Deterrence games and the disruption of information

Abstract

Deterrence is a generic situation where a “Retaliator” (Player R) threatens to bash an “Underminer” (Player U) should he take a stealth threatening move. A typical Underminer is a potential bomb builder, market invader or computer hacker. The Retaliator’s decision whether to bash will depend on a noisy signal her intelligence receives about U’s action. U may or may not have the ability to disrupt R’s signal (type $U^+$ and $U^-,$ respectively). U’s type is his private information. If U can and does disrupt, the signal to R’s intelligence is random, in effect noise. The equilibrium of the game is basically unique. U is better off with the disruption capability than without. More accurate intelligence makes R less likely to bash U. Accordingly, all expected payoffs increase. As R’s belief about U’s ability to disrupt increases, R is more aggressive and U (whether he is able to disrupt or not) is less aggressive. Yet, greater disruption potentially lowers the payoffs of the all players R,  $U^+$ and $U^-.$ Hence a more transparent information system with no potential disruption helps both sides.

Appendix

Proof of Lemma 1

Since $\alpha >\frac{1}{2},$ it is easy to verify from Table 3 that $N{hB_{nh}}$ is strictly dominated by $B_{hN_{nh}}$ for the Retaliator, so $N{hB_{nh}}$ will never be played in equilibrium. Clearly, no equilibrium exists in which R plays pure $N_{hN_{nh}}$ or pure $B{hB_{nh}}.$⁶

Based on the signal, R assigns conditional probabilities over U’s actions, denoted by $P(H|h), P(NH|h)= 1-P(H|h), P(H|nh)$ and $P(NH|nh)= 1-P(H|nh).$ By Bayes’ Rule,

$$\begin{aligned}&P(H|h)=\dfrac{\frac{1}{2}\beta \, p_{+}+(1-\beta )\alpha p_{-}}{\beta (1-p_{+})(1-\alpha )+ \frac{1}{2}\beta p_{+} \, +(1-\beta )(1-p_{-})(1-\alpha ) +(1-\beta ) \alpha p_{-}} \end{aligned}$$

(1)

$$\begin{aligned}&P(H|nh)=\dfrac{\frac{1}{2} \beta \, p_{+} +(1-\beta ) (1-\alpha ) p_{-} }{\beta (1-p_{+})\alpha + \frac{1}{2} \beta p_{+} \, +(1-\beta )(1-p_{-}) \alpha +(1-\beta )p_{-}( 1-\alpha )} \end{aligned}$$

(2)

R’s conditional expected payoffs are

$$\begin{aligned} \Pi _2(B|h)&=P(H|h) w_2+P(NH|h) r_2, \\ \Pi _2(NB|h)&= P(NH|h)\cdot 1+ (1-P(H|h))\cdot 0 =P(NH|h)\\ \Pi _2(B|nh)&=P(H|nh) w_2+P(NH|nh) r_2, \\ \Pi _2(NB|nh)&= P(NH|nh)\cdot 1+ (1-P(NH|nh))\cdot 0=P(NH|nh). \end{aligned}$$

R weakly prefers B to NB given h iff

$$\begin{aligned} \Pi _2(B|h)\ge \Pi _2(NB|h)\longleftrightarrow P(H|h) \ge \frac{1-r_2}{1-r_2+w_2}. \end{aligned}$$

(3)

Similarly, R weakly prefers B to NB given nh iff

$$\begin{aligned} \Pi _2(B|nh)\ge \Pi _2(NB|nh)\longleftrightarrow P(H|nh) \ge \frac{1-r_2}{1-r_2+w_2}. \end{aligned}$$

(4)

Suppose R in equilibrium mixes $N_{hN_{nh}}$ and $B{hB_{nh}}.$ Then $P(H|h) =P(H|nh)= \frac{1-r_2}{1-r_2+w_2}.$ This implies that for $\beta <1$ the signal is uninformative, contradicting $\alpha >\frac{1}{2}.$ $\square$

Proof of Lemma 2

The game $\Gamma _1$ in strategic form is given by Table 3 below.

