Symmetric Bidding in First-Price Auctions

Introduction

Welcome back! In our previous lesson, we set up the core problem of a first-price sealed-bid (FPSB) auction. We formulated a bidder's expected profit by assuming a simple, linear bidding strategy and a uniform distribution of values. The key insight was identifying the fundamental trade-off: bidding higher increases your win probability but reduces your profit margin.

Today, we will solve this problem in its full generality. Your goal is to derive the symmetric Bayesian Nash Equilibrium (BNE) bidding strategy for an FPSB auction with $n$ bidders, where their private values are drawn from any general, continuous probability distribution. This is a cornerstone result in auction theory.

The derivation is mathematically involved and will leverage your skills in calculus. We will progress in three main stages:

Formulating the general expected utility function.
Taking the first-order condition to find the maximum, which results in a differential equation.
Solving this differential equation to find the explicit formula for the optimal bid.

This process of setting up an objective function and using calculus to find the function that optimizes it is a powerful technique you'll encounter frequently in physics and economics.

Step 1: Formulating the General Expected Utility

In the last lesson, we assumed a linear strategy $b (v) = a v$ . Now, let's be more general. We'll look for a symmetric BNE where every bidder uses the same bidding strategy, $b (s)$ , which is a function of their private value (or type) $s$ . We'll assume this function is strictly increasing and differentiable—meaning, higher values lead to higher bids.

Your task is to choose a bid, let's call it $b_{i}$ , to maximize your expected utility, assuming all other $n - 1$ bidders are using this strategy $b (s)$ .

Your expected utility is:

E [u_{i}] = (value - bid) \times Prob (winning with that bid)

E [u_{i} (b_{i}; s_{i})] = (s_{i} - b_{i}) \times Prob (b_{j} \leq b_{i} for all j \neq = i)

The most complex part is calculating the probability of winning. The following video masterfully explains how to construct this general expected utility function.

(AGT10E5) [Game Theory] Solving First Price Auction with N Bidders: Part 1

This video, from Selcuk Ozyurt's 'Advanced Game Theory' series, will guide you through setting up the optimization problem for a general first-price auction. Pay close attention to how the probability of winning is formulated.

Watch the video from the beginning until 12:25. 00:00 - 04:33: Focus on how the problem is defined. Note the assumptions about the symmetric strategy b being strictly increasing, continuous, and differentiable. This allows us to use its inverse, b⁻¹. 04:33 - 12:25: This is the crucial part. Follow the logic for calculating the probability of winning. Understand why the probability of one opponent bidding less than you is F(b⁻¹(bᵢ)) and why this term is raised to the power of n-1 for n-1 opponents.

As the video explains, if bidder $i$ with value $s_{i}$ chooses to bid $b_{i}$ , their expected utility, assuming all other players $j$ use the strategy $b (s_{j})$ , is:

U (b_{i}, s_{i}) = (s_{i} - b_{i}) \cdot Prob (b (s_{j}) \leq b_{i} for all j \neq = i)

Since $b$ is an increasing function, $b (s_{j}) \leq b_{i}$ is equivalent to $s_{j} \leq b^{- 1} (b_{i})$ . The probability that a single opponent's value $s_{j}$ is less than or equal to $b^{- 1} (b_{i})$ is given by the cumulative distribution function (CDF), $F (b^{- 1} (b_{i}))$ .

Because all $n - 1$ opponents' values are drawn independently, the probability that they all bid less than $b_{i}$ is the product of the individual probabilities. This gives us the final expression for player $i$ 's expected utility:

U (b_{i}, s_{i}) = (s_{i} - b_{i}) [F (b^{- 1} (b_{i}))]^{n - 1}

Your goal is to choose the bid $b_{i}$ that maximizes this expression.

Step 2: Finding the Differential Equation

To find the optimal bid, we use calculus. We take the derivative of the utility function $U (b_{i}, s_{i})$ with respect to the variable we control—our bid, $b_{i}$ —and set it to zero. This is the first-order condition for a maximum.

