Common-value all-pay auctions with asymmetric information and bid caps

Abstract

We study a class of two-player common-value all-pay auctions (contests) with asymmetric information under the assumption that one of the players has an information advantage over his opponent and both players are budget-constrained. We extend the results for all-pay auctions with complete information, and show that in our class of all-pay auctions with asymmetric information, sufficiently high (but still binding) bid caps do not change the players’ expected total effort compared to the benchmark auction without any bid cap. Furthermore, we show that there are bid caps that increase the players’ expected total effort compared to the benchmark. Finally, we demonstrate that there are bid caps which may have an unanticipated effect on the players’ expected payoffs—one player’s information advantage may turn into a disadvantage as far as his equilibrium payoff is concerned.

Appendix

Remark 1

As far as equilibrium analysis is concerned, the assumption that player 2 has an information advantage over player 1 can be reduced to (2). Indeed, in the general case of $\Pi _{2}$ being finer than $\Pi _{1}$ note the following. Given $\pi _{1}\in \Pi _{1},$ the event $\pi _{1}$ is common knowledge at any $\omega \in \pi _{1}$. Thus, the equilibrium analysis can be carried out separately for each $\pi _{1}\in \Pi _{1},$ as conditional on the occurrence of $\pi _{1}$ the auction G can be viewed as a distinct common-value all-pay auction $G^{\prime }$, where the set of states of nature is $\Omega ^{\prime }=\pi _{1}$ and the conditional distribution $p(\cdot \mid \pi _{1})$ serves as the common prior distribution $p^{\prime }$. In $G^{\prime }$, player 1 has the trivial information partition, $\Pi _{1}^{\prime }=\left\{ \Omega ^{\prime }\right\} .$

Thus, we may w.l.o.g. assume that $\Pi _{1}=\left\{ \Omega \right\}$ in the original auction G. Now consider another common-value all-pay auction $G^{\prime \prime }$ in which the set of states of nature $\Omega ^{\prime \prime }\equiv \Pi _{2}$ consists of information sets of player 2 in G; the common value function $v^{\prime \prime }$ is given by $v^{\prime \prime }\left( \pi _{2}\right) =E(v\mid \pi _{2})$ for every $\pi _{2}\in \Omega ^{\prime \prime };$ player 2 has the full information partition, consisting of all singleton subsets of $\Omega ^{\prime \prime },$ i.e., $\Pi _{2}^{\prime \prime }=\left\{ \left\{ \pi _{2}\right\} \mid \pi _{2}\in \Omega ^{\prime \prime }\right\} ;$ and player 1 has the trivial information partition $\Pi _{1}^{\prime \prime }=\left\{ \Omega ^{\prime \prime }\right\} .$ It is easy to see that $\Pi _{2}^{\prime \prime }$ is connected w.r.t. $v^{\prime \prime }.$ Naturally, every mixed strategy $F_{2}\left( \cdot ,\cdot \right)$ of player 2 in G corresponds to a unique mixed strategy $F_{2}^{\prime \prime }\left( \cdot ,\cdot \right)$ of 2 in $G^{\prime \prime }$ that is given by $F_{2}^{\prime \prime }\left( \pi _{2},\cdot \right) =F_{2}\left( \omega ,\cdot \right)$ for every $\pi _{2}\in \Omega ^{\prime \prime }$ and every $\omega \in \pi _{2}$ (which is well-defined because of $\Pi _{2}$-measurability of $F_{2}$ in its first coordinate), and the mixed strategies of player 1 in G and $G^{\prime \prime }$ are identical as they are state-independent. The correspondence $\left( F_{1},F_{2}\right) \leftrightarrow \left( F_{1},F_{2}^{\prime \prime }\right)$ between mixed strategy-profiles in G and $G^{\prime \prime }$ obviously preserves both players’ expected payoffs. Thus, G and $G^{\prime \prime }$ are payoff-equivalent, and hence identical in terms of equilibrium analysis. However, $G^{\prime \prime }$ has the property that player 1 has no information on the realized state of nature while player 2 is completely informed of it. Thus, in studying equilibria in two-player common-value all-pay auctions, attention can be confined w.l.o.g. to auctions with the latter property.

Note that when a bid cap is such that a complete characterization of equilibrium in conditional “component” auctions $G^{\prime }$ (which satisfy the first part of (2) and w.l.o.g. satisfy the second part of (2)) is available, then the equilibrium in the original general auction G can also be completely characterized. Thus, our results in Sect. 3 can be stated for general auctions in which player 2 has an information advantage over player 1, provided (3) is satisfied in the “component” auction $G^{\prime }$ obtained by conditioning on $\pi _{1}\in \Pi _{1}$ with the highest values. However, we do not know whether a statement similar to that of Claim 1 can be made a general auction G, as even if the claim and its premises hold for some “component” auction obtained from G,  it remains unclear what properties equilibria may possess in other “component” auctions of G.