Table 3 $\Gamma _1$ in strategic form

By Lemma 1, we can eliminate $N{hB_{nh}}$ of R. Since $\alpha >\frac{1}{2},$ the expected payoff to $U^+$ when playing (NH, ND) is greater than when playing (NH, D) (see Table 3). Similarly, $U^+$’s expected payoff by playing (H, D) is greater than that by playing (H, ND). Hence in equilibrium, $U^+$ must assign zero probability to both (NH, D) and (H, ND). $\square$

Proof of Lemma 3

Suppose R plays pure $B_{hN_{nh}}.$ $U^+$ obtains $\alpha w_1+(1-\alpha )r_1$ by playing NH,  and obtains $\frac{1}{2}$ by playing H. He strictly prefers NH to H iff $\alpha >\frac{\frac{1}{2}-r_1}{ w_1-r_1} .$

$U^-$ obtains $\alpha w_1+(1-\alpha )r_1$ by playing NH,  and obtains $1-\alpha$ by playing H. He strictly prefers NH to H iff $\alpha >\frac{1-r_1}{1+w_1-r_1} .$

Hence R is better off deviating to $N_{hN_{nh}}$ if $\alpha >\max \{\frac{\frac{1}{2}-r_1}{ w_1-r_1}, \frac{1-r_1}{1+w_1-r_1}\}$ and she is better off deviating to $B{hB_{nh}}$ if $\alpha <\min \{\frac{\frac{1}{2}-r_1}{ w_1-r_1}, \frac{1-r_1}{1+w_1-r_1}\}.$

If a pure strategy equilibrium exists, either $\frac{1-r_1}{1 +w_1-r_1}\le \alpha \le \frac{\frac{1}{2}-r_1}{ w_1-r_1}$ or $\frac{\frac{1}{2}-r_1}{ w_1-r_1}\le \alpha \le \frac{1-r_1}{1 +w_1-r_1}.$ Note that the latter is true iff $1-w_1\le r_1,$ but in this case $\frac{\frac{1}{2}-r_1}{ w_1-r_1}\le \frac{1-r_1}{1 +w_1-r_1}\le \frac{1}{2}<\alpha ,$ a contradiction. Hence if a pure strategy equilibrium exists, then

$$\begin{aligned} \frac{1-r_1}{1 +w_1-r_1}\le \alpha \le \frac{\frac{1}{2}-r_1}{ w_1-r_1} \end{aligned}$$

(5)

and $U^+$ chooses H and $U^-$ chooses NH. Note that $\alpha \in (\frac{1}{2},1 )$ satisfies (5) iff

$$\begin{aligned} w_1<1-r_1. \end{aligned}$$

(6)

Using Table 3, it is easy to verify that R has no incentive to deviate from $B_{hN_{nh}}$ given that $U^+$ chooses H and $U^-$ chooses NH if

$$\begin{aligned} (1-\beta )[\alpha +(1-\alpha )r_2] +\frac{1}{2} \beta w_2 \ge \max \{ \beta w_2+(1-\beta )r_2, 1-\beta \}. \end{aligned}$$

Equivalently if

$$\begin{aligned} \alpha \ge \max \bigg \{ \frac{ \frac{1}{2}\beta w_2}{ (1-\beta )(1-r_2)}, 1- \frac{\frac{1}{2}\beta w_2}{ (1-\beta )(1-r_2)}\bigg \}. \end{aligned}$$

(7)

Combining (5) and (7),

$$\begin{aligned} \max \bigg \{ \frac{ \frac{1}{2}\beta w_2}{ (1-\beta )(1-r_2)}, 1- \frac{\frac{1}{2}\beta w_2}{ (1-\beta )(1-r_2)}, \frac{1-r_1}{1 +w_1-r_1}\bigg \}\le \alpha \le \frac{\frac{1}{2}-r_1}{ w_1-r_1}. \end{aligned}$$

(8)

Given $1-w_1>r_1,$ the interval in (8) is not empty iff $\frac{ \frac{1}{2}\beta w_2}{ (1-\beta )(1-r_2)}\le \frac{\frac{1}{2}-r_1}{ w_1-r_1}$ and $1-\frac{ \frac{1}{2}\beta w_2}{ (1-\beta )(1-r_2)}\le \frac{\frac{1}{2}-r_1}{ w_1-r_1}.$ Hence (8) is not void iff

$$\begin{aligned} \frac{(1-r_2)(w_1-\frac{1}{2})}{(1-r_2)(w_1-\frac{1}{2} ) +\frac{1}{2}w_2 (w_1-r_1)} \le \beta \le \frac{(1-r_2)(\frac{1}{2}-r_1)}{(1-r_2)(\frac{1}{2}-r_1) + \frac{1}{2} w_2 (w_1-r_1)}. \end{aligned}$$

