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Let $u_i$ denote player $i$'s expected payoff, so the definition also covers mixed strategies. If $p_i^{\ast }$ is a dominant strategy, thenfor every opponents' profile $p_{-i}$ and every alternative $q_i$. In particular this holds at $p_{-i}=p_{-i}^{\ast }$. Thus no player can improve its payoff by a unilateral deviation from $(p_1^{\ast },\dots ,p_n^{\ast })$, which is exactly the definition of a Nash equilibrium. A profile of dominant strategies is a Nash equilibrium.

For the second-price sealed-bid auction, use the usual private-value auction assumptions: a bidder with value $v$ has quasilinear utility $v$ minus its payment if it wins, and zero if it loses. Fix the highest opposing bid $h$. If $h<v$, winning gives $v-h>0$, and bidding $v$ wins. If $h>v$, winning gives $v-h<0$, and bidding $v$ loses. If $h=v$, winning or losing gives zero, irrespective of the tie rule. Any other bid can only change whether the bidder wins, not the payment $h$ conditional on winning. It therefore cannot improve on the truthful bid in any of these cases. Hence bidding the true value is a weakly dominant strategy. Neither symmetry nor independence of the opposing values is needed for this argument.

For part (a), in a first-price sealed-bid auction with $n\ge 2$, continuous nonnegative bids and $v>0$, no fixed bid is a dominant strategy. A positive bid $b$ can be improved against opponents all bidding zero: replace it by $b/2>0$, still win, and increase the payoff by $b/2$. Bid zero can be improved against an opposing highest bid $h=v/2$: bidding $3v/4$ wins and earns $v/4>0$, whereas zero loses. This proves the no dominant positive-value bid in a first-price auction result.

Allowing mixed strategies does not change that conclusion. Against all-zero opponents, a random bid has expected payoff strictly below $v$ unless it is always zero and the tie rule awards this bidder the item with certainty. Any positive bid incurs a strictly positive payment, and any probability of losing at zero also reduces the payoff below $v$. In the strict case choose a sufficiently small positive fixed bid $\delta$ so that $v-\delta$ exceeds the random bid's expected payoff. In the exceptional case the preceding deviation against $h=v/2$ still improves it. The assumptions matter: for value zero, bid zero is weakly dominant; with only one bidder it is also optimal. Consequently there is no dominant strategy for a positive-value bidder in the standard first-price setting.

Part (b)'s printed nonexistence request is false under the stated symmetric independent private values model. Lack of a dominant strategy does not imply lack of a Nash equilibrium. For incomplete private information the appropriate equilibrium is a Bayesian Nash equilibrium, a Nash equilibrium of the game whose strategies are bidding rules as functions of private values. Here is an explicit counterexample with a full deviation check.

Let the $n\ge 2$ risk-neutral bidders have independent valuations with the uniform distribution on $[0,1]$, and put $a=(n-1)/n$. Suppose every opponent bids $\beta (w)=aw$. A bidder with value $v$ who bids $b\in [0,a]$ wins with probability $(b/a{)}^{n-1}$, since the opposing values must all be below $b/a$. Ties have probability zero, including at either endpoint. Its expected quasilinear utility isFor $b>0$,Thus $U_v$ increases up to $b=av$ and decreases thereafter. For $v=0$, bid zero is optimal directly. A bid above $a$ wins with certainty and pays more than bidding $a$, so it cannot improve on the maximum already found in $[0,a]$. A randomized bid is a mixture of these payoffs and also cannot improve on their maximum. The best response for every value is therefore the proposed bidding rule itself. This proves the uniform private-value first-price bidding equilibrium:The value-$v$ expected equilibrium payoff is $v^n/n$. The bidding rule is a pure strategy in the Bayesian game, even though values are random. Therefore the requested general nonexistence claim in (b) has a counterexample, rather than a valid proof. The defect is present in the original PDF, not introduced by the converted text.

In a second-price sealed-bid auction, fix the other players' bids and let $a$ be their highest bid. A player of valuation $v$ who wins pays $a$, obtaining quasilinear utility $v-a$; losing gives zero. If $v>a$, winning is optimal and bidding $v$ wins. If $v<a$, losing is optimal and bidding $v$ loses. If $v=a$, winning and losing both give zero, regardless of the fixed tie-breaking rule. Thus truthful bidding is a dominant strategy. In particular the profile in which every bidder bids its valuation is a Nash equilibrium, for every valuation profile, not merely on average.

A symmetric independent private values model gives each bidder a private valuation determined by its own type. The types are independent draws from a common distribution, and the bidders have identical roles and preferences. In the standard bidding calculation bidders are risk-neutral with quasilinear utility: they maximize expected valuation minus payment. Strategies map each bidder's privately observed value to a bid. Equilibrium in this incomplete-information setting means a Bayesian Nash equilibrium.

Take $n\ge 2$ with independent uniform values on $[0,1]$. In a first-price sealed-bid auction, if all players bid their valuations, a type $v$ earns zero whenever it wins and zero whenever it loses. But for $v>0$, bidding $v/2$ instead yields positive expected utilityTherefore truthful first-price bidding is not an equilibrium.

For the candidate uniform private-value first-price bidding equilibrium, put $\alpha =(n-1)/n$ and suppose every opponent bids $\alpha$ times its valuation. For a deviation $b\in [0,\alpha ]$, winning means that every opposing value is less than $b/\alpha$. Independence gives winning probability $(b/\alpha {)}^{n-1}$, so the type-$v$ expected utility isFor $b>0$ its derivative isFor $v>0$ it is positive before $b=\alpha v$ and negative after; for $v=0$, every positive bid has nonpositive utility and zero is optimal. Thus the maximum on $[0,\alpha ]$ is $b=\alpha v$. A bid above $\alpha$ wins with probability one but gives $v-b\le v-\alpha$, which is no better than the already considered endpoint $b=\alpha$, hence no better than $b=\alpha v$. Ties have probability zero under the continuous opposing distributions. All admissible deviations are therefore covered.

Each bidder's best response is the proposed rule, provingIts conditional expected utility is $u(v,\alpha v)=v^n/n$. With only one bidder, bidding zero is optimal and agrees with the same formal rule; the derivative argument is needed only for $n\ge 2$.

The SIPV model concerns a fixed number of bidders competing for one object. Each bidder knows their own private valuation; the value of winning depends on that valuation alone. Valuations are independent random variables with the same atomless distribution $F$, known to everyone, and bidders have identical roles and access to the same strategies. The standard revenue equivalence setting also assumes risk neutrality, quasilinear utility, and noncooperative bidding: expected utility is valuation times winning probability minus expected payment.

The revenue equivalence theorem states that auctions with the same equilibrium allocation probability $p(v)$ for each type and the same expected utility for the lowest type have the same expected payment by each type, and therefore the same expected seller revenue. In particular, if the object goes to the highest valuation and the lowest type has zero expected utility, then all such auctions have the same expected revenue. In a strictly increasing symmetric equilibrium, $p(v)=F(v{)}^{n-1}$.

For a bidder of valuation $v$, the profit as a function of winning probability is $vp-e(p)$. At an interior stationary point $p(v)$,The chain rule then gives the first two identities:Integrating from zero and using integration by parts gives the interim payment identityWith the usual zero-payment normalization $e(p(0))=0$, this becomesStationarity alone does not determine the integration constant: a fixed participation fee would add that constant. The stated formula uses the lowest-type normalization from revenue equivalence.

For the lowest-price auction with independent uniform distributions on $[0,1]$, consider the symmetric increasing equilibrium, so that $p(v)=v^{n-1}$. The zero type can bid zero, lose almost surely, and pay nothing. The interim payment identity therefore yieldsSum this expected payment over the $n$ bidders and average over their uniform distributions:This uses $n\ge 2$, as appropriate for this auction format.

For three bidders, assume the common bid is $Av$ with $A>0$. If a bidder deviates to $Az$, $0\le z\le 1$, they win precisely when both opposing valuations $u,w$ are below $z$. On that event they pay $A\min (u,w)$, so their expected payment and profit areFor $0<v<1$, stationarity at $z=v$ gives $2v^2-A{v}^2=0$, henceThis is a global best response, not merely a stationary point: for $A=2$, $\partial U/\partial z=2z(v-z)$ is positive for $0<z<v$ and negative for $v<z\le 1$. Any bid above $2$ gives the same sure-win payment as $z=1$ and cannot improve the payoff.

The uniform lowest-price auction equilibrium also verifies the revenue calculation constructively for any $n\ge 2$. If opponents bid $(n-1)u$, a deviation to $(n-1)z$ has profitThus $z=v$ is optimal and $b(v)=(n-1)v$ is a symmetric Bayesian Nash equilibrium. The realized price is $(n-1)$ times the minimum valuation; this order statistic has expectation ${\int }_0^1(1-t{)}^{n}dt=1/(n+1)$, giving the same expected revenue.

Assume the usual continuous nonnegative valuation distribution, so ties occur only on zero-probability events. Put $t=F(v)$. In a monotone symmetric Bayesian Nash equilibrium, a type $v$ is first with probability $t^{n-1}$ and second with probability $(n-1)(1-t)t^{n-2}$. Its rank-order expected prize allocation in $G_1$ is thereforeThe all-pay effort identity gives $b_1(v)={\int }_{\underline{v}}^{v}s{a}_1^{\prime }(s)ds$. It also verifies equilibrium directly: a type $v$ imitating type $z$ has utility $v{a}_1(z)-b_1(z)$, whose derivative is $(v-z)a_1^{\prime }(z)$, so the true type is a best response.

In the first version of $G_2$, the two contests have expected allocationsThere is no common effort budget, and quasilinear utility makes the two effort choices separable. Since $a_{21}+a_{22}=a_1$, adding their all-pay effort identities yieldsThe equality holds type by type for aggregate effort, rather than only after taking expectations. The within-player correlation of the two efforts does not enter these additive expected payoffs.

For the second version of $G_2$, let $V_{[1]}\ge \cdots \ge V_{[n]}$ denote descending order statistics. The expected effort in a rank-order contest with prize vector $(2,1,0,\dots )$ isEquivalently, decompose the allocation into a unit award to the best player and a unit award to each of the best two players, then use revenue equivalence: the corresponding total auction payments are $V_{[2]}$ and $2V_{[3]}$. Two separate first-place contests with prize values one and two instead generateConsequentlyFor a nondegenerate continuous distribution, the inequality is strict. No regularity of virtual valuations is needed for this comparison. With a uniform distribution on $[0,1]$, the two totals are $(3n-5)/(n+1)$ and $3(n-1)/(n+1)$, giving a difference of $4/(n+1)$.

Use the standard risk-neutral quasilinear utility model: utility is value received minus payment. A bidder may abstain for utility zero, payments to the seller are nonnegative, and the zero-value type has utility zero. The last normalization is necessary for the requested revenue formula; if arbitrary subsidies were allowed, an additive payment constant would remain undetermined. The independent private values model is symmetric, and each bidder has a unit-demand valuation in the two-item part. Participation satisfies individual rationality.

Let $G(\theta )$ be the interim probability of receiving an item and $m(\theta )$ the interim expected payment of a type $\theta$. Write $u(\theta )=\theta G(\theta )-m(\theta )$. By mimicking the Bayesian Nash equilibrium bid of a type $z$, a type $\theta$ could obtain utility $\theta G(z)-m(z)$. Equilibrium therefore gives, for $\theta >z$,The lower inequality uses type $\theta$'s incentive constraint, and the upper uses type $z$'s. Since $G$ is continuous in these auctions, the inequalities give $u^{\prime }(\theta )=G(\theta )$. With $u(0)=0$,This is the interim payment identity underlying revenue equivalence.

For one item, independence and the uniform distribution give $G(\theta )={\theta }^{n-1}$: all other values must be below $\theta$. Hence the expected payment from a bidder conditional on its value isThis is an unconditional-in-winning interim payment, not the amount paid conditional on winning. In the first-price sealed-bid auction, $m(\theta )=b(\theta )G(\theta )$, so the Bayesian Nash equilibrium bid iswith $b(0)=0$. For $n=1$, the seller's revenue is zero and a zero bid suffices under the stipulated allocation rule.

For two items and three unit-demand bidders, a type $\theta$ wins if at most one of the other two values exceeds it. ThereforeThe interim payment identity gives $m(\theta )={\theta }^2-2{\theta }^3/3$. Dividing by the winning probability gives the symmetric Bayesian Nash equilibrium bidTo check that this is an Bayesian Nash equilibrium rather than just a necessary formula, its derivative is $b^{\prime }(\theta )=2(1-\theta )(3-\theta )/[3(2-\theta {)}^2]$, positive for $0\le \theta <1$. A type $\theta$ mimicking a type $z$ receives utility $\theta G(z)-m(z)$, whose derivative in $z$ is $(\theta -z)G^{\prime }(z)$. It is positive before $z=\theta$ and negative after it, so the truthful type-matching bid is globally optimal. Bids above the highest Bayesian Nash equilibrium bid can only increase payment without increasing the winning probability; the usual nonnegative bid range covers the lower boundary.

By symmetry and the law of total expectation, the seller's expected two-item revenue isFor one item and the same three bidders,More generally the single-item revenue with $n$ bidders is $(n-1)/(n+1)$. Allocating a second item lowers competition enough that it adds no expected revenue in this particular three-bidder uniform model.

The final mechanism is a direct revelation mechanism because the message submitted by bidder $i$ is its valuation report, in the same type space $[0,1]$, and allocation and payment are explicit functions of those reports. It is the Clarke pivot mechanism for selecting two unit-demand winners. Truthful reporting is a dominant strategy: bidder $i$'s utility equals its true allocation value plus the other bidders' reported allocation values, minus a term depending only on the others' reports. Reporting its true value makes the efficient allocation maximize the first two terms, while the last is unaffected by its report.

Let the ordered values be ${\theta }_{(1)}\le {\theta }_{(2)}\le {\theta }_{(3)}$. For a winner $i$, the maximum welfare achievable by the other two bidders is their combined value; in the actual allocation only the other winner receives an item. Thus the difference defining its payment is the excluded bidder's value ${\theta }_{(1)}$. For the loser, both other bidders receive items already, so its payment is zero. Consequently both winners pay the lowest valuation, andThis equals the first-price sealed-bid auction revenue above. A conditional check gives the same interim payment identity: the minimum of the other two values has density $2(1-z)$, and bidder $i$ wins exactly when that minimum is below $\theta$. Its conditional expected pivot payment is ${\int }_0^{\theta }2z(1-z)dz={\theta }^2-2{\theta }^3/3=m(\theta )$. The mechanisms have equal expected revenue, even though their realized payments need not coincide.

An equilibrium outcome of a mechanism can be reproduced by asking for types and then sending the messages prescribed by the original equilibrium strategies. Truthful reports are then a Bayesian Nash equilibrium: a profitable false report would induce a profitable original deviation. This reduces optimization over indirect mechanisms to direct revelation mechanisms with Bayesian incentive compatibility, under the same information and participation assumptions.

For $n\ge 2$ risk-neutral bidders with independent uniform distributions on $[0,1]$, the lowest-price auction has a symmetric Bayesian Nash equilibrium $b(v)=(n-1)v$. A deviation bidding $(n-1)z$, with $0\le z\le 1$, wins with probability $z^{n-1}$ and has expected payment $(n-1)z^n/n$. Its expected profit is $v{z}^{n-1}-(n-1)z^n/n$, whose derivative is $(n-1)z^{n-2}(v-z)$, proving optimality at $z=v$. Bids above $n-1$ give the same outcome as $z=1$. The seller receives $(n-1)$ times the minimum of $n$ independent uniform valuations, with expected revenue $(n-1)/(n+1)$.

// Target: mathematical-optimization.bigb

For $n\ge 2$ risk-neutral bidders with independent uniform valuations on $[0,1]$, this rule is a symmetric Bayesian Nash equilibrium. If opponents use $\beta (v)=av$, a bid $b\in [0,a]$ has expected utility $(v-b)(b/a{)}^{n-1}$. Its derivative has the sign of $(n-1)v-nb$, so with $a=(n-1)/n$ the best bid is $av$. Bids above $a$ guarantee winning but only increase payment. The type-$v$ equilibrium utility is $v^n/n$.

With $n$ independent uniform types, $m<n$ equal prizes of scale $w(m/n)$ and unit effort costs, a symmetric Bayesian Nash equilibrium has total expected effort $R_m=m(n-m)w(m/n)/(n+1)$. The all-pay effort identity yields $b(v)=w(m/n){\int }_0^{v}t{q}_m^{\prime }(t)dt$. The allocation derivative is a Beta distribution density with parameters $n-m,m$, so integration gives the formula.