Information acquisition in conflicts

Abstract

This paper considers incentives for information acquisition ahead of conflicts. First, we characterize the (unique) equilibrium of the all-pay auction between two players with one-sided asymmetric information where one player has private information about his valuation. Then, we use our results to study information acquisition prior to an all-pay auction. If the decision to acquire information is observable, but not the information received, one-sided asymmetric information can occur endogenously in equilibrium. Moreover, the cut-off values of the cost of information that determine equilibrium information acquisition are higher than those in the first best. Thus, information acquisition is excessive. In contrast, with open or covert information acquisition, the equilibrium cut-off values are as in the first best.

A Appendix

1.1 A.1 Proof of Lemma 1

• (i) (Continuity) Suppose that $B_{j}$ exhibits a discontinuity at some $\tilde{x}>0$. This implies that a bid of $x_{j}=\tilde{x}$ has strictly positive probability. Thus, there exist $\varepsilon ,\varepsilon ^{\prime }>0$ such that player $i$ strictly prefers $x_{i}=\tilde{x} +\varepsilon$ over all $x_{i}\in \left( \tilde{x}-\varepsilon ^{\prime }, \tilde{x}\right)$: shifting probability mass from $\left( \tilde{x} -\varepsilon ^{\prime },\tilde{x}\right)$ to $\tilde{x}+\varepsilon$ only involves an infinitesimally larger cost of effort, but strictly increases the probability of winning.¹⁷ Since player $i$ will not bid in $\left( \tilde{x}-\varepsilon ^{\prime },\tilde{x}\right)$, player $j$ can strictly increase his payoff by bidding $\tilde{x}-\frac{\varepsilon ^{\prime }}{2}$ instead of $\tilde{x}$.

• (ii) (Support) Let $\bar{b}_{i}$ (b $_{i}$) denote the maximum (minimum) of the support of $B_{i}$. Suppose that $\bar{b}_{i}>\bar{b}_{j}$. Then, $B_{j}\left( x\right) =1$ for all $x\ge \bar{b}_{j}.$ Thus, player $i$ prefers to bid $x_{i}=\left( x_{i}^{\prime }+\bar{b}_{j}\right) /2$ to any bid $x_{i}^{\prime }>\bar{b}_{j},$ contradicting $\bar{b}_{i}>\bar{b}_{j}$. Hence, $\bar{b}_{1}=\bar{b}_{2}=\bar{b}$. Since player 1 can ensure a payoff of zero by bidding zero, we must have $\bar{b}\le v_{1}.$ Suppose that b $_{i}>$ b $_{j}\!>\!0.$ Then, any bid $x_{j}<$ b $_{i}$ loses with probability one; player $j$ could increase his payoff by bidding zero instead, which is a contradiction. Now suppose b $_{i}>$ b $_{j}=0$. Then, player $j$ strictly prefers a bid of zero over all bids in $\left( \text{0},{\underline{\mathrm{b}}}_{i}\right) ,$ thus $B_{j}$ has no probability mass in $\left( \text{0},{\underline{\mathrm{b}}} _{i}\right)$. Since $B_{j}$ has no mass points (except possibly at zero), it follows that $B_{j}$ is constant on $\left( 0,{\underline{\mathrm{b}}}_{i}\right] .$ But then there exists $\varepsilon >0$ such that player $i$ strictly prefers a bid of $\varepsilon$ over any bid in $\left[ {\underline{\mathrm{b}}}_{i},{\underline{\mathrm{b}}} _{i}+\varepsilon \right)$: a bid of $\varepsilon$ has strictly lower costs but only a marginally lower probability of winning. This is a contradiction to the definition of b $_{i}.$ Finally, suppose ${\underline{\mathrm{b}}}_{1}={\underline{\mathrm{b}}}_{2}={\underline{\mathrm{b}}}>0$. By (i), $B_{j}\left( {\underline{\mathrm{b}}}\right) =0$, and there exists an $\varepsilon >0$ such that $x_{i}=0$ is preferred to any bid $x_{i}\in \left[ {\underline{\mathrm{b}}},\ {\underline{\mathrm{b}}} +\varepsilon \right) ,$ which contradicts b $_{i}>0$. Combining these arguments shows that ${\underline{\mathrm{b}}}_{1}={\underline{\mathrm{b}}}_{2}=0$.

• (iii) (Mass points at zero) If $B_{j}\left( 0\right) >0$, there exists an $\varepsilon >0$ such that player $i$ prefers $x_{i}=\varepsilon$ to $x_{i}=0$. Hence, $B_{i}\left(0\right) =0$. This shows that the bid distribution of at most one player can have a mass point at zero.

• (iv) (Monotonicity) Suppose that $B_{j}$ is constant in an interval $\left( x^{\prime },x^{\prime \prime }\right)$ where $0\le x^{\prime }<x^{\prime \prime }\le \bar{b}$, further suppose that $x^{\prime \prime }=\max \left\{ x\left|B_{j}\left( x\right) =B_{j}\left( x^{\prime }\right) \right. \right\} .$ Then $B_{j}\left( x^{\prime }\right) =B_{j}\left( x^{\prime \prime }\right) <1$ since $x^{\prime }<\bar{b}$. There exists an $\varepsilon >0$ such that player $i$ prefers $x_{i}=x^{\prime }$ to all $x_{i}\in \left( x^{\prime },x^{\prime \prime }+\varepsilon \right) :$ by bidding $x^{\prime }$ player $i$ reduces his probability of winning only by (at most) an infinitesimally small amount, but strictly decreases his expected cost of effort. Thus $i$ does not bid in $\left( x^{\prime },x^{\prime \prime }+\varepsilon \right) .$ Since $B_{i}$ has no mass points, we have $B_{i}\left( x^{\prime }\right) =B_{i}\left( x^{\prime \prime }+\varepsilon \right) .$ But then $j$ prefers bidding $x^{\prime }$ over any bid in $\left[ x^{\prime \prime },x^{\prime \prime }+\varepsilon \right]$, and thus, we must have $B_{j}\left( x^{\prime \prime }+\varepsilon \right) =B_{j}\left( x^{\prime }\right) ,$ contradicting $x^{\prime \prime }=\max \left\{ x\left|B_{j}\left( x\right) =B_{j}\left( x^{\prime }\right) \right. \right\} .$

