Inverting First-Price Bids to Recover Values

From Equilibrium Bids Back to Private Values

In a first-price auction, the winner pays their own bid. A bid therefore reflects both a bidder’s private value and the strategic incentive to bid below that value to preserve some surplus when winning. If the equilibrium bidding rule is known, a bid can be mapped back to the value that would generate it. The key result is an inversion formula: it expresses that value using the bid and the distribution of bids.

This mapping is structural, not a direct reading of the bidder’s value from the bid. It depends on a model of how bidders compete. In particular, the symmetric formula below requires independent private values, a fixed number of participating bidders, and a symmetric increasing equilibrium. The formula does not by itself verify that these assumptions describe a given auction.

Definition: Suppose a first-price auction has $N$ bidders, and each bidder’s bid has cumulative distribution function $H$ and density $h$. Under the symmetric, strictly increasing equilibrium assumptions stated below, the private value implied by an interior bid $b$ is $$ v(b)=b+\frac{H(b)}{(N-1)h(b)}. $$ Here $H$ and $h$ describe an individual bidder’s bid, not the maximum of the rivals’ bids. The formula applies where $h(b)>0$.

Model and Assumptions

There are $N\geq 2$ risk-neutral bidders. Their private values are independently drawn from the same distribution $F$ on $[v_L,v_H]$, with density $f(v)>0$ in the interior. Each bidder knows their own value but not the other bidders’ values. Every bidder participates, there is no reserve price, and the highest bidder wins and pays their own bid. Because values have a continuous distribution, ties occur with probability zero.

Assume bidders use the same strictly increasing, differentiable equilibrium bidding function $\beta(v)$. Strict increase matters: observing an opponent’s bid then reveals whether that opponent’s value is above or below the value associated with the bid. The symmetry and independence assumptions mean that, for a bidder with value $v$, each rival’s value is independently below $v$ with probability $F(v)$.

Equilibrium condition: A bidder with value $v$ who submits bid $b$ receives expected utility $$ U(b;v)=(v-b)F\!\left(\beta^{-1}(b)\right)^{N-1} $$ when $b$ is in the interior of the equilibrium bid range. The bidder wins precisely when all $N-1$ rivals have values below the value associated with $b$. At an interior equilibrium bid $b=\beta(v)$, the first-order condition for maximizing this expected utility must hold.

Deriving the Inversion Formula

Let $H(b)$ be the cumulative distribution function of one bidder’s equilibrium bid, and let $h(b)$ be its density. Since $\beta$ is strictly increasing, a bidder’s bid is at most $b$ exactly when their value is at most $\beta^{-1}(b)$. Thus $H(b)=F(\beta^{-1}(b))$. A bidder who submits $b$ wins if all $N-1$ rival bids are below $b$, so the probability of winning is $H(b)^{N-1}$. Their expected utility can therefore also be written as $U(b;v)=(v-b)H(b)^{N-1}$.

\frac{\partial U(b;v)}{\partial b} =-H(b)^{N-1}+(v-b)(N-1)H(b)^{N-2}h(b).

At an interior equilibrium, set this derivative to zero and use $H(b)>0$ and $h(b)>0$. Dividing by $(N-1)H(b)^{N-2}h(b)$ gives the value associated with the equilibrium bid:

\begin{aligned} -H(b)^{N-1}+(v-b)(N-1)H(b)^{N-2}h(b)&=0,\\ (v-b)(N-1)h(b)&=H(b),\\ v&=b+\frac{H(b)}{(N-1)h(b)}. \end{aligned}

This derivation uses a necessary first-order condition for an interior optimum, together with the assumed symmetric increasing equilibrium. The inversion is not a separate proof that a proposed bid rule is an equilibrium: it recovers the value compatible with the equilibrium condition when the model assumptions hold.

