Deriving Symmetric First-Price Equilibrium Bids

Why Bids Are Shaded Below Values

In a first-price auction, the highest bidder wins and pays their own bid. A bidder therefore trades off two consequences of raising a bid: it makes winning more likely, but leaves less surplus if the bidder wins. The equilibrium bid balances these incentives. With independent private values, a bidder’s own value determines how much they gain from winning, while the distribution of rivals’ values determines how likely a particular bid is to win.

This tutorial derives that balance for risk-neutral bidders with the same value distribution. The nonnegative-bid constraint matters: throughout, values are nonnegative. The formula derived here is not automatically valid for distributions with negative values.

Model and assumptions: There are $N\geq 2$ bidders. Bidder $i$ privately observes value $v_i$, and values are independent and identically distributed on $[v_L,v_H]$, where $0\leq v_L<v_H$. Their common cumulative distribution function is $F$, with density $f$ positive and continuous in the interior of the support. The distribution has no atoms. Bidders are risk-neutral, bids must be nonnegative, the highest bid wins and pays their own bid, and ties are resolved by any fixed rule. There is no reserve price or entry decision in this model.

A bidder of value $v$ who wins with bid $b$ earns $v-b$; a bidder who loses earns zero. A symmetric increasing strategy is a common bidding function $b(v)$ that is strictly increasing in value. Because the values are independent and continuously distributed, ties between bidders using such a strategy occur with probability zero.

Equilibrium bidding function: Under the assumptions above, the increasing symmetric Bayesian Nash equilibrium bid for $v\in(v_L,v_H]$ is $$ b(v)=v-\frac{\displaystyle\int_{v_L}^{v} F(t)^{N-1}\,dt}{F(v)^{N-1}}. $$ Its continuous boundary value is $b(v_L)=v_L$. The integral term is the expected amount of shading below value in the equilibrium bid.

Deriving the Bidding Function

Suppose the other $N-1$ bidders use the same strictly increasing strategy $b(\cdot)$. A type $v$ bidder considering the bid $b(z)$ is effectively acting as if their value were $z$: they beat each rival exactly when that rival’s value is below $z$. Independence implies that the probability all rivals have values below $z$ is $F(z)^{N-1}$. Thus the deviating bidder’s expected payoff is

U(v,z)=\bigl(v-b(z)\bigr)F(z)^{N-1}.

In a symmetric equilibrium, type $v$ chooses $z=v$. For an interior type, differentiate the payoff with respect to $z$ and impose the first-order condition at $z=v$:

-b'(v)F(v)^{N-1} +\bigl(v-b(v)\bigr)(N-1)F(v)^{N-2}f(v)=0.

For $v>v_L$, $F(v)>0$, so this condition can be rearranged as

b'(v)F(v)^{N-1} =\bigl(v-b(v)\bigr)(N-1)F(v)^{N-2}f(v).

This differential equation determines how the bid changes with value. To solve it, differentiate the product $(v-b(v))F(v)^{N-1}$. The product rule and the first-order condition give

\begin{aligned} \frac{d}{dv}\left[(v-b(v))F(v)^{N-1}\right] &=(1-b'(v))F(v)^{N-1} +(v-b(v))(N-1)F(v)^{N-2}f(v)\\ &=F(v)^{N-1}. \end{aligned}

Integrate from $v_L$ to $v$. At the lower endpoint, $F(v_L)=0$, and a finite bid makes $(v_L-b(v_L))F(v_L)^{N-1}=0$. Therefore

(v-b(v))F(v)^{N-1} =\int_{v_L}^{v}F(t)^{N-1}\,dt.

Dividing by $F(v)^{N-1}$ yields the stated formula. The condition $v_L\geq 0$ ensures its boundary bid is feasible under the nonnegative-bid constraint. In fact, for $v>v_L$, the integral is positive, so $b(v)<v$: bidders shade their bids below their values. Also, since $F(t)\leq F(v)$ for $t\leq v$, the integral is at most $(v-v_L)F(v)^{N-1}$, which implies $b(v)\geq v_L\geq 0$.

