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Common-value all-pay auctions with asymmetric information

Abstract

We study two-player common-value all-pay auctions in which the players have ex-ante asymmetric information represented by finite connected partitions of the set of states of nature. Our focus is on a family of such auctions in which no player has an information advantage over his opponent. We find sufficient conditions for the existence of equilibrium with monotone strategies, and show that such an equilibrium is unique. We further show that the ex-ante distribution of equilibrium effort is the same for every player (and hence the players’ expected efforts are equal), although their expected payoffs are different and they do not have the same ex-ante probability of winning.

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Notes

  1. 1.

    To mention just a few works, all-pay auctions have been considered by, e.g., Hillman and Riley (1989), Baye et al. (1993, 1996), Amman and Leininger (1996), Krishna and Morgan (1997), Che and Gale (1998), Moldovanu and Sela (2001, 2006), Siegel (2009), and Moldovanu et al. (2012).

  2. 2.

    This partition representation is equivalent to the more common Harsanyi-type formulation of Bayesian games.

  3. 3.

    This framework has been used in several works to analyze common-value second-price auctions [see Einy et al. (2002), Forges and Orzach (2011), and Abraham et al. (2012)], and common-value first-price auctions (see Malueg and Orzach 2009, 2012).

  4. 4.

    See Section 2 of the online appendix to his work.

  5. 5.

    Malueg and Orzach (2012) studied the first-price auction with information partitions of this type.

  6. 6.

    The expected payoff of a player may be higher or lower than that of his opponent, which is natural as by assumption no player has an inherent advantage over his rival in our model. In contrast, in the IA case Siegel (2014) has shown that the uninformed player has a lower equilibrium payoff—he gets zero in expectation, while the expected payoff of his informed rival is positive.

  7. 7.

    Among other conditions, our assumption of connected information partitions plays a role in the emergence of equilibrium with monotone strategies. In Rentschler and Turocy’s (2015) examples of non-monotonicity of equilibrium strategies, the information endowments are non-connected.

  8. 8.

    Ties in information, i.e., the possibility that the two players have identical information sets in some states of nature, is also ruled out by Assumption 2, as can be easily observed.

  9. 9.

    To be precise, consider the following well-defined ranking of the elements of \(\Omega ^{\prime }\): let \(\omega ^{\prime }<\omega ^{\prime \prime }\) if and only if there are indices \(k<l\) such that \(\omega _{k}\in \omega ^{\prime }\) and \(\omega _{l}\in \omega ^{\prime \prime }\). Then index the elements of \(\Omega ^{\prime }\) in accordance with their rank.

  10. 10.

    Note that (12) can be assumed to hold only for \( i=1,2,\ldots ,n-2. \) As for \(i=n-1\), the inequality \(p^{n,n+1}v_{n}-v_{n-1} =v_{n}-v_{n-1}\ge 0\) holds trivially by our assumption that \( v_{i}\ge v_{i-1},1<i\le n\).

  11. 11.

    This can be seen using (13) in the next section.

  12. 12.

    We additionally use the fact that, by definition, \(F_{i(k)}^{*}\left( \pi ^{k},\cdot \right) \) is supported on \([x_{k-1},x_{k+1}]\) if \(1\le k\le n-1\), \(F_{1}^{*}\left( \pi ^{0},\cdot \right) \) is supported on \( [x_{0},x_{1}]\), and \(F_{2}^{*}\left( \pi ^{n},\cdot \right) \) is supported on \([x_{n-1},x_{n}]\).

  13. 13.

    As in the proof of Proposition 2, we use the convention that when \(k=1\), \(v_{k-1}=v_{0}\) is defined as 0.

  14. 14.

    Recall that \(\pi _{i}(\omega _{k})\in \Pi _{i}\) denotes the element of \(\Pi _{i}\) that contains \(\omega _{k}\).

  15. 15.

    This is regardless of whether \(x_{1}>0\) or \(=0\).

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Correspondence to Ori Haimanko.

Additional information

Einy, Haimanko and Sela gratefully acknowledge the support of the Israel Science Foundation Grant 648/13.

Appendix

Appendix

Proof of Proposition 2

For any \(k=0,\ldots ,n\) consider the player i(k) for whom \(\pi ^{k}\in \Pi _{i(k)}\), and assume that his rival uses the strategy from the profile \( F^{*}\). The expected payoff of player i(k), conditional on the event \( \pi ^{k}\), is given as follows. If \(1\le k\le n\), and i(k) exerts effort \(x\in [x_{k-1},x_{k}]\), then

$$\begin{aligned} E_{i(k)}\left( \pi ^{k},x,F_{i(k-1)}^{*}\right)= & {} p^{k,k+1}v_{k}F_{i(k-1)}^{*} \left( \pi ^{k-1}, x\right) -x \nonumber \\= & {} p^{k,k+1}v_{k}\frac{x+p^{k,k+1}v_{k}-x_{k}}{p^{k, k+1}v_{k}}-x \nonumber \\= & {} p^{k,k+1}v_{k}-x_{k}. \end{aligned}$$
(16)

If \(1\le k\le n-1\) and i exerts effort \(x\in [x_{k},x_{k+1}]\), then

$$\begin{aligned} E_{i(k)}\left( \pi ^{k},x,F_{i(k+1)}^{*}\right)= & {} p^{k,k+1}v_{k}+p^{k+1,k}v_{k+1}F_{i(k+1)}^{*}\left( \pi ^{k+1}, x\right) -x \nonumber \\= & {} p^{k,k+1}v_{k}+p^{k+1,k}v_{k+1}\frac{x-x_{k}}{p^{k+1,k}v_{k+1}}-x \nonumber \\= & {} p^{k,k+1}v_{k}-x_{k}. \end{aligned}$$
(17)

Now set \(v_{0}=0\). Then (17) applies also when \(k=0\) (in which case \( i(k)=1\)), i.e. (17) holds for every \(0\le k\le n-1\).

