Mitigation and adaptation: an informational perspective

Abstract

Addressing environmental and public health problems requires concerted efforts from both the government and its citizens. The government typically has more information than the citizens do about the severity of a problem, and spillovers of citizens’ adaptation efforts complicate that asymmetric information problem. This paper presents a model in which an informed government chooses a mitigation policy and uninformed citizens choose adaptation efforts sequentially and non-cooperatively. We find that mitigation and adaptation either strategically complement each other or are substitutes for each other in equilibrium, depending upon whether the government uses an overly stringent or an overly lax mitigation policy to signal the severe state. We then extend our analysis to a two-nation model in which the pollution or public health problem is transboundary. National governments (in this case, two) then have incentives to free-ride in both their mitigation and their information transmission. We find that information asymmetry may not distort mitigation in this case. Meanwhile, there also exists another refined equilibrium in which one national government pools and relies on the other’s stringent policy to signal the severe state.

A Proofs

Proof

(Lemma 1) Suppose $m_{L}\ne m_{L}^{c}$. In equilibrium $\mu \left( m_{L}\right) =0$. We have $\mu \left( m_{L}^{c}\right) \geqslant 0=\mu \left( m_{L}\right)$, and $f\left( m_{L}^{c}\right) +\lambda m_{L}^{c}<f\left( m_{L}\right) +\lambda m_{L}$ as $m_{L}^{c}$ is the unique minimizer of $f\left( m\right) +\lambda m$. Therefore, $S_{L}\left( m_{L}^{c}\right) <S_{L}\left( m_{L}\right)$ given $S_{k}^{\prime }\left( \mu \right) <0$ and $f\left( m_{L}^{c}\right) +\lambda m_{L}^{c}<f\left( m_{L}\right) +\lambda m_{L}$. The type-L government thus has incentive to deviate to $m_{L}^{c}$. $\square$

Proof

(Proposition 1) The belief system is consistent with the equilibrium strategy of the government. The citizen’s response is optimal given the belief system. It remains to show that neither type of the government has any incentive to deviate.

Type-L government has no incentive to deviate to any $m^{\prime }<m_{H}$, as this does not change the citizens’ belief (and behavior), and increases the cost $f\left( m\right) +\lambda m$ as $m_{L}^{c}$ minimizes the strictly convex function $f\left( m\right) +\lambda m$. Any $m^{\prime }>m_{H}$ is dominated by $m_{H}$ as it entails the same belief as $m_{H}$ does while $f\left( m^{\prime }\right) +\lambda m^{\prime }>f\left( m_{H}\right) +\lambda m_{H}$given $m^{\prime }>m_{H}>m_{L}^{c}$ and the fact that $m_{L}^{c}$ minimizes the convex function $f\left( m\right) +\lambda m$. We then show that type-L government has no incentive to deviate to $m_{H}$:

$$\begin{aligned}&f\left( m_{L}^{c}\right) +\lambda m_{L}^{c}+g\left( \left( 1+\varphi \right) a_{L}^{c}\right) +\theta a_{L}^{c}\\&=f\left( \underline{m}\right) +\lambda \underline{m}+g\left( \left( 1+\varphi \right) a_{H}^{c}\right) +\theta a_{H}^{c}\\&\leqslant f\left( m_{H}\right) +\lambda m_{H}+g\left( \left( 1+\varphi \right) a_{H}^{c}\right) +\theta a_{H}^{c}, \end{aligned}$$

where the second line is from the definition of $\underline{m}$, while the third line is from $m_{H}\geqslant \underline{m}>m_{L}^{c}$ and the fact that $m_{L}^{c}$ minimizes the convex function $f\left( m\right) +\lambda m$. The LHS of the inequality is the equilibrium loss of the type-L government while the RHS is its loss if deviating to $m_{H}$.

The type-H government has no incentive to deviate to any $m^{\prime }>m_{H}$, as this does not change citizens’ belief but increases the cost $\delta f\left( m\right) +\lambda m$, due to $m^{\prime }>m_{H}\geqslant m_{H}^{c}$ and the fact that $m_{H}^{c}$ minimizes the convex function $\delta f\left( m\right) +\lambda m$. The type-H government has no incentive to deviate to any $m^{\prime }<m_{H}$ either, because

$$\begin{aligned}&\delta \left[ f\left( m_{H}\right) +g\left( \left( 1+\varphi \right) a_{H}^{c}\right) \right] +\theta a_{H}^{c}+\lambda m_{H}\nonumber \\&\leqslant \delta \left[ f\left( \overline{m}\right) +g\left( \left( 1+\varphi \right) a_{H}^{c}\right) \right] +\theta a_{H}^{c}+\lambda \overline{m}\nonumber \\&=\delta \left[ f\left( m_{H}^{c}\right) +g\left( \left( 1+\varphi \right) a_{L}^{c}\right) \right] +\theta a_{L}^{c}+\lambda m_{H} ^{c}\nonumber \\&\leqslant \delta \left[ f\left( m^{\prime }\right) +g\left( \left( 1+\varphi \right) a_{L}^{c}\right) \right] +\theta a_{L}^{c}+\lambda m^{\prime } \end{aligned}$$

(18)

where the first inequality comes from $m_{H}^{c}\leqslant m_{H}\leqslant \overline{m}$ and the fact that $m_{H}^{c}$ minimizes the convex function $\delta f\left( m\right) +\lambda m$, and the second inequality comes from the fact that $m_{H}^{c}$ minimizes the strictly convex function $\delta f\left( m\right) +\lambda m$. The third line uses the definition of $\overline{m}$. The first line of (18) is the type-H government’s equilibrium loss and the last line is its loss when deviating to $m^{\prime }<m_{H}$. $\square$

Proof

(Proposition 2) The belief system is consistent with the equilibrium strategy of the government. The citizens’ response is optimal given the belief system. It remains to show that neither type of the government has any incentive to deviate.

Note that $m_{L}^{c}$ minimizes the convex function $f\left( m\right) +\lambda m$. Type-L government has no incentive to deviate to any $m^{\prime }$ that satisfies both $m^{\prime }\ne m_{L}^{c}$ and $m^{\prime }>m_{H}$, because this does not change citizens’ beliefs and will incur a larger cost of $f\left( m\right) +\lambda m$. Any deviation to $m^{\prime }<m_{H}$ is dominated by $m=m_{H}$ because the off-equilibrium beliefs are the same while $m^{\prime }<m_{H}\leqslant \widehat{m}<m_{L}^{c}$ so $f\left( m^{\prime }\right) +\lambda m^{\prime }>f\left( m_{H}\right) +\lambda m_{H}$. It remains to show that the type-L government has no incentive to deviate to $m_{H}$. To see this, we have

$$\begin{aligned}&f\left( m_{L}^{c}\right) +\lambda m_{L}^{c}+g\left( \left( 1+\varphi \right) a_{L}^{c}\right) +\theta a_{L}^{c}\\&=f\left( \widehat{m}\right) +\lambda \widehat{m}+g\left( \left( 1+\varphi \right) a_{H}^{c}\right) +\theta a_{H}^{c}\\&\leqslant f\left( m_{H}\right) +\lambda m_{H}+g\left( \left( 1+\varphi \right) a_{H}^{c}\right) +\theta a_{H}^{c}, \end{aligned}$$

where the first line is the equilibrium loss while the third line is the loss when deviating to $m_{H}$. The second line uses the definition of $\widehat{m}$. The inequality comes from that fact that $m_{H}\leqslant \widehat{m}<m_{L}^{c}$ and that $m_{L}^{c}$ minimizes the strictly convex function $f\left( m\right) +\lambda m$, which imply $f\left( m_{H}\right) +\lambda m_{H}\geqslant f\left( \widehat{m}\right) +\lambda \widehat{m}$.

