Corporate governance roles of information quality and corporate takeovers

Abstract

We examine the corporate governance roles of information quality and the takeover market with asymmetric information regarding the value of the target firm. Increasing information quality improves the takeover efficiency however, a highly efficient takeover market also discourages the manager from exerting effort. We find that perfect information quality is not optimal for either current shareholders’ expected payoff maximization or expected firm value maximization. Furthermore, current shareholders prefer a lower level of information quality than the level that maximizes expected firm value, because of a misalignment between current shareholders’ value and total firm value. We also analyze the impact of antitakeover laws, and find that the passage of antitakeover laws may induce current shareholders to choose a higher level of information quality and thus increase expected firm value.

Appendix: Proofs

Proof

Lemma 1 We can rewrite (2) to be ${\Pi }_{r}(p,\,\hat {h})=\left \{ \begin {array}{ll} 0, & \mathit {if} p<p_{l};\\ (1-\hat {h})(v_{0}-p), & \mathit {if} p_{l}\leq p<p_{h};\\ (1-\hat {h})v_{0}+\hat {h}(1+v_{0})-p, & \mathit {if} p\geq p_{h}. \end {array}\right .$

From this equation, $\phantom {\dot {i}\!}{\Pi }_{r}(p,\,\hat {h})$ decreases with p when $\phantom {\dot {i}\!}p\geq p_{h}$ or $\phantom {\dot {i}\!}p_{l}\leq p<p_{h}$. As a result, the optimal p that maximizes $\phantom {\dot {i}\!}{\Pi }_{r}(p,\,\hat {h})$ will be either $\phantom {\dot {i}\!}p_{h}$ or $\phantom {\dot {i}\!}p_{l}$, while all other prices are dominated.

For $\phantom {\dot {i}\!}p_{l}=(1-\gamma )v_{0}$ and $\phantom {\dot {i}\!}p_{h}= 1+(1-\gamma )v_{0}$, we have $\phantom {\dot {i}\!}{\Pi }_{r}(p_{l},\,\hat {h})=(1-\hat {h})\gamma v_{0}$ and $\phantom {\dot {i}\!}{\Pi }_{r}(p_{h},\,\hat {h})=\hat {h}+\gamma v_{0}-1$. That is, $\phantom {\dot {i}\!}{\Pi }_{r}(p_{h},\,\hat {h})\le {\Pi }_{r}(p_{l},\,\hat {h})$ if and only if $\phantom {\dot {i}\!}\hat {h}\le \frac {1}{1+\gamma v_{0}}$, with the equality holds when $\phantom {\dot {i}\!}\hat {h}=\frac {1}{1+\gamma v_{0}}$.

Without accounting signals, the acquirer’s belief about the probability of a high-value realization is her conjectured effort level, $\phantom {\dot {i}\!}\hat {h}=\hat {e}$. Therefore, $\phantom {\dot {i}\!}p^{*}=p_{h}$ when $\phantom {\dot {i}\!}\hat {e}>\frac {1}{1+\gamma v_{0}}$, $\phantom {\dot {i}\!}p^{*}=p_{l}$ when $\phantom {\dot {i}\!}\hat {e}<\frac {1}{1+\gamma v_{0}}$, and the acquirer is indifferent between $\phantom {\dot {i}\!}p_{h}$ and $\phantom {\dot {i}\!}p_{l}$ when $\phantom {\dot {i}\!}\hat {e}=\frac {1}{1+\gamma v_{0}}$. □

Proof

Lemma 2 Upon observing the accounting signal, the acquirer’s belief becomes $\phantom {\dot {i}\!}h(G)$ and $\phantom {\dot {i}\!}h(B)$ as in Eq. 3. It is easy to see that $\phantom {\dot {i}\!}\hat {h}(G)\ge \hat {h}(B)$. From the proof of Lemma 1, we have $\phantom {\dot {i}\!}p^{*}=p_{h}$ when $\phantom {\dot {i}\!}\hat {h}(y)>\frac {1}{1+\gamma v_{0}}$, $\phantom {\dot {i}\!}p^{*}=p_{l}$ when $\phantom {\dot {i}\!}\hat {h}(y)<\frac {1}{1+\gamma v_{0}}$, and the acquirer is indifferent between $\phantom {\dot {i}\!}p_{h}$ and $\phantom {\dot {i}\!}p_{l}$ when $\phantom {\dot {i}\!}\hat {h}(y)=\frac {1}{1+\gamma v_{0}}$.

Given the equations in Eq. 3, we have the following results:

• When $\phantom {\dot {i}\!}\hat {e}>\frac {1}{1+\gamma v_{0}}$:

  • If $\phantom {\dot {i}\!}\frac {1}{2}\leq d<\frac {\hat {e}\gamma v_{0}}{1-\hat {e}(1-\gamma v_{0})}$, we have $\phantom {\dot {i}\!}\hat {h}(G)>\hat {h}(B)>\frac {1}{1+\gamma v_{0}}$. As a result, $\phantom {\dot {i}\!}\alpha _{G}=\alpha _{B}= 1$.

  • If $\phantom {\dot {i}\!}\frac {\hat {e}\gamma v_{0}}{1-\hat {e}(1-\gamma v_{0})}\leq d\leq 1$, we have $\phantom {\dot {i}\!}\hat {h}(G)>\frac {1}{1+\gamma v_{0}}$ and $\phantom {\dot {i}\!}\hat {h}(B)\leq \frac {1}{1+\gamma v_{0}}$ (the equality holds when $\phantom {\dot {i}\!}d=\frac {\hat {e}\gamma v_{0}}{1-\hat {e}(1-\gamma v_{0})}$). Therefore, $\phantom {\dot {i}\!}\alpha _{G}= 1$; $\phantom {\dot {i}\!}\alpha _{B}= 0$ if $\phantom {\dot {i}\!}\frac {\hat {e}\gamma v_{0}}{1-\hat {e}(1-\gamma v_{0})}<d\leq 1$ or $\phantom {\dot {i}\!}\alpha _{B}$ can be any value in (0, 1) if $\phantom {\dot {i}\!}d=\frac {\hat {e}\gamma v_{0}}{1-\hat {e}(1-\gamma v_{0})}$.

• When $\phantom {\dot {i}\!}\hat {e}<\frac {1}{1+\gamma v_{0}}$:

  • If $\phantom {\dot {i}\!}\frac {1}{2}\leq d<\frac {1-\hat {e}}{1-\hat {e}(1-\gamma v_{0})}$, we have $\phantom {\dot {i}\!}\hat {h}(B)<\hat {h}(G)<\frac {1}{1+\gamma v_{0}}$. As a result, $\phantom {\dot {i}\!}\alpha _{G}=\alpha _{B}= 0$.

  • If $\phantom {\dot {i}\!}\frac {1-\hat {e}}{1-\hat {e}(1-\gamma v_{0})}\leq d\leq 1$, we have $\phantom {\dot {i}\!}\hat {h}(B)<\frac {1}{1+\gamma v_{0}}$ and $\phantom {\dot {i}\!}\hat {h}(G)\geq \frac {1}{1+\gamma v_{0}}$ (the equality holds when $\phantom {\dot {i}\!}d=\frac {1-\hat {e}}{1-\hat {e}(1-\gamma v_{0})}$). Therefore, $\phantom {\dot {i}\!}\alpha _{B}= 0$; $\phantom {\dot {i}\!}\alpha _{G}= 1$ if $\phantom {\dot {i}\!}\frac {1-\hat {e}}{1-\hat {e}(1-\gamma v_{0})}<d\leq 1$ or $\phantom {\dot {i}\!}\alpha _{G}$ can be any value in (0, 1) if $\phantom {\dot {i}\!}d=\frac {1-\hat {e}}{1-\hat {e}(1-\gamma v_{0})}$.

• When $\phantom {\dot {i}\!}\hat {e}=\frac {1}{1+\gamma v_{0}}$:

  • If $\phantom {\dot {i}\!}d=\frac {1}{2}$, we have $\phantom {\dot {i}\!}\hat {h}(G)=\hat {h}(B)=\frac {1}{1+\gamma v_{0}}$. Therefore $\phantom {\dot {i}\!}\alpha _{G}$ and $\phantom {\dot {i}\!}\alpha _{B}$ can be any value in (0, 1). In fact, the value of $\phantom {\dot {i}\!}\alpha _{G}$ and $\phantom {\dot {i}\!}\alpha _{B}$ does not affect our results because the signal is uninformative in this case. Without loss of generality, we assume $\phantom {\dot {i}\!}\alpha _{G}= 1$ and $\phantom {\dot {i}\!}\alpha _{B}= 0$.

  • If $\phantom {\dot {i}\!}\frac {1}{2}<d\leq 1$, we have $\phantom {\dot {i}\!}\hat {h}(G)>\frac {1}{1+\gamma v_{0}}>\hat {h}(B)$. Therefore, $\phantom {\dot {i}\!}\alpha _{G}= 1$ and $\phantom {\dot {i}\!}\alpha _{B}= 0$.