1.2 A.2 Proof of Lemma 2

First, we show that no type of player $2$ randomizes. Suppose to the contrary that some type $v_{2}^{\prime }$ of player $2$ does randomize. Let $x_{l}$ ( $x_{h}$) be the infimum (supremum) of the support of the distribution of bids made by type $v_{2}^{\prime }$. For any $x>$ $x_{l},$

$$\begin{aligned} B_{1}\left( \,x_{l}\right) v_{2}^{\prime }-\,x_{l}\ge B_{1}\left( \,x\right) v_{2}^{\prime }-\,x \end{aligned}$$

(18)

for otherwise $v_{2}^{\prime }$ could gain from shifting probability mass to $x$.¹⁸ From (18),

$$\begin{aligned} \,x-\,x_{l}\ge \left( B_{1}\left( \,x\right) -B_{1}\left( \,x_{l}\right) \right) v_{2}^{\prime }. \end{aligned}$$

Since $B_{1}$ is strictly increasing, for any $v_{2}^{\prime \prime }<v_{2}^{\prime }$ we have

$$\begin{aligned} \,x-\,x_{l}>\left( B_{1}\left( \,x\right) -B_{1}\left( \,x_{l}\right) \right) v_{2}^{\prime \prime } \end{aligned}$$

or

$$\begin{aligned} B_{1}\left( \,x_{l}\right) v_{2}^{\prime \prime }-\,x_{l}>B_{1}\left( \,x\right) v_{2}^{\prime \prime }-\,x \end{aligned}$$

that is, type $v_{2}^{\prime \prime }$ strictly prefers to bid $x_{l}$ over bidding $x.$ Therefore, for all $v_{2}^{\prime \prime }<v_{2}^{\prime }$, the supremum of the support of the distribution of bids made by type $v_{2}^{\prime \prime }$ must be weakly smaller than $x_{l}$. Similarly, for all $v_{2}^{\prime \prime \prime }>v_{2}^{\prime }$, the infimum of the support of the distribution of bids made by type $v_{2}^{\prime \prime \prime }$ must be weakly higher than $x_{h}.$ Therefore, only type $v_{2}^{\prime }$ bids in $\left( \,x_{l},\,x_{h}\right) .$ Since the distribution of types, $F,$ is continuous, it follows that $B_{2}$ is constant on $\left( \,x_{l},\,x_{h}\right) ,$ contradicting Lemma 1.

It follows that player $2$ plays a pure strategy $\beta _{2}:\left[ 0,1 \right] \rightarrow \left[ 0,\infty \right)$. Moreover, $\beta _{2}$ is weakly increasing. Now suppose that $v_{2}^{\prime }<v_{2}^{\prime \prime }$ and $\beta _{2}\left( v_{2}^{\prime }\right) =\beta _{2}\left( v_{2}^{\prime \prime }\right)$. Since $\beta _{2}$ is weakly increasing, it follows that $\beta _{2}\left( v_{2}\right) =\beta _{2}\left( v_{2}^{\prime }\right)$ for all $v_{2}\in \left[ v_{2}^{\prime },v_{2}^{\prime \prime }\right]$. Therefore, $B_{2}$ has an atom at $\beta _{2}\left( v_{2}^{\prime }\right)$ (the size of the atom is at least $F\left( v_{2}^{\prime \prime }\right) -F\left( v_{2}^{\prime }\right)$). Since $B_{2}$ is continuous except possibly at zero, this atom can only be at $\beta _{2}\left( v_{2}^{\prime }\right) =0$.

This shows that there is a v $\in \left[ 0,1\right)$ such that, first, for all $v_{2}\le$ v, $\beta _{2}\left( v_{2}\right) =0,$ and second, $\beta _{2}$ is strictly increasing on $\left[ {\underline{\mathrm{v}}},1\right] .$ Since $B_{2}$ is strictly increasing, $\beta _{2}$ has to be continuous as well.

1.3 A.3. Proof of Lemma 3

We first show that $B_{1}$ is differentiable at any $x_{2}\in \left( 0,\bar{b }\right) .$ Let $v_{2}=\beta _{2}^{-1}\left( x_{2}\right)$ and consider a strictly increasing sequence $v_{2}^{n}$ with $v_{2}^{n}\in \left( bad hbox,1\right)$ and $\lim _{n\rightarrow \infty }v_{2}^{n}=v_{2}$. For notational brevity let $x_{2}^{n}=\beta _{2}\left( v_{2}^{n}\right) .$ Then, $x_{2}^{n}$ is strictly increasing and $\lim _{n\rightarrow \infty }x_{2}^{n}=x_{2}.$