Equalities (16) and (17) establish the following fact:

Fact 1

When player i’s opponent uses the strategy from the profile \(F^{*}\), player i is: (i) indifferent between all efforts in \( [x_{k-1},x_{k+1}]\) given the event \(\pi ^{k}\), if \(1\le k\le n-1\) and \( i=i(k);\) (ii) indifferent between all efforts in \([x_{0},x_{1}]\) given \(\pi ^{0}\) (if \(i=1\));  (iii) indifferent between all efforts in \( [x_{n-1},x_{n}] \) given \(\pi ^{n}\) (if \(i=2\)).

It can be shown by induction on k that, for \(k=2,3,\ldots ,n\),

$$\begin{aligned} p^{k,k+1}v_{k}-x_{k}=\sum _{j=1}^{k-1}p^{j,j+1} \left( p^{j+1,j+2} v_{j+1} - v_{j}\right) \ge 0. \end{aligned}$$
(18)

The expression in (18) is non-negative as every summand in \( \sum _{j=1}^{k-1}p^{j,j+1}(p^{j+1,j+2}v_{j+1}-v_{j})\) is non-negative by assumption (12). When \(k=0\) or \(k=1\), equality (18) remains meaningful if the sum is defined as 0. It then follows from (16) and (18), (17) that:

Fact 2

The conditional expected payoffs of player i considered in (16) and (17) are non-negative for the corresponding efforts.

Next consider \(0\le k\le n-2\) and the player i(k). Notice that, given the event \(\pi ^{k}\), if \(y\in [x_{k+1},x_{k+2}]\) then

$$\begin{aligned} E_{i(k)}\left( \pi ^{k},y,F_{i(k+1)}^{*}\right)= & {} p^{k,k+1}v_{k}+p^{k+1,k}v_{k+1}F_{i(k+1)}^{*}\left( \pi ^{k+1},y\right) -y \nonumber \\= & {} p^{k,k+1}v_{k}+p^{k+1,k}v_{k+1}\frac{y+p^{k+2,k+3}v_{k+2}-x_{k+2}}{ p^{k+2,k+3}v_{k+2}}-y \nonumber \\\le & {} p^{k,k+1}v_{k}+p^{k+1,k}v_{k+1}\frac{x_{k+1}+p^{k+2,k+3} v_{k+2}-x_{k+2}}{p^{k+2,k+3}v_{k+2}}-x_{k+1} \nonumber \\= & {} p^{k,k+1}v_{k}+p^{k+1,k}v_{k+1}\frac{x_{k+1}-x_{k}}{p^{k+1,k}v_{k+1}} -x_{k+1} \nonumber \\= & {} p^{k,k+1}v_{k}-x_{k}=E_{i(k)} \left( \pi ^{k}, x_{k}, F_{i(k+1)}^{*}\right) . \end{aligned}$$
(19)

The inequality in (19) holds since, by (12), \( p^{k+2,k+3}v_{k+2}>p^{k+1,k}v_{k+1}\), and the last equality in (19) holds by (17). Since obviously, if \(y>x_{k+2}\) and \( k\le n-2\),

$$\begin{aligned} E_{i(k)}\left( \pi ^{k},y,F_{i(k+1)}^{*}\right) \le E_{i(k)} \left( \pi ^{k},x_{k+2},F_{i(k+1)}^{*}\right) \end{aligned}$$
(20)

and, if \(y>x_{n}\),

$$\begin{aligned} E_{1} \left( \pi ^{n-1},y,F_{2}^{*}\right)\le & {} E_{1} \left( \pi ^{n-1},x_{n},F_{2}^{*}\right) = E_{1} \left( \pi ^{n-1}, x_{n-1},F_{2}^{*}\right) , \nonumber \\ E_{2}\left( \pi ^{n},y,F_{1}^{*}\right)\le & {} E_{2}\left( \pi ^{n}, x_{n},F_{1}^{*}\right) \end{aligned}$$
(21)

Then (20), (21) and (19) establish the following:

Fact 3

When player i’s rival uses the strategy from the profile \( F^{*}\), player i (weakly) prefers effort \(x_{k}\) to any effort above \( x_{\min (k+1,n)}\), given the event \(\pi ^{k}\), if \(0\le k\le n\) and \( i=i(k)\).