Denote $\underline {d}(\hat {e})=\left \{ \begin {array}{ll} \frac {\hat {e}\gamma v_{0}}{1-\hat {e}(1-\gamma v_{0})}, & \mathit {if} \;\;\hat {e}> \frac {1}{1+\gamma v_{0}}\\ \frac {1-\hat {e}}{1-\hat {e}(1-\gamma v_{0})} & \mathit {if}\;\; \hat {e}\le \frac {1}{1+\gamma v_{0}}. \end {array}\right .$ From the above analysis, we have the following results:

• When $\phantom {\dot {i}\!}\frac {1}{2}\le d < \underline {d}(\hat {e})$, the acquirer sticks to her default bidding strategy.

• When $\phantom {\dot {i}\!} \underline {d}(\hat {e}) \le d \le 1$, the acquirer revises her bidding strategy in the following way:

  – If $\phantom {\dot {i}\!}\hat {e}> \frac {1}{1+\gamma v_{0}}$, then $\phantom {\dot {i}\!}\alpha _{G}= 1$, and $\phantom {\dot {i}\!}\alpha _{B}= 0$ or $\phantom {\dot {i}\!}0<\alpha _{B}<1$.

  – If $\phantom {\dot {i}\!}\hat {e}<\frac {1}{1+\gamma v_{0}}$, then $\phantom {\dot {i}\!}\alpha _{G}= 1$ or $\phantom {\dot {i}\!}0<\alpha _{G}<1$, and $\phantom {\dot {i}\!}\alpha _{B}= 0$.

  – If $\phantom {\dot {i}\!}\hat {e}=\frac {1}{1+\gamma v_{0}}$, then $\phantom {\dot {i}\!}\alpha _{G}= 1$ and $\phantom {\dot {i}\!}\alpha _{B}= 0$.

□

Proof

Equilibria in Section 3.3

Given the acquirer’s bidding strategy $\phantom {\dot {i}\!}(\alpha _{G},\alpha _{B})$, the manager’s optimal effort that maximizes his expected payoff is derived in Eq. 6:

$$\begin{array}{@{}rcl@{}} e^{\ast}(\alpha_{G},\,\alpha_{B})=m-m\cdot[\alpha_{G}d+\alpha_{B}(1-d)]. \end{array} $$

In the following, we solve for the equilibrium given the information quality d and the manager’s private benefit m. We need to check that for the bidding strategies as shown in Lemma 2, the manager’s optimal effort in response to the bidding strategies is consistent with the conjectures.

• (1). When the acquirer’s belief satisfies $\phantom {\dot {i}\!}{\displaystyle \hat {e}>\frac {1}{1+\gamma v_{0}}}$:

  (i) If $\phantom {\dot {i}\!}\frac {1}{2}\leq d<\frac {\hat {e}\gamma v_{0}}{1-\hat {e}(1-\gamma v_{0})}$, then $\phantom {\dot {i}\!}\alpha _{G}^{*}=\alpha _{B}^{*}= 1$, which implies $\phantom {\dot {i}\!}e^{*}= 0$ and it does not satisfy the conjecture $\phantom {\dot {i}\!}\hat {e}>\frac {1}{1+\gamma v_{0}}$.

  (ii) If $\phantom {\dot {i}\!}\frac {\hat {e}\gamma v_{0}}{1-\hat {e}(1-\gamma v_{0})}<d\leq 1$, then $\phantom {\dot {i}\!}\alpha _{G}^{*}= 1,\,\alpha _{B}^{*}= 0$, which implies $\phantom {\dot {i}\!}e^{*}=m(1-d)$. For any $\phantom {\dot {i}\!} \frac {1}{2}<d<1$ and $\phantom {\dot {i}\!}0<m<1$, we have $\phantom {\dot {i}\!}m(1-d)<\frac {1}{2}$. In addition, for $\phantom {\dot {i}\!}0<v_{0}<1$ and $\phantom {\dot {i}\!}0<\gamma <1$, we have $\phantom {\dot {i}\!}\frac {1}{1+\gamma v_{0}}>\frac {1}{2}$. Therefore $\phantom {\dot {i}\!}m(1-d)<\frac {1}{1+\gamma v_{0}}$, contradicting with the conjecture. Hence this equilibrium does not hold.

  (iii) If $\phantom {\dot {i}\!}d=\frac {\hat {e}\gamma v_{0}}{1-\hat {e}(1-\gamma v_{0})}$, $\phantom {\dot {i}\!}\alpha _{G}^{*}= 1$ and $\phantom {\dot {i}\!}\alpha _{B}^{*}$ $\phantom {\dot {i}\!}\in (0,1)$. Thus, $\phantom {\dot {i}\!}e^{\ast }=m(1-d)(1-\alpha _{B}^{*})$. Similar to case (ii), it can be proved that $\phantom {\dot {i}\!}m(1-d)(1-\alpha _{B}^{*})<\frac {1}{2}<\frac {1}{1+\gamma v_{0}}$. As a result, this equilibrium does not hold.

• (2). When the acquirer’s belief satisfies $\phantom {\dot {i}\!}0\leq \hat {e}<\frac {1}{1+\gamma v_{0}}$:

  (i) If $\phantom {\dot {i}\!}\frac {1}{2}\leq d<\frac {1-\hat {e}}{1-\hat {e}(1-\gamma v_{0})}$, then $\phantom {\dot {i}\!}\alpha _{G}^{*}=\alpha _{B}^{*}= 0$. Thus, $\phantom {\dot {i}\!}e^{\ast }=m$. This equilibrium exists when $\phantom {\dot {i}\!}m<\frac {1}{1+\gamma v_{0}}$ and $\phantom {\dot {i}\!}\frac {1}{2}\leq d<\frac {1-m}{1-m(1-\gamma v_{0})}$.

  (ii) If $\phantom {\dot {i}\!}\frac {1-\hat {e}}{1-\hat {e}(1-\gamma v_{0})}<d\leq 1$, then $\phantom {\dot {i}\!}\alpha _{G}^{*}= 1$, $\phantom {\dot {i}\!}\alpha _{B}^{*}= 0$. Thus, $\phantom {\dot {i}\!}e^{\ast }=m(1-d)$. We can prove that $\phantom {\dot {i}\!}\frac {1-e^{\ast }}{1-e^{\ast }(1-\gamma v_{0})}=\frac {(1-d)(d m(1-\gamma v_{0})+ 1-m)}{1-(1-d) m (1-\gamma v_{0})}+d>d$, contradicting with the condition. The equilibrium does not hold.

  (iii) If $\phantom {\dot {i}\!}d=\frac {1-\hat {e}}{1-\hat {e}(1-\gamma v_{0})}$, then $\phantom {\dot {i}\!}0\le \alpha _{G}^{*}\le 1$ and $\phantom {\dot {i}\!}\alpha _{B}^{*}= 0$. Therefore, $\phantom {\dot {i}\!}e^{\ast }=m(1-d\alpha _{G}^{*})$.

  In equilibrium, $\phantom {\dot {i}\!}\hat {e}=e^{*}$. Given that $\phantom {\dot {i}\!}d=\frac {1-\hat {e}}{1-\hat {e}(1-\gamma v_{0})}$, we then solve for $\phantom {\dot {i}\!}\alpha _{G}^{*}$, which gives $\phantom {\dot {i}\!}\alpha _{G}^{*}=\frac {1}{d}\frac {\gamma v_{0}dm-(1-d)(1-m)}{(1-d)m+\gamma v_{0}dm}$ and $\phantom {\dot {i}\!}e^{*}=\frac {1-d}{1-d(1-\gamma v_{0})}$.

  Given that $\phantom {\dot {i}\!}\alpha ^{*}_{G}$ needs to satisfy $\phantom {\dot {i}\!}0\le \alpha _{G}^{*}\le 1$, we have:

  • When $\phantom {\dot {i}\!}m>\frac {1}{1+\gamma v_{0}}$ and $\phantom {\dot {i}\!}\frac {1}{2}\leq d<1$, or $\phantom {\dot {i}\!}m\leq \frac {1}{1+\gamma v_{0}}$ and $\phantom {\dot {i}\!}\frac {1-m}{1-m(1-\gamma v_{0})}<d<1$, $\phantom {\dot {i}\!}0<\alpha _{G}^{*}<1$.

  • When $\phantom {\dot {i}\!}m\leq \frac {1}{1+\gamma v_{0}}$ and $\phantom {\dot {i}\!}d=\frac {1-m}{1-m(1-\gamma v_{0})}$, $\phantom {\dot {i}\!}\alpha _{G}^{*}= 0$.

  • When $\phantom {\dot {i}\!}d = 1$, we can derive $\phantom {\dot {i}\!}\alpha _{B}^{*}= 0$ and $\phantom {\dot {i}\!}\alpha _{G}^{*}= 1$. In equilibrium, $\phantom {\dot {i}\!}e^{*}= 0$.