Bidding $x_{2}^{n}$ is at least as good as bidding $x_{2}$ for type $v_{2}^{n},$ thus

$$\begin{aligned} B_{1}\left( x_{2}^{n}\right) v_{2}^{n}-x_{2}^{n}\ge B_{1}\left( x_{2}\right) v_{2}^{n}-x_{2} \end{aligned}$$

or

$$\begin{aligned} 1\ge v_{2}^{n}\frac{B_{1}\left( x_{2}\right) -B_{1}\left( x_{2}^{n}\right) }{x_{2}-x_{2}^{n}}. \end{aligned}$$

Taking $\lim \sup$, we get

$$\begin{aligned} \lim \sup \left( \frac{B_{1}\left( x_{2}\right) -B_{1}\left( x_{2}^{n}\right) }{x_{2}-x_{2}^{n}}\right) \le \frac{1}{v_{2}}. \end{aligned}$$

(19)

Similarly, for type $v_{2},$ bidding $x_{2}$ is at least as good as bidding $x_{2}^{n}.$ Thus,

$$\begin{aligned} B_{1}\left( x_{2}\right) v_{2}-x_{2}\ge B_{1}\left( x_{2}^{n}\right) v_{2}-x_{2}^{n}. \end{aligned}$$

Rearranging and taking $\lim \inf ,$ we get

$$\begin{aligned} \lim \inf \left( \frac{B_{1}\left( x_{2}\right) -B_{1}\left( x_{2}^{n}\right) }{x_{2}-x_{2}^{n}}\right) \ge \frac{1}{v_{2}}. \end{aligned}$$

(20)

From (20) and (19), it follows that

$$\begin{aligned} \lim _{x_{2}^{n}\uparrow x_{2}}\left( \frac{B_{1}\left( x_{2}\right) -B_{1}\left( x_{2}^{n}\right) }{x_{2}-x_{2}^{n}}\right) =\frac{1}{v_{2}}. \end{aligned}$$

A parallel argument, which considers a strictly decreasing sequence $v_{2}^{n}$ with limit $v_{2},$ shows that

$$\begin{aligned} \lim _{x_{2}^{n}\downarrow x_{2}}\left( \frac{B_{1}\left( x_{2}\right) -B_{1}\left( x_{2}^{n}\right) }{x_{2}-x_{2}^{n}}\right) =\frac{1}{v_{2}}. \end{aligned}$$

Thus, $B_{1}$ is differentiable at $v_{2}$, with

$$\begin{aligned} \left. \frac{\text{ d}B_{1}\left( x\right) }{\text{ d}x}\right|_{x=x_{2}}=\lim _{x_{2}^{n}\rightarrow x_{2}}\left( \frac{B_{1}\left( x_{2}\right) -B_{1}\left( x_{2}^{n}\right) }{x_{2}-x_{2}^{n}}\right) =\frac{1 }{v_{2}}. \end{aligned}$$

We next show that the bid distribution $B_{2}$ is differentiable. Since $B_{1}$ is strictly increasing on $\left( 0,\bar{b}\right)$, player $1$ must be indifferent between all bids $x\in \left( 0,\bar{b}\right) .$ Fix one $x_{1}\in \left( 0,\bar{b}\right) .$ Consider a sequence $x_{1}^{n}$ with limit $x_{1}$ and with $x_{1}^{n}\in \left( 0,\bar{b}\right)$ for all $n.$ For all $n,$ player $1$ is indifferent between bidding $x_{1}^{n}$ and bidding $x_{1}:$

$$\begin{aligned} B_{2}\left( x_{1}^{n}\right) v_{1}-x_{1}^{n}=B_{2}\left( x_{1}\right) v_{1}-x_{1} \end{aligned}$$

Rearranging,

$$\begin{aligned} \frac{B_{2}\left( x_{1}\right) -B_{2}\left( x_{1}^{n}\right) }{ x_{1}-x_{1}^{n}}=\frac{1}{v_{1}} \end{aligned}$$

Thus,

$$\begin{aligned} \lim _{n\rightarrow \infty }\left( \frac{B_{2}\left( x_{1}\right) -B_{2}\left( x_{1}^{n}\right) }{x_{1}-x_{1}^{n}}\right) =\frac{1}{v_{1}} \end{aligned}$$

and therefore $B_{2}$ is differentiable.

Since $F$ is differentiable by assumption, it follows that $\beta _{2}$ must be differentiable as well.

1.4 A.4. Proof of Lemma 4

(i) Suppose to the contrary that $\alpha _{1}>0.$ Then $\alpha _{2}=0$ by (6) and thus

$$\begin{aligned} B_{1}\left( \beta _{2}\left( 1\right) \right) =\int _{0}^{1}\frac{v_{1}}{v} \text{ d}F\left( v\right) +\alpha _{1}>1, \end{aligned}$$

contradiction. Thus, $\alpha _{1}=0.$ Inserting $\alpha _{1}=0$ in (7), we get (9). The left-hand side of (9) is strictly greater than one for $\alpha _{2}=0,$ it strictly decreases in $\alpha _{2},$ and is equal to zero for $\alpha _{2}=1.$ By continuity, there is a unique $\alpha _{2}\in \left( 0,1\right)$ such that (9) holds. Part (ii) can be proven similarly. From (i) and (ii), it follows that $\alpha _{1}$ and $\alpha _{2}$ are uniquely determined.