Now consider \(2\le k\le n\) and the player i(k). Given the event \(\pi ^{k}\), if \(y\in [x_{k-2},x_{k-1}]\) then

$$\begin{aligned} E_{i(k)}\left( \pi ^{k},y,F_{i(k-1)}^{*}\right)= & {} p^{k,k+1}v_{k}F_{i(k-1)}^{*} \left( \pi ^{k-1},y\right) -y \nonumber \\= & {} p^{k,k+1}v_{k}\frac{y-x_{k-2}}{p^{k-1,k-2}v_{k-1}}-y \nonumber \\\le & {} p^{k,k+1}v_{k}\frac{x_{k-1}-x_{k-2}}{p^{k-1,k-2}v_{k-1}}-x_{k-1} \nonumber \\= & {} p^{k,k+1}v_{k}-x_{k} = E_{i(k)} \left( \pi ^{k}, x_{k}, F_{i(k-1)}^{*}\right) . \end{aligned}$$
(22)

The inequality in (22) holds since by (12) \( p^{k,k+1}v_{k}\ge p^{k-1,k-2}v_{k-1}\), and the last equality in (22) holds by (16). Note also that when \(k \ge 2\) and \( 0\le y\le x_{k-2}\),

$$\begin{aligned} E_{i(k)} \left( \pi ^{k},y,F_{i(k-1)}^{*}\right) \le 0. \end{aligned}$$
(23)

Then (22), (23), and Fact 2 lead to the following:

Fact 4

When player i’s rival uses the strategy from the profile \( F^{*}\), player i (weakly) prefers effort \(x_{k}\) to any effort below \( x_{k-1}\) given the event \(\pi ^{k}\), if \(2\le k\le n\) and \(i=i(k)\).

Facts 13, and 4 show that for any \(0\le k\le n\) with \(i=i(k)\), conditional on the event \(\pi ^{k}\) any effort in the support of \( F_{i}^{*}\left( \pi ^{k},\cdot \right) \) is a best response of player i against the mixed strategy of his rival in the profile \(F^{*}\). Thus \( F_{i}^{*}\) is also an unconditional best response for each player i, which means that \(F^{*}\) is indeed an equilibrium of G.

Proof of Proposition 3

Assume that \(F=(F_{1},F_{2})\) is an equilibrium, and that both strategies are monotone. We will show that \((F_{1},F_{2})=(F_{1}^{*},F_{2}^{*})\).

For any \(i=1,2\) and k satisfying \(i(k)=i\), denote by \(\mu _{F_{i}(\pi ^{k},\cdot )}\) the probability distribution (measure) with a c.d.f. \( F_{i}(\pi ^{k},\cdot )\), and by \(suppF_{i}(\pi ^{k},\cdot )\) the support of that distribution (i.e., the closure of set of all \(x\in [0,\infty )\) such that for any \(\varepsilon >0\), \(\mu _{F_{i}(\pi ^{k},\cdot )}((x-\varepsilon ,x+\varepsilon )>0\)). Since equilibrium strategies’ supports are necessarily bounded (as the marginal cost of effort is positive and fixed at 1), each \(suppF_{i}(\pi ^{k},\cdot )\) is a compact set.

Lemma 1

For any \(k=1,2,\ldots ,n\), the c.d.f. \(F_{i(k)}(\pi ^{k},\cdot )\) is continuous on \((0,\infty )\), i.e. \(\mu _{F_{i(k)}(\pi ^{k},\cdot )}\) has no atoms in \((0,\infty )\). Hence all conditional expected payoffs of each player are continuous at every positive effort (chosen as that player’s pure strategy), provided his rival sticks to his strategy in F.

Proof of Lemma 1

Suppose that, for some k, \(F_{i(k)}(\pi ^{k},\cdot )\) is discontinuous at some \(a>0\), i.e. a is an atom of \(\mu _{F_{i(k)}(\pi ^{k},.)}\). Since i(k)’s rival best-responds to \(F_{i(k)}\), there exists \(\delta >0\) such that \(\mu _{F_{i(k-1)}(\pi ^{k-1},\cdot )}((a-\delta ,a])=0\) if \(k\ge 1\), and \(\mu _{F_{i(k+1)}(\pi ^{k+1},\cdot )}((a-\delta ,a])=0\) if \(k\le n-1\). Thus, conditional on \(\pi ^{k}\), player i(k) can improve his expected payoff by shifting mass from a to an effort slightly below a. This contradicts the assumption that, conditional on \(\pi ^{k}\), a is chosen by i in equilibrium with positive probability. \(\square \)

Lemma 2

For each \(k=1,2,\ldots ,n\), denote \(x_{k}\equiv \max \left[ suppF_{i(k-1)}(\pi ^{k-1},\cdot )\right] \), and let \(x_{0}\equiv 0\). Then \( x_{0}\le x_{1}\le \cdots \le x_{n}\). Furthermore, for every \(1\le k\le n-1\),

$$\begin{aligned} suppF_{i(k)}(\pi ^{k},\cdot )=[x_{k-1},x_{k+1}]; \end{aligned}$$
(24)

additionally,

$$\begin{aligned} suppF_{1}(\pi ^{0},\cdot )=[x_{0},x_{1}] \end{aligned}$$
(25)

and

$$\begin{aligned} suppF_{2}(\pi ^{n},\cdot )=[x_{n-1},x_{n}]. \end{aligned}$$
(26)

Proof of Lemma 2

We start by showing (25). \(\square \)

Claim 1

\([x_{0},x_{1}]\subset suppF_{1}(\pi ^{0},\cdot )\).