Theorem: Under the stated symmetric first-price auction assumptions, for an interior equilibrium bid $b=\beta(v)$, $$ v=b+\frac{H(b)}{(N-1)h(b)}. $$ The adjustment $H(b)/((N-1)h(b))$ is the implied difference between value and bid. It depends on both the bid distribution and the number of rivals.

Two Equivalent Ways to Write the Condition

The probability of winning against the rivals is $W(b)=H(b)^{N-1}$. Its density is $w(b)=(N-1)H(b)^{N-2}h(b)$. The first-order condition can be expressed using these quantities as $v=b+W(b)/w(b)$. These are the cumulative distribution function and density of the maximum rival bid, not those of a single rival’s bid.

The two expressions agree because $$ \frac{W(b)}{w(b)} =\frac{H(b)^{N-1}}{(N-1)H(b)^{N-2}h(b)} =\frac{H(b)}{(N-1)h(b)}. $$ This distinction prevents a common bidder-count error: if using the distribution of an individual bid, include the factor $N-1$ in the denominator. If using the maximum rival bid distribution, that factor is already incorporated in its density.

Worked Example: Uniform Values and Three Bidders

Suppose three bidders have independent values uniformly distributed from 0 to 100, and consider the symmetric increasing equilibrium with no reserve. A hypothetical bid of 40 is observed. Derive the bid distribution under the equilibrium rule and use it to recover the implied value.

For this benchmark, $F(v)=v/100$ on $[0,100]$. The symmetric equilibrium bid is $\beta(v)=2v/3$, which is strictly increasing. A bid is at most $b$ exactly when the value is at most $3b/2$, so, for $0\leq b\leq 200/3$, $$ H(b)=F(3b/2)=\frac{3b}{200}, \qquad h(b)=\frac{3}{200}. $$

There are $N=3$ bidders, so the inversion formula gives $$ v(40)=40+\frac{H(40)}{(3-1)h(40)} =40+\frac{3(40)/200}{2(3/200)} =40+\frac{0.6}{0.03} =60. $$

The recovered value is 60 currency units. This matches the equilibrium bid rule because $\beta(60)=2(60)/3=40$. The example checks that the distribution-based inversion reverses the equilibrium bidding rule under the stated model.

Applying the Formula to a Bid Distribution

In an application, $H(b)$ describes the share of individual bids at or below $b$, while $h(b)$ describes how concentrated bids are near $b$. The value adjustment is larger when $H(b)$ is large relative to the density $h(b)$. Intuitively, when a bid is high relative to the distribution’s local density, the first-order condition requires a larger gap between value and bid to balance the marginal cost of raising the bid against the gain in winning probability.

The expression requires a density, so the value recovery is local: it depends not just on the fraction of bids below $b$, but also on how quickly that fraction changes around $b$. If the density is very small, the adjustment can be large. This can make inversion sensitive to the accuracy of the density near the bid being evaluated. The formula defines the model-implied mapping; applying it to a finite data set requires a way to obtain a bid distribution and density, which is a separate estimation problem.

Worked Example: A Quadratic Bid Distribution

Consider a hypothetical symmetric auction with $N=3$ bidders. Suppose an individual bidder’s bid distribution on $[0,100]$ is $H(b)=(b/100)^2$, so its density is $h(b)=2b/10000$. For a bid of 50 currency units, calculate the implied private value.

First evaluate the distribution and density at the bid: $$ H(50)=\left(\frac{50}{100}\right)^2=0.25, \qquad h(50)=\frac{2(50)}{10000}=0.01. $$ The density is measured per currency unit. Since there are two rivals, the adjustment is $$ \frac{H(50)}{(N-1)h(50)} =\frac{0.25}{2(0.01)} =\frac{0.25}{0.02} =12.5. $$

Therefore, $$ v(50)=50+12.5=62.5. $$ Under this model, a bid of 50 currency units corresponds to an implied private value of 62.5 currency units. The recovered value exceeds the bid, as expected here: the formula’s adjustment is positive wherever $H(b)>0$ and $h(b)>0$.