1.5 A.5. Proof of Proposition 1

Uniqueness follows from the discussion in the main text. It remains to establish that the strategies are an equilibrium. Consider player $1$ and suppose player $2$ follows (12). The expected payoff of player $1$ for a bid $x_{1}\in \left( 0,\left( 1-\alpha _{2}\right) v_{1} \right]$ is equal to

$$\begin{aligned} E\left[ u_{1}\left( x_{1}\right) \right] =F\left( \beta _{2}^{-1}\left( x_{1}\right) \right) v_{1}-x_{1} \end{aligned}$$

since $\beta _{2}^{-1}$ exists on $\left( 0,\left( 1-\alpha _{2}\right) v_{1} \right]$. Inserting (12), we get $E\left[ u_{1}\left( x_{1}\right) \right] =\alpha _{2}v_{1}$ for all $x_{1}\in \left( 0,\left( 1-\alpha _{2}\right) v_{1}\right] .$ Moreover, if (8) does not hold, then $\alpha _{2}=0$ and player $1$ has a payoff of zero; thus, in this case he is indifferent between all $x_{1}\in \left[ 0,\left( 1-\alpha _{2}\right) v_{1}\right] .$ Bidding more than $\left( 1-\alpha _{2}\right) v_{1}$ is always suboptimal. Thus (11) is a best response.

Now consider player $2$ and suppose he has a valuation $v_{2}.$ Given $B_{1},$ his payoff $B_{1}\left( x\right) v_{2}-x$ is strictly concave in his bid $x$ since

$$\begin{aligned} B_{1}^{\prime \prime }\left( x\right) =\frac{\partial ^{2}}{\partial x^{2}} \left( \int _{0}^{x}\frac{1}{F^{-1}\left( \frac{z+\alpha _{2}v_{1}}{v_{1}} \right) }\text{ d}z\right) =\frac{\partial }{\partial x}\frac{1}{F^{-1}\left( \frac{ x+\alpha _{2}v_{1}}{v_{1}}\right) }<0. \end{aligned}$$

If $v_{2}>F^{-1}\left( \alpha _{2}\right)$, then the first-order condition (3) describes the unique maximum. If $v_{2}\le F^{-1}\left( \alpha _{2}\right) ,$ then for all $x_{2}>0,$

$$\begin{aligned} B_{1}^{\prime }\left( x_{2}\right) v_{2}-1=\frac{1}{F^{-1}\left( \frac{ x_{2}+\alpha _{2}v_{1}}{v_{1}}\right) }v_{2}-1<0. \end{aligned}$$

Therefore, (12) is a best response.

1.6 A.6. Proof of Proposition 2

Suppose player $j$ does not acquire information. If $i$ does not acquire information either, he gets an expected payoff of zero by Fact 1; if $i$ acquires information, his payoff is described by (17). Hence, $i$’s best response is to acquire information if and only if $c$ is smaller than

$$\begin{aligned} \bar{c}:=\int _{{\underline{\mathrm{v}}}}^{1}\int _{{\underline{\mathrm{v}}}}^{v_{i}}\left( \frac{ E\left( V\right) }{v_{j}}v_{i}-E\left( V\right) \right) \text{ d}F\left( v_{j}\right) \text{ d}F\left( v_{i}\right) \end{aligned}$$

(21)

where, from (15), v $=F^{-1}\left( \alpha _{i}\right) >0$, and v is defined by

$$\begin{aligned} \int _{{\underline{\mathrm{v}}}}^{1}\frac{E\left( V\right) }{v}\text{ d}F\left( v\right) =1. \end{aligned}$$

(22)

Note that from (22), it follows that v $<E\left( V\right)$.

Now suppose that $j$ acquires information. If $i$ remains uninformed, he gets $F\left({\underline{\mathrm{v}}}\right) E\left( V\right)$, as in (16). If $i$ acquires information, his payoff is described by (14). Thus, $i$’s best response is to acquire information if and only if $c$ is smaller than

$$\begin{aligned} {\underline{\mathrm{c}}}:=\int _{0}^{1}\int _{0}^{v_{i}}\left( v_{i}-v_{j}\right) \text{ d}F\left( v_{j}\right) \text{ d}F\left( v_{i}\right) -\int _{0}^{{\underline{\mathrm{v}}}}E\left( V\right) \text{ d}F\left( v\right) \end{aligned}$$

(23)

where again v is defined by (22).

Let

$$\begin{aligned} c^{\prime }:=E_{v_{i},v_{j}}\left[ \max \left\{ v_{i},v_{j}\right\} \right] -E_{v_{j}}\left[ \max \left\{ E\left( V\right) ,v_{j}\right\} \right] . \end{aligned}$$

(In Appendix A.7, we will show that in the first best, both players acquire information if and only if $c<c^{\prime }.$) The following lemmas will be used repeatedly below.

Lemma 5

$$\begin{aligned} c^{\prime }=\int _{0}^{1}\int _{0}^{v_{i}}\left( v_{i}-v_{j}\right) \text{ d}F\left( v_{j}\right) \text{ d}F\left( v_{i}\right) -\int _{0}^{E\left( V\right) }\left( E\left( V\right) -v_{j}\right) \text{ d}F\left( v_{j}\right) >0. \end{aligned}$$

Proof

For the equality,

$$\begin{aligned} c^{\prime }&= E_{v_{i},v_{j}}\left[ \max \left\{ v_{i},v_{j}\right\} \right] -E_{v_{j}}\left[ \max \left\{ E\left( V\right) ,v_{j}\right\} \right] \\&= \int _{0}^{1}\int _{0}^{v_{i}}v_{i}\text{ d}F\left( v_{j}\right) \text{ d}F\left( v_{i}\right) +\int _{0}^{1}\int _{v_{i}}^{1}v_{j}\text{ d}F\left( v_{j}\right) \text{ d}F\left( v_{i}\right) \\&-\int _{0}^{E\left( V\right) }E\left( V\right) \text{ d}F\left( v_{j}\right) -\int _{E\left( V\right) }^{1}v_{j}\text{ d}F\left( v_{j}\right) \\&= \int _{0}^{1}\int _{0}^{v_{i}}\left( v_{i}-v_{j}\right) \text{ d}F\left( v_{j}\right) \text{ d}F\left( v_{i}\right) -\int _{0}^{E\left( V\right) }\left( E\left( V\right) -v_{j}\right) \text{ d}F\left( v_{j}\right) . \end{aligned}$$

