# Auction with independent private values - An example from Game Theory: Analysis of Conflict by Roger Myerson

I have difficulties understanding the equilibrium analysis of the following auction game:

Suppose that there are $$n$$ bidders in an auction for a single indivisible object. Each player knows privately how much the object would be worth to him, if he won the auction. Suppose that there is some positive number $$M$$ and some increasing differentiable function $$F(\cdot)$$ such that each player considers the values of the object to the other $$n-1$$ players to be independent random variables drawn from the interval $$[0,M]$$ with cumulative distribution $$F$$. (So $$F(v)$$ is the probability that any given player has a value for the object that is less than $$v$$.) In the auction, each player $$i$$ simultaneously submits a sealed bid $$b_i$$, which must be a nonnegative number. The object is delivered to the player whose bid is highest, and this player must pay the amount for his bid. No other player pays anything.

The analysis is presented as follows:

Let $$b = (b_1,\dots,b_n)$$ denote the profile of bids and letting $$v = (v_1,\dots,v_n)$$ denote the profile of values of the object to the $$n$$ players, the expected payoff to player $$i$$ is \begin{align*} \begin{alignedat}{4} &u_i(b,v) &&= v_i - b_i\quad &&\text{if}\quad \{i\} = \text{argmax}_{j \in \{1,\dots,n\}} b_j,\\ & &&= 0\quad &&\text{if}\quad i \notin \text{argmax}_{j \in \{1,\dots,n\}} b_j. \end{alignedat} \end{align*} This is a Bayesian game where each player's type is his value for the object. We now show how to find a Bayesian equilibrium in which every player chooses his bid according to some function $$\beta$$ that is differentiable and increasing.

In the equilibrium, player $$i$$ expects the other players to be choosing bids between 0 and $$\beta(M)$$, so player $$i$$ would never want to submit a bid higher than $$\beta(M)$$. Suppose that, when $$i$$'s value is actually $$v_i$$, he submits a bid equal to $$\beta(w_i)$$. Another player $$j$$ will submit a bid that is less than $$\beta(w_i)$$ if and only if $$j$$'s value $$\tilde{v}_j$$ satisfies $$\beta(\tilde{v}_j) < \beta(w_i)$$, which happens if and only if $$\tilde{v}_j < w_i$$, because $$\beta$$ is increasing. So the probability that the bid $$\beta(w_i)$$ will win the auction is $$F(w_i)^{n-1}$$, because the types of the other $$n-1$$ players are independently distributed according to $$F$$. Thus, the expected payoff to player $$i$$ from bidding $$\beta(w_i)$$ with value $$v_i$$ would be \begin{align*} (v_i - \beta(w_i))F(w_i)^{n-1}. \end{align*} However, by the definition of an equilibrium, the optimal bid for $$i$$ with value $$v_i$$ should be $$\beta(v_i)$$. So the derivative of this expected payoff with respect to $$w_i$$ should equal 0 when $$w_i$$ equals $$v_i$$. That is, \begin{align*} 0 = (v_i - \beta(v_i))F'(v_i)(n-1)F(v_i)^{n-2} - \beta'(v_i)F(v_i)^{n-1}. \end{align*} This differential equation implies that, for any $$x$$ in $$[0,M]$$ \begin{align*} \beta(x)F(x)^{n-1} = \int_0^x y(n-1)F(y)^{n-2}F'(y)dy. \end{align*}

(1) The analysis begins with claiming that every player chooses his bid acording to some function $$\beta$$. I don't quite understand why this is necessarily ture. That is, why is there necessarily a function $$\beta$$ for each player to place his bid?

(2) In the analysis, it claims that "by the definition of an equilibrium, the optimal bid for $$i$$ with value $$v_i$$ should be $$\beta(v_i)$$". I cannot see how can we make this conclusion from the definition of an equilibrium. (The author defines a (Nash) equilibrium to be a randomized-strategy profile where no player could increase his expected payoff by unilaterally deviating from the prediction of the randomized-strategy profile.)

(3) Finally, how is the differential equation solved? (That is, how to derive the last expression?)

• "by unilaterally deviating from the prediction of the randomized-strategy profile" should be "by unilaterally deviating from his strategy". Commented Nov 23, 2023 at 8:53
• In (1), are you asking why the equilibrium is symmetric, why $\beta$ is the same for all players, or are you asking why a player's strategy can be described by a function? Commented Nov 23, 2023 at 12:00
• @Giskard Actually, I was gonna ask why a player's strategy can be described by a function. Commented Nov 23, 2023 at 16:18
• @VARulle You are right. The author's description is indeed a little misleading. Basically, as for my second question, I am not quite sure why he put a $\beta(w_i)$ at first, and then how would the definition of an equilibrium imply that $\beta(w_i)=\beta(v_i)$. Commented Nov 23, 2023 at 16:23

(2a) The game is symmetric, so a natural approach is to try to find a symmetric Bayesian Nash equilibrium, i.e. an equilibrium where all players use the same strategy (function). This common function is just denoted by $$\beta(.)$$ here. In equilibrium, type $$v_i$$ therefore bids $$\beta(v_i)$$.
(2b) The "natural" way to begin the proof would be to assume that type $$v_i$$ bids $$b_i$$. But then, since $$\beta$$ is increasing, there exists a unique value $$w_i$$ such that $$\beta(w_i)=b_i$$, viz. $$w_i=\beta^{-1}(b_i)$$. The introduction of $$b_i$$ can be shortcut, however, by assuming right from the outset that the player with value $$v_i$$ bids $$\beta(w_i)$$, as is done in the proof. Then the equilibrium condition is invoked and since by definition type $$v_i$$ actually bids $$\beta(v_i)$$ in equilibrium, it is enough to go on by setting $$\beta(w_i)=\beta(v_i)$$.
(3) Rearrange the terms of the differential equation to get \begin{align*} \beta'(v_i)F(v_i)^{n-1}+\beta(v_i)F'(v_i)(n-1)F(v_i)^{n-2} = v_i(n-1)F(v_i)^{n-2}F'(v_i) \end{align*} and note that the LHS is just the derivative of $$\beta(v_i)F(v_i)^{n-1}$$. Integrate from $$0$$ to $$x$$, using $$F(0)=0$$.
• In (1), why can we assume that the strategies of the equilibrium will be deterministic, in the since that a bid is assigned deterministically by $\beta$ to a valuation? Commented Nov 24, 2023 at 10:28