Type-H government has no incentive to deviate to $m^{\prime }<m_{H}$, as beliefs do not change and $m^{\prime }<m_{H}\leqslant \widehat{m}<m_{L} ^{c}<m_{H}^{c}$ implying $\delta f\left( m^{\prime }\right) +\lambda m^{\prime }>\delta f\left( m_{H}\right) +\lambda m_{H}$. Any deviation to $m^{\prime }>m_{H}$ is weakly dominated by $m=m_{H}^{c}$ as beliefs are the same, while $m=m_{H}^{c}$ minimizes $\delta f\left( m\right) +\lambda m$. It remains to show that it has no incentive to deviate to $m_{H}^{c}$. We have

$$\begin{aligned}&\delta f\left( m_{H}\right) +\lambda m_{H}+\delta g\left( \left( 1+\varphi \right) a_{H}^{c}\right) +\theta a_{H}^{c}\\&\leqslant \delta f\left( \widetilde{m}\right) +\lambda \widetilde{m}+\delta g\left( \left( 1+\varphi \right) a_{H}^{c}\right) +\theta a_{H}^{c}\\&=\delta f\left( m_{H}^{c}\right) +\lambda m_{H}^{c}+\delta g\left( \left( 1+\varphi \right) a_{L}^{c}\right) +\theta a_{L}^{c} \end{aligned}$$

where the first line is the equilibrium loss while the third line is the loss when deviating to $m_{H}^{c}$ (with $\mu \left( m_{H}^{c}\right) =0$ given $m_{H}^{c}>m_{H}$). The inequality comes from that fact that $\widetilde{m} \leqslant m_{H}\leqslant \widehat{m}<m_{L}^{c}<m_{H}^{c}$ implying $\delta f\left( m_{H}\right) +\lambda m_{H}\leqslant \delta f\left( \widetilde{m} \right) +\lambda \widetilde{m}$ given that $m_{H}^{c}$ minimizes the strictly convex function $\delta f\left( m\right) +\lambda m$. The third line uses the definition of $\widetilde{m}$. $\square$

Proof

(Remark 2) Since $m_{H}<m_{L}^{c}<m_{H}^{c}$, we have

$$\begin{aligned}&f\left( m_{H}\right)>f\left( m_{L}^{c}\right) \\&\Rightarrow \left( \delta -1\right) f\left( m_{H}\right)>\left( \delta -1\right) f\left( m_{L}^{c}\right) \\&\Rightarrow \delta f\left( m_{H}\right) +\lambda m_{H}>f\left( m_{H}\right) +\lambda m_{H}+\left( \delta -1\right) f\left( m_{L} ^{c}\right) \\&\Rightarrow \delta f\left( m_{H}\right) +\lambda m_{H}-\left( \delta f\left( m_{H}^{c}\right) +\lambda m_{H}^{c}\right) \\& >f\left( m_{H}\right)+\lambda m_{H}+\left( \delta -1\right) f\left( m_{L}^{c}\right) -\left( \delta f\left( m_{L}^{c}\right) +\lambda m_{L}^{c}\right) \\&\Rightarrow \left[ \delta f\left( m_{H}\right) +\lambda m_{H}\right] -\left[ \delta f\left( m_{H}^{c}\right) +\lambda m_{H}^{c}\right] >\left[ f\left( m_{H}\right) +\lambda m_{H}\right] -\left[ f\left( m_{L} ^{c}\right) +\lambda m_{L}^{c}\right] , \end{aligned}$$

where the second line follows from $\delta >1$ and the fourth line follows from the fact that $m_{H}^{c}$ uniquely minimizes $\delta f\left( m\right) +\lambda m$. $\square$

Proof

(Corollary 1) Let a be the adaptation level undertaken by every citizen. We have $A=\left( 1+\varphi \right) a$. By (2),

$$\begin{aligned} \alpha \left( a_{0}-\left( 1+\varphi \right) a_{L}^{c}\right)&=\theta \\&\Rightarrow a_{L}^{c}=\frac{a_{0}-\frac{\theta }{\alpha }}{1+\varphi }.\\ \delta \alpha \left( a_{0}-\left( 1+\varphi \right) a_{H}^{c}\right)&=\theta \\&\Rightarrow a_{H}^{c}=\frac{a_{0}-\frac{\theta }{\alpha \delta }}{1+\varphi }. \end{aligned}$$

Thus

$$\begin{aligned}&g\left( \left( 1+\varphi \right) a_{L}^{c}\right) +\theta a_{L} ^{c}-g\left( \left( 1+\varphi \right) a_{H}^{c}\right) -\theta a_{H}^{c}\\&=\left[ \frac{\alpha }{2}\left( a_{0}-\left( 1+\varphi \right) a_{L} ^{c}\right) ^{2}+\theta a_{L}^{c}\right] -\left[ \frac{\alpha }{2}\left( a_{0}-\left( 1+\varphi \right) a_{H}^{c}\right) ^{2}+\theta a_{H}^{c}\right] \\&=\left[ \frac{\alpha }{2}\left( \frac{\theta }{\alpha }\right) ^{2} +\theta \frac{a_{0}-\frac{\theta }{\alpha }}{1+\varphi }\right] -\left[ \frac{\alpha }{2}\left( \frac{\theta }{\alpha \delta }\right) ^{2}+\theta \frac{a_{0}-\frac{\theta }{\alpha \delta }}{1+\varphi }\right] \\&=\frac{\theta ^{2}}{2\alpha }\left( 1-\frac{1}{\delta ^{2}}\right) +\frac{\frac{\theta ^{2}}{\alpha }\left( \frac{1}{\delta }-1\right) }{1+\varphi }\\&=\frac{\theta ^{2}}{\alpha }\left( 1-\frac{1}{\delta }\right) \left[ \frac{1}{2}\left( 1+\frac{1}{\delta }\right) -\frac{1}{1+\varphi }\right] , \end{aligned}$$

and

$$\begin{aligned}&\delta g\left( \left( 1+\varphi \right) a_{L}^{c}\right) +\theta a_{L}^{c}-\delta g\left( \left( 1+\varphi \right) a_{H}^{c}\right) -\theta a_{H}^{c}\\&=\left[ \frac{\alpha \delta }{2}\left( a_{0}-\left( 1+\varphi \right) a_{L}^{c}\right) ^{2}+\theta a_{L}^{c}\right] -\left[ \frac{\alpha \delta }{2}\left( a_{0}-\left( 1+\varphi \right) a_{H}^{c}\right) ^{2}+\theta a_{H}^{c}\right] \\&=\left[ \frac{\alpha \delta }{2}\left( \frac{\theta }{\alpha }\right) ^{2}+\theta \frac{a_{0}-\frac{\theta }{\alpha }}{1+\varphi }\right] -\left[ \frac{\alpha \delta }{2}\left( \frac{\theta }{\alpha \delta }\right) ^{2} +\theta \frac{a_{0}-\frac{\theta }{\alpha \delta }}{1+\varphi }\right] \\&=\frac{\delta \theta ^{2}}{2\alpha }\left( 1-\frac{1}{\delta ^{2}}\right) +\frac{\frac{\theta ^{2}}{\alpha }\left( \frac{1}{\delta }-1\right) }{1+\varphi }\\&=\frac{\theta ^{2}}{\alpha }\left( 1-\frac{1}{\delta }\right) \left[ \frac{\delta }{2}\left( 1+\frac{1}{\delta }\right) -\frac{1}{1+\varphi }\right] . \end{aligned}$$