(9)

Note that the LHS of (9) imposes no restriction on $\beta$ if $w_1<\frac{1}{2}.$ If the denominator is positive then the LHS of (9) is negative and if the denominator is negative then LHS$>1$ and the sign of the inequality changes to $\beta \le$ LHS. Also since $w_1- \frac{1}{2}<\frac{1}{2}-r_1 ,$ (9) defines a non-empty interval.

Hence a pure strategy equilibrium exists iff (8) and (9) hold and in this case R plays pure $B_{hN_{nh}}$ and obtains $(1-\beta )[\alpha +(1-\alpha )r_2] +\frac{1}{2} \beta w_2,$ $U^+$ plays pure H and obtains $\frac{1}{2},$ and $U^-$ plays pure NH and obtains $\alpha w_1+(1-\alpha )r_1.$ $\square$

Proof of Lemma 4

Consider an equilibrium where R mixes $B_{hN_{nh}}$ and $N_{hN_{nh}}$ with probability q and $1-q,$ respectively, $0<q<1.$

Given q,  $U^+$’s and $U^-$’s payoffs when playing pure strategies are

$$\begin{aligned}&\Pi ^+_1(NH|q)=\Pi ^-_1(NH|q)= q[\alpha w_1+(1-\alpha )r_1]+(1-q) w_1 \end{aligned}$$

(10)

$$\begin{aligned}&\Pi ^+_1(H|q) = \frac{1}{2}\, q +(1-q) > \Pi ^-_1(H|q) = q (1-\alpha ) +(1-q). \end{aligned}$$

(11)

Suppose the equilibrium is separating. Then it has to be $p_+=1$ and $p_-=0$ s.t. $\Pi ^+_1(H|q)>\Pi ^+_1(NH|q)=\Pi ^-_1(NH|q)> \Pi ^-_1(H|q).$⁷ Since R is indifferent with $B_{hN_{nh}}$ and $N_{hN_{nh}},$ by (3) and (4),

$$\begin{aligned} P(H|h)= \frac{1-r_2}{1-r_2+w_2} \quad \text{and}\quad P(H|nh)\le \frac{1-r_2}{1-r_2+w_2}. \end{aligned}$$

(12)

Plugging in $p_+=1$ and $p_-=0$ into (1) and setting it equal $\frac{1-r_2}{1-r_2+w_2},$ we have

$$\alpha = \dfrac{ (1-r_2) (1-\beta )-\frac{1}{2} \beta w_2}{(1-r_2)(1-\beta )}$$

For other parameters, by (10) and (11), if $U^+$ randomizes H and NH,  $U^-$ must play pure NH;  if $U^-$ randomizes H and NH,  $U^+$ must play pure H.

Next consider an equilibrium where R mixes $B_{hN_{nh}}$ and $B{hB_{nh}}$ with probability q and $1-q,$ respectively, $0<q<1.$

Given q,  Player $U^+$’s and $U^-$’s payoffs if playing pure strategies are

$$\begin{aligned}&\Pi ^+_1(NH|q)=\Pi ^-_1(NH|q)=q[\alpha w_1+(1-\alpha )r_1]+(1-q) r_1 \end{aligned}$$

(13)

$$\begin{aligned}&\Pi ^+_1(H|q) = q\cdot \frac{1}{2}+(1-q)\cdot 0>\Pi ^-_1(H|q) = q (1-\alpha ). \end{aligned}$$

(14)

Suppose the equilibrium is separating. Then it has to be $p_+=1$ and $p_-=0.$ Since R is indifferent with $B_{hN_{nh}}$ and $B{hB_{nh}},$ by (3) and (4),

$$\begin{aligned} P(H|h)\ge \frac{1-r_2}{1-r_2+w_2} \text{ and } P(H|nh)= \frac{1-r_2}{1-r_2+w_2}. \end{aligned}$$

(15)

Plugging in $p_+=1$ and $p_-=0$ into (2) and setting it equal $\frac{1-r_2}{1-r_2+w_2},$ we have

$$\alpha = \dfrac{ \frac{1}{2} \beta w_2}{(1-r_2)(1-\beta )}$$

For other parameters, by (13) and (14), if $U^+$ randomizes H and NH,  $U^-$ must play pure NH;  if $U^-$ randomizes H and NH,  $U^+$ must play pure H. $\square$

Proof of Proposition 1

Case 1. Pure Strategy Equilibrium.