Proof of the uniqueness of equilibrium in Proposition 1

Fix an equilibrium $(F_{1},F_{2})$ in a common-value all pay auction G satisfying (3). We will prove that $(F_{1},F_{2})=(F_{1}^{*},F_{2}^{*}).$

In what follows, for $k=1,2$ and $\omega \in \Omega ,\,F_{k}\left( \omega ,\cdot \right)$ will be treated either as a probability measure on $\left[ 0,d\right] ,$ or as the corresponding c.d.f., sometimes in the same context. Accordingly, for $A\subset [0,d]$, $F_{k}\left( \omega ,A\right)$ will stand for the probability that the equilibrium effort of player k,  conditional on $\omega ,$ belongs of the set A,  and for $x\in {\mathbb {R}},\,F_{k}\left( \omega ,x\right)$ will stand for the probability that the equilibrium effort of player k,  conditional on $\omega ,$ is less or equal to x. Also, as $F_{1}$ is state-independent, $F_{1}\left( \omega ,\cdot \right)$ will be shortened to $F_{1}\left( \cdot \right) ,$ whenever convenient.

We will start the proof by listing the properties and partially characterizing the structure of the strategies in $(F_{1},F_{2})$ in a sequence of lemmas.

Lemma 1

$F_{1}\left( \cdot \right)$ and $F_{2}\left( \omega ,\cdot \right)$ are non-atomic on (0, d) for every $\omega \in \Omega .$

Proof

If $F_{k}\left( \omega ,\{c\}\right) >0$ for some $c\in (0,d),\,k$ and $\omega ,$ then $F_{-k}\left( \omega ^{\prime },\left( c-\varepsilon ,c\right] \right) =0$ for the other player $-k$ and every $\omega ^{\prime }\in \Omega ,$ and some sufficiently small $\varepsilon >0.$ But then k would be strictly better off by shifting the probability from c to $c-\frac{\varepsilon }{2}$, a contradiction to $F_{k}$ being an equilibrium strategy. $\square$

Lemma 2

There is no interval $(a,b)\subset (0,d)$ on which in some state of nature only one player places a positive probability according to his equilibrium strategy.

Proof

If such (a, b) existed, there would also exist a positive $a^{\prime }$ such that only one player (k) places a positive probability on $(a^{\prime },b)$, and it would then be profitable for k to deviate (in at least one state of nature, if this is the informed player 2) by shifting a positive probability from $(a^{\prime },b)$ to $a^{\prime }$, a contradiction to $F_{k}$ being an equilibrium strategy. $\square$

Lemma 3

There exist $0\le \beta \le d$ such that $F_{1}\left( [0,\beta ]_{-}\cup \{d\}\right) \,=1$ (where $[0,\beta ]_{-}$ denotes the interval $[0,\beta )$ if $\beta >0$ and $[0,0]_{-}=\{0\}$),  and either the restriction of $F_{1}\left( \cdot \right)$ to $[0,\beta ]_{-}$ has full support on it, or $F_{1}\left( \cdot \right)$ is concentrated on d (i.e., $F_{1}\left( \{d\}\right) =1$).

Proof

Suppose first that there is a (non-degenerate) interval $(a,b)\subset (0,d)$ such that $F_{1}\left( (a,b)\right) =0$ (and thus $F_{2}\left( \omega ,(a,b)\right) =0$ for every $\omega \in \Omega ,$ by Lemma 2), but $F_{1}\left( \left[ b,d\right) \right) >0$. By increasing b if necessary, it can also be assumed that b is maximal with respect to this property, i.e., that $F_{1}\left( [b,b+\varepsilon )\right) >0$ for every small enough $\varepsilon >0.$ However, the expected payoff of player 1 at $\frac{a+b}{2}$ is strictly bigger than his payoff for any effort in $[b,b+\varepsilon )$, if $\varepsilon >0$ is small enough. This contradicts the assumption that $F_{1}$ is an equilibrium strategy, and shows that there exists no interval (a, b) as above. Non-existence of such (a, b) together with Lemma 1 now imply the claim of Lemma 3. $\square$

Lemma 4

For every $\omega \in \Omega ,\,F_{2}\left( \omega ,[0,\beta ]_{-}\cup \{d\}\right) \,=1$ if $F_{1}\left( \{d\}\right) <1$, and $F_{2}\left( \omega ,\{0\}\cup \{d\}\right) \,=1$ if $F_{1}\left( \{d\}\right) =1.$

Proof

In the contrary case there would exist an open subinterval of (0, d) where only player 2 places positive probability, which would contradict Lemma 2. $\square$

Lemma 5

$F_{1}(\cdot )$ cannot be concentrated on d,  and thus (by Lemma 3) the restriction of $F_{1}\mathrm{(} \cdot \mathrm{)}$ to $[0,\beta ]_{-}$ has full support on $[0,\beta ]_{-}$.