Proof of Claim 1

By definition \(x_{1}\in suppF_{1}(\pi ^{0},\cdot )\). Now suppose by the way of contradiction that there exists \( x\in [0,x_{1})\cap (suppF_{1}(\pi ^{0},\cdot ))^{c}\). This implies the existence of \(\delta >0\) such that \([x,x+\delta )\subseteq [0,x_{1})\cap (suppF_{1}(\pi ^{0},\cdot ))^{c}\). Using the first part of Lemma 1, it follows that \(\mu _{F_{1}(\pi ^{0},\cdot )}([x,x+\delta ])=0\), and w.l.o.g. we can assume that \(x+\delta \in suppF_{1}(\pi ^{0},\cdot )\). Now two possible cases emerge.

Case 1: \(x+\delta \in suppF_{2}(\pi ^{1},\cdot )\).

Given the monotonicity of \(F_{1}\) and the first part of Lemma 1, under efforts x and \(x+\delta \) player 2 has the same expected value of winning (conditional on \(\pi ^{1}\)), and so \(E_{2}(\pi ^{1},F_{1},x)>E_{2}(\pi ^{1},F_{1},x+\delta )\). By the second part of Lemma 1, \(E_{2}(\pi ^{1},F_{1},x)>E_{2}(\pi ^{1},F_{1},z)\) for any z in some open neighborhood \((x+\delta -\delta _{1},x+\delta +\delta _{1})\) of \(x+\delta \). As \(F_{2}\) is an equilibrium strategy, \(\mu _{F_{2}(\pi ^{1},\cdot )}((x+\delta -\delta _{1},x+\delta +\delta _{1}))=0\). Hence, \(x+\delta \not \in suppF_{2}(\pi ^{1},\cdot )\), a contradiction.

Case 2: \(x+\delta \in (suppF_{2}(\pi ^{1},\cdot ))^{c}\).

In this case there exists \(\delta _{1}>0\) such that \([x+\delta -\delta _{1},x+\delta ]\subset (0,\infty )\cap (suppF_{2}(\pi ^{1},\cdot ))^{c}\). Given the monotonicity of \(F_{2}\) and the first part of Lemma 1, this means that under efforts \(x+\delta -\delta _{1}\) and \(x+\delta \) player 1 has the same expected value of winning (conditional on \(\pi ^{0}\)), hence \( E_{1}(\pi ^{0},x+\delta -\delta _{1},F_{2})>E_{1}(\pi ^{0},x+\delta ,F_{2})\) . Arguing as in Case 1 above, we conclude that \(x+\delta \not \in suppF_{1}(\pi ^{0},\cdot )\), a contradiction.

Since contradiction arises in both cases, the claim is established. \(\square \)

By definition, \(suppF_{1}(\pi ^{0},\cdot )\subset [x_{0},x_{1}]\), and so Claim 1 implies (25). The following claim is needed to establish the equality (24) for \(k=1\).

Claim 2

If \(x_{1}>0\), \([x_{0},x_{1}]\subset suppF_{2}(\pi ^{1},\cdot )\). In particular, \(x_{1}\le x_{2}\).

Proof of Claim 2

Since \([x_{0},x_{1}]\) is assumed to have a non-empty interior, it suffices to show that \((x_{0},x_{1})\subset suppF_{2}(\pi ^{1},\cdot )\). Suppose by the way of contradiction that there exists \(x\in (0,x_{1})\cap (suppF_{2}(\pi ^{1},\cdot ))^{c}\). This implies the existence of \(\delta >0\) such that \([x-\delta ,x]\subset (0,x_{1})\cap (suppF_{2}(\pi ^{1},\cdot ))^{c}\). By the monotonicity of \(F_{2}\) and the first part of Lemma 1, player 1 has the same expected value of winning (conditional on \(\pi ^{0}\)) under efforts \(x-\delta \) and x , and hence, by the second part of Lemma 1, \(E_{1}(\pi ^{0},x-\delta ,F_{2})>E_{1}(\pi ^{0},z,F_{2})\) for any z in some open neighborhood \((x-\epsilon ,x+\epsilon )\) of x. As \(F_{1}\) is an equilibrium strategy, \(\mu _{F_{1}(\pi ^{0},\cdot )}((x-\epsilon ,x+\epsilon ))=0\), in contradiction to the assumption that \(x\in [x_{0},x_{1}]=suppF_{1}(\pi ^{0},\cdot )\).

By recycling the arguments used to prove Claim 1 above, it can be shown that \([x_{1},x_{2}]\subset suppF_{2}(\pi ^{1},\cdot )\). In conjunction with Claim 2 this impliesFootnote 15 \([x_{0},x_{2}]\subset suppF_{2}(\pi ^{1},\cdot )\). Since the reverse inclusion holds by the definition of \(x_{2}\), we have established the equality (24 ) for \(k=1\).

