What We're Solving

We need to find the optimal bidding strategy in a first-price sealed-bid auction where all bidders draw their private values uniformly from $[0, 1]$. The key insight is that bidding your true value is never optimal here. We'll derive the Bayesian Nash Equilibrium bid function using the principle that each bidder maximizes expected profit, taking into account that the highest bid wins.

Setting Up the Problem

Let's fix the scenario. There are $n$ bidders. Each bidder $i$ privately observes their value $v_i \sim \text{Uniform}[0, 1]$, independently drawn. In a first-price auction, the winner pays their own bid. So if you win with bid $b$, your profit is $v - b$.

We're looking for a symmetric equilibrium: a single bid function $b(v)$ that every bidder uses. Symmetric means we assume all opponents also play $b(\cdot)$, and we ask: what is the best response for a bidder with value $v$?

Also, let's assume $b(v)$ is strictly increasing and differentiable. This is a natural assumption (higher values should lead to higher bids), and we'll verify it holds at the end.

Expected Payoff

Suppose you bid $b$ and all opponents use strategy $b(\cdot)$. You win if and only if your bid beats every opponent's bid. Since $b$ is strictly increasing, bidding $b$ is equivalent to "acting like" a player with value $b^{-1}(b)$. Let's denote your effective type as $z = b^{-1}(b)$, meaning you win if all $n-1$ opponents have values below $z$.

The probability you win is:

$$P(\text{win}) = P(\text{all opponents bid less than } b) = z^{n-1}$$

because each opponent has value below $z$ with probability $z$, and there are $n-1$ opponents acting independently.

Your expected payoff as a function of your bid $b$ (or equivalently, $z$) is:

$$\Pi(z; v) = (v - b(z)) \cdot z^{n-1}$$

We want to maximize this over $z$, and in equilibrium, the optimal choice must be $z = v$ (you shouldn't want to pretend to be a different type).

Deriving the Equilibrium Bid Function

Take the first-order condition with respect to $z$ and set it equal to zero:

$$\frac{d}{dz} \left[(v - b(z)) z^{n-1}\right] = 0$$ $$-b'(z) z^{n-1} + (v - b(z))(n-1)z^{n-2} = 0$$

Now apply the equilibrium condition $z = v$:

$$-b'(v) v^{n-1} + (v - b(v))(n-1)v^{n-2} = 0$$

Multiply through by $v$ and rearrange:

$$v^{n-1} b'(v) + (n-1) v^{n-2} b(v) = (n-1) v^{n-1}$$

Notice that the left side is exactly the derivative of $v^{n-1} b(v)$:

$$\frac{d}{dv}\left[v^{n-1} b(v)\right] = (n-1) v^{n-1}$$

Integrate both sides from $0$ to $v$, using the boundary condition $b(0) = 0$ (a bidder with zero value bids zero):

$$v^{n-1} b(v) = \int_0^v (n-1) t^{n-1} \, dt = \frac{n-1}{n} v^n$$

Solving for $b(v)$:

$$\boxed{b(v) = \frac{n-1}{n} v}$$

Intuition Behind the Result

This is a beautifully clean result. Let's unpack what it means.

You always shade your bid below your value. With $n$ bidders, you bid a fraction $(n-1)/n$ of your true value. The gap is your "information rent" for holding a private value.
Competition closes the gap. As $n \to \infty$, the bid fraction approaches 1, so bidders bid closer and closer to their true values. With more competitors, shading becomes increasingly risky (you might lose to someone close to your value), so equilibrium bids rise.
With $n = 2$, you bid exactly half your value. With $n = 10$, you bid $90\%$ of your value.

Revenue Equivalence Check

A useful sanity check: the Revenue Equivalence Theorem says that under standard conditions (symmetric, risk-neutral bidders with i.i.d. private values), all standard auctions yield the same expected revenue. Let's verify this for the first-price and second-price auctions.

In a second-price auction, the winner pays the second-highest value. The expected payment of the highest-value bidder (with value $v$) equals the expected second-highest order statistic given that $v$ is the highest. For $n$ uniform bidders, this is:

$$E[\text{2nd highest} \mid \text{highest} = v] = \frac{n-1}{n} v$$

Which is exactly $b(v)$ from our first-price solution. The expected payments match, confirming our answer is consistent with Revenue Equivalence.

An Alternative Way to See It

There's a neat shortcut. In equilibrium, the expected payment from a bidder with value $v$ must equal the expected value of the second-highest order statistic among $n$ draws, conditional on $v$ being the maximum. For $n$ uniform bidders:

$$b(v) = E[Y_{(n-1)} \mid Y_{(n)} = v]$$

where $Y_{(n-1)}$ is the second-highest of $n-1$ uniform draws on $[0, v]$. The expected maximum of $n-1$ uniform$[0,v]$ draws is $\frac{n-1}{n} v$. So we recover $b(v) = \frac{n-1}{n} v$ again, this time from a more probabilistic angle.

Natural Extensions to Think About

A few follow-up directions worth anticipating:

Non-uniform distributions: The same ODE approach works for any distribution $F(v)$ with density $f(v)$. The equilibrium bid becomes $b(v) = v - \int_0^v F(t)^{n-1} dt / F(v)^{n-1}$, which is the expected value of the highest opponent's value, conditional on you winning.
Risk aversion: Risk-averse bidders shade less, because the guaranteed gain from winning is more valuable relative to the uncertain gain from aggressive shading. Revenue Equivalence breaks down here, and first-price auctions raise more revenue than second-price ones.
Reserve prices: A seller can optimally set a reserve price to exclude low-value bidders and increase revenue, at the cost of sometimes failing to sell the good.

5496. Uniform Value Bidding in First-Price Auction

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