• (3). When the acquirer’s belief is $\phantom {\dot {i}\!}\hat {e}={\displaystyle \frac {1}{1+\gamma v_{0}}}$:

  (i) If $\phantom {\dot {i}\!}\frac {1}{2}<d\leq 1$, the acquirer’s optimal bidding strategy is $\phantom {\dot {i}\!}\alpha _{G}^{*}= 1$, $\phantom {\dot {i}\!}\alpha _{B}^{*}= 0$. The manager’s best response is $\phantom {\dot {i}\!}e^{\ast }=m(1-d)$. However, it can be proved that $\phantom {\dot {i}\!}m(1-d)<\frac {1}{2}$<$\frac {1}{1+\gamma v_{0}}$. As a result, the equilibrium does not hold.

  (ii) If $\phantom {\dot {i}\!}d=\frac {1}{2}$, $\phantom {\dot {i}\!}\alpha _{G}^{*}$ and $\phantom {\dot {i}\!}\alpha _{B}^{*}$ can be any value in [0,1]. Moreover, $\phantom {\dot {i}\!}\alpha _{G}^{*}$ and $\phantom {\dot {i}\!}\alpha _{B}^{*}$ are essentially the same, as the signal is completely uninformative. The manager’s optimal response is $\phantom {\dot {i}\!}e^{\ast }=m\left (1-\frac {\alpha _{G}^{*}+\alpha _{B}^{*}}{2}\right )$.

  The equilibrium exists when $\phantom {\dot {i}\!}m\geq \frac {1}{1+\gamma v_{0}}$ and $\phantom {\dot {i}\!}\alpha _{G}^{*}+\alpha _{B}^{*}= 2\left (1-\frac {1}{m(1+\gamma v_{0})}\right )$. $\phantom {\dot {i}\!}\alpha _{G}^{*}$ and $\phantom {\dot {i}\!}\alpha _{B}^{*}$ cannot be uniquely identified. However, in order to be consistent with the other equilibrium, and to make the bidding strategy continuous, we assume that, in this case $\phantom {\dot {i}\!}\alpha _{B}^{*}= 0$ and $\phantom {\dot {i}\!}\alpha _{G}^{*}= 2\left (1-\frac {1}{m(1+\gamma v_{0})}\right )$ (which can also be written as $\phantom {\dot {i}\!}\alpha ^{*}_{G}=\frac {1}{d}\frac {\gamma v_{0}dm-(1-d)(1-m)}{(1-d)m+\gamma v_{0}dm}$ with $\phantom {\dot {i}\!}d=\frac {1}{2}$).¹³

To summarize, we have the following equilibria:

• When $\phantom {\dot {i}\!}d = 1$, the manager chooses $\phantom {\dot {i}\!}e^{*}= 0$ and the acquirer chooses $\phantom {\dot {i}\!}\alpha _{G}^{*}= 1,\,\alpha _{B}^{*}= 0$ in the equilibrium.

• When $\phantom {\dot {i}\!}m\leq \frac {1}{1+\gamma v_{0}}$ and $\phantom {\dot {i}\!}\frac {1}{2}\leq d\leq \frac {1-m}{1-m(1-\gamma v_{0})}$ (for our convenience in later proofs, we define this condition as C1), the manager chooses $\phantom {\dot {i}\!}e^{*}=m$ and the acquirer chooses $\phantom {\dot {i}\!}\alpha _{G}^{*}=\alpha _{B}^{*}= 0$ in the equilibrium.

• When $\phantom {\dot {i}\!}m\leq \frac {1}{1+\gamma v_{0}}$ and $\phantom {\dot {i}\!}\frac {1-m}{1-m(1-\gamma v_{0})}<d<1$, or $\phantom {\dot {i}\!}m>\frac {1}{1+\gamma v_{0}}$ and $\phantom {\dot {i}\!}\frac {1}{2}\leq d<1$ (for our convenience, we define this condition as C2), the manager chooses $\phantom {\dot {i}\!}e^{*}=\frac {1-d}{1-d(1-\gamma v_{0})}$ and the acquirer chooses $\phantom {\dot {i}\!}\alpha _{G}^{*}=\frac {1}{d}\frac {\gamma v_{0}dm-(1-d)(1-m)}{(1-d)m+\gamma v_{0}dm}$ and $\phantom {\dot {i}\!}\alpha _{B}^{*}= 0$ in the equilibrium.

Notice that C1, C2, and d = 1 are mutually exclusive conditions about m and d. A unique equilibrium exists under each condition. □

Proof

Proposition 1

The manager’s optimal effort is $\phantom {\dot {i}\!}e^{\ast }(\alpha _{G},\alpha _{B})=m-m\cdot [\alpha _{G}d+\alpha _{B}(1-d)]$.

Given the equilibrium derived above, we can obtain that in equilibrium, $e^{\ast }(\alpha _{y}^{*}) =\left \{ \begin {array}{ll} 0, & \mathit {given}~d = 1;\\ m, & \mathit {given}~C1;\\ \frac {1-d}{1-d(1-\gamma v_{0})}, & \mathit {given}~C2. \end {array}\right .$ $\alpha _{G}^{*}=\left \{ \begin {array}{ll} \frac {1}{d}\frac {\gamma v_{0}dm-(1-d)(1-m)}{(1-d)m+\gamma v_{0}dm}, & \mathit {given~ C2~or}~d = 1;\\ 0, & \mathit {given}~C1. \end {array}\right .$

Therefore $\frac {\partial e^{*}}{\partial d}=\left \{ \begin {array}{ll} -\frac {\gamma v_{0}}{(1-d(1-\gamma v_{0}))^{2}}<0, & \mathit {given~C2};\\ 0, & \mathit {given~C1~or~d = 1,} \end {array}\right .$ and $\frac {\partial \alpha ^{*}_{G}}{\partial d}=\left \{ \begin {array}{ll} \frac {(1-\gamma v_{0})[1-m(1-\gamma v_{0})]\left (d-\frac {1-m}{1-(1-\gamma v_{0})m}\right )^{2}+\frac {(1-m)\gamma v_{0}}{1-(1-\gamma v_{0})m}}{m(d-d^{2}+\gamma v_{0}d^{2})^{2}}>0, & \mathit {given ~C2~or~d = 1;}\\ 0, & \mathit {given}~C1. \end {array}\right .$□

Proof

Lemma 3

• 1) The expected current firm value $\phantom {\dot {i}\!}E[v|e^{*},\,\alpha _{G}^{*},\,\alpha _{B}^{*}]=e^{\ast }(\alpha _{y}^{*})$. Hence $\phantom {\dot {i}\!}E[v|e^{*},\,\alpha _{G}^{*},\,\alpha _{B}^{*}]$ weakly decreases in d.

• 2) The probability of takeover success is, $Prob(T|e^{*},\,\alpha _{G}^{*},\,\alpha _{B}^{*})= 1-e^{*}+e^{*}[\alpha ^{*}_{G}d+\alpha ^{*}_{B}(1-d)] =\left \{ \begin {array}{ll} 1, & \mathit {given }~d = 1;\\ 1-m, & \mathit {given }~C1;\\ 1-\frac {(1-d)^{2}}{m(1-d(1-\gamma v_{0}))^{2}}, & \mathit {given }~C2. \end {array}\right .$

  We can prove that $\frac {d\,Prob(T|e^{*},\,\alpha _{G}^{*},\,\alpha _{B}^{*})}{d\,d}=\left \{ \begin {array}{ll} 0, & \mathit {given }~d = 1;\\ 0, & \mathit {given }~C1;\\ \frac {2(1-d)v_{0}\gamma }{m(1-d(1-\gamma v_{0}))^{3}}, & \mathit {given }~C2. \end {array}\right .$

  Therefore, $\phantom {\dot {i}\!}Prob(T|e^{*},\,\alpha _{G}^{*},\,\alpha _{B}^{*})$ weakly increases in d.

• 3) The probability of overbidding is, $Prob(OT|e^{*},\,\alpha _{G}^{*},\,\alpha _{B}^{*})=(1-e^{*})[(1-d)\alpha _{G}^{*}+d \alpha ^{*}_{B} ]\ =\left \{ \begin {array}{ll} 0, & \mathit {given~d = 1~or~C1};\\ \frac {(1-d)v_{0}\gamma [dmv_{0}\gamma -(1-m)(1-d)]}{m(1-d(1-\gamma v_{0}))^{2}}, & \mathit {given~C2}. \end {array}\right .$

  We can prove that given $\phantom {\dot {i}\!}C2$, $\phantom {\dot {i}\!}Prob(OT|e^{*},\,\alpha _{G}^{*},\,\alpha _{B}^{*})$ is concave and has a unique maximum point at $\phantom {\dot {i}\!}d=\frac {2-m}{2-m+m\gamma v_{0}}$.