The analogues of Claims 12 can be used to iteratively extend the proof in order to establish (24) for \(k>1\), and to show that \( x_{2}\le x_{3}\le \cdots \le x_{n}\). The analogue of Claim 2 would also show that \([x_{n-1},x_{n}]\subset suppF_{2}(\pi ^{n},\cdot )\), if \(x_{n-1}<x_{n}\). But, obviously, \(E_{2}(\pi ^{n},F_{1},x)<E_{2}(\pi ^{n},F_{1},x^{n})\) for every \(x>x_{n}\), and hence the inclusion holds as equality, establishing (26). \(\square \)

Lemma 3

If \(x_{1}>0\) and \(F_{2}(\pi ^{1},\cdot )\) is continuous at 0, then for every \(1\le k\le n\),

$$\begin{aligned}&x_{k}=\sum _{j=1}^{k}p^{j,j+1}v_{j}\left[ 1-\frac{p^{j-1,j}}{p^{j-1,j-2}} \left[ 1- \frac{p^{j-2,j-1}}{p^{j-2,j-3}} \bigg [1-\cdots \right. \right. \nonumber \\&\qquad \qquad \left. \left. \left. \times \left[ 1-\frac{p^{0,1}}{p^{0,-1}} F_{1}(\pi ^{0},0)\right] \cdots \right] \right] \right] \end{aligned}$$
(27)

(with the convention that \(p^{0,-1}=p^{0,1} =p^{1,0}= 1\)), and

$$\begin{aligned}&F_{i(k)}(\pi ^{k},x_{k}) \nonumber \\&\quad =\frac{p^{k,k+1}}{p^{k,k-1}}\left[ 1-\frac{p^{k-1,k}}{p^{k-1,k-2}} \left[ 1-\frac{p^{k-2,k-1}}{p^{k-2,k-3}}\left[ 1-\cdots \left[ 1-\frac{p^{0,1}}{ p^{0,-1}}F_{1}(\pi ^{0},0)\right] \cdots \right] \right] \right] .\nonumber \\ \end{aligned}$$
(28)

Proof of Lemma 3

The proof is by induction on k. Notice that, since \([0,x_{1}]\subset suppF_{2}(\pi ^{1},\cdot )\), \(x_{1}>0\), and \( E_{2}(\pi ^{1},F_{1},\cdot )\) is continuous on \((0,\infty )\),

$$\begin{aligned} p^{1,2}v_{1}-x_{1}= & {} E_{2} \left( \pi ^{1},F_{1},x_{1}\right) \\= & {} \lim _{\varepsilon \rightarrow 0+}E_{2} \left( \pi ^{1}, F_{1}, \varepsilon \right) = p^{1,2}v_{1}F_{1}(\pi ^{0},0). \end{aligned}$$

Thus \(x_{1}=p^{1,2}v_{1}[1-F_{1}(\pi ^{0},0)]\), i.e. (27) holds for \(k=1\). Similarly, using additionally the assumption that \( F_{2}(\pi ^{1},\cdot )\) is continuous at 0,

$$\begin{aligned} v_{1}F_{2}(\pi ^{1},x_{1})-x_{1}= & {} E_{1}(\pi ^{0},x_{1},F_{2}) \\= & {} E_{1}(\pi ^{0},0,F_{2})=0, \end{aligned}$$

and hence \(F_{2}(\pi ^{1},x_{1})=p^{1,2}[1-F_{1}(\pi ^{0},0)]\), which establishes (28) for \(k=1\).

Now, given \(1<k\le n\), assume that (27), (28) have been shown to hold for \(k-1\). Since \(\left[ x_{k-1},x_{k}\right] \subset suppF_{i(k)}\left( \pi ^{k},\cdot \right) \), \(x_{k-1}>0\), and \( E_{i(k)}(\pi ^{k},\cdot , F_{i(k-1)})\) is continuous on \((0,\infty )\),

$$\begin{aligned} p^{k,k+1}v_{k}-x_{k}= & {} E_{i(k)}\left( \pi ^{k},x_{k},F_{i(k-1)}\right) \\= & {} E_{i(k)} \left( \pi ^{k},x_{k-1},F_{i(k-1)}\right) = p^{k,k+1}v_{k}F_{i(k-1)}\left( \pi ^{k-1},x_{k-1}\right) -x_{k-1}, \end{aligned}$$

and hence

$$\begin{aligned} x_{k}= & {} p^{k,k+1}v_{k}\left[ 1-F_{i(k-1)}\left( \pi ^{k-1},x_{k-1}\right) \right] +x_{k-1} \nonumber \\= & {} p^{k,k+1}v_{k}\left[ 1-\frac{p^{k-1,k}}{p^{k-1,k-2}} \left[ 1-\frac{p^{k-2,k-1}}{ p^{k-2,k-3}} \left[ 1-\cdots \left[ 1-\frac{p^{0,1}}{p^{0,-1}} F_{1}(\pi ^{0},0) \right] \cdots \right] \right] \right] +x_{k-1} \nonumber \\= & {} \sum _{j=1}^{k}p^{j,j+1}v_{j}\left[ 1-\frac{p^{j-1,j}}{p^{j-1,j-2}} \left[ 1- \frac{p^{j-2,j-1}}{p^{j-2,j-3}} \left[ 1-\cdots \left[ 1-\frac{p^{0,1}}{p^{0,-1}} F_{1}(\pi ^{0},0)\right] \cdots \right] \right] \right] ,\nonumber \\ \end{aligned}$$
(29)

where the last two equalities hold by the induction hypothesis on (27) and (28). This completes the induction step for (27).