This case is shown in Lemma 3.

Next we deal with Case 2 and 3. Consider an equilibrium where R mixes $B_{hN_{nh}}$ and $N_{hN_{nh}}$ with probability q and $1-q,$ respectively, $0<q<1.$ Suppose $U^+$ chooses H with probability $p_{+};$ $U^-$ chooses H with probability $p_{-}.$

By (12) and (1), $P(H|h)= \frac{1-r_2}{1-r_2+w_2},$ that is

$$\begin{aligned} \alpha = \dfrac{ (1-r_2)[\beta (1-p_{+})+(1-\beta )(1-p_{-})]-\frac{1}{2} \beta w_2 p_{+} }{(1-r_2)[\beta (1-p_{+})+(1-\beta )(1-p_{-})]+ w_2 (1-\beta )p_{-}}. \end{aligned}$$

(16)

R ’s equilibrium payoff is the same whether she plays $B_{hN_{nh}}$ or $N_{hN_{nh}},$ that is

$$\begin{aligned} \Pi _2 (p_{+},p_{-})=\beta (1-p_{+})+(1-\beta )(1-p_{-}). \end{aligned}$$

(17)

Case 2. (Region 2) Suppose $p_+^*=1$ and $0<p^*_-<1.$

Since $\Pi ^-_1(H|q)=\Pi ^-_1(NH|q),$ by (10) and (11), we have

$$\begin{aligned} q^*=\frac{1-w_1}{ \alpha -(1-\alpha )(w_1-r_1)} \end{aligned}$$

and $q^*\in (0,1)$ iff $\alpha >\frac{1-r_1}{1 +w_1-r_1}.$ This imposes no restriction if $1-w_1<r_1$ since then $\frac{1-r_1}{1 +w_1-r_1}<\frac{1}{2}.$

$$\begin{aligned} \Pi ^{+*}_1 =\frac{(1+w_1)(\alpha -\frac{1}{2})+r_1(1-\alpha )}{ \alpha -(1-\alpha )(w_1-r_1)}, \quad \Pi ^{-*}_1 = \frac{(2w_1-r_1)\alpha -(w_1-r_1)}{\alpha -(1-\alpha )(w_1-r_1)}. \end{aligned}$$

(18)

Since $p_{+}=1,$ by (16) we have

$$\begin{aligned} p_{+}^*=1,\quad p^*_{-}=\frac{(1-\alpha ) (1-r_2)(1-\beta )-w_2\frac{1}{2} \beta }{(1-\beta ) [\alpha w_2 +(1-\alpha )(1-r_2) ]} \end{aligned}$$

and $0<p_{-}^*<1$ iff $\beta <\frac{(1-\alpha )(1-r_2)}{(1-\alpha )(1-r_2)+\frac{1}{2}w_2}$ By (17),

$$\begin{aligned} \Pi _2^*=\frac{(1-\beta ) \alpha w_2 +w_2 \frac{1}{2} \beta }{ \alpha w_2 +(1-\alpha )(1-r_2)}. \end{aligned}$$

(19)

Case 3. (Region 3) Suppose $0<p^*_+<1$ and $p^*_-=0.$

Since $\Pi ^+_1(H|q)=\Pi ^+_1(NH|q),$ by (10) and (11), we have

$$\begin{aligned} q^*=\frac{1-w_1}{\frac{1}{2}-(1-\alpha )(w_1-r_1)} \end{aligned}$$

and $q^*\in (0,1)$ iff $\alpha >\frac{\frac{1}{2}-r_1}{w_1-r_1}.$ Note that this region is not empty iff $w_1>\frac{1}{2}.$ The equilibrium payoffs of $U^+$ and $U^-$ satisfy

$$\begin{aligned} \Pi ^{+*}_1= \Pi ^{-*}_1 = \frac{w_1 \frac{1}{2}-(1-\alpha )(w_1-r_1)}{\frac{1}{2}-(1-\alpha )(w_1-r_1)}. \end{aligned}$$