Proof

If, contrary to the assertion, $F_{1}(\cdot )$ is concentrated on d,  then for each $1\le i\le n,\,F_{2}\left( \omega _{i},\{0\}\cup \{d\}\right) \,=1$ by Lemma 4. It cannot be that all $F_{2}\left( \omega _{i},\cdot \right)$ are concentrated 0,  or all $F_{2}\left( \omega _{i},\cdot \right)$ are concentrated on d [in the first case, player 1 would have a profitable deviation to a positive effort near 0,  and in the second case player 1 would have a profitable deviation to 0 since his expected payoff under $F_{1}$ would be negative, as follow from assumption (3)]. It is easy to see that then there exists $i^{\prime }$ such that $F_{2}\left( \omega _{i^{\prime }},\{0\}\cup \{d\}\right) =1,$ whereas $F_{2}\left( \omega _{i},\cdot \right)$ is concentrated on 0 for $i<i^{\prime }$ and $F_{2}\left( \omega _{i},\cdot \right)$ is concentrated on d for $i>i^{\prime }.$ If $i^{\prime }\ne n,$ however, $F_{1}$ (which is concentrated on d) gives player 1 a negative expected payoff by assumption (3) on d,  and thus cannot be a best response. We conclude that $i^{\prime }=n.$ But then, by lowering his effort from d to a positive effort near zero,  player 1 will save almost the entire cost of effort d, while his loss of utility from winning will not exceed $\frac{1}{2}p_{n}v_{n}.$ By assumption (3), this will be a profitable deviation, in contradiction to $F_{1}$ being an equilibrium strategy. $\square$

Lemma 6

The interval $\left[ 0,\beta \right] _{-}$ is non-degenerate, i.e., $0<\beta$ and hence $\left[ 0,\beta \right] _{-}=[0,\beta ).$

Proof

If $\beta =0$ then both equilibrium strategies prescribe mixtures of effort 0 and effort d. It follows from Lemma 5 that $F_{1}(\{0\}\cup \{d\})=1$ and $F_{1}(\{0\})>0$. Then the only strategy $F_{2}$ of 2 that may constitute a best response to $F_{1}$ would prescribe d with probability 1 at every $\omega \in \Omega ,$ and then $F_{1}$ must be concentrated on 0 [since choosing d will give player 1 a negative expected payoff by assumption (3) on d]. However, no strategy of player 2 can be a best response to such $F_{1}$, a contradiction. $\square$

Lemma 7

If $1\le i<j\le n$ and $x\in [0,d]$ (respectively, $y\in [0,d]$) is a pure strategy best response of player 2 given $\left\{ \omega _{i}\right\}$ (respectively, given $\{ \omega _{j}\}$) against strategy $F_{1},$ then $x\le y.$

Proof

Assume to the contrary that $x>y,$ and let $F_{1}^{-}$ be the function defined by $F_{1}^{-}\left( z\right) \equiv \lim _{z^{\prime }\nearrow z}F_{1}\left( z^{\prime }\right)$ for every real z. Note that for every $k=1,\ldots ,n$ and every $z\in [0,d],$

$$\begin{aligned} E_{2}\left( \left\{ \omega _{i}\right\} ,F_{1},z\right) =v_{i}\cdot \frac{ F_{1}^{-}(z)+F_{1}(z)}{2}-z. \end{aligned}$$

By assumption on x, y,  we thus have

$$\begin{aligned} v_{i}\cdot \frac{F_{1}^{-}(x)+F_{1}(x)}{2}-x= & {} E_{2}\left( \left\{ \omega _{i}\right\} ,F_{1},x\right) \end{aligned}$$

(22)

$$\begin{aligned}\ge & {} E_{2}\left( \left\{ \omega _{i}\right\} ,F_{1},y\right) =v_{i}\cdot \frac{F_{1}^{-}(y)+F_{1}(y)}{2}-y \end{aligned}$$

(23)

and

$$\begin{aligned} v_{j}\cdot \frac{F_{1}^{-}(x)+F_{1}(x)}{2}= & {} E_{2}\left( \left\{ \omega _{j}\right\} ,F_{1},x\right) \end{aligned}$$

(24)

$$\begin{aligned}\le & {} E_{2}\left( \left\{ \omega _{j}\right\} ,F_{1},y\right) =v_{j}\cdot \frac{F_{1}^{-}(y)+F_{1}(y)}{2}-y. \end{aligned}$$

(25)