This differentiation is non-trivial, involving the product rule, the chain rule, and the rule for differentiating an inverse function. The next video segment walks through this calculation in detail.

(AGT10E6) [Game Theory] Solving First Price Auction with N Bidders: Part 2

This is the second part of the derivation. It starts exactly where the previous one left off and executes the differentiation to arrive at the first-order condition.

Watch from the beginning to 07:31. 00:00 - 03:04: Follow the application of the product rule and chain rule. The explanation of how to find the derivative of the inverse function b⁻¹ is particularly important. 03:04 - 07:31: This is the key step. After finding the derivative, the video applies the symmetric equilibrium condition: in equilibrium, player i's optimal bid bᵢ must be the one prescribed by the strategy function b(sᵢ). By substituting bᵢ = b(sᵢ) into the first-order condition, the complex expression simplifies into a differential equation.

Let's summarize the result. The first-order condition $\frac{d}{d b _{i}} U (b_{i}, s_{i}) = 0$ is:

- [F (b^{- 1} (b_{i}))]^{n - 1} + (s_{i} - b_{i}) \cdot (n - 1) [F (b^{- 1} (b_{i}))]^{n - 2} f (b^{- 1} (b_{i})) \frac{1}{b ^{'} ( b ^{- 1} ( b _{i} ) )} = 0

Here, $f$ is the probability density function (the derivative of $F$ ) and $b^{'}$ is the derivative of the bid function.

Now, we impose the symmetric equilibrium condition: the optimal bid $b_{i}$ must be $b (s_{i})$ . This means $b^{- 1} (b_{i}) = s_{i}$ . Substituting this into the equation simplifies it immensely. After rearranging, we get the first-order linear differential equation that the equilibrium bid function $b (s)$ must solve:

b^{'} (s) F (s)^{n - 1} = (s - b (s)) (n - 1) F (s)^{n - 2} f (s)

Notice that $(n - 1) F (s)^{n - 2} f (s)$ is the derivative of $F (s)^{n - 1}$ by the chain rule. So we can write:

b^{'} (s) F (s)^{n - 1} = (s - b (s)) \frac{d}{d s} [F (s)^{n - 1}]

Step 3: Solving for the Bidding Strategy

The videos have provided us with the differential equation. Now, we must solve it to find the function $b (s)$ . For this, we'll turn to a written resource that provides a clear and rigorous proof.

First-Price Sealed-Bid Auctions

The following document from Brown University provides a theorem and a full proof that solves the exact differential equation we just derived. We will walk through the key steps of this proof together.

Refer to the proof of Theorem 1.1, starting from Equation (12). You don't need to read the whole document, but use it to follow the steps outlined below.

Let's solve the differential equation from the proof. Starting from equation (12) in the text, we have:

b^{'} (s) F (s)^{n - 1} = (s - b (s)) \frac{d}{d s} [F (s)^{n - 1}]

Rearranging terms gives:

b^{'} (s) F (s)^{n - 1} + b (s) \frac{d}{d s} [F (s)^{n - 1}] = s \frac{d}{d s} [F (s)^{n - 1}]

The left-hand side is now exactly the derivative of the product $b (s) F (s)^{n - 1}$ , according to the product rule. This is a common technique for solving differential equations.

\frac{d}{d s} [b (s) F (s)^{n - 1}] = s \frac{d}{d s} [F (s)^{n - 1}]

Now, we integrate both sides from the lowest possible valuation (let's call it $s$ , often 0) up to some value $v$ .

\int_{s}^{v} \frac{d}{d s} [b (s) F (s)^{n - 1}] d s = \int_{s}^{v} s \frac{d}{d s} [F (s)^{n - 1}] d s

By the Fundamental Theorem of Calculus, the left side is simply:

[b (s) F (s)^{n - 1}]_{s}^{v} = b (v) F (v)^{n - 1} - b (s) F (s)^{n - 1}

A bidder with the lowest possible value $s$ knows they can't win against anyone (as values are continuous), so their optimal bid $b (s)$ will be such that their profit is 0. We also know $F (s) = 0$ , so the second term vanishes. This leaves $b (v) F (v)^{n - 1}$ .