Since \(\left[ x_{k-1},x_{k}\right] \subset suppF_{i(k-1)}\left( \pi ^{k-1},\cdot \right) \), \(x_{k-1}>0\), and \(E_{i(k-1)}(\pi ^{k-1}, \cdot , F_{i(k)})\) is continuous on \((0,\infty )\),

$$\begin{aligned} p^{k-1,k}v_{k-1}+p^{k,k-1}v_{k}F_{i(k)}\left( \pi ^{k},x_{k}\right) -x_{k}= & {} E_{i(k-1)}\left( \pi ^{k-1},x_{k},F_{i(k)}\right) \\= & {} E_{i(k-1)}\left( \pi ^{k-1},x_{k-1},F_{i(k)}\right) \\= & {} p^{k-1,k}v_{k-1}-x_{k-1}, \end{aligned}$$

and hence

$$\begin{aligned}&F_{i(k)}\left( \pi ^{k},x_{k}\right) \nonumber \\&\quad = \frac{1}{p^{k,k-1}v_{k}} \left( x_{k}-x_{k-1}\right) \nonumber \\&\quad = \frac{p^{k,k+1}}{p^{k,k-1}}\left[ 1 - \frac{p^{k-1,k}}{p^{k-1,k-2}} \left[ 1-\frac{ p^{k-2,k-1}}{p^{k-2,k-3}} \left[ 1-\cdots \left[ 1-\frac{p^{0,1}}{p^{0,-1}}F_{1}(\pi ^{0},0) \right] \cdots \right] \right] \right] .\nonumber \\ \end{aligned}$$
(30)

This completes the induction step for (28) and finishes the proof. \(\square \)

Corollary 4

If \(x_{1}>0\) and \(F_{2}(\pi ^{1},\cdot )\) is continuous at 0, then \(F_{1}(\pi ^{0},0)=0\) (i.e., \(F_{1}(\pi ^{0},\cdot )\) is also continuous at 0). Hence, by Lemma 3, for every \(1\le k\le n\),

$$\begin{aligned} x_{k}=\sum _{j=1}^{k}p^{j,j-1}p^{j,j+1}v_{j}, \end{aligned}$$
(31)

and

$$\begin{aligned} F_{i(k)}\left( \pi ^{k},x_{k}\right) =p^{k,k+1}. \end{aligned}$$
(32)

Proof of Corollary 4

Since \(F_{2}(\pi ^{n},x_{n})=1\), by (28) in Lemma 3 taken for \(k=n\),

$$\begin{aligned} 1-\frac{p^{n-1,n}}{p^{n-1,n-2}} \left[ 1-\frac{p^{n-2,n-1}}{p^{n-2,n-3}} \left[ 1-\cdots \left[ 1- \frac{p^{0,1}}{p^{0,-1}} F_{1}(\pi ^{0},0)\right] \cdots \right] \right] =p^{n,n-1}, \end{aligned}$$

which can only occur when \(F_{1}(\pi ^{0},0)=0\). Substituting \(F_{1}(\pi ^{0},0)=0\) into (27) and (28) yields (31) and (32). \(\square \)

Lemma 5

If \(F_{2}(\pi ^{1},\cdot )\) is continuous at 0 then \( x_{1}>0\).

Proof of Lemma 5

Assume to the contrary that \(F_{2}(\pi ^{1},\cdot )\) is continuous at 0 (and hence \(x_{2}>0\)) but \(x_{1}=0\). Then \( [0,x_{2}]\subset suppF_{1}(\pi ^{2},\cdot )\). Since \(E_{1}(\pi ^{2},F_{2},\cdot )\) is continuous on \([0,\infty )\),

$$\begin{aligned} p^{2,3}v_{2}-x_{2}= & {} E_{1}\left( \pi ^{2},F_{2},x_{2}\right) \\= & {} E_{1}\left( \pi ^{2},F_{2},0\right) =0. \end{aligned}$$

Thus

$$\begin{aligned} x_{2}=p^{2,3}v_{2}. \end{aligned}$$

Similarly, since \([0,x_{2}]=suppF_{2}(\pi ^{1},\cdot )\), \(x_{2}>0\), and \( E_{2}(\pi ^{1},F_{1},\cdot )\) is continuous on \((0,\infty )\),

$$\begin{aligned} p^{1,2}v_{1}+p^{2,1}v_{2}F_{1} \left( \pi ^{2},x_{2}\right) -x_{2}= & {} E_{2}\left( \pi ^{1},F_{1},x_{2}\right) \\= & {} \lim _{\varepsilon \rightarrow 0+}E_{2} \left( \pi ^{1}, F_{1}, \varepsilon \right) \\= & {} p^{1,2}v_{1}+p^{2,1}v_{2}F_{1} \left( \pi ^{2},0\right) . \end{aligned}$$

This yields

$$\begin{aligned} F_{1}(\pi ^{2},x_{2})=\frac{1}{p^{2,1}}\left[ p^{2,3} + p^{2,1}F_{1}(\pi ^{2},0)\right] . \end{aligned}$$
(33)