□

Proof

Proposition 2

The expected firm value is $\phantom {\dot {i}\!}{\Pi }_{v}=E[v|e^{*},\,\alpha _{G}^{*},\,\alpha _{B}^{*}]+Prob(T|e^{*},\,\alpha _{G}^{*},\,\alpha _{B}^{*})v_{0}$. Substituting $\phantom {\dot {i}\!}E[v|e^{*},\,\alpha _{G}^{*},\,\alpha _{B}^{*}]$ and $\phantom {\dot {i}\!}Prob(T|e^{*},\,\alpha _{G}^{*},\,\alpha _{B}^{*})$ into $\phantom {\dot {i}\!}{\Pi }_{v}$, we have

$$ {\Pi}_{v}=\left\{ \begin{array}{ll} m+(1-m)v_{0}, & \mathit{given }~C1;\\ v_{0}+\frac{1-d}{1-d(1-\gamma v_{0})}-\frac{(1-d)^{2}v_{0}}{m(1-d(1-v_{0}\gamma))^{2}}, & \mathit{given ~C2,~or~d = 1.} \end{array}\right. $$

(9)

It is easy to show that $\phantom {\dot {i}\!}{\Pi }_{v}$ is continuous in both d and m. We consider two cases separately: (i) when $\phantom {\dot {i}\!}m\le \frac {1}{1+\gamma v_{0}}$, and (ii) when $\phantom {\dot {i}\!}m>\frac {1}{1+\gamma v_{0}}$. In each case, the choice of information quality d also affects which equilibrium holds. Therefore, we need to compare the payoffs in each possible equilibrium and determine the optimal information quality for firm value.

• (1). When $\phantom {\dot {i}\!}m\le \frac {1}{1+\gamma v_{0}}$:

  • If $\phantom {\dot {i}\!}\frac {1}{2}\leq d\leq \frac {1-m}{1-(1-\gamma v_{0})m}$, $\phantom {\dot {i}\!}C1$ holds, and expected firm value is $\phantom {\dot {i}\!}m+(1-m)v_{0}$ for any information quality in this range.

  • If $\phantom {\dot {i}\!}\frac {1-m}{1-(1-\gamma v_{0})m}<d<1$, $\phantom {\dot {i}\!}C2$ holds. We need to solve a constrained optimization problem for $\phantom {\dot {i}\!}{\Pi }_{v}$.

    Taking the first-order derivative of $\phantom {\dot {i}\!}{\Pi }_{v}$ with respect to d, we have $\phantom {\dot {i}\!}\frac {d{\Pi }_{v}}{d\,d}=\frac {v_{0}\gamma [(1-d)(2v_{0}-m)-dmv_{0}\gamma ]}{m(1-d(1-v_{0}\gamma ))^{3}}$.

    The first-order condition gives the unconstrained optimal-information quality level in this equilibrium, denoted as $\phantom {\dot {i}\!}d_{v}\equiv \frac {2v_{0}-m}{2v_{0}-m+m\gamma v_{0}}$. We verify that the second-order condition holds.

    Next, we need to check whether the interior solution exists for the optimization problem. Notice that $\phantom {\dot {i}\!}\frac {d{\Pi }_{v}}{d\,d}<0$ at $\phantom {\dot {i}\!}d = 1$. Therefore, $\phantom {\dot {i}\!}d = 1$ is never optimal for firm value maximization.

    • If $\phantom {\dot {i}\!}\frac {1}{2}<v_{0}<1$, $\phantom {\dot {i}\!}d_{v}$ satisfies $\phantom {\dot {i}\!}\frac {1-m}{1-(1-\gamma v_{0})m}<d_{v}<1$. The optimal information quality maximizes $\phantom {\dot {i}\!}{\Pi }_{v}$, $\phantom {\dot {i}\!}d_{v}^{*}=d_{v}\equiv \frac {2v_{0}-m}{2v_{0}-m+m\gamma v_{0}}$.

    • If $\phantom {\dot {i}\!}v_{0}\leq \frac {1}{2}$, combined with $\phantom {\dot {i}\!}m\le \frac {1}{1+\gamma v_{0}}$, we can prove that $\phantom {\dot {i}\!}d_{v}\leq \frac {1-m}{1-(1-\gamma v_{0})m}$.

      Notice that $\phantom {\dot {i}\!}{\Pi }_{v}$ is continuous at $\phantom {\dot {i}\!}d=\frac {1-m}{1-(1-\gamma v_{0})m}$. Therefore, the optimal information quality maximizes $\phantom {\dot {i}\!}{\Pi }_{v}$, $\phantom {\dot {i}\!}d_{v}^{*}\in \left [\frac {1}{2},\,\frac {1-m}{1-(1-\gamma v_{0})m}\right ]$.

• (2). When $\phantom {\dot {i}\!}m>\frac {1}{1+\gamma v_{0}}$:

  With the assumption $\phantom {\dot {i}\!}m<v_{0}$, we can prove that $\phantom {\dot {i}\!}m>\frac {1}{1+\gamma v_{0}}$ holds only when $\phantom {\dot {i}\!}v_{0} >\frac {\sqrt {4\gamma + 1}-1}{2\gamma } \left (>\frac {1}{2}\right )$.

  For any $\phantom {\dot {i}\!}\frac {1}{2}\le d<1$, $\phantom {\dot {i}\!}C2$ holds. We have the same unconstrained maximum point $\phantom {\dot {i}\!}d_{v}$ that maximizes $\phantom {\dot {i}\!}{\Pi }_{v}$. We can prove that $\phantom {\dot {i}\!}\frac {1}{2}<d_{v}<1$. In addition, $\phantom {\dot {i}\!}{\Pi }_{v}$ is continuous at $\phantom {\dot {i}\!}d = 1$ and $\phantom {\dot {i}\!}\frac {d{\Pi }_{v}}{d\,d}<0$ at $\phantom {\dot {i}\!}d = 1$. As a result, $\phantom {\dot {i}\!}d_{v}^{*}=d_{v}\equiv \frac {2v_{0}-m}{2v_{0}-m+m\gamma v_{0}}$.

Combining (1) and (2), we conclude that $\left \{ \begin {array}{ll} d_{v}^{*}=\frac {2v_{0}-m}{2v_{0}-m+m\gamma v_{0}}, & \mathit {if }~ v_{0}>\frac {1}{2},\\ d_{v}^{*}\in \left [\frac {1}{2},\,\frac {1-m}{1-(1-\gamma v_{0})m}\right ], & \mathit {if }~ v_{0}\leq \frac {1}{2}. \end {array}\right .$□

Proof

Proposition 3

The expected payoff for current shareholders is $\phantom {\dot {i}\!}{\Pi }_{s}=E[v|e^{*},\,\alpha _{G}^{*},\,\alpha _{B}^{*}]+Prob(T|e^{*},\,\alpha _{G}^{*},\,\alpha _{B}^{*})(1-\gamma )v_{0}+Prob(OT|e^{*},\,\alpha _{G}^{*},\,$ $\phantom {\dot {i}\!}\alpha _{B}^{*})$.

Substituting $\phantom {\dot {i}\!}E[v|e^{*},\,\alpha _{G}^{*},\,\alpha _{B}^{*}]$, $\phantom {\dot {i}\!}Prob(T|e^{*},\,\alpha _{G}^{*},\,\alpha _{B}^{*})$ and $\phantom {\dot {i}\!}Prob(OT|e^{*},\,\alpha _{G}^{*},\,\alpha _{B}^{*})$ into $\phantom {\dot {i}\!}{\Pi }_{s}$, we have

$${\Pi}_{s}=\left\{ \begin{array}{ll} m+(1-m)(1-\gamma)v_{0}, & \mathit{given }~C1,\\ \frac{m[1-d(1-\gamma v_{0})][1+v_{0}-d(1+v_{0}-{v_{0}^{2}}\gamma(1-\gamma))]-(1-d)^{2}v_{0}}{m(1-d(1-v_{0}\gamma))^{2}}, & \mathit{given~C2,~or~d = 1.} \end{array}\right. $$

It’s easy to prove that $\phantom {\dot {i}\!}{\Pi }_{s}$ is continuous in both d and m.

Under the condition $\phantom {\dot {i}\!}C2$, we solve the constrained optimization problem for $\phantom {\dot {i}\!}{\Pi }_{s}$. Taking the first-order derivative of $\phantom {\dot {i}\!}{\Pi }_{s}$ with respect to d, we have $\phantom {\dot {i}\!}\frac {d{\Pi }_{s}}{d\,d}= 0$ when $\phantom {\dot {i}\!}d=d_{s}\equiv \frac {2v_{0}-m(1+\gamma v_{0})}{2v_{0}-m(1-\gamma ^{2}{v_{0}^{2}})}$. We verify that the second-order condition holds.

Notice that $\phantom {\dot {i}\!}\frac {d{\Pi }_{s}}{d\,d}<0$ at $\phantom {\dot {i}\!}d = 1$. Therefore, $\phantom {\dot {i}\!}d = 1$ is never optimal for current shareholders.