Both $F_{1}^{-}$ and $F_{1}$ are non-decreasing functions, that are strictly increasing on $[0,\beta )$ by Lemmas 5 and 6; moreover, $F_{1}(d)>F_{1}(z)$ for any $z<\beta$ if $F_{1}(\left\{ d\right\} )>0.$ Thus, if either $y<\beta ,$ or $F_{1}(\left\{ d\right\} )>0$ and $x=d,$ since $x>y$ and $v_{j}>v_{i}$ we also have

$$\begin{aligned} (v_{j}-v_{i})\frac{F_{1}^{-}(x)+F_{1}(x)}{2}>(v_{j}-v_{i})\frac{ F_{1}^{-}(y)+F_{1}(y)}{2}. \end{aligned}$$

(26)

But adding (26) to the inequality in (22)–(23) contradicts the inequality obtained in (24)–(25), and so it is impossible for x to exceed y when either $y<\beta$ or $F_{1}(\left\{ d\right\} )>0$ and $x=d$. In the remaining case either $F_{1}(\left\{ d\right\} )=0$ or $x<d,$ and ($x>$)$y\ge \beta .$ But then x cannot be a best response of player 2 given $\left\{ \omega _{i}\right\}$, as lowering his bid to $\beta$ will guarantee him $v_{i}$ with the same probability as bidding x but at a lesser cost, contrary to the assumption on x. $\square$

In keeping with our earlier notation, given $0\le a\le b\le d$ denote by $\left[ a,b\right] _{-}$ the interval [a, b) if $b>0$ and the set $\{0\}$ if $a=b=0.$

Lemma 8

For each $i=1,\ldots ,n$ there exists a (possibly empty or degenerate) subinterval $\left[ a_{i},b_{i}\right] _{-}$ of $[0,\beta )$ such that $F_{2}\left( \omega _{i},\left[ a_{i},b_{i}\right] _{-}\cup \{d\}\right) \,=1,$ and, if $F_{2}\left( \omega _{i},\{d\}\right) <1,$ the restriction of $F_{2}\left( \omega _{i},\cdot \right)$ to $[0,\beta )$ has full support on $\left[ a_{i},b_{i}\right] _{-}$.

Proof

If $F_{2}\left( \omega _{i},\{0\}\cup \{d\}\right) \,=1,$ the claim obviously holds for $0<a_{i}=b_{i}$ (an empty $\left[ a_{i},b_{i} \right] _{-}$), or $a_{i}=b_{i}=0.$ Assume now that $F_{2}\left( \omega _{i},\{0\}\cup \{d\}\right) \,<1$, but the claim does not hold. Then there are $0<a<b<\beta$ such that $F_{2}\left( \omega _{i},(a,b)\right) =0$, but $F_{2}\left( \omega _{i},[0,a]\right) >0$ and $F_{2}\left( \omega _{i},[b,\beta )\right) >0.$ Since $F_{1}\left( (a,b)\right) >0$, by Lemma 2 there must be $j\ne i$ such that $F_{2}\left( \omega _{j},(a,b)\right) >0.$ Assume that $i<j$ (the opposite case is treated similarly). Then there are $x\in [b,\beta )$ and $y\in (a,b)$ which are player 2’s pure strategy best responses to $F_{1}$ given $\left\{ \omega _{i}\right\}$ and $\left\{ \omega _{j}\right\}$ respectively. But, since $x>y,$ this contradicts Lemma 7. Consequently, the restriction of each $F_{2}\left( \omega _{i},\cdot \right)$ to $[0,\beta )$ has full support on some $\left[ a_{i},b_{i}\right] _{-}$. $\square$

Lemma 9

For each $i=1,\ldots ,n-1,$ the set $\left[ a_{i},b_{i}\right] _{-}$ is non-empty and $F_{2}\left( \omega _{i},\cdot \right)$ has full support on $\left[ a_{i},b_{i}\right] _{-}$. The set $\left[ a_{n},b_{n} \right] _{-}$ is also non-empty, and the restriction of $F_{2}\left( \omega _{n},\cdot \right)$ to $[0,\beta )$ has full support on $\left[ a_{n},b_{n} \right] _{-}$.

Proof

If there is $1\le i\le n$ such that $F_{2}\left( \omega _{i},\{d\}\right) >0$, denote by $i^{\prime }$ the smallest index with this property. By Lemma 7, $F_{2}\left( \omega _{i},\cdot \right)$ is concentrated on d for every $i>i^{\prime },$ and hence, if $i^{\prime }<n,\,F_{2}\left( \omega _{n},\cdot \right)$ is concentrated on d. Thus

$$\begin{aligned} E_{1}(d^{\prime },F_{2})\le \sum _{j=1}^{n-1}p_{j}v_{j}+\frac{1}{2} p_{n}v_{n}-d^{\prime }<0 \end{aligned}$$

