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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.. :/

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    $\begingroup$ 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 '14 at 23:00
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(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).

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