I have a question regarding the below situation

Alice and Bob are two partners in a joint project. Simultaneously, Alice and Bob put efforts $e_A$ and $e_B \in [0,1]$ respectively, to the project, and the project succeeds with probability $e_A^{1/2}e_B^{1/2}$. The payoffs of Alice and Bob are $\theta_A-e_A^2$ and $\theta_B-e_B^2$, respectively, if the project succeeds; the payoffs are $-e_A^2$ and $-e_B^2$ otherwise. Here, $\theta_A$ and $\theta_B$ are privately known by Alice and Bob, and respectively, and they are independently and uniformly distributed on $[0,1]$.

How can I represent this game as a Bayesian game and determine a symmetric Nash equilibrium in increasing strategies ?


1 Answer 1


A pure strategy profile consists of functions $e_A(\theta_A)$ and $e_B(\theta_B)$ by which the players choose their efforts depending on their valuations.

First, note that the effort $1$ is dominated by the effort $0$, and the effort $0$ is dominated by very small efforts unless the valuation is $0$. Thus, at non-zero valuations the efforts will be in the interior of $[0,1]$, so equilibrium implies that the derivatives of the expected payoffs with respect to the efforts vanish.

The expected payoff for $A$, given $\theta_A$, is

$$ \int_0^1e_B^{1/2}\mathrm d\theta_Be_A^{1/2}\theta_A-e_A^2\;. $$

Setting the derivative with respect to $e_A$ to zero yields

$$ e_A=\left(\frac{\theta_A}4\int_0^1e_B^{1/2}\mathrm d\theta_B\right)^{\frac23}\;. $$

Thus $e_A=\lambda\theta_A^{2/3}$, and by symmetry $e_B=\lambda\theta_B^{2/3}$, and we can determine $\lambda$ by substituting above:

$$ \lambda\theta_A^{2/3}=\left(\frac{\theta_A}4\int_0^1\lambda^{1/2}\theta_B^{1/3}\mathrm d\theta_B\right)^{\frac23}\;, $$

with solution $\lambda=\frac3{16}$. Thus, the symmetric equilibrium strategy profile is $e_A(\theta_A)=\frac3{16}\theta_A^{2/3}$ and $e_B(\theta_B)=\frac3{16}\theta_B^{2/3}$.

The expected total payoff is

$$ \int_0^1\mathrm d\theta_A\int_0^1\mathrm d\theta_B\left(\frac3{16}\theta_A^{1/3}\theta_B^{1/3}(\theta_A+\theta_B)-\left(\frac3{16}\theta_A^{4/3}\right)^2-\left(\frac3{16}\theta_B^{4/3}\right)^2\right)=\frac{999}{9856}\approx0.101\;, $$

whereas if the players were to cooperate to maximize the expected payoff, they would each make the effort $e=\left(\frac{\theta_A+\theta_B}4\right)^{\frac23}$, resulting in a expected total payoff of

$$ \int_0^1\mathrm d\theta_A\int_0^1\mathrm d\theta_B\left(\left(\frac{\theta_A+\theta_B}4\right)^{\frac23}(\theta_A+\theta_B)-2\left(\frac{\theta_A+\theta_B}4\right)^{\frac23}\right)=\frac{2916\cdot2^{1/3}-1899\cdot2^{2/3}}{6160}\approx0.107\;. $$

Thus the players come surprisingly close to the cooperative result.


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