The inequality $c^{\prime }>0$ follows from Jensen’s inequality. To see this, define

$$\begin{aligned} g\left( v_{i}\right) :=\int _{0}^{1}\max \left\{ v_{i},v_{j}\right\} \text{ d}F\left( v_{j}\right) . \end{aligned}$$

Since $g$ is strictly convex in $v_{i}$,

$$\begin{aligned} E_{v_{i}}\left[ g\left( v_{i}\right) \right] >g\left( E_{v_{i}}\left( v_{i}\right) \right) \end{aligned}$$

or equivalently

$$\begin{aligned} E_{v_{i},v_{j}}\left[ \max \left\{ v_{i},v_{j}\right\} \right] >E_{v_{j}} \left[ \max \left\{ E\left( V\right) ,v_{j}\right\} \right] . \end{aligned}$$

$\square$

Lemma 6

(i) c $>c^{\prime }$ and (ii) $\bar{c}>$ c.

Proof

(i) Using Lemma 5,

$$\begin{aligned} {\underline{\mathrm{c}}}-c^{\prime }&= \int _{0}^{E\left( V\right) }\left( E\left( V\right) -v_{j}\right) \text{ d}F\left( v_{j}\right) -\int _{0}^{{\underline{\mathrm{v}}}}E\left( V\right) \text{ d}F\left( v_{j}\right) \\&= \int _{{\underline{\mathrm{v}}}}^{E\left( V\right) }\left( E\left( V\right) -v_{j}\right) \text{ d}F\left( v_{j}\right) -\int _{0}^{{\underline{\mathrm{v}}}}v_{j}\text{ d}F\left( v_{j}\right) . \end{aligned}$$

Adding and subtracting both $\int _{0}^{E\left( V\right) }{\underline{\mathrm{v}}} \text{ d}F\left( v_{j}\right)$ and $\int _{{\underline{\mathrm{v}}}}^{E\left( V\right) }{\underline{\mathrm{v}}}\frac{E\left( V\right) }{v_{j}}\text{ d}F\left( v_{j}\right)$ yields

$$\begin{aligned} {\underline{\mathrm{c}}}-c^{\prime }&= \int _{0}^{{\underline{\mathrm{v}}}}\left( {\underline{\mathrm{v}}} -v_{j}\right) \text{ d}F\left( v_{j}\right) +\int _{{\underline{\mathrm{v}}}}^{E\left( V\right) }\left( E\left( V\right) -v_{j}+{\underline{\mathrm{v}}}-{\underline{\mathrm{v}}}\frac{E\left( V\right) }{v_{j}}\right) \text{ d}F\left( v_{j}\right) \\&-\int _{0}^{E\left( V\right) }{\underline{\mathrm{v}}}\text{ d}F\left( v_{j}\right) +\int _{ {\underline{\mathrm{v}}}}^{E\left( V\right) }{\underline{\mathrm{v}}}\frac{E\left( V\right) }{v_{j}} \text{ d}F\left( v_{j}\right) . \end{aligned}$$

First observe that

$$\begin{aligned} \int _{{\underline{\mathrm{v}}}}^{E\left( V\right) }{\underline{\mathrm{v}}}\frac{E\left( V\right) }{ v_{j}}\text{ d}F\left( v_{j}\right)&= {\underline{\mathrm{v}}}\left[ \int _{{\underline{\mathrm{v}}}}^{1} \frac{E\left( V\right) }{v_{j}}\text{ d}F\left( v_{j}\right) -\int _{E\left( V\right) }^{1}\frac{E\left( V\right) }{v_{j}}\text{ d}F\left( v_{j}\right) \right] \\&= {\underline{\mathrm{v}}}\left[ 1-\int _{E\left( V\right) }^{1}\frac{E\left( V\right) }{ v_{j}}\text{ d}F\left( v_{j}\right) \right] \end{aligned}$$

where the second equality uses (22). Therefore,

$$\begin{aligned} {\underline{\mathrm{c}}}-c^{\prime }&= \int _{0}^{{\underline{\mathrm{v}}}}\left( {\underline{\mathrm{v}}} -v_{j}\right) \text{ d}F\left( v_{j}\right) +\int _{{\underline{\mathrm{v}}}}^{E\left( V\right) } \frac{\left( E\left( V\right) -v_{j}\right) \left( v_{j}-{\underline{\mathrm{v}}}\right) }{v_{j}}\text{ d}F\left( v_{j}\right) \\&+\,{\underline{\mathrm{v}}}\left[ 1-\int _{0}^{E\left( V\right) }\text{ d}F\left( v_{j}\right) -\int _{E\left( V\right) }^{1}\frac{E\left( V\right) }{v_{j}}\text{ d}F\left( v_{j}\right) \right] \end{aligned}$$

which is strictly positive.