by assumption (3) for all efforts $d^{\prime }\le d$ that are sufficiently close to d. Consequently, the support of the equilibrium strategy $F_{1}$ of player 1 will exclude some left-hand neighborhood of d. Thus player 2 will not exert effort d at $\omega _{n}$ with positive probability, a contradiction to $F_{2}\left( \omega _{n},\cdot \right)$ being concentrated on d. Thus, even if $i^{\prime }$ is defined, it must be equal to n,  implying that $F_{2}\left( \omega _{i},\{d\}\right) =0$ for every $i=1,\ldots ,n-1.$ Thus, by Lemma 8 the distribution $F_{2}\left( \omega _{i},\cdot \right)$ has full support on $\left[ a_{i},b_{i}\right] _{-}\subset [0,\beta )$ for $i=1,\ldots ,n-1$ (and in particular each $\left[ a_{i},b_{i}\right] _{-}$ is non-empty). As $F_{2}\left( \omega _{n},\cdot \right)$ cannot be concentrated on d by the argument above$,\, \left[ a_{n},b_{n}\right] _{-}$ is also non-empty, and the restriction of $F_{2}\left( \omega _{n},\cdot \right)$ to $[0,\beta )$ has full support on $\left[ a_{n},b_{n}\right] _{-}$ by Lemma 8. $\square$

Corollary 10

The intervals $\left\{ [a_{i},b_{i}]_{\_}\right\} _{i=1}^{n}$ are disjoint (barring the set $\{0\}$), and “ordered” according to the index i on the interval $[0,\beta ),$ i.e., if $i<j,$ then $[a_{i},b_{i}]_{\_}$ lies below $[a_{j},b_{j}]_{\_}$ (or coincides with $[a_{j},b_{j}]_{\_},$ if $[a_{j},b_{j}]_{\_}=\{0\}$). Moreover, $\cup _{i=1}^{n}\left[ a_{i},b_{i}\right] _{-}=[0,\beta ).$

Proof

It follows immediately from Lemmas 7 and 9 that the intervals are “ordered”. And, if $\cup _{i=1}^{n}\left[ a_{i},b_{i}\right] _{-}\varsubsetneq [0,\beta )$ then there must be a “gap” (a, b) on which only player 1 places positive probability, which contradicts Lemma 2. $\square$

Notation

It follows from Corollary 10 that there are points $0=y_{0}\le y_{1}\le \cdots <y_{n}\equiv \beta$ such that $[a_{i},b_{i}]_{ \_}=[y_{i-1},y_{i}]_{\_}$ for every $i=1,2,\ldots ,n,$ i.e., for $i=1,\ldots ,n-1,\, F_{2}\left( \omega _{i},\cdot \right)$ has full support on $[y_{i-1},y_{i}]_{\_},$ and the restrictions of $F^{1}\left( \cdot \right)$ and $F_{2}\left( \omega _{n},\cdot \right)$ to $[0,\beta )$ have full support on $[0,y_{n})$ and $[y_{n-1},y_{n})$, respectively. We denote by $i_{0}$ the smallest integer with $y_{i_{0}}>0.$ ¹⁸

Since $F^{1}\left( \cdot \right)$ has full support on $[0,y_{n})$ (when restricted to $[0,y_{n})$) and $F_{2}\left( \omega ,\cdot \right)$ has no atoms in $(0,y_{n}),$ player 1 is indifferent between any two efforts in $(0,y_{n})$. Thus, after denoting

$$\begin{aligned} e_{1}\equiv \lim _{y\searrow 0}E_{1}(y,F_{2})\ge 0, \end{aligned}$$

the following equality must hold for every $i=i_{0},\ldots ,n$ and every positive $x\in [y_{i-1},y_{i}):$

$$\begin{aligned} \sum _{j=1}^{i-1}p_{j}v_{j}+p_{i}v_{i}F_{2}(\omega _{i},x)-x=E_{1}(x,F_{2})=e_{1}. \end{aligned}$$

In particular,

$$\begin{aligned} F_{2}(\omega _{i},x)=\frac{x-\sum _{j=1}^{i-1}p_{j}v_{j}+e_{1}}{p_{i}v_{i}} \end{aligned}$$

(27)

for every $i=i_{0},\ldots ,n$ and every $x\in [y_{i-1},y_{i}).$ Since, for $i=i_{0}+1,\ldots ,n,\,F_{2}\left( \omega _{i},\cdot \right)$ has full support (after restriction to $[0,\beta )$ for $i=n$) on $[y_{i-1},y_{i})\subset (0,\beta ),$ we have $F_{2}(\omega _{i},y_{i-1})=0,$ and thus

$$\begin{aligned} y_{i}=\sum _{j=1}^{i}p_{j}v_{j}-e_{1} \end{aligned}$$

(28)

for every $i=i_{0},\ldots ,n-1.$

Since, for $i=i_{0},\ldots ,n,\,F_{2}\left( \omega _{i},\cdot \right)$ has full support (after restriction to $[0,\beta )$ for $i=n$) on $[y_{i-1},y_{i})$ and $F_{1}\left( \cdot \right)$ has no atoms (except, possibly, at 0) on $[0,\beta ),$ player 2 is indifferent between all positive efforts in $[y_{i-1},y_{i})$. Thus, the following equality must hold for every positive $x\in [y_{i-1},y_{i}):$

