= Solution
Use the usual continuous-distribution convention $F(0)=0$, $F(1)=1$, and write $f=F'$. 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 $v$ maximizes
$$
\boxed{h_v(b)=vF(b)-b,\qquad 0\leq b\leq1.}
$$
Bids 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,
$$
h_v'(q)=vf(q)-1.
$$
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> is
$$
\boxed{\mathbb P(1\text{ wins}\mid v_1=v)>\frac12
\quad\Longleftrightarrow\quad v>\frac1{F'(F^{-1}(1/2))}.}
$$
Strict <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 is
$$
F(t)=2t-\frac{15}{4}\left[(t-\tfrac1{10})_+^2-(t-\tfrac15)_+^2\right]
-\frac{145}{144}(t-\tfrac25)_+^2.
$$
Its 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 satisfies
$$
vF(b(v))-b(v)\geq0\quad\Longrightarrow\quad b(v)\leq vF(b(v))\leq v.
$$
Consequently $F(b(v))\leq F(v)$. 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>:
$$
\boxed{\mathbb P(1\text{ wins})=\mathbb E[F(b(V))]\leq\frac12
\leq\mathbb P(2\text{ wins}).}
$$
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.
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