(ii) With (21) and (23), $\bar{c}-$ c is equal to

$$\begin{aligned}&\int _{{\underline{\mathrm{v}}}}^{1}\int _{{\underline{\mathrm{v}}}}^{v_{i}}\left( \frac{E\left( V\right) v_{i}}{v_{j}}-E\left( V\right) \right) \text{ d}F\left( v_{j}\right) \text{ d}F\left( v_{i}\right) +\int _{0}^{1}\int _{0}^{{\underline{\mathrm{v}}}}E\left( V\right) \text{ d}F\left( v_{j}\right) \text{ d}F\left( v_{i}\right) \\&~~~~~~~~~-\int _{0}^{1}\int _{0}^{v_{i}}\left( v_{i} -v_{j}\right) \text{ d}F\left( v_{j}\right) \text{ d}F\left( v_{i}\right) =\int _{0}^{1}h\left( v_{i}\right) \text{ d}F\left( v_{i}\right) \end{aligned}$$

where

$$\begin{aligned} h\left( v_{i}\right) =\int _{0}^{{\underline{\mathrm{v}}}}E\left( V\right) \text{ d}F\left( v_{j}\right) -\int _{0}^{v_{i}}\left( v_{i}-v_{j}\right) \text{ d}F\left( v_{j}\right) \end{aligned}$$

if $v_{i}\le \,$ v, and

$$\begin{aligned} h\left( v_{i}\right) \!=\!\int _{{\underline{\mathrm{v}}}}^{v_{i}}\left( \frac{E\left( V\right) v_{i}}{v_{j}}-E\left( V\right) \right) \text{ d}F\left( v_{j}\right) \!+\!\int _{0}^{{\underline{\mathrm{v}}}}E\left( V\right) \text{ d}F\left( v_{j}\right) \!-\!\int _{0}^{v_{i}}\left( v_{i}-v_{j}\right) \text{ d}F\left( v_{j}\right) \end{aligned}$$

if $v_{i}>$ v. Then, it is sufficient to show that $h\left( v_{i}\right) >0$ for all $v_{i}\in \left[ 0,1\right]$.

Case 1: :

$v_{i}\le$ v $.$ From (22), it follows that v $<E\left( V\right)$, and thus

$$\begin{aligned} \int _{0}^{{\underline{\mathrm{v}}}}E\left( V\right) \text{ d}F\left( v_{j}\right) >\int _{0}^{v_{i}}v_{i}\text{ d}F\left( v_{j}\right) >\int _{0}^{v_{i}}\left( v_{i}-v_{j}\right) \text{ d}F\left( v_{j}\right) . \end{aligned}$$

Case 2: :

$v_{i}\in \left( {\underline{\mathrm{v}}},E\left( V\right) \right] .$ Here, $h\left( v_{i}\right)$ is equal to

$$\begin{aligned} \int _{{\underline{\mathrm{v}}}}^{v_{i}}\frac{\left( v_{i}-v_{j}\right) \left( E\left( V\right) -v_{j}\right) }{v_{j}}\text{ d}F\left( v_{j}\right) +\int _{0}^{{\underline{\mathrm{v}}} }\left( E\left( V\right) -v_{i}+v_{j}\right) \text{ d}F\left( v_{j}\right) . \end{aligned}$$

The first term is strictly positive because $v_{j}\le v_{i}\le E\left( V\right)$ and $v_{i}>$ v. The second term is strictly positive as $v_{i}\le E\left( V\right)$ and, by (15), v $>0$.

Case 3: :

$v_{i}\in \left( E\left( V\right) ,1\right] .$ Since v is independent of $v_{i}$, we get

$$\begin{aligned} h^{\prime }\left( v_{i}\right)&= \int _{{\underline{\mathrm{v}}}}^{v_{i}}\frac{E\left( V\right) }{v_{j}}\text{ d}F\left( v_{j}\right) -\int _{0}^{v_{i}}\text{ d}F\left( v_{j}\right) , \\ h^{\prime \prime }\left( v_{i}\right)&= \frac{E\left( V\right) }{v_{i}} F^{\prime }\left( v_{i}\right) -F^{\prime }\left( v_{i}\right) , \end{aligned}$$

hence, $h$ is strictly concave for $v_{i}>E\left( V\right)$. Moreover, as $v_{i}\rightarrow 1$, $h^{\prime }$ converges to

$$\begin{aligned} \int _{{\underline{\mathrm{v}}}}^{1}\frac{E\left( V\right) }{v_{j}}\text{ d}F\left( v_{j}\right) -\int _{0}^{1}\text{ d}F\left( v_{j}\right) =1-1=0. \end{aligned}$$

(The first integral is one by (22).) Thus, $h^{\prime }$ must be positive for all $v_{i}\in \left( E\left( V\right) ,1\right)$ and thus $h\left( v_{i}\right) >h\left( E\left( V\right) \right) >0$ where the last inequality follows from case 2.$\square$

We are now in a position to prove Proposition 2. From Lemmas 5 and , it follows directly that $\bar{c}>$ c $>0$. Thus, (i) if $c<$ c, information acquisition is strictly dominant. (ii) If c $<c<\bar{c}$, a player invests in information only if the opponent remains uninformed, and there exist two asymmetric equilibria where exactly one player invests. Moreover, there is a symmetric equilibrium where both players invest in information with probability $p=\left( \bar{c}-c\right) /\left( \bar{c}- {\underline{\mathrm{c}}}\right) :$ if player $i$ acquires information, he gets

$$\begin{aligned} \left( 1-p\right) \left( \bar{c}-c\right) +p\left( F\left( {\underline{\mathrm{v}}} \right) E\left( V\right) +{\underline{\mathrm{c}}}-c\right) =pF\left( {\underline{\mathrm{v}}} \right) E\left( V\right) \end{aligned}$$

which is equal to his payoff if he remains uninformed. Thus, $i$ is indifferent between investing and not investing in information. Moreover, for all $p$ that are strictly smaller (greater) than this critical value, $i$ strictly prefers (not) to acquire information. Finally, (iii) if $c>\bar{c}$ , not investing is strictly dominant.