$$\begin{aligned} v_{i}F_{1}\left( x\right) -x= & {} E_{2}(\{\omega _{i}\},F_{1},x) \\= & {} \lim _{y\searrow y_{i-1}}E_{2}(\{\omega _{i}\},F_{1},y)=v_{i}F_{1}\left( y_{i-1}\right) -y_{i-1}, \end{aligned}$$

and in particular

$$\begin{aligned} F_{1}\left( x\right) =\frac{x}{v_{i}}+F_{1}\left( y_{i-1}\right) -\frac{ y_{i-1}}{v_{i}}. \end{aligned}$$

(29)

The rest of the proof will separately consider the following cases.

Case 1 Assume that $e_{1}>0$ or $i_{0}>1.$

The assumption of Case 1 implies that either $F_{2}\left( \omega _{i_{0}-1},\cdot \right)$ is concentrated on 0 (if $i_{0}>1$) or $F_{2}\left( \omega _{1},\{0\}\right) >0$ by (27) (if $i_{0}=1$ but $e_{1}>0$). Thus $F_{1}\left( \cdot \right)$ has no atom at 0 (otherwise shifting mass from 0 to a sightly higher effort would constitute a profitable deviation), and hence $F_{1}\left( 0\right) =0.$ Using this, (28) and (29),  we obtain

$$\begin{aligned} F_{1}\left( x\right) =\frac{x+e_{1}-\sum _{j=1}^{i}p_{j}v_{j}}{v_{i}} +\sum _{j=i_{0}+1}^{i}p_{j}+\frac{\sum _{j=1}^{i_{0}}p_{j}v_{j}-e_{1}}{ v_{i_{0}}} \end{aligned}$$

(30)

for every $i=i_{0},\ldots ,n,$ and every positive $x\in [y_{i-1},y_{i}).$

Case 1.1 Assume that $F_{1}\left( \{d\}\right) >0.$

It is a corollary of this assumption that $F_{2}\left( \omega _{n},\{d\}\right) >0$ (since otherwise for some small $\varepsilon >0$ and every i we would have $F_{2}\left( \omega _{i},(d-\varepsilon ,d]\right) =0,$ and then player 1 could profitably shift probability from d to a lower effort). Note further that having an atom at d by both $F_{1}\left( \cdot \right)$ and $F_{2}\left( \omega _{n},\cdot \right)$ implies that these distributions place no probability on an open interval $(d-\varepsilon ,d)$ for some small $\varepsilon >0,$ and hence

$$\begin{aligned} \beta <d, \end{aligned}$$

(31)

and $F_{2}\left( \omega _{n},\{d\}\right) =1-F_{2}(\omega _{n},\beta ).$ ¹⁹ As $F_{2}(\omega _{n},\cdot )$ has no atom at $\beta ,\, F_{2}(\omega _{n},\beta )$ is given by (27) for $x=\beta$ and $i=n.$ Since $F_{1}\left( \{d\}\right) >0,$ the expected payoff of player 1 from choosing d must be equal to the expected payoff from choosing effort levels in $(0,\beta ),$ i.e.,

$$\begin{aligned} e_{1}= & {} \lim _{y\searrow 0}E_{1}(y,F_{2})=E_{1}(d,F_{2}) \\= & {} \sum _{j=1}^{n-1}p_{j}v_{j}+p_{n}v_{n}\left[ F_{2}(\omega _{n},\beta )+ \frac{1}{2}\left( 1-F_{2}(\omega _{n},\beta )\right) \right] -d \\= & {} \sum _{j=1}^{n-1}p_{j}v_{j}+\frac{1}{2}p_{n}v_{n}\left[ F_{2}(\omega _{n},\beta )+1\right] -d \\= & {} \sum _{j=1}^{n-1}p_{j}v_{j}+\frac{1}{2}p_{n}v_{n}\left[ \frac{\beta -\sum _{j=1}^{n-1}p_{j}v_{j}+e_{1}}{p_{n}v_{n}}+1\right] -d \\= & {} \frac{1}{2}\sum _{j=1}^{n}p_{j}v_{j}+\frac{1}{2}\beta +\frac{1}{2}e_{1}-d. \end{aligned}$$

Therefore

$$\begin{aligned} \beta =2d-\sum _{j=1}^{n}p_{j}v_{j}+e_{1}. \end{aligned}$$

(32)