For the right side, we use integration by parts, $\int u d v = uv - \int v d u$ . Let $u = s$ and $d v = \frac{d}{d s} [F (s)^{n - 1}] d s$ . Then $d u = d s$ and $v = F (s)^{n - 1}$ .

[s F (s)^{n - 1}]_{s}^{v} - \int_{s}^{v} F (s)^{n - 1} d s = v F (v)^{n - 1} - \int_{s}^{v} F (s)^{n - 1} d s

Equating the two sides gives:

b (v) F (v)^{n - 1} = v F (v)^{n - 1} - \int_{s}^{v} F (s)^{n - 1} d s

Finally, dividing by $F (v)^{n - 1}$ yields the symmetric Bayesian Nash Equilibrium bidding strategy:

b (v) = v - \frac{\int _{s}^{v} F ( s ) ^{n - 1} d s}{F ( v ) ^{n - 1}}

Step 4: Interpreting the Equilibrium Strategy

This formula is a remarkable result. It tells you exactly how much to bid for any given valuation $v$ and any distribution $F$ .

Optimal Bidding Strategy and Expected Payoff in Auctions — The left graph shows the derived bidding strategy $b(v)$ as a function of valuation $v$. It is an increasing function below the 45-degree line $b(v)=v$. The right graph shows that for a given valuation $v_i$, the expected payoff is maximized at the specific bid $b(v_i)$ prescribed by the strategy.

The formula confirms our intuition about "bid shading." Your optimal bid is your true value, $v$ , minus a discount term:

Shading Amount = \frac{\int _{s}^{v} F ( s ) ^{n - 1} d s}{F ( v ) ^{n - 1}}

This term represents the expected gain you give up to increase your probability of winning.

Connection to Previous Lesson:
Let's check this for the uniform distribution on $[0, 1]$ , where $F (s) = s$ and $s = 0$ .
The integral becomes $\int_{0}^{v} s^{n - 1} d s = [\frac{s ^{n}}{n}]_{0}^{v} = \frac{v ^{n}}{n}$ .
The denominator is $F (v)^{n - 1} = v^{n - 1}$ .
So, the bidding function is:

b (v) = v - \frac{v ^{n} / n}{v ^{n - 1}} = v - \frac{v}{n} = \frac{n - 1}{n} v

This is exactly the linear strategy that was assumed in the simplified example from our last lesson! For $n = 2$ , the strategy is $b (v) = v /2$ . For $n = 10$ , it's $b (v) = 0.9 v$ . As the number of bidders increases, you shade your bid less.

Conclusion

Today, we accomplished a major goal in our study of game theory by deriving the symmetric BNE for a first-price sealed-bid auction. We moved from an intuitive understanding of bid shading to a rigorous mathematical derivation that holds for any continuous distribution of player types.

Key Takeaways:

The problem is solved by maximizing a bidder's expected utility, assuming all other bidders follow the same strategy function $b (s)$ .
The maximization process involves taking the first-order condition (derivative equals zero), which leads to a first-order linear differential equation.
Solving this differential equation using integration by parts yields the explicit equilibrium bidding strategy: $b (v) = v - \frac{\int _{s}^{v} F ( s ) ^{n - 1} d s}{F ( v ) ^{n - 1}}$
This general formula confirms that the optimal strategy is to shade your bid below your true value, and it quantifies the precise amount of shading based on the number of bidders and their value distribution.

This concludes our module on static games of incomplete information. We have seen how to analyze situations where players have private information that is crucial to their strategy. In our next module, we will explore Dynamic Games of Incomplete Information, where players' actions over time can reveal some of their private information, leading to fascinating concepts like signaling and screening.

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