By (3),

$$\begin{aligned}&-m_{0}+m_{L}^{c}+\lambda =0\\&\Rightarrow m_{L}^{c}=m_{0}-\lambda \text {.}\\&-\delta m_{0}+\delta m_{H}^{c}+\lambda =0\\&\Rightarrow m_{H}^{c}=m_{0}-\frac{\lambda }{\delta }\text {.}\\& \frac{1}{2}\left( m_{0}-m_{L}^{c}\right) ^{2}+\lambda m_{L}^{c}=\frac{1}{2}\lambda ^{2}+\lambda m_{0}-\lambda ^{2}=\lambda m_{0}-\frac{1}{2}\lambda ^{2}.\\&\frac{\delta }{2}\left( m_{0}-m_{H}^{c}\right) ^{2}+\lambda m_{H}^{c} =\frac{1}{2}\frac{\lambda ^{2}}{\delta }+\lambda m_{0}-\frac{\lambda ^{2}}{\delta }=\lambda m_{0}-\frac{1}{2}\frac{\lambda ^{2}}{\delta }. \end{aligned}$$

Given the definitions of $\underline{m}$, $\widehat{m}$, $\widetilde{m}$ and $\overline{m}$, $\underline{m}$ and $\widehat{m}$ solve

$$\begin{aligned}&\frac{(m_{0}-m)^{2}}{2}+\lambda m-\left( \lambda m_{0}-\frac{\lambda ^{2} }{2}\right) =\frac{\theta ^{2}}{\alpha }\left( 1-\frac{1}{\delta }\right) \left( \frac{\delta +1}{2\delta }-\frac{1}{1+\varphi }\right) \\&\Leftrightarrow \left( m_{0}-\lambda -m\right) ^{2}=\left( \frac{\theta }{\delta }\right) ^{2}\frac{\delta -1}{\alpha }\left( \delta +1-\frac{2\delta }{1+\varphi }\right) >0, \end{aligned}$$

while $\overline{m}$ and $\widetilde{m}$ solve

$$\begin{aligned}&\frac{\delta (m_{0}-m)^{2}}{2}+\lambda m-\left( \lambda m_{0}-\frac{\lambda ^{2}}{2\delta }\right) =\frac{\theta ^{2}}{\alpha }\left( 1-\frac{1}{\delta }\right) \left( \frac{\delta +1}{2}-\frac{1}{1+\varphi }\right) \\&\Leftrightarrow \left( m_{0}-\frac{\lambda }{\delta }-m\right) ^{2}=\left( \frac{\theta }{\delta }\right) ^{2}\frac{\delta -1}{\alpha }\left( \delta +1-\frac{2}{1+\varphi }\right) >0, \end{aligned}$$

where both inequalities follow from $1+\varphi>\delta >1$. The solutions are:

$$\begin{aligned}&\underline{m}=m_{0}-\lambda +\frac{\theta }{\delta }\sqrt{\frac{\delta -1}{\alpha }\left( \delta +1-\frac{2\delta }{1+\varphi }\right) }>m_{L} ^{c}=m_{0}-\lambda ,\nonumber \\&\widehat{m}=m_{0}-\lambda -\frac{\theta }{\delta }\sqrt{\frac{\delta -1}{\alpha }\left( \delta +1-\frac{2\delta }{1+\varphi }\right) }<m_{L}^{c} =m_{0}-\lambda ,\nonumber \\&\overline{m}=m_{0}-\frac{\lambda }{\delta }+\frac{\theta }{\delta }\sqrt{\frac{\delta -1}{\alpha }\left( \delta +1-\frac{2}{1+\varphi }\right) } >m_{H}^{c}=m_{0}-\frac{\lambda }{\delta },\nonumber \\&\widetilde{m}=m_{0}-\frac{\lambda }{\delta }-\frac{\theta }{\delta }\sqrt{\frac{\delta -1}{\alpha }\left( \delta +1-\frac{2}{1+\varphi }\right) } <m_{H}^{c}=m_{0}-\frac{\lambda }{\delta }. \end{aligned}$$

(19)

Because $\delta >1$, we have $\overline{m}>\underline{m}$. We also have $\widehat{m}\ge \widetilde{m}$ if and only if

$$\begin{aligned}&m_{0}-\lambda -\frac{\theta }{\delta }\sqrt{\frac{\delta -1}{\alpha }\left( \delta +1-\frac{2\delta }{1+\varphi }\right) }\ge m_{0}-\frac{\lambda }{\delta }-\frac{\theta }{\delta }\sqrt{\frac{\delta -1}{\alpha }\left( \delta +1-\frac{2}{1+\varphi }\right) }\\&\Leftrightarrow \lambda \left( 1-\frac{1}{\delta }\right) \le \frac{\theta }{\delta }\sqrt{\frac{\delta -1}{\alpha (1+\varphi )}}\left( -\sqrt{\varphi \delta +\varphi +\delta +1-2\delta }+\sqrt{\varphi \delta +\varphi +\delta +1-2}\right) \\&\Leftrightarrow \frac{\lambda }{\theta }\le \frac{\sqrt{\varphi \delta +\varphi +\delta -1}-\sqrt{\varphi \delta +\varphi -\delta +1}}{\sqrt{\alpha (\delta -1)(1+\varphi )}}. \end{aligned}$$

Moreover, $\underline{m}>m_{H}^{c}$ if and only if

$$\begin{aligned}&m_{0}-\lambda +\frac{\theta }{\delta }\sqrt{\frac{\delta -1}{\alpha }\left( \delta +1-\frac{2\delta }{1+\varphi }\right) }>m_{0}-\frac{\lambda }{\delta }\\&\Leftrightarrow \lambda \left( 1-\frac{1}{\delta }\right)<\frac{\theta }{\delta }\sqrt{\frac{\delta -1}{\alpha }\cdot \frac{\varphi \delta +\varphi +\delta +1-2\delta }{1+\varphi }}\\&\Leftrightarrow \frac{\lambda }{\theta }<\sqrt{\frac{\varphi -\delta +\varphi \delta +1}{\alpha (\varphi +1)(\delta -1)}}. \end{aligned}$$