1.7 A.7. Proof of Proposition 3

If the social planner does not acquire information, welfare equals $E\left( V\right)$. If she acquires information about the valuation of one player, welfare is equal to

$$\begin{aligned} E_{v_{j}}\left[ \max \left\{ E\left( V\right) ,v_{j}\right\} \right] -c. \end{aligned}$$

If the social planner acquires information about both players, welfare equals

$$\begin{aligned} E_{v_{i},v_{j}}\left[ \max \left\{ v_{i},v_{j}\right\} \right] -2c. \end{aligned}$$

As above, let

$$\begin{aligned} c^{\prime }=E_{v_{i},v_{j}}\left[ \max \left\{ v_{i},v_{j}\right\} \right] -E_{v_{j}}\left[ \max \left\{ E\left( V\right) ,v_{j}\right\} \right] . \end{aligned}$$

(24)

Moreover, let

$$\begin{aligned} c^{\prime \prime }&= E_{v_{i}}\left[ \max \left\{ E\left( V\right) ,v_{i}\right\} \right] -E\left( V\right) \nonumber \\&= \int _{E\left( V\right) }^{1}\left( v_{i}-E\left( V\right) \right) \text{ d}F\left( v_{i}\right) . \end{aligned}$$

(25)

Lemma 7

(i) $0<c^{\prime }<c^{\prime \prime }$ and (ii) $c^{\prime }<$ c and $c^{\prime \prime }<\bar{c}.$

Proof

(i) In Lemma 5, we have already shown that $c^{\prime }>0$. Moreover, using Lemma ,

$$\begin{aligned} c^{\prime }&= \int _{0}^{1}\int _{0}^{v_{i}}\left( v_{i}-v_{j}\right) \text{ d}F\left( v_{j}\right) \text{ d}F\left( v_{i}\right) -\int _{0}^{E\left( V\right) }\left( E\left( V\right) -v_{j}\right) \text{ d}F\left( v_{j}\right) \\&= \int _{0}^{1}\int _{0}^{v_{i}}\left( v_{i}-E\left( V\right) \right) \text{ d}F\left( v_{j}\right) \text{ d}F\left( v_{i}\right) +\int _{0}^{1}\int _{0}^{v_{i}}\left( E\left( V\right) -v_{j}\right) \text{ d}F\left( v_{j}\right) \text{ d}F\left( v_{i}\right) \\&-\int _{0}^{1}\int _{0}^{E\left( V\right) }\left( E\left( V\right) -v_{j}\right) \text{ d}F\left( v_{j}\right) \text{ d}F\left( v_{i}\right) \\&= \int _{0}^{1}\int _{0}^{v_{i}}\left( v_{i}-E\left( V\right) \right) \text{ d}F\left( v_{j}\right) \text{ d}F\left( v_{i}\right) +\int _{E\left( V\right) }^{1}\int _{E\left( V\right) }^{v_{i}}\left( E\left( V\right) -v_{j}\right) \text{ d}F\left( v_{j}\right) \text{ d}F\left( v_{i}\right) \\&-\int _{0}^{E\left( V\right) }\int _{v_{i}}^{E\left( V\right) }\left( E\left( V\right) -v_{j}\right) \text{ d}F\left( v_{j}\right) \text{ d}F\left( v_{i}\right) \end{aligned}$$

which is strictly smaller than

$$\begin{aligned} \int _{0}^{1}\int _{0}^{v_{i}}\left( v_{i}-E\left( V\right) \right) \text{ d}F\left( v_{j}\right) \text{ d}F\left( v_{i}\right)&< \int _{E\left( V\right) }^{1}\int _{0}^{v_{i}}\left( v_{i}-E\left( V\right) \right) \text{ d}F\left( v_{j}\right) \text{ d}F\left( v_{i}\right) \\&< \int _{E\left( V\right) }^{1}\int _{0}^{1}\left( v_{i}-E\left( V\right) \right) \text{ d}F\left( v_{j}\right) \text{ d}F\left( v_{i}\right) =c^{\prime \prime }. \end{aligned}$$

(ii) The first inequality is Lemma , part (i). Moreover, by (21) and (25), $\bar{c}>c^{\prime \prime }$ is equivalent to

$$\begin{aligned} \int _{{\underline{\mathrm{v}}}}^{1}\int _{{\underline{\mathrm{v}}}}^{v_{i}}\left( \frac{E\left( V\right) v_{i}}{v_{j}}-E\left( V\right) \right) \text{ d}F\left( v_{j}\right) \text{ d}F\left( v_{i}\right) >\int _{E\left( V\right) }^{1}\left( v_{i}-E\left( V\right) \right) \text{ d}F\left( v_{i}\right) . \end{aligned}$$