(20)

Since $p_{-}=0,$ by (16) we have

$$\begin{aligned}&p_{+}^*=\frac{(1-\alpha ) (1-r_2)}{w_2 \frac{1}{2}\beta +(1-r_2) (1-\alpha ) \beta },\quad p^*_{-}=0 \end{aligned}$$

and $0<p_{+}^*<1$ iff $\beta >\frac{(1-\alpha )(1-r_2)}{(1-\alpha )(1-r_2)+\frac{1}{2}w_2} .$ By (17),

$$\begin{aligned} \Pi _2^*= \frac{w_2 \frac{1}{2}}{w_2 \frac{1}{2} +(1-\alpha )(1-r_2) }. \end{aligned}$$

(21)

Finally we analyze Case 4 and 5. Consider an equilibrium where R mixes $B_{hN_{nh}}$ and $B{hB_{nh}}$ with probability q and $1-q,$ respectively, $0<q<1.$ In this case, R is indifferent between playing $B_{hN_{nh}}$ and $B{hB_{nh}}.$ By (1), (2) and (15), setting $P(H|nh)= \frac{1-r_2}{1-r_2+w_2}$ we have

$$\begin{aligned} \alpha = \dfrac{ \frac{1}{2}\beta w_2 p_{+}+ w_2 (1-\beta ) p_{-} }{(1-r_2)[\beta (1-p_{+})+(1-\beta )(1-p_{-})]+ w_2 (1-\beta )p_{-}}. \end{aligned}$$

(22)

R obtains the same expected payoff whether she plays $B_{hN_{nh}}$ or $B{hB_{nh}},$

$$\begin{aligned} \Pi _2 (p_{+},p_{-})=r_2[\beta (1-p_{+})+(1-\beta )(1-p_{-})]+w_2[\beta p_{+}+(1-\beta )p_{-}]. \end{aligned}$$

(23)

Case 4. (Region 4) Suppose $0<p^*_-<0$ and $p^*_+=1.$

Since $\Pi ^-_1(H|q)=\Pi ^-_1(NH|q),$ by (13) and (14), we have

$$\begin{aligned} q^*=\frac{r_1}{1-\alpha -\alpha (w_1-r_1)}. \end{aligned}$$

Clearly, $q^*\in (0,1)$ iff $\alpha <\frac{1-r_1}{1 +w_1-r_1}.$ Note that $\frac{1-r_1}{1 +w_1-r_1}>\frac{1}{2}$ iff $1-w_1>r_1$ and this region of $\alpha$ is not empty.

Since $p_{+}=1,$ by (22), we have

$$\begin{aligned} p_{+}^*=1,\quad p^*_{-}=\frac{ (1-r_2)(1-\beta )\alpha - \frac{1}{2}\beta w_2 }{(1-\beta ) [w_2+(1-r_2-w_2) \alpha ]} \end{aligned}$$

and $0<p_{-}^*<1$ iff $\beta <\frac{\alpha (1-r_2)}{\alpha (1-r_2)+\frac{1}{2}w_2} .$

By (23),

$$\begin{aligned} \Pi _2 ^*=w_2\cdot \frac{(w_2-r_2)\beta (\frac{1}{2}-\alpha )+(1-2\alpha )r_2+ \alpha }{w_2+(1-r_2-w_2) \alpha }. \end{aligned}$$

(24)

The equilibrium payoffs of $U^+$ and $U^-$ are

$$\begin{aligned} \Pi ^{+*}_1 =\frac{r_1\frac{1}{2}}{1-\alpha -\alpha (w_1-r_1)},\quad \Pi ^{-*}_1= \frac{r_1(1-\alpha )}{1-\alpha -\alpha (w_1-r_1)}. \end{aligned}$$

(25)

Case 5. (Region 5) Suppose $0<p^*_+<1$ and $p^*_-=0.$

Since $\Pi ^+_1(H|q)=\Pi ^+_1(NH|q),$ by (13) and (14), we have

$$\begin{aligned} q^*=\frac{r_1}{\frac{1}{2}-\alpha (w_1-r_1)} \end{aligned}$$

and $q^*\in (0,1)$ iff $\alpha <\frac{\frac{1}{2}-r_1}{ w_1-r_1}.$ Since $\alpha >\frac{1}{2},$ we must have $1-w_1>r_1$ in this case.