For \(k>2\), the first equalities in (29) and (30) in the proof of Lemma 3 hold in the current context as well, and they imply that

$$\begin{aligned}&x_{k}-x_{k-1} \end{aligned}$$
(34)
$$\begin{aligned}&\quad = p^{k,k+1}v_{k}\left[ 1-F_{i(k-1)}\left( \pi ^{k-1},x_{k-1}\right) \right] \end{aligned}$$
(35)
$$\begin{aligned}&\quad = p^{k,k+1}v_{k}\left[ 1-\frac{1}{p^{k-1,k-2}v_{k-1}}\left( x_{k-1}-x_{k-2}\right) \right] , \end{aligned}$$
(36)

and also that

$$\begin{aligned} \text {(}1=\text {) }F_{2}\left( \pi ^{n},x_{n}\right) =\frac{1}{p^{n,n-1}v_{n} }\left( x_{n}-x_{n-1}\right) . \end{aligned}$$
(37)

It follows from (34), (35) and (33) that

$$\begin{aligned} x_{3}-x_{2}=v_{3}\left[ 1-\frac{1}{p^{2,1}}\left[ p^{2,3}+p^{2,1}F_{1}(\pi ^{2},0)\right] \right] \end{aligned}$$

if \(n=3\), and, by induction based on (34)–(36), that

$$\begin{aligned} x_{n}-x_{n-1}= & {} v_{n}\left[ 1-\frac{p^{n-1,n}}{p^{n-1,n-2}} \left[ 1-\frac{p^{n-2,n-1}}{p^{n-2,n-3}} \bigg [ 1-\cdots \bigg [1 \right. \right. \nonumber \\&\left. \left. \left. \left. -\frac{p^{3,4}}{p^{3,2}} \left[ 1-\frac{1}{p^{2,1}}\left[ p^{2,3}+p^{2,1} F_{1}(\pi ^{2},0)\right] \right] \right] \cdots \right] \right] \right] \end{aligned}$$
(38)

if \(n>3\). From (37) we have \(x_{n} - x_{n-1} = p^{n,n-1}v_{n}\), which is equal to the expression in (38) only if

$$\begin{aligned} 1-\frac{1}{p^{2,1}}\left[ p^{2,3}+p^{2,1}F_{1}(\pi ^{2},0)\right] =p^{3,2}. \end{aligned}$$

Equivalently,

$$\begin{aligned} -p^{2,3}p^{1,2}=p^{2,1}F_{1}(\pi ^{2},0). \end{aligned}$$

However, the left-hand side is negative while the right-hand side is non-negative, a contradiction. \(\square \)

Lemma 6

If \(F_{1}\left( \pi ^{0},\cdot \right) \) is continuous at 0, then \(F_{2}\left( \pi ^{1},\cdot \right) \) is continuous at 0.

Proof of Lemma 6

Since \(F_{1}\left( \pi ^{0},\cdot \right) \) is continuous at 0, \(x_{1}>0\). Notice that, since \([0,x_{1}]\subset suppF_{2}(\pi ^{1},\cdot )\) and \(E_{2}(\pi ^{1},F_{1},\cdot )\) is continuous on \([0,\infty )\),

$$\begin{aligned} p^{1,2}v_{1}-x_{1}= & {} E_{2}\left( \pi ^{1},F_{1},x_{1}\right) \\= & {} E_{2}\left( \pi ^{1},F_{1},0\right) =0. \end{aligned}$$

Thus \(x_{1}=p^{1,2}v_{1}\). And, since \([0,x_{1}]=suppF_{1}(\pi ^{0},\cdot )\), \(x_{1}>0\), and \(E_{1}(\pi ^{0},\cdot ,F_{2})\) is continuous on \((0,\infty )\),

$$\begin{aligned} v_{1}F_{2}\left( \pi ^{1},x_{1}\right) -x_{1}= & {} E_{1} \left( \pi ^{0},x_{1},F_{2}\right) \\= & {} \lim _{\varepsilon \rightarrow 0+}E_{1}\left( \pi ^{0},\varepsilon , F_{2}\right) =v_{1}F_{2}\left( \pi ^{1},0\right) , \end{aligned}$$

and hence

$$\begin{aligned} F_{2}\left( \pi ^{1},x_{1}\right) =p^{1,2}+F_{2}\left( \pi ^{1},0\right) . \end{aligned}$$
(39)

For \(k>1\), the first equalities in (29) and (30) in the proof of Lemma 3 hold in the current context as well, and they imply that

$$\begin{aligned}&x_{k}-x_{k-1} \end{aligned}$$
(40)
$$\begin{aligned}&\quad = p^{k,k+1}v_{k}\left[ 1-F_{i(k-1)}\left( \pi ^{k-1},x_{k-1}\right) \right] \end{aligned}$$
(41)
$$\begin{aligned}&\quad = p^{k,k+1}v_{k}\left[ 1-\frac{1}{p^{k-1,k-2}v_{k-1}}\left( x_{k-1}-x_{k-2}\right) \right] , \end{aligned}$$
(42)

and also that

$$\begin{aligned} \text {(}1=\text {) }F_{2}\left( \pi ^{n},x_{n}\right) =\frac{1}{p^{n,n-1}v_{n} }\left( x_{n}-x_{n-1}\right) . \end{aligned}$$
(43)