Similar to the proof of Proposition 2, we consider all possible cases for the values of m, $\phantom {\dot {i}\!}v_{0}$, and $\phantom {\dot {i}\!}\gamma$. In each case, the choice of information quality d also affects which equilibrium holds. We need to compare the payoffs in each possible equilibrium and determine the optimal information quality for expected firm value. In each case that we consider below, if $\phantom {\dot {i}\!}C1$ holds, then $\phantom {\dot {i}\!}{\Pi }_{s}=m+(1-m)(1-\gamma )v_{0}$ for any information quality in this range. If $\phantom {\dot {i}\!}C2$ holds, we then need to check whether the the unconstrained maximum point $\phantom {\dot {i}\!}d_{s}$ is an interior solution.

Given the assumption $\phantom {\dot {i}\!}m<v_{0}$, we can prove that $\phantom {\dot {i}\!}v_{0}>\frac {1}{1+\gamma v_{0}}$ if and only if $\phantom {\dot {i}\!}v_{0}>\frac {\sqrt {4\gamma + 1}-1}{2\gamma } \left (>\frac {1}{2}\right )$. Therefore we consider the following cases:

• When $\phantom {\dot {i}\!}0<v_{0}\leq \frac {\sqrt {4\gamma + 1}-1}{2\gamma }$, then $\phantom {\dot {i}\!}m<v_{0}$ implies $\phantom {\dot {i}\!}m<\frac {1}{1+\gamma v_{0}}$.

  • If $\phantom {\dot {i}\!}0<\gamma \leq 2-\sqrt {2}$ and $\phantom {\dot {i}\!}\frac {1}{2-\gamma }<v_{0}\leq \frac {\sqrt {4\gamma + 1}-1}{2\gamma }$, $\phantom {\dot {i}\!}d_{s}$ satisfies $\phantom {\dot {i}\!}\frac {1-m}{1-(1-\gamma v_{0})m}<d_{s}<1$. The optimal information quality maximizes $\phantom {\dot {i}\!}{\Pi }_{s}$ is $\phantom {\dot {i}\!}d_{s}^{*}=d_{s}\equiv \frac {2v_{0}-m(1+\gamma v_{0})}{2v_{0}-m(1-\gamma ^{2}{v_{0}^{2}})}$. We then compare $\phantom {\dot {i}\!}d_{s}^{*}$ with $\phantom {\dot {i}\!}d_{v}^{*}$. Because $\phantom {\dot {i}\!}\frac {1}{2-\gamma }>\frac {1}{2}$, the optimal information quality that maximizes firm value is $\phantom {\dot {i}\!}d_{v}^{*}=d_{v}\equiv \frac {2v_{0}-m}{2v_{0}-m+m\gamma v_{0}}$. We can prove that $\phantom {\dot {i}\!}d_{s}^{*}<d_{v}^{*}$.

  • If $\phantom {\dot {i}\!}0<\gamma \leq 2-\sqrt {2}$ and $\phantom {\dot {i}\!}0<v_{0}\leq \frac {1}{2-\gamma }$, or $\phantom {\dot {i}\!}2-\sqrt {2}<\gamma <1$ and $\phantom {\dot {i}\!}0<v_{0}\leq \frac {\sqrt {4\gamma + 1}-1}{2\gamma }$, the unconstrained optimal information quality $\phantom {\dot {i}\!}d_{s}\leq \frac {1-m}{1-(1-\gamma v_{0})m}$. Therefore, the optimal information quality maximizes $\phantom {\dot {i}\!}{\Pi }_{s}$ is $\phantom {\dot {i}\!}d_{s}^{*}\in \left [\frac {1}{2},\,\frac {1-m}{1-(1-\gamma v_{0})m}\right ]$. Under this condition, if $\phantom {\dot {i}\!}v_{0}\leq \frac {1}{2}$, then $\phantom {\dot {i}\!}d_{s}^{*},\,d_{v}^{*}\in \left [\frac {1}{2},\,\frac {1-m}{1-(1-\gamma v_{0})m}\right ]$; if $\phantom {\dot {i}\!}v_{0}>\frac {1}{2}$, it is then easy to show that $\phantom {\dot {i}\!}d_{s}^{*}\in \left [\frac {1}{2},\,\frac {1-m}{1-(1-\gamma v_{0})m}\right ]<d_{v}^{*}=d_{v}$.

• When $\phantom {\dot {i}\!}\frac {\sqrt {4\gamma + 1}-1}{2\gamma }<v_{0}<1$ and $\phantom {\dot {i}\!}m<\frac {1}{1+\gamma v_{0}}$, the case is similar to the one above. We need to compare the interior solution $\phantom {\dot {i}\!}d_{s}$ with $\phantom {\dot {i}\!}\frac {1-m}{1-(1-\gamma v_{0})m}$ for the optimization problem of the shareholder. Because $\phantom {\dot {i}\!}\frac {\sqrt {4\gamma + 1}-1}{2\gamma }>\frac {1}{2}$, in this case $\phantom {\dot {i}\!}d_{v}^{*}=d_{v}$.

  • If $\phantom {\dot {i}\!}0<\gamma \leq 2-\sqrt {2}$ and $\phantom {\dot {i}\!}\frac {\sqrt {4\gamma + 1}-1}{2\gamma }<v_{0}<1$, or $\phantom {\dot {i}\!}2-\sqrt {2}<\gamma <1$ and $\phantom {\dot {i}\!}\frac {1}{2-\gamma }<v_{0}<1$, $\phantom {\dot {i}\!}d_{s}$ satisfies $\phantom {\dot {i}\!}\frac {1-m}{1-(1-\gamma v_{0})m}<d_{s}<1$. The optimal information quality maximizes $\phantom {\dot {i}\!}{\Pi }_{s}$ is $\phantom {\dot {i}\!}d_{s}^{*}=d_{s}\equiv \frac {2v_{0}-m(1+\gamma v_{0})}{2v_{0}-m(1-\gamma ^{2}{v_{0}^{2}})}$.

  • If $\phantom {\dot {i}\!}2-\sqrt {2}<\gamma <1$ and $\phantom {\dot {i}\!}\frac {\sqrt {4\gamma + 1}-1}{2\gamma }<v_{0}<\frac {1}{2-\gamma }$, we can prove $\phantom {\dot {i}\!}d_{s}\leq \frac {1-m}{1-(1-\gamma v_{0})m}$. Therefore, the optimal information quality maximizes $\phantom {\dot {i}\!}{\Pi }_{s}$ is $\phantom {\dot {i}\!}d_{s}^{*}\in \left [\frac {1}{2},\,\frac {1-m}{1-(1-\gamma v_{0})m}\right ]$.

  In both cases, $\phantom {\dot {i}\!}d_{s}^{*} <d_{v}^{*}$.

• When $\phantom {\dot {i}\!}\frac {\sqrt {4\gamma + 1}-1}{2\gamma }<v_{0}<1$ and $\phantom {\dot {i}\!}v_{0}>m>\frac {1}{1+\gamma v_{0}}$, $\phantom {\dot {i}\!}C2$ holds. In this case $\phantom {\dot {i}\!}d_{v}^{*}=d_{v}$.

  We have the same unconstrained maximum point $\phantom {\dot {i}\!}d_{s}$ that maximizes $\phantom {\dot {i}\!}{\Pi }_{s}$. By checking whether the interior solution exists for the optimization problem, we have:

  • If (i) $\phantom {\dot {i}\!}0<\gamma \leq 2-\sqrt {2}$, $\phantom {\dot {i}\!}\frac {\sqrt {4\gamma + 1}-1}{2\gamma }<v_{0}<\frac {\sqrt {2}-1}{\gamma }$ and $\phantom {\dot {i}\!}v_{0}>m>\frac {1}{1+\gamma v_{0}}$ or (ii) $\phantom {\dot {i}\!}0<\gamma \leq 2-\sqrt {2}$, $\phantom {\dot {i}\!}\frac {\sqrt {2}-1}{\gamma }<v_{0}<1$ and $\phantom {\dot {i}\!}\frac {1}{1+\gamma v_{0}}<m<\frac {2v_{0}}{(1+\gamma v_{0})^{2}}$, or (iii) $\phantom {\dot {i}\!}2-\sqrt {2}<\gamma <1$, $\phantom {\dot {i}\!}\frac {1}{2-\gamma }<v_{0}<1$ and $\phantom {\dot {i}\!}\frac {1}{1+\gamma v_{0}}<m<\frac {2v_{0}}{(1+\gamma v_{0})^{2}}$, $\phantom {\dot {i}\!}d_{s}$ satisfies $\phantom {\dot {i}\!}\frac {1}{2}<d_{s}<1$. The optimal information quality maximizes $\phantom {\dot {i}\!}{\Pi }_{s}$, $\phantom {\dot {i}\!}d_{s}^{*}=d_{s}\equiv \frac {2v_{0}-m(1+\gamma v_{0})}{2v_{0}-m(1-\gamma ^{2}{v_{0}^{2}})}$.

