Consider a game in which 2 players transmit packets in a network with a selected power $x ∈ [1, A]$ and $y∈ [1, A]$, respectively. The utility of the players can be expressed as:

$$u_{i} (x,y) = \log(x) + \log(y)$$ for $i = 1, 2$

(a) Does this game have a pure strategy Nash equilibrium? If yes, what is the Nash equilibrium?

(b) If we apply the transformation $g = \exp(u)$ to the utility function, does the new utility function

$$U_{i} (x,y) = \exp(\log(x) + \log(y))$$ for $i = 1, 2$

guarantee a pure Nash equilibrium solution?

thank you.. :/

  • 1
    $\begingroup$ Could you demonstrate some work you've done on the problem so we could maybe point you in the right direction? $\endgroup$
    – JustAskin
    Mar 11, 2014 at 23:00

1 Answer 1


(a) Assume player II chooses $y \in [1,A]$. Then what is player I's best replay? Since his payoff function in monotone increasing in $x$, his best reply is $x_0=A$ regardless of player II's strategy. Due to symmetry the same holds for player II. Therefore the only Nash equilibrium of this game is already in pure strategies and it is $$(x_0, y_0)=(A, A)$$

(b) The transformed payoff function remains monotone in $x$. So player I will choose $A$ regardless of II's choise and similarly for player II.

(The reason for this invariance is that the function $\exp$ (that was applied for the transformation) is monotone and does not affect the characteristic of monotonicity of the initial payoff function).


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .