Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 39 (Designing Online Contests)

Write $S=R+\delta$ and let $C_i=\delta +{\sum }_{j\ne i}{b}_j$. Player $i$'s payoff in the proportional contest with outside effort is $u_i(b)=v_{i}b/(b+C_i)-b$. When $C_i>0$, this is strictly concave andThe best response is zero when $v_i\le C_i$, and otherwise is $\sqrt{v_i{C}_i}-C_i$. Thus the nonnegative-effort Karush-Kuhn-Tucker conditions at a pure strategy Nash equilibrium giveFor an active player the first-order equation is $v_i(S-b_i)=S^2$. For an inactive player the derivative at zero is $v_i/S-1$. Strict concavity makes these conditions sufficient as well as necessary.

If $\delta =0$, an equilibrium cannot have just one active player: that player would win with certainty and could lower its positive effort. An all-zero profile also cannot be an equilibrium under the usual completion of proportional allocation at zero: at least one player can gain by investing an arbitrarily small amount. Consequently there are at least two active players, so every $C_i$ is positive. If $\delta >0$ this positivity is automatic. The allocation rule's otherwise undefined all-zero value at $\delta =0$ is therefore immaterial to the equilibrium calculation.

Let $k=\hat{n}>0$, $H_k={\sum }_{i=1}^{k}1/v_i$, and use the harmonic mean ${\bar{v}}_k=k/H_k$. Summing the active efforts givesThe positive root supplies the total-effort formula for a proportional contest with outside effort:At $\delta =0$ this reduces to $R=(k-1){\bar{v}}_k/k$. There is also a genuine zero-active case: if $\delta \ge v_1$, then $k=0$ and $R=0$; no harmonic mean of an empty family is needed.

For a fully explicit active-set rule, setThe equilibrium equation is $G(S)=0$. For $\delta >0$, $G$ is strictly decreasing from $+\infty$ to $-1$. For $\delta =0$ its limit at zero is $n-1>0$, and it is strictly decreasing wherever a zero could occur. Hence the positive root is unique. Moreover $S\ge \delta$, since its equation gives $1-\delta /S\ge 0$, and $b_i=S(1-S/v_i{)}_{+}$ supplies an equilibrium with total effort $S-\delta$.

Player $j$ is active exactly when $v_j>S$, equivalently $G(v_j)<0$. Ordering the valuations, including ties, givesTherefore the active-set threshold for a proportional contest with outside effort isThe qualifying indices form a prefix because $G$ is decreasing. Equality excludes the marginal player, as required by strict positivity of effort. Equivalently, the positive root $S_k$ of the displayed quadratic must satisfy $v_k>S_k\ge v_{k+1}$, with $v_{n+1}=0$. These formulas include one active player when $\delta >0$ and exclude that case when $\delta =0$.

Let $w_m=w(m/n)$, so a type $v$ values a prize at $v{w}_m$. In a symmetric increasing Bayes-Nash equilibrium of this rank-order contest, a player reporting type $t$ wins when at most $m-1$ opponents have larger types. Its winning probability isHere the count of opponents above $t$ has a binomial distribution. Differentiation, or the associated order statistic density, givesDefine the effort by the all-pay effort identityThis also verifies equilibrium globally. The derivative of a type $v$'s payoff from reporting $t$ is $w_m(v-t)q_m^{\prime }(t)$, positive before $v$ and negative after $v$. Thus truthful reporting is a best response. Bidding above the maximal equilibrium effort gains no additional winning probability. Type zero chooses zero effort.

By exchanging the two integrations, the expected value of total effort isThe last integral is the moment $\mathbb{E}[X(1-X)]$ for a Beta distribution with parameters $n-m$ and $m$, namely $m(n-m)/(n(n+1))$. This is the uniform-value multi-prize all-pay effort formula.

Put $x=m/n$ and $h(x)=w(x)x(1-x)$. The positive constant $n^2/(n+1)$ multiplying $h(x)$ does not affect the maximizing $m$. Sincethe assumed inequality makes $h$ nonincreasing on the feasible interval. Hence one prize maximizes expected total effort.

For the power family, the PDF gives $w(x)=x^{-\alpha }$. ThenIf $\alpha \ge 1$, this derivative is strictly negative for $0<x<1$: the bracket is affine and its values at the endpoints are $1-\alpha \le 0$ and $-1$. Thus $m=1$ is optimal.

If $0<\alpha <1$, the unique continuous maximizer isThe objective strictly increases before $x_{\ast }$ and decreases after it. Therefore its discrete maximizer lies amongCompare the surviving candidates using $m^{1-\alpha }(n-m)$, since $R_m=n^{\alpha }{m}^{1-\alpha }(n-m)/(n+1)$. If $n{x}_{\ast }$ is an integer, that integer is the unique maximizer; if its floor is zero, the only feasible candidate is $1$. This proves the discrete prize-count optimization for a power-valued contest including the endpoint cases.

The sufficient condition need only hold on $[1/n,(n-1)/n]$. The power family is undefined at zero, so the printed endpoint $x=0$ cannot apply to it. More generally, a positive finite value $w(0)$ would make the displayed inequality fail at zero. The design argument uses only positive feasible prize fractions and needs no value there.

Use the usual continuous-distribution convention $F(0)=0$, $F(1)=1$, and write $f=F^{\prime }$. After observing $b$, the follower can win by matching it, because ties favor the follower. Its best response is to match when $v_2>b$ and choose zero when $v_2<b$; the indifferent equality has probability zero. Thus the leader wins with probability $F(b)$ and a leader of type $v$ maximizesBids above $1$ are dominated by bidding $1$. This is the leader optimization in a sequential private-value all-pay contest. Since $F$ is a concave function, $h_v$ is concave. An interior optimum satisfies $vf(b)=1$, with the usual endpoint conditions when that equation has no interior solution.

For precision, select the smallest maximizer when the leader is indifferent. This defines a Stackelberg equilibrium and supplies the printed strict conditional comparison at the threshold type. Let $q=F^{-1}(1/2)$. The density $f(q)$ is positive: if it vanished there, its nonnegative nonincreasing continuation would force $F$ to remain $1/2$ up to $1$, which is impossible. At the median,If this is positive, every maximizer is strictly greater than $q$, so the leader wins with probability greater than $1/2$. If it is negative, every maximizer is strictly smaller than $q$. If it is zero, $q$ is a maximizer, and the smallest maximizer is at most $q$. Therefore the median-density threshold for the leader in an all-pay contest isStrict concavity of $F$ would make the optimum unique, removing the selection convention. With mere concavity, the printed assertion needs that convention at equality. For example, $F(b)=b$ and $v=1$ make all bids optimal, and choosing $b=3/4$ makes the leader more likely to win even though the strict threshold is not exceeded.

The same issue can occur at an interior type, rather than just an endpoint. A continuously differentiable concave distribution isIts density decreases from $2$ to $5/4$, is constant on $[1/5,2/5]$, and then decreases to $1/24$; integration gives $F(1)=1$. Its median is $31/100$. At $v=4/5=1/f(q)$, every bid in $[1/5,2/5]$ maximizes $h_v$, and choosing $2/5$ gives winning probability $49/80>1/2$. This confirms the genuine best-response selection at a flat leader objective issue. The smallest-maximizer convention avoids it; the threshold type has zero ex ante probability.

The unconditional comparison is valid for every optimal selection. Zero effort guarantees the leader payoff zero, so an optimal bid satisfiesConsequently $F(b(v))\le F(v)$. If $V$ has the continuous distribution $F$, the probability integral transform makes $F(V)$ uniform on $[0,1]$. Taking expectations proves the ex ante follower advantage in a sequential all-pay contest:For strictly increasing atom-free $F$, optimal bids in fact satisfy $b(v)<v$ for almost every $v\in (0,1)$, so the first inequality is strict. Conditional advantage for unusually high leader types is therefore compatible with an unconditional follower advantage.

We derive the probabilities for a sequential elimination all-pay contest from a discounted subgame perfect equilibrium, using backward induction, and only then take $\delta \uparrow 1$. This preserves the selection supplied by discounting.

First consider a two-player all-pay auction with effective prizes $A_1\ge A_2>0$, so the incremental payoff is $A_i$ times winning probability minus effort. For $0\le b\le A_2$, the independent mixed strategies have effort cumulative distribution functionsThe second player has an atom $1-A_2/A_1$ at zero. For positive bids on this support, the first player's payoff is $A_1{G}_2(b)-b=A_1-A_2$ and the second's is $A_2{G}_1(b)-b=0$. Bids above $A_2$ cannot improve either payoff. The first has no zero atom, so the second's zero bid also earns zero. If the first deviates to zero, it can win only when the second bids zero; even with every such tie resolved in its favor, its payoff is at most $A_1(1-A_2/A_1)=A_1-A_2$. The two-player complete-information all-pay equilibrium therefore has winning probabilitiesThese follow by integrating $G_2$ against the uniform $G_1$. A third player with effective prize at most $A_2$ cannot profit by entering: for $0<b\le A_2$, its winning probability is $G_1(b)G_2(b)\le b/A_2$, giving payoff at most zero; above $A_2$ its effort exceeds its prize.