$\square$

Proof

(Proposition 3) For any separating equilibrium in Proposition 1 where $m_{H}>\max \{m_{H}^{c},\underline{m}\}$, consider a deviation $m^{\prime }=m_{H}-\varepsilon >\max \{m_{H}^{c},\underline{m}\}$ where $\varepsilon$ is a small enough positive number. This deviation is not attractive to Type-L government even if it is believed to have type-H (the best possible belief for it) under the deviation:

$$\begin{aligned}&f\left( m_{L}^{c}\right) +\lambda m_{L}^{c}+g\left( \left( 1+\varphi \right) a_{L}^{c}\right) +\theta a_{L}^{c}\\&=f\left( \underline{m}\right) +\lambda \underline{m}+g\left( \left( 1+\varphi \right) a_{H}^{c}\right) +\theta a_{H}^{c}\\&<f\left( m^{\prime }\right) +\lambda m^{\prime }+g\left( \left( 1+\varphi \right) a_{H}^{c}\right) +\theta a_{H}^{c}, \end{aligned}$$

where the equality uses the definition of $\underline{m}$ and the inequality is from $m_{L}^{c}<\underline{m}<m^{\prime }$, which implies $f\left( \underline{m}\right) +\lambda \underline{m}<f\left( m^{\prime }\right) +\lambda m^{\prime }$ given that $m_{L}^{c}$ uniquely minimizes the strictly convex function $f\left( m\right) +\lambda m$. However, this deviation is attractive to type-H government if it is considered to have type-H (its true type) as

$$\begin{aligned} \delta f\left( m_{H}\right) +\lambda m_{H}+\delta g\left( \left( 1+\varphi \right) a_{H}^{c}\right) +\theta a_{H}^{c}>\delta f\left( m^{\prime }\right) +\lambda m^{\prime }+\delta g\left( \left( 1+\varphi \right) a_{H}^{c}\right) +\theta a_{H}^{c}, \end{aligned}$$

given $m_{H}^{c}<m^{\prime }<m_{H}$ and the fact that $m_{H}^{c}$ uniquely minimizes the strictly convex function $\delta f\left( m\right) +\lambda m$. Therefore, the type- H government has incentive to deviate to $m^{\prime }$, and thus any separating equilibrium in Proposition 1 with $m_{H}>\max \{m_{H}^{c},\underline{m}\}$ does not survive the intuitive criterion.

For any separating equilibrium in Proposition 2 where $m_{H}<\widehat{m}$, consider a deviation $m^{\prime }=m_{H}+\varepsilon <\widehat{m}$ where $\varepsilon$ is a small enough positive number. This deviation is not attractive to Type-L government even if it is believed to have type-H (the best possible belief for it) under the deviation:

$$\begin{aligned}&f\left( m_{L}^{c}\right) +\lambda m_{L}^{c}+g\left( \left( 1+\varphi \right) a_{L}^{c}\right) +\theta a_{L}^{c}\\&=f\left( \widehat{m}\right) +\lambda \widehat{m}+g\left( \left( 1+\varphi \right) a_{H}^{c}\right) +\theta a_{H}^{c}\\&\leqslant f\left( m^{\prime }\right) +\lambda m^{\prime }+g\left( \left( 1+\varphi \right) a_{H}^{c}\right) +\theta a_{H}^{c}, \end{aligned}$$

where the equality uses the definition of $\widehat{m}$ and the inequality is from $m^{\prime }<\widehat{m}<m_{L}^{c}$, where $m_{L}^{c}$ uniquely minimizes the strictly convex function $f\left( m\right) +\lambda m$. However, this deviation is attractive to type-H government if it is considered to have type-H (its true type) as

$$\begin{aligned} \delta f\left( m_{H}\right) +\lambda m_{H}+\delta g\left( \left( 1+\varphi \right) a_{H}^{c}\right) +\theta a_{H}^{c}>\delta f\left( m^{\prime }\right) +\lambda m^{\prime }+\delta g\left( \left( 1+\varphi \right) a_{H}^{c}\right) +\theta a_{H}^{c}, \end{aligned}$$

given $m_{H}<m^{\prime }<m_{H}^{c}$, where $m_{H}^{c}$ uniquely minimizes the strictly convex function $\delta f\left( m\right) +\lambda m$. Therefore, the type- H government has incentive to deviate to $m^{\prime }$, and thus any separating equilibrium in Proposition 2 with $m_{H}<\widehat{m}$ does not survive the intuitive criterion. $\square$

Proof

(Lemma 3) Suppose at least one government does not choose $m_{L}^{tc}$. Then there exists $n\in \{1,2\}$ such that $m_{n}\ne m_{L} ^{BR}\left( m_{-n}\right)$. Then government n has incentive to deviate to $m_{L}^{BR}\left( m_{-n}\right)$, which gives a lower loss of $f\left( m_{n}+\beta m_{-n}\right) +\lambda m_{n}$ and possibly higher belief $\mu$ and so a lower loss of $g\left( (1+\varphi +\gamma )a(\mu )\right) +\theta a(\mu )$. $\square$

Proof

(Proposition 4) (i) Under this equilibrium, a government with type-L has no incentive to unilaterally deviate to any $m^{\prime }\ne m_{L}^{tc}$ because this unilateral deviation does not change beliefs and $m^{\prime }\ne m_{L}^{BR}\left( m_{L}^{tc}\right)$. Meanwhile, the type-H governments do not have incentive to unilaterally deviate because any deviation is not a best response to the other’s choice $m_{H}^{tc}$ and possibly reduces $\mu$.

(ii) Under this equilibrium, a government with type-L has no incentive to unilaterally deviate to any $m^{\prime }\ne m_{L}^{tc}$ because this unilateral deviation does not change beliefs and $m^{\prime }\ne m_{L} ^{BR}\left( m_{L}^{tc}\right)$. We then show that no type-H government has incentive to unilaterally deviate to any $m^{\prime }>m_{H}$. To see this, note that $m_{H}^{tc}=m_{H}^{BR}\left( m_{H}^{tc}\right)$. By Lemma 2, we have

$$\begin{aligned} m^{\prime }>m_{H}>m_{H}^{tc}>m_{H}^{BR}\left( m_{H}\right) . \end{aligned}$$

Since $m_{H}^{BR}\left( m_{H}\right)$ uniquely minimizes the convex function $\delta f\left( m+\beta m_{H}\right) +\lambda m$, the deviation to $m^{\prime }$ causes a larger loss in $\delta f\left( m+\beta m_{H}\right) +\lambda m$ than $m_{H}$ but does not change beliefs.