Why the Solution Is a Best Response

A first-order condition alone would establish only a local optimum. Here the candidate strategy also gives a direct global incentive check. Differentiate the payoff from choosing the cutoff $z$, using the equilibrium differential equation for $b'(z)$:

\begin{aligned} \frac{\partial U(v,z)}{\partial z} &=-b'(z)F(z)^{N-1} +(v-b(z))(N-1)F(z)^{N-2}f(z)\\ &=(N-1)f(z)F(z)^{N-2}(v-z). \end{aligned}

For interior $z$, the factors $(N-1)$, $f(z)$, and $F(z)^{N-2}$ are nonnegative. Thus the payoff increases with $z$ when $z<v$ and decreases when $z>v$. Its maximum over the strategy’s bid range is attained at $z=v$. A bid below $b(v_L)$ cannot win against any rival using the strategy, so it yields zero; the equilibrium bid gives nonnegative expected utility. A bid above $b(v_H)$ wins for certain but costs more than $b(v_H)$, and is no better than that boundary bid for any type $v\leq v_H$. Hence no bid outside the strategy’s range improves on the prescribed bid. This verifies global optimality within the model.

Worked Example: Uniform Values on Zero to One

Suppose $N=3$ bidders have independent values uniformly distributed on $[0,1]$. Bids are nonnegative, as required, and $F(t)=t$ on this support. For $v>0$, the equilibrium formula gives

\begin{aligned} b(v) &=v-\frac{\int_0^v t^2\,dt}{v^2}\\ &=v-\frac{v^3/3}{v^2} =\frac{2}{3}v. \end{aligned}

A bidder with value $v=0.75$ therefore bids $b(0.75)=0.50$. The expected surplus conditional on winning is $0.75-0.50=0.25$, and the probability of winning is $F(0.75)^2=0.75^2=0.5625$. Expected utility is therefore $0.25(0.5625)=0.140625$.

For a check, consider instead bidding $0.40$. Since $b(z)=2z/3$, that bid corresponds to $z=0.60$. Its win probability is $0.60^2=0.36$, and its expected utility is $(0.75-0.40)(0.36)=0.126$, below $0.140625$. The calculation illustrates the global incentive result: shading more to $0.40$ reduces the chance of winning enough to lower expected utility.

Uniform Values on a General Nonnegative Interval

For a uniform distribution on $[a,c]$, with $0\leq a<c$, the CDF is $F(t)=(t-a)/(c-a)$. Substitution into the general formula gives a particularly simple expression. For $v>a$,

\begin{aligned} \frac{\int_a^v F(t)^{N-1}\,dt}{F(v)^{N-1}} &=\frac{\int_a^v \left(\frac{t-a}{c-a}\right)^{N-1}dt} {\left(\frac{v-a}{c-a}\right)^{N-1}}\\ &=\frac{(v-a)^N/[N(c-a)^{N-1}]} {(v-a)^{N-1}/(c-a)^{N-1}} =\frac{v-a}{N},\\ b(v)&=v-\frac{v-a}{N} =a+\frac{N-1}{N}(v-a). \end{aligned}

The minimum bid is $a$, and the fraction of the distance above the lower endpoint that is bid is $(N-1)/N$. As the number of bidders increases, the bid approaches value at any fixed $v>a$: stronger competition makes winning more valuable and reduces the incentive to shade.

Worked Example: Four Bidders With Values From Two to Ten

Suppose four bidders have independent uniform values on $[2,10]$, measured in dollars. The support is nonnegative and the model’s distributional assumptions hold. Here $a=2$ and $N=4$, so

b(v)=2+\frac{3}{4}(v-2).

A bidder with value $v=8$ bids

b(8)=2+\frac{3}{4}(8-2)=2+4.5=\$6.50.