Since $p_{-}=0,$ by (22),

$$\begin{aligned}&p_{+}^*=\frac{\alpha (1-r_2)}{[\frac{1}{2}w_2 +(1-r_2)\alpha ] \beta },\quad p^*_{-}=0 \end{aligned}$$

(26)

and $0<p_{+}^*<1$ iff $\beta >\frac{\alpha (1-r_2)}{\alpha (1-r_2)+\frac{1}{2}w_2} .$

By (23) and (26),

$$\begin{aligned} \Pi _2^*=w_2\cdot \frac{ \frac{1}{2} r_2+\alpha (1-r_2)}{ \frac{1}{2} w_2+ \alpha (1-r_2)}. \end{aligned}$$

(27)

In this case, U of both types has expected payoff

$$\begin{aligned} \Pi ^{+*}_1= \Pi ^{-*}_1= \frac{\frac{1}{2}r_1}{\frac{1}{2}-\alpha (w_1-r_1)}. \end{aligned}$$

(28)

$\square$

Table 4 Equilibrium strategies in the five regions

Table 5 Equilibrium (ex ante) payoffs in the five regions

Proof of Proposition 2

(i) and (ii) can be verified by Table 4.

(iii) By Table 4, the strategies of $U^+$ and $U^-$ are continuous in $\beta$ within each region. By Table 5, the payoff of R is continuous in $\beta$ within each region. If we plug in the value of $\beta$ on the border line, it can be verified that they are also continuous on the border. By Table 4, R’s strategy is constant in $\beta$ within each region. By Table 5, the payoffs of $U^+$ and $U^-$ are constant in $\beta$ within each region. It can be verified that the value of $\beta$ on the border line supports any equilibrium of the adjacent regions. Hence the equilibrium is upper-semi-continuous in $\beta .$

The continuity of the equilibrium in $\beta =0$ follows immediately from Jelnov et al. (2017) and Table 4.

Suppose $\beta =1.$ That is, it is common knowledge that U is of type $U^+.$ Consider an equilibrium where R mixes $B_{hN_{nh}}$ and $B{hB_{nh}}$ with probability q and $1-q,$ respectively, $0<q<1.$ In this case, R is indifferent between playing $B_{hN_{nh}}$ and $B{hB_{nh}}.$ By (1), (2) and (15), setting $P(H|h)= \frac{1-r_2}{1-r_2+w_2}$ we have

$$\alpha = \dfrac{ w_2 \frac{1}{2} p_{+} }{(1-r_2)(1-p_{+})}$$

equivalently, $p^*_{+}=\frac{ \alpha (1-r_2)}{\frac{1}{2} w_2+ \alpha (1-r_2)},$ and

$$\begin{aligned} \Pi ^*_2=r_2(1-p_{+}) +w_2 p_{+} = w_2\cdot \frac{ \frac{1}{2}r_2+\alpha (1-r_2)}{\frac{1}{2} w_2+ \alpha (1-r_2)}. \end{aligned}$$

Since $U^+$ is indifferent with NH and H.

$$\begin{aligned} \Pi ^+_1(NH|q)=q[\alpha w_1+(1-\alpha )r_1]+(1-q) r_1 =\Pi ^+_1(H|q) =\frac{1}{2} q. \end{aligned}$$

Hence, $q^*=\frac{r_1}{\frac{1}{2}-\alpha (w_1-r_1)}$ and $\,Pr_2(B{hB_{nh}})=1-q^*.$ Note that $q^*\in (0,1)$ iff $\alpha <\frac{\frac{1}{2}-r_1}{+w_1-r_1}.$ U’s expected payoff is $\Pi ^*_1=\frac{\frac{1}{2} r_1 }{\frac{1}{2}-\alpha (w_1-r_1)}.$