Next, for all deviations to $m^{\prime }<m_{H}$, the best one is $m_{H} ^{BR}\left( m_{H}\right)$. For a type-H government to have no incentive to deviate to $m_{H}^{BR}\left( m_{H}\right)$, we need

$$\begin{aligned}&\delta f\left( m_{H}^{BR}\left( m_{H}\right) +\beta m_{H}\right) +\lambda m_{H}^{BR}\left( m_{H}\right) +\delta g\left( (1+\varphi +\gamma )a_{L}^{tc}\right) +\theta a_{L}^{tc}\\&\ge \delta f\left( (1+\beta )m_{H}\right) +\lambda m_{H}+\delta g\left( (1+\varphi +\gamma )a_{H}^{tc}\right) +\theta a_{H}^{tc}, \end{aligned}$$

which is equivalent to

$$\begin{aligned}&\left[ \delta g\left( (1+\varphi +\gamma )a_{L}^{tc}\right) +\theta a_{L}^{tc}\right] -\left[ \delta g\left( (1+\varphi +\gamma )a_{H} ^{tc}\right) +\theta a_{H}^{tc}\right] \nonumber \\&\ge \left[ \delta f\left( (1+\beta )m_{H}\right) +\lambda m_{H}\right] -\left[ \delta f\left( m_{H}^{BR}\left( m_{H}\right) +\beta m_{H}\right) +\lambda m_{H}^{BR}\left( m_{H}\right) \right] , \end{aligned}$$

(20)

where the LHS is the loss of being considered to have type-L due to the deviation and the RHS is the gain in best responding to the other government’s $m_{H}$. The LHS is independent of $m_{H}$. The derivative of the RHS with respect to $m_{H}$ is

$$\begin{aligned}&(1+\beta )\delta f^{\prime }\left( (1+\beta )m_{H}\right) +\lambda -\delta f^{\prime }\left( m_{H}^{BR}\left( m_{H}\right) +\beta m_{H}\right) \left( m_{H}^{BR\prime }\left( m_{H}\right) +\beta \right) -\lambda m_{H}^{BR\prime }\left( m_{H}\right) \\&=(1+\beta )\delta f^{\prime }\left( (1+\beta )m_{H}\right) +\lambda +\lambda \left( m_{H}^{BR\prime }\left( m_{H}\right) +\beta \right) -\lambda m_{H}^{BR\prime }\left( m_{H}\right) \\&=(1+\beta )\left[ \delta f^{\prime }\left( (1+\beta )m_{H}\right) +\lambda \right] \end{aligned}$$

where the second line uses $\delta f^{\prime }\left( m_{H}^{BR}\left( m_{H}\right) +\beta m_{H}\right) +\lambda =0$ by the definition of $m_{H}^{BR}\left( m_{H}\right)$. The derivative equals 0 when $m_{H} =m_{H}^{tc}$ and is positive for all $m_{H}>m_{H}^{tc}$ due to $f^{\prime \prime }>0$, implying that the RHS of (20) is increasing for $m_{H}>m_{H}^{tc}$. Also note that the RHS of (20) equals 0 when $m_{H}=m_{H}^{tc}$. Define $m^{+}$ by the following equation:

$$\begin{aligned}&\left[ \delta g\left( (1+\varphi +\gamma )a_{L}^{tc}\right) +\theta a_{L}^{tc}\right] -\left[ \delta g\left( (1+\varphi +\gamma )a_{H} ^{tc}\right) +\theta a_{H}^{tc}\right] \nonumber \\&\equiv \left[ \delta f\left( (1+\beta )m^{+}\right) +\lambda m^{+}\right] -\left[ \delta f\left( m_{H}^{BR}\left( m^{+}\right) +\beta m^{+}\right) +\lambda m_{H}^{BR}\left( m^{+}\right) \right] . \end{aligned}$$

(21)

Then for all $m_{H}\in \left( m_{H}^{tc},m^{+}\right]$, (20) is satisfied and so the type-H government has no incentive to unilaterally deviate to any $m^{\prime }<m_{H}$. $\square$

Proof

(Proposition 6) Part (i). According to Lemma 3, if such an equilibrium exists, both governments choose $m_{L}=m_{L}^{tc}$. There are two possibilities under state H for an asymmetric equilibrium: (a) For all $n\in \{1,2\}$, $m_{nH}\ne m_{L}^{tc}$ and $m_{1H}\ne m_{2H}$; and (b) there exists $n\in \{1,2\}$, such that $m_{nH}\ne m_{L}^{tc}$ while $m_{-nH}=m_{L}^{tc}$. We discuss the two cases respectively.

(a) Since $m_{1H}\ne m_{2H}$, there exists $n\in \{1,2\}$ such that $m_{nH}\ne m_{H}^{BR}\left( m_{-nH}\right)$. Then type-H government n would have incentive to unilaterally deviate to an action $m^{\prime } =m_{H}^{BR}\left( m_{-nH}\right) +\varepsilon \ne m_{L}^{tc}$, where $\varepsilon$ is equal or very close to 0, because by observing the off-equilibrium mitigation pair $\left( m^{\prime },m_{-nH}\right)$, citizens would form the belief that the state is H and a single deviation occurred, by the minimality rule. Therefore, no equilibrium in case (i), if exists, is supported by unprejudiced beliefs.

(b) If $m_{L}^{tc}\ne m_{H}^{BR}\left( m_{nH}\right)$, then this equilibrium, if exists, is not supported by unprejudiced beliefs. This is because by observing off-equilibrium mitigation pair $\left( m_{H} ^{BR}\left( m_{H}\right) ,m_{H}\right)$ citizens would form a belief that the state is H and one deviation occurred by the minimality rule, and so type-H government $-n$ has incentive to deviate to $m_{H}^{BR}\left( m_{nH}\right)$. However, if $m_{L}^{tc}=m_{H}^{BR}\left( m_{nH}\right)$, then this equilibrium, if exists, is supported by unprejudiced beliefs. This is because, first, type-H government $-n$ has no incentive to deviate, and second, if type-H government n or any type-L government deviates, the minimality rule generates a tie and so citizens will not be able to infer the state. Since $m_{L}^{tc}=m_{H}^{BR}\left( m_{nH}\right)$, we have

$$\begin{aligned} -f^{\prime }\left( m_{L}^{tc}+\beta m_{nH}\right) =\frac{\lambda }{\delta }. \end{aligned}$$

Combining this equation with (10), we have $m_{L}^{tc}+\beta m_{nH}=\left( 1+\beta \right) m_{H}^{tc}$, implying $m_{nH}=\frac{\left( 1+\beta \right) m_{H}^{tc}-m_{L}^{tc}}{\beta }=m_{H}^{tc}+\frac{m_{H} ^{tc}-m_{L}^{tc}}{\beta }>m_{H}^{tc}>m_{L}^{tc}$.