  • If (i) $\phantom {\dot {i}\!}0<\gamma \leq 2-\sqrt {2}$, $\phantom {\dot {i}\!}\frac {\sqrt {2}-1}{\gamma }<v_{0}<1$ and $\phantom {\dot {i}\!}\frac {2v_{0}}{(1+\gamma v_{0})^{2}}<m<v_{0}$ or (ii) $\phantom {\dot {i}\!}2-\sqrt {2}<\gamma <1$, $\phantom {\dot {i}\!}\frac {1}{2-\gamma }<v_{0}<1$ and $\phantom {\dot {i}\!}\frac {2v_{0}}{(1+\gamma v_{0})^{2}}<m<v_{0}$, or (iii)$2-\sqrt {2}<\gamma <1$, $\phantom {\dot {i}\!}\frac {\sqrt {4\gamma + 1}-1}{2\gamma }<v_{0}<\frac {1}{2-\gamma }$, and $\phantom {\dot {i}\!}v_{0}>m>\frac {1}{1+\gamma v_{0}}$, then $\phantom {\dot {i}\!}d_{s}<\frac {1}{2}$. The optimal information quality maximizes $\phantom {\dot {i}\!}{\Pi }_{s}$, $\phantom {\dot {i}\!}d_{s}^{*}=\frac {1}{2}$.

  In both cases, $\phantom {\dot {i}\!}d_{s}^{*}<d_{v}^{*}$.

To summarize the above discussions, we have the following:

when $\phantom {\dot {i}\!}v_{0}\leq \frac {1}{2}$, $\phantom {\dot {i}\!}d_{s}^{*},\,d_{v}^{*}\in \left [\frac {1}{2},\,\frac {1-m}{1-(1-\gamma v_{0})m}\right ]$; when $\phantom {\dot {i}\!}v_{0}>\frac {1}{2}$, $\phantom {\dot {i}\!}d_{s}^{*}<d_{v}^{*} <1$. □

Proof

Proposition 4

From the proof of Proposition 3 we can obtain the following three scenarios:

• Scenario 1: when (i) $\phantom {\dot {i}\!}0<\gamma \leq 2-\sqrt {2}$, $\phantom {\dot {i}\!}0<v_{0}\leq \frac {1}{2-\gamma }$, and $\phantom {\dot {i}\!}m<v_{0}$, or (ii) $\phantom {\dot {i}\!}2-\sqrt {2}<\gamma <1$, $\phantom {\dot {i}\!}0<v_{0}\leq \frac {\sqrt {4\gamma + 1}-1}{2\gamma }$, and $\phantom {\dot {i}\!}m<v_{0}$, or (iii) $\phantom {\dot {i}\!}2-\sqrt {2}<\gamma <1$, $\phantom {\dot {i}\!}\frac {\sqrt {4\gamma + 1}-1}{2\gamma }<v_{0}\leq \frac {1}{2-\gamma }$, and $\phantom {\dot {i}\!}m\le \frac {1}{1+\gamma v_{0}}$. (i)-(iii) can be rewritten as: $\phantom {\dot {i}\!}\frac {1}{2}<v_{0}<1$, $\phantom {\dot {i}\!}2-\frac {1}{v_{0}}<\gamma <1$, $\phantom {\dot {i}\!}0<m<\frac {1}{1+\gamma v_{0}}$, and $\phantom {\dot {i}\!}0<m<v_{0}$; or $\phantom {\dot {i}\!}0<v_{0}<\frac {1}{2}$ and $\phantom {\dot {i}\!}0<m<v_{0}$. In this case, $\phantom {\dot {i}\!}d_{s}^{*}\in [\frac {1}{2},\,\frac {1-m}{1-(1-\gamma v_{0})m}]$. The upper bound of the range of the optimal information quality $\phantom {\dot {i}\!}\frac {1-m}{1-(1-\gamma v_{0})m}$ is decreasing in $\phantom {\dot {i}\!}\gamma$.

• Scenario 2 (A1): when (i) $\phantom {\dot {i}\!}0<\gamma \leq 2-\sqrt {2}$, $\phantom {\dot {i}\!}\frac {\sqrt {2}-1}{\gamma }<v_{0}<1$ and $\phantom {\dot {i}\!}\frac {2v_{0}}{(1+\gamma v_{0})^{2}}<m<v_{0}$, or (ii) $\phantom {\dot {i}\!}2-\sqrt {2}<\gamma <1$, $\phantom {\dot {i}\!}\frac {1}{2-\gamma }<v_{0}<1$ and $\phantom {\dot {i}\!}\frac {2v_{0}}{(1+\gamma v_{0})^{2}}<m<v_{0}$, or (iii)$2-\sqrt {2}<\gamma <1$, $\phantom {\dot {i}\!}\frac {\sqrt {4\gamma + 1}-1}{2\gamma }<v_{0}<\frac {1}{2-\gamma }$, and $\phantom {\dot {i}\!} \frac {1}{1+\gamma v_{0}}<m<v_{0}$. (i)-(iii) can be rewritten as: $\phantom {\dot {i}\!}\frac {\sqrt {2v_{0}}-\sqrt {m}}{v_{0}\sqrt {m}}<\gamma <1$, $\phantom {\dot {i}\!}\frac {1}{1+\gamma v_{0}}<m<v_{0}$, and $\phantom {\dot {i}\!}\frac {1}{2}<v_{0}<1$. In this case, $\phantom {\dot {i}\!}d_{s}^{*}=\frac {1}{2}$.

• Scenario 3 (A2): when (i) $\phantom {\dot {i}\!}0<\gamma \leq 2-\sqrt {2}$, $\phantom {\dot {i}\!}\frac {1}{2-\gamma }<v_{0}<\frac {\sqrt {2}-1}{\gamma }$ and $\phantom {\dot {i}\!}m<v_{0}$ or (ii) $\phantom {\dot {i}\!}0<\gamma \leq 2-\sqrt {2}$, $\phantom {\dot {i}\!}\frac {\sqrt {2}-1}{\gamma }<v_{0}<1$ and $\phantom {\dot {i}\!}0<m<\frac {2v_{0}}{(1+\gamma v_{0})^{2}}$, or (iii) $\phantom {\dot {i}\!}2-\sqrt {2}<\gamma <1$, $\phantom {\dot {i}\!}\frac {1}{2-\gamma }<v_{0}<1$ and $\phantom {\dot {i}\!}0<m<\frac {2v_{0}}{(1+\gamma v_{0})^{2}}$. (i)-(iii) can be rewritten as: $\phantom {\dot {i}\!}0<\gamma <\frac {\sqrt {2v_{0}}-\sqrt {m}}{v_{0}\sqrt {m}}$ and $\phantom {\dot {i}\!}\frac {1}{1+\gamma v_{0}}<m<v_{0}$, or $\phantom {\dot {i}\!}0<\gamma <2-\frac {1}{v_{0}}$ and $\phantom {\dot {i}\!}0<m<\frac {1}{1+\gamma v_{0}}$, and $\phantom {\dot {i}\!}\frac {1}{2}<v_{0}<1$, $\phantom {\dot {i}\!}0<m<v_{0}$. In this case, $\phantom {\dot {i}\!}d_{s}^{*}=d_{s}\equiv \frac {2v_{0}-m(1+\gamma v_{0})}{2v_{0}-m(1-\gamma ^{2}{v_{0}^{2}})}$. Furthermore, the interior optimal information quality $\phantom {\dot {i}\!}d_{s}^{*}$ is decreasing in $\phantom {\dot {i}\!}\gamma$.

Let $\phantom {\dot {i}\!}\gamma$ and $\phantom {\dot {i}\!}\gamma ^{\prime }$ be the acquirer’s shares of the synergy value before and after the adoption of anti-takeover laws, respectively, and $\phantom {\dot {i}\!}0<\gamma ^{\prime }<\gamma <1$. We then consider how the adoption of antitakeover laws affects the conditions for each scenario to hold. If the scenario still holds, it is clear from the above results, that antitakeover laws then increase the optimal information quality. If the scenario does not hold, we need to find a new scenario and compare it with the old scenario.

• If scenario 1 holds initially, then the optimal information quality after antitakeover laws either remains in the same range, or jumps to scenario 3 when $\phantom {\dot {i}\!}\gamma$ decreases to $\phantom {\dot {i}\!}\gamma ^{\prime }$. For example, suppose initially $\phantom {\dot {i}\!}0<\gamma \leq 2-\sqrt {2}$, $\phantom {\dot {i}\!}\frac {1}{2}<v_{0}\leq \frac {1}{2-\gamma }$, and $\phantom {\dot {i}\!}m<v_{0}$ holds. If $\phantom {\dot {i}\!}\gamma ^{\prime }$ is not too small such that $\phantom {\dot {i}\!}\frac {1}{2}<v_{0}\leq \frac {1}{2-\gamma ^{\prime }}$ remains, then scenario 1 still holds with $\phantom {\dot {i}\!}\gamma ^{\prime }$. If $\phantom {\dot {i}\!}\gamma ^{\prime }$ is small enough such that $\phantom {\dot {i}\!}\frac {1}{2-\gamma ^{\prime }}<v_{0}$, then scenario 1 does not hold, but scenario 3 holds. We can show a similar result for all of the conditions under scenario 1 and prove that $\phantom {\dot {i}\!}d_{s}^{\ast \ast }\geq d_{s}^{\ast }$.