Consider next an equilibrium where R mixes $B_{hN_{nh}}$ and $N_{hN_{nh}}$ with probability q and $1-q,$ respectively. By (1), (2) and (12), setting $P(H|nh)= \frac{1-r_2}{1-r_2+w_2}$ we have

$$\begin{aligned} \alpha = \dfrac{ (1-r_2)(1-p_{+}) -\frac{1}{2} p_{+} w_2}{(1-r_2) (1-p_{+})} \end{aligned}$$

and $p_{+}^*=\frac{(1-\alpha )(1-r_2)}{\frac{1}{2}w_2+(1-\alpha )(1-r_2)}.$ R obtains $\Pi ^*_2=1-p^*_{+}=\frac{ \frac{1}{2} w_2}{\frac{1}{2} w_2+(1-\alpha )(1-r_2)}.$

Since U is indifferent between H and NH,  $\Pi ^+_1(NH)=q[\alpha w_1+(1-\alpha )r_1]+(1-q) w_1 =\Pi ^+_1(H) = 1-\frac{1}{2}q.$ Hence $q^*=\frac{1-w_1}{\frac{1}{2}-(1-\alpha )(w_1-r_1)}.$ Finally, $q^*\in (0,1)$ iff $\alpha >\frac{\frac{1}{2} -r_1}{w_1-r_1},$ and U’s expected payoff is $\Pi ^*_1=1-\frac{\frac{1}{2}(1-w_1)}{\frac{1}{2}-(1-\alpha )(w_1-r_1)}.$ This is the same outcome obtained if we substitute $\beta =1$ in Tables 4 and 5. $\square$

Proof of Proposition 3

(i) First note that by Table 5, $\frac{\partial ^2 \Pi _1^{t*}}{\partial \alpha \, \partial \beta } = 0$ and $\,\frac{\partial \Pi _1^{t*}}{ \partial \beta } = 0,$ for $t\in \{+,-\}.$

By Table 5, we can calculate the derivatives of the equilibrium payoffs with respect to $\alpha$ and $\beta ,$ respectively, as well as the cross partial derivatives of the equilibrium payoffs with respect to $\alpha$ and $\beta .$ We present these derivatives in Tables 6 and 7.

Table 6 Impact of $\alpha$ on U’s equilibrium (ex ante) payoff

Table 7 Impact of $\alpha$ and $\beta$ on R’s equilibrium (ex ante) payoff

(ii) Since Player R’s payoff within every region is weakly decreasing in $\beta$ and it is continuous in $\beta$ in all regions, we conclude that R’s payoff is weakly decreasing in $\beta$ for all $\beta \in [0,1].$

$U^+$’s and $U^-$’s payoffs are independent of $\beta$ within each of the five regions. By Table 5 and using Maple, we can show that for a fixed $\alpha ,$ $U^+$’s and $U^-$’s payoffs are both decreasing across regions as $\beta$ increases.

(iii) It can be verified from Tables 4 and 5. $\square$

Proof of Claim 1

In such equilibrium, both $U^+$ and $U^-$ assign a probability $\delta +(1-\delta )q$ on R playing $B{hB_{nh}}$ and a probability $(1-\delta )(1-q)$ on R playing $B_{hN_{nh}}.$ The payoffs of $U^+$ and $U^-$ if playing pure strategies are

$$\begin{aligned}&\Pi ^+_1(NH|q,\delta )=\Pi ^-_1(NH|q,\delta )=(1-\delta )(1-q)[\alpha w_1+(1-\alpha )r_1]+(\delta +(1-\delta )q) r_1 \\ &\Pi ^+_1(H|q,\delta ) = (1-\delta )(1-q)\cdot \frac{1}{2} >\Pi ^-_1(H|q,\delta ) = (1-\delta )(1-q) (1-\alpha ) \end{aligned}$$

$U^-$ mixes H and NH,  then $\Pi ^-_1(H|q,\delta )=\Pi ^-_1(NH|q,\delta ).$ The solution in q is

$$\begin{aligned} q^*=1-\frac{r_1}{(1-\delta )[1-\alpha -\alpha (w_1-r_1)]} \end{aligned}$$

R’s best reply (as a function of the signal and $\beta )$ is the same as in $\Gamma _0.$ Since $R^{low}$ mixes $B{hB_{nh}}$ and $B_{hN_{nh}},$ we have (see (15))

$$\begin{aligned} P_{low}(H|nh)=\frac{1-r_2}{1-r_2+w_2} \end{aligned}$$

(29)

where $P_{low}(H|nh)$ is $R^{low}$’s conditional probability on U playing H,  given nh. By (2), replacing $\beta$ with $\beta _{low}$ and substituting $p_{+}=1,$ we have

$$\begin{aligned} P_{low}(H|nh)=\dfrac{\frac{1}{2} \beta _{low} +(1-\beta _{low}) (1-\alpha ) p_{-} }{ \frac{1}{2} \beta _{low} \, +(1-\beta _{low})(1-p_{-}) \alpha +(1-\beta _{low})p_{-}( 1-\alpha )}. \end{aligned}$$