It follows from (40), (41 ) and (39) that

$$\begin{aligned} x_{2}-x_{1}=p^{2,3}v_{2}\left[ p^{2,1}-F_{2}(\pi ^{1},0)\right] , \end{aligned}$$

and, by induction based on (40)–(42), that

$$\begin{aligned} x_{n}-x_{n-1}= & {} v_{n}\left[ 1-\frac{p^{n-1,n}}{p^{n-1,n-2}} \left[ 1-\frac{p^{n-2,n-1} }{p^{n-2,n-3}} \bigg [1-\cdots \bigg [1 \right. \right. \nonumber \\&\left. \left. \left. \left. -\frac{p^{2,3}}{p^{2,1}} \left[ p^{2,1}-F_{2} (\pi ^{1},0)\right] \right] \cdots \right] \right] \right] . \end{aligned}$$
(44)

From (43), also \(x_{n}-x_{n-1} = p^{n,n-1}v_{n}\), which is equal to the expression in (44) only if \(F_{2}(\pi ^{1},0)=0\). Thus \(F_{2}(\pi ^{1},\cdot )\) is continuous at 0. \(\square \)

Proof of Proposition 3

Note first that at most one of the functions \(F_{1}\left( \pi ^{0},\cdot \right) \) and \( F_{2}\left( \pi ^{1},\cdot \right) \) can have a discontinuity at zero, since otherwise player 1 would have had a profitable deviation from \(F_{1}\), consisting of a shift of mass from 0 to an effort slightly above zero (conditional on \(\pi _{0}\)). Note that \(F_{2}\left( \pi ^{1},\cdot \right) \) cannot be discontinuous at 0, since then \(F_{1}\left( \pi ^{0},\cdot \right) \) is continuous at 0 and therefore so is \(F_{2}\left( \pi ^{1},\cdot \right) \), by Lemma 6, a contradiction.

Accordingly, \(F_{2}\left( \pi ^{1},\cdot \right) \) must be continuous at 0. Then by Lemma 5 and Corollary 4 \(F_{1}\left( \pi ^{0},\cdot \right) \) is also continuous at 0, and the sequences \(\left( x^{k}\right) _{k=1}^{n}, \left( F_{i(k)}(\pi ^{k},x_{k}) \right) _{k=1}^{n}\) are determined by (31) and (32) in a unique fashion (and are equal to the corresponding parameters for the strategy profile \((F_{1}^{*}, F_{2}^{*})\)).

Note now that, if \(1<k\le n\), then as \(\left[ x_{k-1},x_{k}\right] \subset suppF_{i(k-1)}\left( \pi ^{k-1},\cdot \right) \) and \(E_{i(k-1)}(\pi ^{k-1},\cdot ,F_{i(k)})\) is continuous on \([0,\infty )\), for every \(x\in \left[ x_{k-1},x_{k}\right] \):

$$\begin{aligned} p^{k-1,k}v_{k-1}+p^{k,k-1}v_{k}F_{i(k)}\left( \pi ^{k},x\right) -x= & {} E_{i(k-1)}\left( \pi ^{k-1},x,F_{i(k)}\right) \\= & {} E_{i(k-1)}\left( \pi ^{k-1},x_{k-1},F_{i(k)}\right) \\= & {} p^{k-1,k}v_{k-1}-x_{k-1}. \end{aligned}$$

Similarly, if \(0\le k<n\), then as \(\left[ x_{k},x_{k+1}\right] \subset supp F_{i(k+1)}\left( \pi ^{k+1},\cdot \right) \) and \(E_{i(k+1)}(\pi ^{k+1},\cdot ,F_{i(k)})\) is continuous on \([0,\infty )\), for every \(x\in \left[ x_{k},x_{k+1}\right] \) :

$$\begin{aligned} p^{k+1,k+2}v_{k+1}-x_{k+1}= & {} E_{i(k+1)}\left( \pi ^{k+1},x_{k+1}, F_{i(k)}\right) \\= & {} E_{i(k+1)}\left( \pi ^{k+1},x,F_{i(k)}\right) =p^{k+1,k+2}v_{k+1}F_{i(k)}\left( \pi ^{k},x\right) -x. \end{aligned}$$

It follows that each \(F_{i(k)}\left( \pi ^{k},\cdot \right) \) is linear on the interval \(\left[ x_{k-1},x_{k}\right] \) (if \(k>1\)) and on the interval \(\left[ x_{k},x_{k+1}\right] \) (if \(k<n\)). Since, as has been shown, \(F_{i(k)}\left( \pi ^{k},\cdot \right) \) is a continuous function, with \(F_{i(k)}\left( \pi ^{k},x_{k-1}\right) =0\) if \(k>1\) and \( F_{i(k)}\left( \pi ^{k},x_{k+1}\right) =1\) if \(k<n\) (and \(F_{1}\left( \pi ^{0},0\right) =0\), \(F_{2}\left( \pi ^{n},x_{n}\right) =1\)), the values of \( \left( F_{i(k)}(\pi ^{k},x_{k})\right) _{k=1}^{n}\) determine the c.d.f.s \( \left( F_{i(k)}(\pi ^{k},\cdot )\right) _{k=1}^{n}\) in a unique fashion. Thus \((F_{1},F_{2})=(F_{1}^{*},F_{2}^{*})\). \(\square \)

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Einy, E., Goswami, M.P., Haimanko, O. et al. Common-value all-pay auctions with asymmetric information. Int J Game Theory 46, 79–102 (2017). https://doi.org/10.1007/s00182-015-0524-4

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Keywords

  • Common-value all-pay auctions
  • Asymmetric information

JEL Classification

  • C72
  • D44
  • D82