The bid is $\$1.50$ below value. The probability all three rivals have values below $8$ is $F(8)^3=(6/8)^3=0.421875$. Thus expected utility is $(\$8-\$6.50)(0.421875)=\$0.6328125$, or about $\$0.633$. This is expected monetary surplus before the auction outcome is known, not the surplus conditional on winning.

A Different Common Distribution: Power Values

The formula also applies when the common distribution is not uniform. For example, suppose values have CDF $F(v)=v^k$ on $[0,1]$, where $k>0$. This is a valid continuous distribution on a nonnegative support. The integrand is $F(t)^{N-1}=t^{k(N-1)}$, so

\begin{aligned} \frac{\int_0^v F(t)^{N-1}\,dt}{F(v)^{N-1}} &=\frac{v^{k(N-1)+1}/[k(N-1)+1]}{v^{k(N-1)}}\\ &=\frac{v}{k(N-1)+1},\\ b(v)&=\frac{k(N-1)}{k(N-1)+1}v. \end{aligned}

Worked Example: Three Bidders With a Power Distribution

Let $F(v)=v^2$ on $[0,1]$, so $k=2$, and let $N=3$. The equilibrium bidding function is

b(v)=\frac{2(3-1)}{2(3-1)+1}v=\frac{4}{5}v.

A bidder with value $0.80$ bids $b(0.80)=0.64$. The probability of winning is $F(0.80)^2=(0.80^2)^2=0.4096$, while the surplus if the bidder wins is $0.80-0.64=0.16$. Expected utility is $0.16(0.4096)=0.065536$. The arithmetic differs from the uniform example because the distribution of rival values—and therefore the probability of beating them at a given cutoff—differs.

Boundary Conditions and Common Pitfalls

The lower endpoint is not a minor technical detail. The derivation assumes a feasible nonnegative strategy, and the formula has the boundary value $b(v_L)=v_L$. If values can be negative while bids must remain nonnegative, this expression may prescribe infeasible negative bids. For example, with two bidders and uniform values on $[-1,1]$, direct substitution would give $b(v)=(v-1)/2$, which is negative for $v<1$. That is not an equilibrium bidding function under a nonnegative-bid constraint. Such a model requires a separate analysis that accounts for the constraint; the formula in this tutorial does not cover it.

Another common mistake is to use the probability of beating one rival instead of all $N-1$ rivals. With independent values, a cutoff $z$ beats one rival with probability $F(z)$, but wins the auction only if every rival’s value is below $z$, giving $F(z)^{N-1}$. A second mistake is to treat the bid as a function of value alone and ignore the distribution of rivals: the integral term shows that equilibrium shading depends on both $F$ and the number of bidders.

These are private-value auctions: a bidder’s value is known to that bidder and does not depend on rivals’ information. A setting in which bidders receive signals about a shared object value is a common-value auction and does not satisfy this model’s private-value assumptions. The distinction is substantive, not just terminology.

Key takeaway: For risk-neutral bidders with independent, identically distributed nonnegative private values and nonnegative bids, the increasing symmetric first-price bid equals value minus an integral adjustment determined by the probability of beating all rivals. Always check the support and bid constraint before applying the formula.

Check Your Understanding

Use the assumptions and derivation above to answer the following questions.

In a first-price auction with $N$ bidders, why is the probability of winning at cutoff $z$ equal to $F(z)^{N-1}$, rather than $F(z)$?
For $N=5$ bidders with independent uniform values on $[0,1]$, derive the symmetric equilibrium bid function.
With two bidders and $F(v)=v^2$ on $[0,1]$, calculate the equilibrium bid for a bidder with value $0.60$.
Explain how the sign of the derivative of $U(v,z)$ shows that $z=v$ is a global optimum over the bid range.
Why can the formula derived here not simply be applied to uniform values on $[-1,1]$ when bids must be nonnegative?