(30)

By (29) and (30), it can be verified that

$$\begin{aligned} p^*_{-}=\frac{ (1-r_2)(1-\beta _{low})\alpha - \frac{1}{2}\beta _{low} w_2 }{(1-\beta _{low}) [w_2+(1-r_2-w_2) \alpha ]} \end{aligned}$$

$R^{low}$ obtains

$$\begin{aligned} \Pi _2^{low} (1,p^*_{-})&=r_2[(1-\beta _{low})(1-p^*_{-})]+w_2[\beta _{low} +(1-\beta _{low})p^*_{-}] \end{aligned}$$

(31)

$$\begin{aligned}&= r_2 (1-p^*_{-}) + w_2 p^*_{-} +( w_2 -r_2)(1-p^*_-) \beta _{low} \end{aligned}$$

(32)

and since $R^{high}$ plays purely $B{hB_{nh}},$

$$\begin{aligned} \Pi _2^{high} (1,p^*_{-}) = r_2 (1-p^*_{-}) + w_2 p^*_{-} +( w_2 -r_2)(1-p^*_-) \beta _{high}. \end{aligned}$$

(33)

Next we need to ensure that $R^{high}$ has no incentive to deviate from $B{hB_{nh}}.$ By (3) and (4), $P_{high}(H|h)\ge \frac{1-r_2}{1-r_2+w_2}$ and $P_{high}(H|nh)\ge \frac{1-r_2}{1-r_2+w_2}$ must hold. By (1) and (2), for $\beta = \beta _{high}$ and $p_{+}=1,$ we have

$$\begin{aligned} P_{high}(H|h)&=\dfrac{\frac{1}{2}\beta _{high}\, +(1-\beta _{high})\alpha p_{-}}{ \frac{1}{2}\beta _{high} \, +(1-\beta _{high})(1-p_{-})(1-\alpha ) +(1-\beta _{high}) \alpha p_{-}} \ge \frac{1-r_2}{1-r_2+w_2} \end{aligned}$$

(34)

$$\begin{aligned} P_{high}(H|nh)&=\dfrac{\frac{1}{2} \beta _{high}\, +(1-\beta _{high}) (1-\alpha ) p_{-} }{ \frac{1}{2} \beta _{high} \, +(1-\beta _{high})(1-p_{-}) \alpha +(1-\beta _{high})p_{-}( 1-\alpha )} \ge \frac{1-r_2}{1-r_2+w_2}. \end{aligned}$$

(35)

To guarantee the existence of this equilibrium, we need to show that there exist parameters $(\alpha , \beta _{high}, \beta _{low}, \delta , r_1, w_1, r_2, w_2)$ s.t. $p^*_{-} \in (0,1),$ $q^* \in (0,1)$ and (34) and (35) are satisfied. Namely, $\alpha <\frac{1-\delta -r_1}{(1-\delta )(1+w_1-r_1)},$ $\beta _{low}<\frac{\alpha (1-r_2)}{\alpha (1-r_2)+\frac{1}{2}w_2}$ and $\beta _{high}$ satisfies (34) and (35). It can be shown that such a parameter set is non-empty. For example, if $\alpha =0.55,$ $\beta _{high}=0.9,$ $\beta _{low}=0.1,$ $\delta =0.1,$ $r_1=0.125,$ $w_1=0.625,$ $r_2=0.5,$ $w_2=0.75,$ such equilibrium exists. $U^+$ plays pure H,  $U^-$ plays (0.38, 0.62) over (H, NH),  $R^{high}$ plays pure $B{hB_{nh}},$ and $R^{low}$ plays (0.21, 0.79) over $(B{hB_{nh}},B_{hN_{nh}}).$

Comparing $R^{high}$’s payoff (33) to that of $R^{low}$ (31), we show that in the equilibrium of this game R is better off with a higher belief of U being able to disrupt. $\square$