• If scenario 2 holds initially, then either scenario 2 remains, or scenario 1 or scenario 3 holds as $\phantom {\dot {i}\!}\gamma$ decreases to $\phantom {\dot {i}\!}\gamma ^{\prime }$. For example, suppose initially (i) $\phantom {\dot {i}\!}0<\gamma \leq 2-\sqrt {2}$, $\phantom {\dot {i}\!}\frac {\sqrt {2}-1}{\gamma }<v_{0}<1$ and $\phantom {\dot {i}\!}\frac {2v_{0}}{(1+\gamma v_{0})^{2}}<m<v_{0}$ holds. If $\phantom {\dot {i}\!}\gamma ^{\prime }$ becomes very small such that $\phantom {\dot {i}\!}v_{0}<\frac {\sqrt {2}-1}{\gamma ^{\prime }}$ or $\phantom {\dot {i}\!}m<\frac {2v_{0}}{(1+\gamma ^{\prime }v_{0})^{2}}$, then scenario 3 holds. In the same way, we can show that when (ii) $\phantom {\dot {i}\!}2-\sqrt {2}<\gamma <1$, $\phantom {\dot {i}\!}\frac {1}{2-\gamma }<v_{0}<1$ and $\phantom {\dot {i}\!}\frac {2v_{0}}{(1+\gamma v_{0})^{2}}<m<v_{0}$, or (iii)$2-\sqrt {2}<\gamma <1$, $\phantom {\dot {i}\!}\frac {\sqrt {4\gamma + 1}-1}{2\gamma }<v_{0}<\frac {1}{2-\gamma }$, and $\phantom {\dot {i}\!}v_{0}>m>\frac {1}{1+\gamma v_{0}}$ initially holds, then after the anti-takeover law, either scenario 2 remains, or scenario 1 or scenario 3 holds. Again, we have $\phantom {\dot {i}\!}d_{s}^{\ast \ast }\geq d_{s}^{\ast }$.

• If scenario 3 holds initially, then the condition in scenario 3 is relaxed as $\phantom {\dot {i}\!}\gamma$ decreased to $\phantom {\dot {i}\!}\gamma ^{\prime }$. Therefore, scenario 3 still holds after the anti-takeover law. From the above result, the interior solution is decreasing in $\phantom {\dot {i}\!}\gamma$, therefore we have $\phantom {\dot {i}\!}d_{s}^{\ast \ast }>d_{s}^{\ast }$.

□

Proof

Proposition 5

We focus on the cases of A1 and A2, where condition C2 holds. In the proofs of Propositions 2 and 3, we have derived shareholder’s expected payoff and expected firm value given d and $\phantom {\dot {i}\!}\gamma$ as

$$\begin{array}{@{}rcl@{}} {\Pi}_{s}(d,\,\gamma)\!&=&\!\frac{m[1-d(1-\gamma v_{0})][1+v_{0}-d(1+v_{0}-{v_{0}^{2}}\gamma(1-\gamma))]-(1-d)^{2}v_{0}}{m(1-d(1-v_{0}\gamma))^{2}},\\ {\Pi}_{v}(d,\,\gamma)\!&=&\!v_{0}+\frac{1-d}{1-d(1-\gamma v_{0})}-\frac{(1-d)^{2}v_{0}}{m(1-d(1-v_{0}\gamma))^{2}}. \end{array} $$

Recall that the manager’s equilibrium effort is $\phantom {\dot {i}\!}e^{*}(d)=\frac {1-d}{1-d(1-\gamma v_{0})}$, and the probability of takeover is $\phantom {\dot {i}\!}PT^{*}(d)= 1-\frac {(1-d)^{2}}{m(1-d(1-\gamma v_{0}))^{2}}$.

We denote $\phantom {\dot {i}\!}e^{*}$ (and $\phantom {\dot {i}\!}e^{**}$) and $\phantom {\dot {i}\!}PT^{*}$ (and $\phantom {\dot {i}\!}PT^{**}$) to be the manager’s equilibrium effort and the probability of takeover before (and after) the passage of antitakeover laws given the optimal information quality levels, $\phantom {\dot {i}\!}d_{s}^{*}$ and $\phantom {\dot {i}\!}d_{s}^{**}$.

For simplicity, we assume after the antitakeover laws, the prevailing scenario does not change. If the scenario changes after the antitakeover laws, for example, if A1 holds with $\phantom {\dot {i}\!}\gamma$ while A2 holds with $\phantom {\dot {i}\!}\gamma ^{\prime }$ after the adoption of antitakeover laws, we can then consider it as a combination of A1 and A2, and show that our results remain.

• Under A1, $\phantom {\dot {i}\!}d_{s}^{\ast }=d_{s}^{**}=\frac {1}{2}$. Substitute $\phantom {\dot {i}\!}d_{s}^{\ast }$ and $\phantom {\dot {i}\!}d_{s}^{**}$ into $\phantom {\dot {i}\!}{\Pi }_{s}(d,\,\gamma )$ and $\phantom {\dot {i}\!}{\Pi }_{v}(d,\,\gamma$), we obtain that $\phantom {\dot {i}\!}{\Pi }_{s}^{*}(\gamma )= 1+v_{0}-\frac {v_{0}}{m(1+v_{0}\gamma )^{2}}-v_{0}\gamma$; $\phantom {\dot {i}\!}{\Pi }_{v}^{*}(\gamma )=v_{0}-\frac {v_{0}}{m(1+v_{0}\gamma )^{2}}+\frac {1}{1+\gamma v_{0}}$.

  Taking the first-order derivatives, we have $\phantom {\dot {i}\!}\frac {\partial {\Pi }_{s}^{*}}{\partial \gamma }=v_{0}\left (\frac {2v_{0}}{m(1+v_{0}\gamma )^{3}}-1\right )<0$, and $\phantom {\dot {i}\!}\frac {\partial {\Pi }_{v}^{*}}{\partial \gamma }=\frac {v_{0}(2v_{0}-m\gamma v_{0}-m)}{m(1+v_{0}\gamma )^{3}}>0$.

  As a result, we have $\phantom {\dot {i}\!}{\Pi }_{s}^{*}(\gamma )<{\Pi }_{s}{~}^{*}(\gamma ^{\prime })$ and $\phantom {\dot {i}\!}{\Pi }_{v}^{*}(\gamma )>{\Pi }_{v}{~}^{*}(\gamma ^{\prime })$.

  Substituting $\phantom {\dot {i}\!}d_{s}^{\ast }$ and $\phantom {\dot {i}\!}d_{s}^{**}$ into $\phantom {\dot {i}\!}e^{*}(d)$ and $\phantom {\dot {i}\!}PT^{*}(d)$, we obtain $\phantom {\dot {i}\!}e^{*}=\frac {1}{1+\gamma v_{0}}$, and $\phantom {\dot {i}\!}e^{**}=\frac {1}{1+\gamma ^{\prime }v_{0}};$ $\phantom {\dot {i}\!}PT^{*}= 1-\frac {1}{m(1+\gamma v_{0})^{2}}$ and $\phantom {\dot {i}\!}PT^{**}= 1-\frac {1}{m(1+\gamma ^{\prime }v_{0})^{2}}$. It is obvious that $\phantom {\dot {i}\!}e^{**}>e^{*}$ and $\phantom {\dot {i}\!}PT^{**}<PT^{*}$.

• Under A2, we have $\phantom {\dot {i}\!}d_{s}^{\ast }=\frac {2v_{0}-m-m\gamma v_{0}}{2v_{0}-m+m\gamma ^{2}{v_{0}^{2}}}$, $\phantom {\dot {i}\!}d_{s}^{\ast \ast }=\frac {2v_{0}-m-m\gamma ^{\prime }v_{0}}{2v_{0}-m+m\gamma ^{'2}{v_{0}^{2}}}$. Substituting $\phantom {\dot {i}\!}d_{s}^{\ast }$ and $\phantom {\dot {i}\!}d_{s}^{**}$ into $\phantom {\dot {i}\!}{\Pi }_{s}(d,\,\gamma )$ and $\phantom {\dot {i}\!}{\Pi }_{v}(d,\,\gamma$), we obtain that, $\phantom {\dot {i}\!}{\Pi }_{s}^{*}(\gamma )=\frac {4{v_{0}^{2}}(1-\gamma )+m(1+\gamma v_{0})^{2}}{4v_{0}}$, $\phantom {\dot {i}\!}{\Pi }_{v}^{*}(\gamma )=\frac {m}{4v_{0}}+v_{0}-\frac {mv_{0}\gamma ^{2}}{4}$.