By (17), the left-hand side is $i$’s ex ante expected payoff if $i$ acquired information and $j$ remained uninformed. Since, in this case, $j$ never bids higher than his expected value, the LHS must be weakly higher than the RHS, because the latter is the payoff $i$ could ensure by bidding $E\left( V\right)$ for all types $v_{i}\ge E\left( V\right)$ and bidding zero otherwise. It remains to show that for some realizations of $v_{i}$, $i$ can do strictly better. Note first that $F^{-1}\left( \alpha _{i}\right) = {\underline{\mathrm{v}}}>0$, that is, $j$’s maximum bid is $\bar{ b}=\left( 1-\alpha _{i}\right) E\left( V\right) <E\left( V\right)$. Hence, for all realizations $v_{i}\in \left( \left( 1-\alpha _{i}\right) E\left( V\right) ,E\left( V\right) \right)$, $i$ can ensure a strictly positive payoff by bidding $\left( 1-\alpha _{i}\right) E\left( V\right)$, and hence, the LHS must be strictly larger than the RHS. $\square$

The inequalities in (i) allow us to characterize first best information acquisition: if $c<c^{\prime }$, both should acquire information; if $c\in \left( c^{\prime },c^{\prime \prime }\right) ,$ exactly one player should acquire information; finally, if $c>c^{\prime \prime },$ no one should. With (ii), we can compare equilibrium investments and first best investments (see Fig. 2 in the main text). If $c<c^{\prime },$ both players invest as in the first best. If $c\in \left( c^{\prime },\min \left\{ {\underline{\mathrm{c}}} ,c^{\prime \prime }\right\} \right) ,$ both players acquire information although exactly one player should. If $c\in \left( \min \left\{ {\underline{\mathrm{c}}},c^{\prime \prime }\right\} ,c^{\prime \prime }\right) ,$ in the asymmetric equilibria exactly one player acquires information, as in the first best. If $c\in \left( c^{\prime \prime },\bar{c}\right) ,$ at least one player acquires information, but neither of the players should. Finally, if $c>\bar{c},$ no player invests, as in the first best. Therefore, the number of players investing in information is higher than the first best.

1.8 A.8. Proof of Proposition 4

If no player invests in information, both get an expected payoff of zero. If only player $i$ invests, $i$’s expected payoff is $E_{v_{i}}\left[ \max \left\{ v_{i}-E\left( V\right) ,0\right\} \right] -c$, while $j$ gets $E_{v_{i}}\left[ \max \left\{ E\left( V\right) -v_{i},0\right\} \right]$. If both players acquire information, each of them gets $E_{v_{i},v_{j}}\left[ \max \left\{ v_{i}-v_{j},0\right\} \right] -c$.

Now suppose that $j$ remains uninformed. Player $i$’s best response is to acquire information whenever $c$ is smaller than $E_{v_{i}}\left[ \max \left\{ v_{i}-E\left( V\right) ,0\right\} \right]$ which, with (25), is equal to $c^{\prime \prime }$. If $j$ acquires information, $i$ invests whenever $c$ is smaller than

$$\begin{aligned} E_{v_{i},v_{j}}\left[ \max \left\{ v_{i}-v_{j},0\right\} \right] -E_{v_{i}} \left[ \max \left\{ E\left( V\right) -v_{i},0\right\} \right] \end{aligned}$$

which, by Lemma 5, is equal to $c^{\prime }$. Since $0<c^{\prime }<c^{\prime \prime }$, both players (no player) acquire information if $c<c^{\prime }$ ($c>c^{\prime \prime }$). If $c\in \left(c^{\prime },c^{\prime \prime }\right)$, there are two equilibria where exactly one player acquires information, and a mixed strategy equilibrium where players acquire information with probability $\left( c^{\prime \prime }-c\right) /\left( c^{\prime \prime }-c^{\prime }\right)$.

1.9 A.9. Proof of Proposition 5

• (i) We first analyze whether there can be an equilibrium where both players acquire information with probability $1$. If this is the case, then they bid as in Fact 2 and both get a payoff of

  $$\begin{aligned} \int _{0}^{1}\int _{0}^{v_{i}}\left( v_{i}-v_{j}\right) \text{ d}F\left( v_{j}\right) \text{ d}F\left( v_{i}\right) -c. \end{aligned}$$

  Now suppose that $i$ deviates and remains uninformed. Then, his optimal bid is as if he had a value of $E\left( V\right)$ which leads to a deviation payoff of

  $$\begin{aligned} \int _{0}^{E\left( V\right) }\left( E\left( V\right) -v_{j}\right) \text{ d}F\left( v_{j}\right) . \end{aligned}$$

  Hence, it pays off to save the cost of information whenever $c$ is larger than

  $$\begin{aligned} \int _{0}^{1}\int _{0}^{v_{i}}\left( v_{i}-v_{j}\right) \text{ d}F\left( v_{j}\right) \text{ d}F\left( v_{i}\right) -\int _{0}^{E\left( V\right) }\left( E\left( V\right) -v_{j}\right) \text{ d}F\left( v_{j}\right) \end{aligned}$$

  which, by Lemma 5, is equal to $c^{\prime }$. Thus, if and only if $c<c^{\prime }$, an equilibrium exists where both players acquire information.

• (ii) Now suppose that both players do not invest in information with probability $1$. Then, both get zero payoff. If $i$ deviates and acquires information, his optimal bid is zero if $v_{i}\le E\left( V\right)$ and $E\left( V\right)$ if $v_{i}>E\left( V\right)$. (The type $v_{i}=E\left( V\right)$ is exactly indifferent. Thus, lower types prefer a bid of zero, and higher types prefer a bid at the upper bound of the support of $j$’s bids.) The deviation payoff is

  $$\begin{aligned} \int _{E\left( V\right) }^{1}\left( v_{i}-E\left( V\right) \right) \text{ d}F\left( v_{i}\right) -c. \end{aligned}$$

  Therefore, if and only if $c$ is larger than $c^{\prime \prime }$ (from (25)), there is an equilibrium where no player acquires information.