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.
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Notes
The discussion paper version of this paper contains a much more extensive literature review. See Ma et al. (2020), Harvard Kennedy School Discussion Paper.
We are indebted to a referee for making this point.
The model in JTZ (2017) is slightly different than the one we have. There U has the option to open his facility for public inspection. Yet, the conclusion in Proposition 2(i) and Proposition 3(i) remain true if this option is eliminated. In line with JTZ (2017), we show that the probability of either type of U harming R is decreasing in \(\alpha\) in Regions 2 and 3, independent of \(\alpha\) in Region 1, and increasing in \(\alpha\) in Regions 4 and 5.
This result is based on \(\alpha\) being common knowledge. For the analysis of the case where \(\alpha\) is private information of R, see Biran and Ma (2023).
We thank the associate editor and a referee for suggesting the two extensions of the model.
If R plays in equilibrium \(N_{hN_{nh}},\) both \(U^+\) and \(U^-\) are better off playing pure H in which case R is better off deviating to \(B{hB_{nh}}.\) Similarly, if R plays in equilibrium \(B{hB_{nh}},\) both \(U^+\) and \(U^-\) are better off playing pure NH, and R is better off deviating to \(N_{hN_{nh}}.\)
If it is to the contrary that \(p_+=0\) and \(p_-=1,\) then it suggests \(\Pi ^-_1(H|q)>\Pi ^-_1(NH|q)=\Pi ^+_1(NH|q)> \Pi ^+_1(H|q),\) contradicting with \(\Pi ^+_1(H|q)>\Pi ^-_1(H|q).\)
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Appendix
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 \(B{hB_{nh}}.\)Footnote 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,
R’s conditional expected payoffs are
R weakly prefers B to NB given h iff
Similarly, R weakly prefers B to NB given nh iff
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.
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
and \(U^+\) chooses H and \(U^-\) chooses NH. Note that \(\alpha \in (\frac{1}{2},1 )\) satisfies (5) iff
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
Equivalently if
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
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
Suppose the equilibrium is separating. Then it has to be \(p_+=1\) and \(p_-=0\) s.t. \(\Pi ^+_1(H|q)>\Pi ^+_1(NH|q)=\Pi ^-_1(NH|q)> \Pi ^-_1(H|q).\)Footnote 7 Since R is indifferent with \(B_{hN_{nh}}\) and \(N_{hN_{nh}},\) by (3) and (4),
Plugging in \(p_+=1\) and \(p_-=0\) into (1) and setting it equal \(\frac{1-r_2}{1-r_2+w_2},\) we have
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
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),
Plugging in \(p_+=1\) and \(p_-=0\) into (2) and setting it equal \(\frac{1-r_2}{1-r_2+w_2},\) we have
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
R ’s equilibrium payoff is the same whether she plays \(B_{hN_{nh}}\) or \(N_{hN_{nh}},\) that is
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
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 \(\frac{1-r_1}{1 +w_1-r_1}<\frac{1}{2}.\)
Since \(p_{+}=1,\) by (16) we have
and \(0<p_{-}^*<1\) iff \(\beta <\frac{(1-\alpha )(1-r_2)}{(1-\alpha )(1-r_2)+\frac{1}{2}w_2}\) By (17),
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
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
Since \(p_{-}=0,\) by (16) we have
and \(0<p_{+}^*<1\) iff \(\beta >\frac{(1-\alpha )(1-r_2)}{(1-\alpha )(1-r_2)+\frac{1}{2}w_2} .\) By (17),
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
R obtains the same expected payoff whether she plays \(B_{hN_{nh}}\) or \(B{hB_{nh}},\)
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
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
and \(0<p_{-}^*<1\) iff \(\beta <\frac{\alpha (1-r_2)}{\alpha (1-r_2)+\frac{1}{2}w_2} .\)
By (23),
The equilibrium payoffs of \(U^+\) and \(U^-\) are
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
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),
and \(0<p_{+}^*<1\) iff \(\beta >\frac{\alpha (1-r_2)}{\alpha (1-r_2)+\frac{1}{2}w_2} .\)
In this case, U of both types has expected payoff
\(\square\)
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
equivalently, \(p^*_{+}=\frac{ \alpha (1-r_2)}{\frac{1}{2} w_2+ \alpha (1-r_2)},\) and
Since \(U^+\) is indifferent with NH and H.
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
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.
(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
\(U^-\) mixes H and NH, then \(\Pi ^-_1(H|q,\delta )=\Pi ^-_1(NH|q,\delta ).\) The solution in q is
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))
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
By (29) and (30), it can be verified that
\(R^{low}\) obtains
and since \(R^{high}\) plays purely \(B{hB_{nh}},\)
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
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. \(\square\)
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Ma, S., Tauman, Y. & Zeckhauser, R. Deterrence games and the disruption of information. Int J Game Theory 53, 261–287 (2024). https://doi.org/10.1007/s00182-023-00870-3
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DOI: https://doi.org/10.1007/s00182-023-00870-3