  Taking the first-order derivatives, we have $\phantom {\dot {i}\!}\frac {\partial {\Pi }_{s}^{*}}{\partial \gamma }=\frac {1}{2}(m-2v_{0}+mv_{0}\gamma )<0$, and $\phantom {\dot {i}\!}\frac {\partial {\Pi }_{v}^{*}}{\partial \gamma }=-\frac {1}{2}mv_{0}\gamma <0$. As a result, we have $\phantom {\dot {i}\!}{\Pi }_{s}^{*}(\gamma )<{\Pi }_{s}^{*}(\gamma ^{\prime })$ and $\phantom {\dot {i}\!}{\Pi }_{v}^{*}(\gamma )<{\Pi }_{v}^{*}(\gamma ^{\prime })$.

  Substituting $\phantom {\dot {i}\!}d_{s}^{\ast }$ and $\phantom {\dot {i}\!}d_{s}^{**}$ into $\phantom {\dot {i}\!}e^{*}(d)$ and $\phantom {\dot {i}\!}PT^{*}(d)$, we obtain $\phantom {\dot {i}\!}e^{*}=\frac {m(1+\gamma v_{0})}{2v_{0}}$ and $\phantom {\dot {i}\!}e^{**}=\frac {m(1+\gamma ^{\prime }v_{0})}{2v_{0}}$; $\phantom {\dot {i}\!}PT^{*}= 1-\frac {m(1+\gamma v_{0})^{2}}{4{v_{0}^{2}}}$ and $\phantom {\dot {i}\!}PT^{**}= 1-\frac {m(1+\gamma ^{\prime }v_{0})^{2}}{4{v_{0}^{2}}}$. It is obvious that $\phantom {\dot {i}\!}e^{**}<e^{*}$ and $\phantom {\dot {i}\!}PT^{**}>PT^{*}$.

Combining all the cases, we get Proposition 5. □

Proof

Proposition 6

Under the generalized information system, and analogous to the equilibrium in the main setting, denoting $\phantom {\dot {i}\!}h_{0}\equiv \frac {1}{1+\gamma v_{0}}$, we can derive the equilibrium as follows:

• (separating-price-bidding equilibrium) When $\phantom {\dot {i}\!}L_{G}\geq \frac {h_{0}(1-m)}{m(1-h_{0})}$ and $\phantom {\dot {i}\!}1\geq L_{B}\geq \frac {L_{G}h_{0}}{h_{0}(1-m)+mL_{G}(2h_{0}+(1-h_{0})L_{G}-1)}$, the manager chooses $\phantom {\dot {i}\!}e^{*}=\frac {mL_{B}(L_{G}-1)}{L_{G}-L_{B}}$ and the acquirer chooses $\phantom {\dot {i}\!}\alpha _{G}^{*}= 1$, $\phantom {\dot {i}\!}\alpha _{B}^{*}= 0$.

• (low-price-bidding equilibrium) When $\phantom {\dot {i}\!}L_{G}<\frac {h_{0}(1-m)}{m(1-h_{0})}$(Condition C1’), the manager chooses $\phantom {\dot {i}\!}e^{*}=m$ and the acquirer chooses $\phantom {\dot {i}\!}\alpha _{G}^{*}= 0$, $\phantom {\dot {i}\!}\alpha _{B}^{*}= 0$.

• (mixed-price-bidding equilibrium) When $\phantom {\dot {i}\!}L_{G}\geq \frac {h_{0}(1-m)}{m(1-h_{0})}$ and $\phantom {\dot {i}\!}0<L_{B}<\frac {L_{G}h_{0}}{h_{0}(1-m)+mL_{G}(2h_{0}+(1-h_{0})L_{G}-1)}$ (Condition C2’), the manager’s effort is $\phantom {\dot {i}\!}e^{*}=\frac {h_{0}}{h_{0}+L_{G}(1-h_{0})}$. The acquirer chooses $\phantom {\dot {i}\!}\alpha _{G}^{*}=\frac {(L_{G}-L_{B})(mh_{0}+mL_{G}(1-h_{0})-h_{0})}{m(1-L_{B})L_{G}((1-h_{0})L_{G}+h_{0})}$, $\phantom {\dot {i}\!}\alpha _{B}^{*}= 0$.

The shareholder’s expected payoff is

$$\begin{array}{@{}rcl@{}} {\Pi}_{s}(L_{G},L_{B})\,=\,e^{*}\,+\,(1-e^{*}\,+\,e^{*}\alpha_{G}^{*}\frac{L_{G}(1-L_{B})}{L_{G}-L_{B}})(1\,-\,\gamma)v_{0}+(1\,-\,e^{*})\alpha_{G}^{*}\frac{L_{G}\,-\,1}{L_{G}\,-\,L_{B}}. \end{array} $$

The expected firm value is $\phantom {\dot {i}\!}{\Pi }_{v}(L_{G},L_{B})=e^{*}+(1-e^{*}+e^{*}\alpha _{G}^{*}\frac {L_{G}(1-L_{B})}{L_{G}-L_{B}})v_{0}$.

When the separating-price-bidding equilibrium sustains, we substitute $\phantom {\dot {i}\!}\alpha _{G}^{*}= 1$, $\phantom {\dot {i}\!}\alpha _{B}^{*}= 0$ and $\phantom {\dot {i}\!}e^{*}=\frac {mL_{B}(L_{G}-1)}{L_{G}-L_{B}}$ into $\phantom {\dot {i}\!}{\Pi }_{s}(L_{G},L_{B})$ and solve for the optimal $\phantom {\dot {i}\!}L_{G}$ and $\phantom {\dot {i}\!}L_{B}$. We find that the optimal level of likelihood ratios for current shareholders under the separating-price-bidding equilibrium is a corner solution: $\phantom {\dot {i}\!}L_{Gs}=\frac {2-m}{m\gamma v_{0}}$ and $\phantom {\dot {i}\!}L_{Bs}=\frac {L_{Gs}h_{0}}{h_{0}(1-m)+mL_{Gs}(2h_{0}+(1-h_{0})L_{Gs}-1)}$. Notice that this optimal corner solution in the separating-price-bidding equilibrium also approaches the conditions for the mixed-price-bidding equilibrium. Therefore, the separating-price-bidding equilibrium is weakly dominated by the mixed-price-bidding equilibrium. Similarly, we can also prove that the separating-price-bidding equilibrium is a dominated equilibrium for firm value maximization as well.

When the low-price-bidding equilibrium sustains, $\phantom {\dot {i}\!}\alpha _{G}^{*}= 0$, $\phantom {\dot {i}\!}\alpha _{B}^{*}= 0$ and $\phantom {\dot {i}\!}e^{*}=m$. Both shareholder’s expected payoff and expected firm value are independent of the information system as long as the condition for sustaining the low-price-bidding equilibrium sustains $\phantom {\dot {i}\!}L_{G}<\frac {h_{0}(1-m)}{m(1-h_{0})}$. We do not include this case in the proposition, but even with this case, in general we still have $\phantom {\dot {i}\!}L_{Gs}^{*}\le L_{Gv}^{*}$.

When the mixed-price-bidding equilibrium is the optimal equilibrium, substituting $\phantom {\dot {i}\!}e^{*}=\frac {h_{0}}{h_{0}+L_{G}(1-h_{0})}$, $\phantom {\dot {i}\!}\alpha _{G}^{*}=\frac {(L_{G}-L_{B})(mh_{0}+mL_{G}(1-h_{0})-h_{0})}{m(1-L_{B})L_{G}((1-h_{0})L_{G}+h_{0})},$ and $\phantom {\dot {i}\!}\alpha _{B}^{*}= 0$ into $\phantom {\dot {i}\!}{\Pi }_{s}(L_{G},L_{B})$ and $\phantom {\dot {i}\!}{\Pi }_{v}(L_{G},L_{B})$, we obtain that $\phantom {\dot {i}\!}{\Pi }_{s}(L_{G},L_{B})$ and $\phantom {\dot {i}\!}{\Pi }_{v}(L_{G},L_{B})$ are independent of $\phantom {\dot {i}\!}L_{B}$. Taking the first-order derivative with respect to $\phantom {\dot {i}\!}L_{G}$, we then obtain the optimal likelihood ratios $\phantom {\dot {i}\!}L_{Gs}^{*}$ and $\phantom {\dot {i}\!}L_{Gv}^{*}$ that satisfy the first-order conditions: $\phantom {\dot {i}\!}L_{Gs}^{*}=\frac {2v_{0}-m(1+\gamma v_{0})}{m\gamma v_{0}(1+\gamma v_{0})}$ and $\phantom {\dot {i}\!}L_{Gv}^{*}=\frac {2v_{0}-m}{m\gamma v_{0}}$. The second-order conditions hold. It is easy to show that $\phantom {\dot {i}\!}L_{Gs}^{*}<L_{Gv}^{*}$. □