $x_2^*$ which are unique Nash equilibrium.

As for the slope of the vest response function which is equal to $-{z_1\over z_2}$

I can only say these about this question. I cannot write specific answer with using this payoff function z. What is your ideas? How can I solve this question expect for what I did. I am stuck at this point. I am waiting for your helps please. Thanks.

Best Answer

You derivation of the first order condition is correct and this is indeed a necessary and sufficient condition for optimal since the objective function is concave. One detail is that $u_1(x_1+x_2)=u_2(x_1+x_2)=u'(x_1+x_2)$ and the same for $v$.

To obtain the slope of the best response function for player $1$, for instance, you can, then, use the implicit function theorem. Let

$$ F(x_1,x_2)=u'(x_1+x_2)-v'(2b-x_1-x_2)=0, $$

Then,

$$\frac{dx_1}{dx_2}=-\frac{F_{x_2}(x_1,x_2)}{F_{x_1}(x_1,x_2)}=-\frac{u''(x_1+x_2)+v''(2b-x_1-x_2)}{u''(x_1+x_2)+v''(2b-x_1-x_2)}=-1.$$

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