Part (ii). Under the belief system, government $-n$’s action does not affect beliefs; moreover, it chooses best response actions under both states: $m_{L}^{tc}=m_{H}^{BR}\left( m_{H}^{tc}+\frac{m_{H}^{tc}-m_{L}^{tc}}{\beta }\right)$ and $m_{L}^{tc}=m_{L}^{BR}\left( m_{L}^{tc}\right)$. So government $-n$ has no incentive to deviate. For type-L government n to have no incentive to deviate, we need

$$\begin{aligned}&f\left( \left( 1+\beta \right) m_{L}^{tc}\right) +\lambda m_{L} ^{tc}+g\left( \left( 1+\varphi +\gamma \right) a_{L}^{tc}\right) +\theta a_{L}^{tc}\le \nonumber \\&f\left( m_{H}^{tc}+\frac{m_{H}^{tc}-m_{L}^{tc}}{\beta }+\beta m_{L} ^{tc}\right) +\lambda \left( m_{H}^{tc}+\frac{m_{H}^{tc}-m_{L}^{tc}}{\beta }\right) +g\left( \left( 1+\varphi +\gamma \right) a_{H}^{tc}\right) +\theta a_{H}^{tc}, \end{aligned}$$

(22)

where the LHS is its equilibrium loss and the RHS is the loss when mimicking the type-H government n by choosing $m_{H}^{tc}+\frac{m_{H}^{tc} -m_{L}^{tc}}{\beta }$. Condition (22) can be simplified to

$$\begin{aligned}&g\left( \left( 1+\varphi +\gamma \right) a_{L}^{tc}\right) -g\left( \left( 1+\varphi +\gamma \right) a_{H}^{tc}\right) -f\left( m_{H}^{tc} +\frac{m_{H}^{tc}-m_{L}^{tc}}{\beta }+\beta m_{L}^{tc}\right) \nonumber \\&\le -f\left( \left( 1+\beta \right) m_{L}^{tc}\right) +\frac{\lambda \left( \beta +1\right) }{\beta }\left( m_{H}^{tc}-m_{L}^{tc}\right) +\theta \left( a_{H}^{tc}-a_{L}^{tc}\right) . \end{aligned}$$

(23)

For type-H government n to have no incentive to deviate, we need

$$\begin{aligned}&\delta f\left( m_{nH}+\beta m_{L}^{tc}\right) +\lambda m_{nH}+\delta g\left( \left( 1+\varphi +\gamma \right) a_{H}^{tc}\right) +\theta a_{H}^{tc}\nonumber \\&=\delta f\left( m_{H}^{tc}+\frac{m_{H}^{tc}-m_{L}^{tc}}{\beta }+\beta m_{L}^{tc}\right) +\lambda \left( m_{H}^{tc}+\frac{m_{H}^{tc}-m_{L}^{tc} }{\beta }\right) +\delta g\left( \left( 1+\varphi +\gamma \right) a_{H} ^{tc}\right) +\theta a_{H}^{tc}\nonumber \\&\le \delta f\left( m_{H}^{BR}\left( m_{L}^{tc}\right) +\beta m_{L} ^{tc}\right) +\lambda m_{H}^{BR}\left( m_{L}^{tc}\right) +\delta g\left( \left( 1+\varphi +\gamma \right) a_{L}^{tc}\right) +\theta a_{L} ^{tc}\nonumber \\&=\delta f\left( \left( 1+\beta \right) m_{H}^{tc}\right) +\lambda \left( \left( 1+\beta \right) m_{H}^{tc}-\beta m_{L}^{tc}\right) +\delta g\left( \left( 1+\varphi +\gamma \right) a_{L}^{tc}\right) +\theta a_{L} ^{tc} \end{aligned}$$

(24)

where the LHS of the inequality is its equilibrium loss, the RHS is the loss when deviating to its best response $m_{H}^{BR}\left( m_{L}^{tc}\right)$, and the last equality follows from the fact $m_{H}^{BR}\left( m\right) +\beta m=\left( 1+\beta \right) m_{H}^{tc}$ for all m due to (9) and (10). Condition (24) can be simplified as

$$\begin{aligned}&g\left( \left( 1+\varphi +\gamma \right) a_{L}^{tc}\right) -g\left( \left( 1+\varphi +\gamma \right) a_{H}^{tc}\right) -f\left( m_{H}^{tc} +\frac{m_{H}^{tc}-m_{L}^{tc}}{\beta }+\beta m_{L}^{tc}\right) \nonumber \\&\ge -f\left( \left( 1+\beta \right) m_{H}^{tc}\right) +\frac{\lambda \left( 1-\beta ^{2}\right) }{\delta \beta }\left( m_{H}^{tc} -m_{L}^{tc}\right) +\frac{\theta }{\delta }\left( a_{H}^{tc}-a_{L} ^{tc}\right) . \end{aligned}$$

(25)

Combining (23) and (25) and after some rearrangement, we have

$$\begin{aligned}&-f\left( \left( 1+\beta \right) m_{H}^{tc}\right) +\frac{\lambda \left( 1-\beta ^{2}\right) }{\delta \beta }\left( m_{H}^{tc}-m_{L}^{tc}\right) +\frac{\theta }{\delta }\left( a_{H}^{tc}-a_{L}^{tc}\right) \nonumber \\&\le g\left( \left( 1+\varphi +\gamma \right) a_{L}^{tc}\right) -g\left( \left( 1+\varphi +\gamma \right) a_{H}^{tc}\right) -f\left( \frac{\left( \beta +1\right) \left( m_{H}^{tc}+\left( \beta -1\right) m_{L}^{tc}\right) }{\beta }\right) \nonumber \\&\le -f\left( \left( 1+\beta \right) m_{L}^{tc}\right) +\frac{\lambda \left( 1+\beta \right) }{\beta }\left( m_{H}^{tc}-m_{L}^{tc}\right) +\theta \left( a_{H}^{tc}-a_{L}^{tc}\right) . \end{aligned}$$

(26)

$\square$

Proof

(Corollary 3) Given the functional forms and by symmetry, we have the following first order conditions.

$$\begin{aligned} -[m_{0}-(1+\beta )m_{L}^{tc}]+\lambda&=0~~~~~~\Leftrightarrow ~~~~~~m_{L} ^{tc}=\frac{m_{0}-\lambda }{1+\beta },\\ -\delta [m_{0}-(1+\beta )m_{H}^{tc}]+\lambda&=0~~~~~~\Leftrightarrow ~~~~~~m_{H}^{tc}=\frac{m_{0}-\frac{\lambda }{\delta }}{1+\beta },\\ -\alpha [a_{0}-(1+\varphi +\gamma )a_{L}^{tc}]+\theta&=0~~~~~~\Leftrightarrow ~~~~~~a_{L}^{tc}=\frac{a_{0}-\frac{\theta }{\alpha } }{1+\varphi +\gamma },\\ -\delta \alpha [a_{0}-(1+\varphi +\gamma )a_{H}^{tc}]+\theta&=0~~~~~~\Leftrightarrow ~~~~~~a_{H}^{tc}=\frac{a_{0}-\frac{\theta }{\alpha \delta }}{1+\varphi +\gamma }. \end{aligned}$$

Equation (13) can be written as

$$\begin{aligned} f\left( (1+\beta )m_{L}^{tc}\right) -f(m_{H}^{tc}+\beta m_{L}^{tc})+g\left( (1+\varphi +\gamma )a_{L}^{tc}\right) -g\left( (1+\varphi +\gamma )a_{H} ^{tc}\right) >\lambda \left( m_{H}^{tc}-m_{L}^{tc}\right) +\theta \left( a_{H}^{tc}-a_{L}^{tc}\right) . \end{aligned}$$

(27)