Since $F_{2}(\omega _{n},\cdot )$ has full support on $[y_{n-1},\beta )\cup \{d\}$ (as we have seen, d is chosen with positive probability), the conditional expected payoff of player 2 at $\omega _{n}$ from choosing d must be equal to his conditional expected payoff from choosing effort levels close to (but smaller than) $\beta ,$ i.e.,

$$\begin{aligned} v_{n}F_{1}\left( \beta \right) -\beta= & {} \lim _{x\nearrow \beta }E_{2}(\{\omega _{n}\},F_{1},x) \\= & {} E_{2}(\{\omega _{n}\},F_{1},d)=v_{n}\left[ F_{1}\left( \beta \right) + \frac{1}{2}\left( 1-F_{1}\left( \beta \right) \right) \right] -d \\= & {} \frac{v_{n}}{2}\left[ 1+F_{1}\left( \beta \right) \right] -d. \end{aligned}$$

It follows, by using (30) and (32) for $x=\beta$ and $i=n$,²⁰ that

$$\begin{aligned} 0= & {} \frac{v_{n}}{2}\left[ 1-F_{1}\left( \beta \right) \right] +\beta -d \\= & {} \frac{v_{n}}{2}\left[ 1-\frac{\left( 2d-\sum _{j=1}^{n}p_{j}v_{j}+e_{1}\right) +e_{1}-\sum _{j=1}^{n}p_{j}v_{j}}{ v_{n}}\right. \nonumber \\&\left. -\sum _{j=i_{0}+1}^{n}p_{j}-\frac{\sum _{j=1}^{i_{0}}p_{j}v_{j}-e_{1}}{ v_{i_{0}}}\right] +\left( 2d-\sum _{j=1}^{n}p_{j}v_{j}+e_{1}\right) -d \\= & {} \frac{v_{n}}{2}\left[ 1-\sum _{j=i_{0}+1}^{n}p_{j}-\frac{ \sum _{j=1}^{i_{0}}p_{j}v_{j}-e_{1}}{v_{i_{0}}}\right] \\= & {} \frac{v_{n}}{2}\left[ \sum _{j=1}^{i_{0}}p_{j}\left( 1-\frac{v_{j}}{ v_{i_{0}}}\right) +\frac{e_{1}}{v_{i_{0}}}\right] . \end{aligned}$$

But the expression in brackets is positive under the assumptions of Case 1, a contradiction. Thus, the properties assumed in Case 1 and Case 1.1 cannot hold jointly in equilibrium.

Case 1.2 Assume that $F_{1}\left( \{d\}\right) =0.$

This assumption implies that $F_{2}(\omega _{n},\{d\})=0$ as well (since otherwise for some small $\varepsilon >0$ we would have $F_{1}\left( (d-\varepsilon ,d]\right) =0,$ and then player 2 could profitably shift mass from d to a lower effort at $\omega _{n}$). Thus, $F_{1}\left( [0,\beta )\right) =\,F_{2}\left( \omega _{i},[0,\beta )\right) \,=1,$ and $F_{1}\left( \cdot \right) ,\,F_{2}\left( \omega _{i},\cdot \right)$ are non-atomic on $(0,\beta ),$ for each $i=i_{0},\ldots ,n.$ In particular, formulas (27) and (30) hold for $x=y_{n}=\beta$ and $i=n,$ and

$$\begin{aligned} F_{2}(\omega _{n},\beta )=F_{1}(\beta )=1. \end{aligned}$$

(33)

It follows from (27) and (33) that

$$\begin{aligned} \beta =y_{n}=\sum _{j=1}^{n}p_{j}v_{j}-e_{1}. \end{aligned}$$

Thus

$$\begin{aligned} 1= & {} F_{1}(\beta )=\frac{\beta +e_{1}-\sum _{j=1}^{n}p_{j}v_{j}}{v_{i}} +\sum _{j=i_{0}+1}^{n}p_{j}+\frac{\sum _{j=1}^{i_{0}}p_{j}v_{j}-e_{1}}{ v_{i_{0}}} \\= & {} \frac{\left( \sum _{j=1}^{n}p_{j}v_{j}-e_{1}\right) +e_{1}-\sum _{j=1}^{n}p_{j}v_{j}}{v_{i}}+\sum _{j=i_{0}+1}^{n}p_{j}+\frac{ \sum _{j=1}^{i_{0}}p_{j}v_{j}-e_{1}}{v_{i_{0}}} \\= & {} \sum _{j=i_{0}+1}^{n}p_{j}+\frac{\sum _{j=1}^{i_{0}}p_{j}v_{j}-e_{1}}{ v_{i_{0}}} \\= & {} \sum _{j=i_{0}+1}^{n}p_{j}+\sum _{j=1}^{i_{0}}p_{j}\frac{v_{j}}{v_{i_{0}}}- \frac{e_{1}}{v_{i_{0}}}. \end{aligned}$$

The last expression is smaller than 1, however, under the assumptions of Case 1, a contradiction.