Worked Example: Using the Maximum Rival Bid Distribution

In a hypothetical four-bidder auction, suppose the individual bid distribution is $H(b)=(b/120)^3$ over the relevant range, with density $h(b)=3b^2/120^3$. Recover the implied value for a bid of 60 using first the individual bid distribution and then the distribution of the highest rival bid.

At $b=60$, $$ H(60)=\left(\frac{60}{120}\right)^3=0.125, \qquad h(60)=\frac{3(60)^2}{120^3}=0.00625. $$ Using the individual bid distribution, with $N-1=3$ rivals, gives $$ v(60)=60+\frac{0.125}{3(0.00625)} =60+\frac{0.125}{0.01875} =60+6.6667 \approx 66.67. $$

Alternatively, the maximum rival bid has cumulative distribution $W(b)=H(b)^3=(b/120)^9$ and density $w(b)=9b^8/120^9$. At 60, $W(60)=1/512$ and $w(60)=0.00029296875$ per currency unit. Thus $$ 60+\frac{W(60)}{w(60)} =60+\frac{1/512}{0.00029296875} \approx 66.67. $$ Both calculations recover the same value. Their agreement follows from $W/w=H/(3h)$, not from treating $H$ as the maximum rival bid distribution.

Conditions, Interpretation, and Boundaries

Inversion is meaningful only under the model that produced the first-order condition. Independent private values rule out dependence between bidders’ values; symmetry imposes the same bid distribution and strategy for each bidder; and strict monotonicity ensures the ranking of bids matches the ranking of values. A fixed number of participants is also essential because the winning probability depends on the number of rivals. If bidders face different value distributions or use different strategies, the symmetric inversion need not recover their values.

The formula applies to interior bids where the distribution has positive density. At the bottom of the bid support, $H(b)$ may be zero, and the ratio can be an indeterminate $0/0$; the value at that boundary must be handled through a limit or the model’s boundary condition, not by direct substitution. In practice, very small estimated densities can also produce unstable implied values. These are limitations of the mapping, not evidence that a particular recovered value is known with certainty.

A recovered value is an implied value conditional on the auction assumptions and the bid distribution used. It is not a direct observation of a bidder’s private information, and the formula alone does not establish that those assumptions fit the data. The direction of the adjustment is informative: where $H(b)>0$ and $h(b)>0$, the implied value exceeds the bid, reflecting the incentive to shade bids below values in a first-price auction.

Common Mistake

Wrong reasoning: Use $v=b+H(b)/h(b)$ when $H$ and $h$ are the distribution and density of one bidder’s bid, because the bidder wins whenever their bid exceeds a rival’s bid.
Correction: A bidder must exceed all $N-1$ rivals. The winning probability is $H(b)^{N-1}$, whose derivative is $(N-1)H(b)^{N-2}h(b)$. The correct individual-bid formula is $v=b+H(b)/((N-1)h(b))$. The shorter ratio $W(b)/w(b)$ is valid only when $W$ and $w$ refer to the maximum rival bid.

Key takeaway: Under a symmetric, strictly increasing first-price equilibrium, invert a bid using the individual bid distribution as $v=b+H(b)/((N-1)h(b))$. The distribution must correspond to one bidder’s bid, and the formula is valid only where the model assumptions hold and the density is positive.

Check Your Understanding

Use the equilibrium condition and keep track of which bid distribution is being used.

State two assumptions that allow the probability of winning after bidding $b$ to be written as $H(b)^{N-1}$.
For $N=5$, $H(b)=0.4$, $h(b)=0.02$ per currency unit, and $b=30$, what value does the individual-bid inversion imply?
Explain why the density of the maximum rival bid is not the same as the density $h$ of one bidder’s bid.
Why does the inversion formula require positive density at the bid being evaluated?
Does a recovered value establish that the symmetric independent-private-values model is correct for the auction? Explain.