For the dynamic induction, relabel any remaining subgame's valuations as $v_1>\cdots >v_k$, with $r<k$ prizes left. Define the backward-induction threshold in an elimination all-pay contestFor $r=1$ the sum is empty and $C_1=v_2$. The discounted continuation value in an elimination contest is obtained from the following net utilities:The base case is the one-prize all-pay auction: only the two highest valuations need positive effort, with prizes $v_1,v_2$ and payoffs $v_1-v_2,0$. Every lower player has a nonprofitable deviation by the preceding calculation.

For $r\ge 2$, if either of the top two wins, the other becomes the highest player in a subgame with $r-1$ prizes. The continuation threshold in either such subgame is the same numberThe remaining top player's continuation payoff is $v_i-D$. Its effective prize in a sequential contest, net of the discounted payoff from losing, is thereforeThus $A_1>A_2$ for $\delta <1$, and the two-player distributions above apply. Adding the losing-state baselines giveswhich agree with the proposed formulas.

For a player of rank $i\ge 3$, the inductive continuation utility after either top player wins is identical: its new rank is $i-1$, and the utility expression depends only on its own value and the lower-valued tail. Thus its current zero-effort payoff is that common continuation value multiplied by $\delta$, exactly the stated $J_i$. Its effective prize for deviating to win now is $A_i=v_i-J_i$. For $3\le i\le r$, direct subtraction givesFor $i\ge r+1$, $A_i=v_i\le v_{r+1}\le C_r$, since the coefficients in $C_r$ are nonnegative and sum to one. Such a player cannot gain by entering against the two active players. There is one additional zero-bid deviation to check for the highest player. If both active bids become zero, the tie could award a prize to any remaining player. Losing to any player below rank $2$ gives no greater continuation utility than losing to rank $2$: deleting rank $2$ makes the ordered remaining rival list componentwise smallest, and its continuation threshold is a nonnegative weighted sum of that list. Thus even resolving every all-zero tie in the highest player's favor gives payoff at most its usual losing baseline plus $A_1(1-A_2/A_1)=A_1-A_2$. No tie rule can improve this deviation. This verifies all best responses and the continuation-utility formulas. Applying the construction to every remaining-player set proves a subgame perfect equilibrium of the discounted contest, not merely an on-path prescription.

Now take the vanishing-discount limit of an elimination all-pay contest. At every nonfinal subgame,Thus the two highest remaining players win with equal probabilities in a nonfinal stage. In the final stage the effective prizes are their actual valuations, so its lower-valued participant beats the higher one with probability equal to half their valuation ratio.

Return to the original ranks. Only the original top $m+1$ can ever win; at most $m-1$ higher players can have left before the final stage, so players of rank $m+2$ or worse never enter its top pair. Player $1$ must lose $m-1$ fair nonfinal contests to remain unawarded until the final stage, where its opponent has value $v_{m+1}$. HenceA player $2\le i\le m$ first enters the top pair at stage $i-1$, after $i-2$ higher-ranked winners have left. To remain unawarded, it must lose the $m-i+1$ nonfinal contests from then on, followed by the final contest against rank $m+1$. ThereforeThere are exactly $m$ distinct winners, so the expected value of their indicator sum is $m$. Combining these calculations givesThis proves the ranked winning probabilities in an undiscounted elimination contest. When $m=1$, it reduces to the ordinary two-player all-pay winning probabilities; when $m>1$, the repeated fair stages explain each power of $1/2$.

The Bradley-Terry model assigns comparison probability $q_i={\theta }_i/({\theta }_i+{\theta }_{i+1})$ to each observed edge. Up to factors independent of the parameters, its likelihood function isThe comparison graph is a path graph, hence a tree. After fixing ${\theta }_n=1$, its edge ratios are unconstrained positive coordinates: every collection $r_i={\theta }_i/{\theta }_{i+1}>0$ determines uniquely ${\theta }_i={\prod }_{k=i}^{n-1}{r}_k$.

For a convenient strict-concavity calculation, use the edge log-ratios in a Bradley-Terry comparison tree ${\eta }_i=\log r_i$. The log-likelihood separates asEach derivative is $m(p_i-q_i)$, where $q_i$ is the logistic function of ${\eta }_i$, and each second derivative is $-m{q}_i(1-q_i)<0$. Because $0<p_i<1$, there is a unique finite global maximum atConverting back gives the Bradley-Terry maximum-likelihood estimate on a path:Equivalently, recurse backwards using ${\hat{\theta }}_i={\hat{\theta }}_{i+1}{p}_i/(1-p_i)$. All estimates are finite and positive. The sample size $m$ changes the curvature of the likelihood but not this maximizer; absence of comparison cycles is what permits every empirical edge proportion to be fitted simultaneously.