Using the quadratic functional forms and the solutions above, Equation (27) becomes

$$\begin{aligned}&\frac{1}{2}\left[ m_{0}-(1+\beta )\frac{m_{0}-\lambda }{1+\beta }\right] ^{2}-\frac{1}{2}\left[ m_{0}-\frac{m_{0}-\frac{\lambda }{\delta }}{1+\beta }-\beta \frac{m_{0}-\lambda }{1+\beta }\right] ^{2}+\frac{\alpha }{2}\left[ a_{0}-\left( a_{0}-\frac{\theta }{\alpha }\right) \right] ^{2}\\&-\frac{\alpha }{2}\left[ a_{0}-\left( a_{0}-\frac{\theta }{\alpha \delta }\right) \right] ^{2}>\lambda \left( \frac{m_{0}-\frac{\lambda }{\delta } }{1+\beta }-\frac{m_{0}-\lambda }{1+\beta }\right) +\theta \left( \frac{a_{0}-\frac{\theta }{\alpha \delta }}{1+\varphi +\gamma }-\frac{a_{0}-\frac{\theta }{\alpha }}{1+\varphi +\gamma }\right) \\&\Leftrightarrow \frac{\lambda ^{2}}{2}-\frac{1}{2}\left[ \frac{\lambda }{(1+\beta )\delta }+\frac{\beta \lambda }{1+\beta }\right] ^{2}+\frac{\alpha }{2}\left( \frac{\theta }{\alpha }\right) ^{2}-\frac{\alpha }{2}\left( \frac{\theta }{\alpha \delta }\right) ^{2}>\lambda \left( \frac{\lambda -\frac{\lambda }{\delta }}{1+\beta }\right) +\frac{\theta \left( \frac{\theta }{\alpha }-\frac{\theta }{\alpha \delta }\right) }{1+\varphi +\gamma }\\&\Leftrightarrow \frac{\lambda ^{2}}{2}\left[ 1-\left( \frac{1+\beta \delta }{(1+\beta )\delta }\right) ^{2}\right] +\frac{\theta ^{2}}{2\alpha }\left( 1-\frac{1}{\delta ^{2}}\right)>\frac{\lambda ^{2}}{1+\beta }\left( 1-\frac{1}{\delta }\right) +\frac{\theta ^{2}}{(1+\varphi +\gamma )\alpha }\left( 1-\frac{1}{\delta }\right) \\&\Leftrightarrow \frac{\lambda ^{2}}{2}\frac{(\delta -1)(2\beta \delta +\delta +1)}{[(1+\beta )\delta ]^{2}}+\frac{\theta ^{2}}{2\alpha }\frac{(\delta -1)(\delta +1)}{\delta ^{2}}>\frac{\delta -1}{\delta }\left[ \frac{\lambda ^{2}}{1+\beta }+\frac{\theta ^{2}}{(1+\varphi +\gamma )\alpha }\right] \\&\Leftrightarrow \frac{\lambda ^{2}(2\beta \delta +\delta +1)}{(1+\beta )^{2}\delta }+\frac{\theta ^{2}(\delta +1)}{\alpha \delta }>\frac{2\lambda ^{2} }{1+\beta }+\frac{2\theta ^{2}}{(1+\varphi +\gamma )\alpha }\\&\Leftrightarrow \frac{\lambda ^{2}}{1+\beta }\left[ \frac{2\beta \delta +\delta +1}{(1+\beta )\delta }-\frac{2(1+\beta )\delta }{(1+\beta )\delta }\right]>\frac{\theta ^{2}}{\alpha }\left( \frac{2}{1+\varphi +\gamma }-\frac{\delta +1}{\delta }\right) \\&\Leftrightarrow -\frac{\lambda ^{2}(\delta -1)}{(1+\beta )^{2}\delta } >-\frac{\theta ^{2}}{\alpha }\left( \frac{\delta +1}{\delta }-\frac{2}{1+\varphi +\gamma }\right) \\&\Leftrightarrow \frac{\lambda ^{2}}{(1+\beta )^{2}}\left( \frac{\alpha }{\theta ^{2}}\right)<\frac{\delta +1}{\delta -1}-\frac{\delta }{\delta -1}\left( \frac{2}{1+\varphi +\gamma }\right) \\&\Leftrightarrow \alpha \left( \frac{\lambda }{\theta }\right) ^{2} <(1+\beta )^{2}\left[ 1+\frac{2}{\delta -1}\left( 1-\frac{\delta }{1+\varphi +\gamma }\right) \right] , \end{aligned}$$

which is (16).

• (i) Using Proposition 5, Part (i) is proved.

• (ii) Proposition 6 shows that if (26) is satisfied, there exists such an equilibrium. Condition (26) is derived by combining Eqs. (24) and (22).

Equation (24) can be rewritten as

$$\begin{aligned}&\delta f\left( \left( 1+\frac{1}{\beta }\right) m_{H}^{tc}+\left( \beta -\frac{1}{\beta }\right) m_{L}^{tc}\right) -\delta f\left( (1+\beta )m_{H}^{tc}\right) +\lambda \left( \frac{1}{\beta }-\beta \right) (m_{H}^{tc}-m_{L}^{tc})\\&\le \delta g\left( (1+\varphi +\gamma )a_{L}^{tc}\right) -\delta g\left( (1+\varphi +\gamma )a_{H}^{tc}\right) +\theta (a_{L}^{tc}-a_{H}^{tc}). \end{aligned}$$

Plugging the solutions at the beginning of this proof to this inequality, we have the following:

$$\begin{aligned}&\frac{\delta }{2}\left[ m_{0}-\left( 1+\frac{1}{\beta }\right) \frac{m_{0}-\frac{\lambda }{\delta }}{1+\beta }-\left( \beta -\frac{1}{\beta }\right) \frac{m_{0}-\lambda }{1+\beta }\right] ^{2}-\frac{\delta }{2}\left[ m_{0}-\left( m_{0}-\frac{\lambda }{\delta }\right) \right] ^{2}\\&+\lambda \left( \frac{1-\beta ^{2}}{\beta }\right) \left( \frac{m_{0} -\frac{\lambda }{\delta }}{1+\beta }-\frac{m_{0}-\lambda }{1+\beta }\right) \\&\le \frac{\alpha \delta }{2}\left[ a_{0}-\left( a_{0}-\frac{\theta }{\alpha }\right) \right] ^{2}-\frac{\alpha \delta }{2}\left[ a_{0}-\left( a_{0}-\frac{\theta }{\alpha \delta }\right) \right] ^{2}+\theta \left( \frac{a_{0}-\frac{\theta }{\alpha }}{1+\varphi +\gamma }-\frac{a_{0}-\frac{\theta }{\alpha \delta }}{1+\varphi +\gamma }\right) \\&\Rightarrow \frac{\delta }{2}\left[ m_{0}-\frac{1}{\beta }\left( m_{0} -\frac{\lambda }{\delta }\right) -\frac{\beta -1}{\beta }(m_{0}-\lambda )\right] ^{2}-\frac{\delta }{2}\left( \frac{\lambda }{\delta }\right) ^{2} +\lambda \left( \frac{1-\beta }{\beta }\right) \left( \lambda -\frac{\lambda }{\delta }\right) \\&\le \frac{\alpha \delta }{2}\left( \frac{\theta }{\alpha }\right) ^{2} -\frac{\alpha \delta }{2}\left( \frac{\theta }{\alpha \delta }\right) ^{2} +\frac{\theta }{1+\varphi +\gamma }\left( \frac{\theta }{\alpha \delta } -\frac{\theta }{\alpha }\right) \\&\Rightarrow \frac{\delta }{2}\left\{ \left[ \frac{\lambda }{\beta \delta }+\frac{\lambda (\beta -1)}{\beta }\right] ^{2}-\left( \frac{\lambda }{\delta }\right) ^{2}\right\} +\frac{\lambda ^{2}(1-\beta )}{\beta }\left( 1-\frac{1}{\delta }\right) \\&\le \frac{\delta \theta ^{2}}{2\alpha }\left( 1-\frac{1}{\delta ^{2}}\right) -\frac{\theta ^{2}}{(1+\varphi +\gamma )\alpha }\left( 1-\frac{1}{\delta }\right) \\&\Rightarrow \frac{\lambda ^{2}}{2\delta }\left[ \frac{1-(1-\beta )\delta }{\beta }+1\right] \left[ \frac{1-(1-\beta )\delta }{\beta }-1\right] +\frac{\beta (1-\beta )\lambda ^{2}}{\beta ^{2}}\left( 1-\frac{1}{\delta }\right) \\&\le \frac{\theta ^{2}}{\alpha }\left( 1-\frac{1}{\delta }\right) \left[ \frac{\delta }{2}\left( 1+\frac{1}{\delta }\right) -\frac{1}{1+\varphi +\gamma }\right] \\&\Rightarrow \frac{\lambda ^{2}}{\delta }\cdot \frac{1+\beta -(1-\beta )\delta }{\beta }\cdot \frac{(1-\beta )(1-\delta )}{\beta }+\frac{2\beta (1-\beta )\lambda ^{2}}{\beta ^{2}}\left( 1-\frac{1}{\delta }\right) \\&\le \frac{2\theta ^{2}}{\alpha }\left( \frac{\delta +1}{2}-\frac{1}{1+\varphi +\gamma }\right) \left( 1-\frac{1}{\delta }\right) \\&\Rightarrow \frac{\lambda ^{2}(1-\beta )}{\beta ^{2}}\left( 1-\frac{1}{\delta }\right) [2\beta -(1+\beta -(1-\beta )\delta )]\le \frac{\theta ^{2}}{\alpha }\left( \delta +1-\frac{2}{1+\varphi +\gamma }\right) \left( 1-\frac{1}{\delta }\right) \\&\Rightarrow \frac{\lambda ^{2}(1-\beta )^{2}(\delta -1)}{\beta ^{2}}\le \frac{\theta ^{2}}{\alpha }\left[ \delta -1+2\left( 1-\frac{1}{1+\varphi +\gamma }\right) \right] \\&\Rightarrow \alpha \left( \frac{\lambda }{\theta }\right) ^{2}\le \left( \frac{\beta }{1-\beta }\right) ^{2}\left[ 1+\frac{2}{\delta -1}\left( 1-\frac{1}{1+\varphi +\gamma }\right) \right] , \end{aligned}$$

where the last line uses $0<\beta <1$.

Condition (22) can be rewritten as

$$\begin{aligned}&f\left( (1+\beta )m_{L}^{tc}\right) -f\left( \frac{1+\beta }{\beta } m_{H}^{tc}-\frac{(1-\beta )(1+\beta )}{\beta }m_{L}^{tc}\right) -\lambda \left( \frac{1+\beta }{\beta }\right) (m_{H}^{tc}-m_{L}^{tc})\\&\le g\left( (1+\varphi +\gamma )a_{H}^{tc}\right) -g\left( (1+\varphi +\gamma )a_{L}^{tc}\right) +\theta (a_{H}^{tc}-a_{L}^{tc}). \end{aligned}$$

Plugging the solutions at the beginning of this proof to this inequality, we have the following:

$$\begin{aligned}&\frac{\left[ m_{0}-(m_{0}-\lambda )\right] ^{2}}{2}-\frac{1}{2}\left\{ m_{0}-\left[ \frac{m_{0}-\frac{\lambda }{\delta }}{\beta }-\frac{1-\beta }{\beta }(m_{0}-\lambda )\right] \right\} ^{2}-\frac{\lambda (1+\beta )}{\beta }\left( \frac{m_{0}-\frac{\lambda }{\delta }}{1+\beta }-\frac{m_{0}-\lambda }{1+\beta }\right) \\&\le \frac{\alpha }{2}\left[ a_{0}-\left( a_{0}-\frac{\theta }{\alpha \delta }\right) \right] ^{2}-\frac{\alpha }{2}\left[ a_{0}-\left( a_{0} -\frac{\theta }{\alpha }\right) \right] ^{2}+\theta \left[ \frac{a_{0} -\frac{\theta }{\alpha \delta }}{1+\varphi +\gamma }-\frac{a_{0}-\frac{\theta }{\alpha }}{1+\varphi +\gamma }\right] \\&\Rightarrow \frac{\lambda ^{2}}{2}-\frac{1}{2}\left[ \frac{\lambda }{\beta \delta }-\frac{\lambda (1-\beta )}{\beta }\right] ^{2}-\frac{\lambda }{\beta }\left( \lambda -\frac{\lambda }{\delta }\right) \le \frac{\alpha }{2}\left[ \left( \frac{\theta }{\alpha \delta }\right) ^{2}-\left( \frac{\theta }{\alpha }\right) ^{2}\right] +\theta \left( \frac{\frac{\theta }{\alpha }-\frac{\theta }{\alpha \delta }}{1+\varphi +\gamma }\right) \\&\Rightarrow \lambda ^{2}-\lambda ^{2}\left( \frac{1-\delta +\beta \delta }{\beta \delta }\right) ^{2}-\frac{2\lambda ^{2}}{\beta }\left( 1-\frac{1}{\delta }\right) \le \frac{\theta ^{2}}{\alpha }\left( \frac{1}{\delta ^{2} }-1\right) +\frac{\theta ^{2}}{\alpha }\left( 1-\frac{1}{\delta }\right) \frac{2}{1+\varphi +\gamma }\\&\Rightarrow \lambda ^{2}\left[ \left( 1-\frac{1}{\delta }\right) \frac{2\beta \delta +1-\delta }{\beta ^{2}\delta }-\left( 1-\frac{1}{\delta }\right) \frac{2}{\beta }\right] \le \frac{\theta ^{2}}{\alpha }\left( 1-\frac{1}{\delta }\right) \left[ \frac{2}{1+\varphi +\gamma }-\left( \frac{1}{\delta }+1\right) \right] \\&\Rightarrow \alpha \left( \frac{\lambda }{\theta }\right) ^{2}\frac{1-\delta }{\beta ^{2}\delta }\le \frac{2}{1+\varphi +\gamma }-\frac{\delta +1}{\delta }\\&\Rightarrow \alpha \left( \frac{\lambda }{\theta }\right) ^{2}\ge \beta ^{2}\left[ 1+\frac{2}{\delta -1}\left( 1-\frac{\delta }{1+\varphi +\gamma }\right) \right] , \end{aligned}$$

where the last inequality uses $\delta >1$. Combining the above analysis, we derive (17). $\square$