We conclude that the assumptions of Case 1 cannot hold in equilibrium, and consider next the complementary Case 2:

Case 2 Assume that $i_{0}=1$ and $e_{1}=0.$

Given the assumptions, it follows that (27), (28), and (29) can be rewritten in the following way:

$$\begin{aligned} y_{i}=\sum _{j=1}^{i}p_{j}v_{j}\text { }(=x_{i}) \end{aligned}$$

(34)

for every $i=1,\ldots ,n-1$ (where $x_{i}$ was defined in (4)), and

$$\begin{aligned} F_{2}(\omega _{i},x)= & {} \frac{x-\sum _{j=1}^{i-1}p_{j}v_{j}}{p_{i}v_{i}}, \end{aligned}$$

(35)

$$\begin{aligned} F_{1}\left( x\right)= & {} \frac{x}{v_{i}}+F_{1}\left( y_{i-1}\right) -\frac{ y_{i-1}}{v_{i}} \end{aligned}$$

(36)

for every $i=1,\ldots ,n$ and every $x\in [y_{i-1},y_{i}).$

Denote $e_{2}\equiv F_{1}\left( y_{0}\right) \ge 0.$ From (36) and (34) we obtain

$$\begin{aligned} F_{1}\left( x\right) =\frac{x}{v_{i}}+\sum _{j=1}^{i-1}p_{j}\left[ 1-\frac{ v_{j}}{v_{i}}\right] +e_{2} \end{aligned}$$

(37)

for every $i=1,\ldots ,n$ and every $x\in [y_{i-1},y_{i}).$

Case 2.1 Assume that $F_{1}\left( \{d\}\right) >0.$

Arguing as in Case 1.1 (but substituting $e_{1}=0$), we obtain

$$\begin{aligned} y_{n}=\beta =2d-\sum _{j=1}^{n}p_{j}v_{j}\text { }(=\widetilde{x}), \end{aligned}$$

(38)

where $\widetilde{x}$ was defined in (5), and

$$\begin{aligned} v_{n}F_{1}\left( \beta \right) -\beta =\frac{v_{n}}{2}\left[ 1+F_{1}\left( \beta \right) \right] -d. \end{aligned}$$

Using these equalities and (37), it follows that

$$\begin{aligned} 0= & {} \frac{v_{n}}{2}\left[ 1-F_{1}\left( \beta \right) \right] +\beta -d \\= & {} \frac{v_{n}}{2}\left[ 1-\frac{2d-\sum _{j=1}^{n}p_{j}v_{j}}{v_{n}} -\sum _{j=1}^{n-1}p_{j}\left[ 1-\frac{v_{j}}{v_{n}}\right] -e_{2}\right] \\&+\left( 2d-\sum _{j=1}^{n}p_{j}v_{j}\right) -d=-\frac{v_{n}}{2}e_{2}, \end{aligned}$$

and we conclude that $e_{2}=0.$ This turns (37) into

$$\begin{aligned} F_{1}\left( x\right) =\frac{x}{v_{i}}+\sum _{j=1}^{i-1}p_{j}\left[ 1-\frac{ v_{j}}{v_{i}}\right] \end{aligned}$$

(39)

for every $i=1,\ldots ,n$ and every $x\in [y_{i-1},y_{i}).$

Due to (34), (35), (38) and (39), the equilibrium strategy profile $(F_{1},F_{2})$ is identical to $(F_{1}^{*},F_{2}^{*})$ described in Sect. 3.

Case 2.2 Assume that $F_{1}\left( \{d\}\right) =0.$

Arguing as in Case 1.2 (but substituting $e_{1}=0$ and $i_{0}=1$), we obtain

$$\begin{aligned} F_{2}(\omega _{n},\beta )=F_{1}(\beta )=1 \end{aligned}$$

(40)

and

$$\begin{aligned} \beta =y_{n}=\sum _{j=1}^{n}p_{j}v_{j}. \end{aligned}$$

(41)

(Since $\beta \le d\le \sum _{j=1}^{n}p_{j}v_{j},$ (41) shows that Case 2.2 can only occur when $d=\sum _{j=1}^{n}p_{j}v_{j}.$) It follows from (37), (40), and (41) that

$$\begin{aligned} 1=\frac{\sum _{j=1}^{n}p_{j}v_{j}}{v_{n}}+\sum _{j=1}^{n-1}p_{j}\left[ 1-\frac{ v_{j}}{v_{n}}\right] +e_{2}=1+e_{2}, \end{aligned}$$

and hence $e_{2}=0.$ Thus, (39) also holds in Case 2.2, for every $i=1,\ldots ,n$ and every $x\in [y_{i-1},y_{i}).$ It follows from this, together with (34), (35), and (41), that the equilibrium strategy profile $(F_{1},F_{2})$ is identical to $(F_{1}^{*},F_{2}^{*})$ described in Sect. 3 in the case of $d=\sum _{j=1}^{n}p_{j}v_{j}$. Q.E.D.