# Equilibrium of Perturbed Dollar Auction Game - An Example from Game Theory: Analysis of Conflict by Roger Myerson

I am studying game theory using Myerson's textbook (Chapter 3 - Equilibria of Strategic-Form Games, Section 3.6 - The Decision-Analytic Approach to Games). I have difficulties understanding and deriving his equilibrium of the following perturbed Dollar Auction game:

There are two risk-neutral players, each of whom must choose a bid that can be any real number between $$0$$ and $$1$$. Suppose that for each player $$j$$, there is an independent probability $$0.1$$ that $$j$$'s bid will be determined, not by a rational intelligent decision-maker, but by a naive agent who chooses the bid from a uniform distribution over the interval from $$0$$ to $$1$$. The high bidder pays the amount of his bid and then wins $$1$$ dollar. (In case of a tie, each has a probability $$0.5$$ of winning and buying the dollar for his bid.)

Let us then interpret $$\hat{u}_i(c_1,c_2)$$ as the conditionally expected payoff that $$i$$ would get in this perturbed game given that player $$i$$'s bid is not being determined by such a naive agent and given that, for each player $$j$$, $$c_j$$ is the bid that player $$j$$ would make if his bid were not being otherwise determined by such a naive agent. Then the utility functions in this perturbed game are \begin{align*} \hat{u}_i(c_1,c_2) & = 0.1c_i(1-c_i)\quad\quad\quad\quad\quad\quad\quad\quad \text{if}\quad i \notin \text{argmax}_{j \in \{1,2\}}c_j,\\ & = 0.1c_i(1-c_i) + 0.9(1-c_i)\quad\hspace{0.5cm} \text{if}\quad \{i\} = \text{argmax}_{j \in \{1,2\}}c_j,\\ & = 0.1c_i(1-c_i) + \frac{0.9(1-c_i)}{2} \quad\hspace{0.4cm} \text{if}\quad c_1 = c_2. \end{align*}

I have no problem understanding everything so far. But then, Myerson stated that

There is an equilibrium of this perturbed game in which each player randomly chooses a bid from the interval between 0.5 and 0.975 in such a way that, for any number $$x$$ in this interval, the cumulative probability of his bidding below $$x$$ is $$\frac{0.025 + 0.1x^2 - 0.1x}{0.9(1-x)}$$. The median bid for a player under this distribution is 0.954.

Let us describe the mixed strategy played by both players 1 and 2 using the probability density function $$f$$. We say that a pure strategy $$x$$ is in the support $$S$$ of a mixed strategy $$f$$ if $$f(x)>0$$.

Edit:
Sorry, the below notation and lemma only holds for when the players are not a naive agents. Expressing this rigorously would require me to write out the equations, and I am currently not willing to do that. The logic still holds for the perturbed game, the perturbation just makes the equations more messy.

We will denote the expected utility of player 1 in the perturbed game given to mixed strategies with: $$E\left( \hat{u}_1(f_1,f_2) \right) = \int_{x_1 \in S_1} \int_{x_2 \in S_2} \hat{u}_1(x_1,x_2) \ \text{d}x_1\text{d}x_2.$$

Lemma For pure strategies $$x_1$$ in the support (aside from a zero-measure set) of a best-response $$f_1$$ to a strategy $$f_2$$ we have $$E\left( \hat{u}_1(x_1,f_2) \right) = E\left( \hat{u}_1(f_1,f_2) \right) \hskip 10pt \blacksquare$$ The intuition for the lemma is that if a player played a set of pure strategies that have a lower payoff than the mixed strategy on average with positive probability, the player could improve their payoff by shifting probability away from the lower payoff pure strategies to the higher payoff pure strategies.

You want to find the symmetric equilibrium mentioned in your answer. Assuming it exists, for all $$x_1 \in S_1$$ it will fulfill $$E\left( \hat{u}_1(x_1,f) \right) = E\left( \hat{u}_1(f,f) \right)$$ since $$f$$ is a best response to $$f$$.
It helps a lot if you assume that the support set is a continuous and closed interval $$[x_l,x_u]$$, because the above equation is 'special' for both $$x_l$$ and $$x_u$$, you know the probability of winning the auction in these cases.

In my experience from here on it is just calculus busy work, separating the integrals into cases, e.g.; $$x_1 < x_2$$, etc., then doing calculations.