Terrorism deterrence in a two country framework: strategic interactions between R&D, defense and pre-emption

Abstract

In this paper, we analyze the equilibrium responses (in terms of defense, R&D and preemption) to a potential terrorist attack in a two-country framework (Home and Foreign) using a multi-stage game with imperfect information. We highlight three different types of strategic interactions: (a) how the choice of defense, R&D and pre-emption affects the choice of the same in the other country (strategic interdependence across countries); (b) the strategic interaction between the instruments of terrorism deterrence, namely, defense, R&D and pre-emption in a given country and (c) the strategic interaction between the terrorist and the defender. Our main results are as follows: (i) defense effort in Home is a strategic complement to the defense effort in Foreign. (ii) Even without R&D sharing between countries, we find that R&D effort in one country is a strategic substitute to that in the other; (iii) similar results hold for pre-emption because of its public good nature; (iv) for a given country, defense and R&D efforts may be strategic substitutes or complements depending on the magnitude of the ratio of weighted expected damage between Foreign and Home; (v) R&D and pre-emption may be strategic substitutes or complements depending on the magnitude of the elasticity of damage and (vi) an increase in the likelihood of the terrorist being weak reduces defense effort, may increase or decrease R&D depending on the magnitude of elasticity of damage but increases pre-emptive effort in both countries.

Appendices

Appendix A: Terrorist effort

Lemma 3:

• (a) $\frac{\partial t_{H}^{*}}{\partial d_{H}}<0$

  $$\begin{aligned} V_{T} =&\sum_{i={H,F}} p_{i}b_{i} \theta-\underline {c}(1+\alpha_{i}r_{i})t_{i} \end{aligned}$$

  (6)
  $$\begin{aligned} \frac{\partial V_{T}}{\partial t_{H}} =&p_{H}kt_{H}^{k-1}e^{m} \theta -\underline{c}(1+\alpha_{H}r_{H})=0 \end{aligned}$$

  (7)
  $$\begin{aligned} t_{H}^{*} =& \biggl[\frac{\underline{c}(1+\alpha _{H}r_{H})}{p_{H}e^{m}\theta k} \biggr]^{\frac{1}{k-1}} \end{aligned}$$

  (8)

  or

  $$\begin{aligned} t_{H}^{*} =& \biggl[\frac{\underline{c}(1+\alpha _{H}r_{H})(d_{H}+d_{F})}{d_{F}e^{m}\theta k} \biggr]^{\frac {1}{k-1}} \end{aligned}$$

  (9)
  $$\begin{aligned} =& \biggl[\frac{\underline{c}}{\theta k}(1+\alpha_{H}r_{H}) \frac {(d_{H}+d_{F})}{d_{F}}e^{-m} \biggr]^{\frac{1}{k-1}} \end{aligned}$$

  (10)
  $$\begin{aligned} =&\overline{\gamma_{H}}h_{1H}(\cdot)h_{2H}( \cdot)h_{3H}(\cdot) \end{aligned}$$

  (11)

  Where,

  $$\begin{aligned} \overline{\gamma_{H}} =& \biggl(\frac{\underline{c}}{\theta k} \biggr)^{\frac {1}{k-1}} \end{aligned}$$

  (12)
  $$\begin{aligned} h_{1H} =&(1+\alpha_{H}r_{H})^{\frac{1}{k-1}} \end{aligned}$$

  (13)
  $$\begin{aligned} h_{2H} =&\biggl(\frac{d_{H}+d_{F}}{d_{F}}\biggr)^{\frac{1}{k-1}} \end{aligned}$$

  (14)
  $$\begin{aligned} h_{3H} =&e^{\frac{-m}{k-1}} \\ =&\overline{\gamma_{H}}H_{H} \end{aligned}$$

  (15)

  Where,

  $$ H_{H}=h_{1H}h_{2H}h_{3H} $$

  (16)

  Differentiating (15) with respect to d _H , we get

  $$\begin{aligned} \frac{\partial t_{H}^{*}}{\partial d_{H}} =&\overline{\gamma_{H}}\frac {\partial H_{H}}{\partial d_{H}} \end{aligned}$$

  (17)
  $$\begin{aligned} \ln H_{H} =&\ln h_{1H}+\ln h_{2H}+\ln h_{3H} \end{aligned}$$

  (18)
  $$\begin{aligned} \frac{1}{H_{H}}\frac{\partial H_{H}}{\partial d_{H}} =&\frac {1}{(k-1)}\frac{d_{F}}{(d_{H}+d_{F})} \frac{1}{d_{F}}=\frac{1}{k-1}\frac {1}{(d_{H}+d_{F})} \end{aligned}$$

  (19)

  Or,

  $$ \frac{\partial H_{H}}{\partial d_{H}}=\frac{H_{H}}{k-1}\frac {1}{(d_{H}+d_{F})} $$

  (20)

  Or,

  $$ \frac{\partial t_{H}^{*}}{\partial d_{H}}=\frac{-\overline{\gamma _{H}}H_{H}}{(1-k)}\frac{1}{(d_{H}+d_{F})}=\frac{-t_{H}^{*}}{(1-k)} \frac {1}{(d_{H}+d_{F})}<0 $$

  (21)

• (b) $\frac{\partial t_{H}^{*}}{\partial d_{F}}>0$. Similarly,

  $$ \frac{\partial t_{H}^{*}}{\partial d_{F}}=\frac {t_{H}^{*}}{(1-k)}\frac{d_{H}}{(d_{H}+d_{F})d_{F}}>0, $$

  (22)

• (c) $\frac{\partial t_{H}^{*}}{\partial r_{H}}<0$. In the same vein,

  $$ \frac{\partial t_{H}^{*}}{\partial r_{H}}=-\frac{t_{H}^{*}}{(1-k)}\frac{\alpha_{H}}{(1+\alpha _{H}r_{H})}<0, $$

  (23)

• (d) $\frac{\partial t_{H}^{*}}{\partial e_{H}}<0$. As before,

  $$ \frac{\partial t_{H}^{*}}{\partial e_{H}}=\frac {mt_{H}^{*}}{e(1-k)}<0, $$

  (24)

Appendix B: Defense effort

Lemma 5:

• (a) $d_{H}^{*}=\frac{1}{p_{H2}X_{H}}$.

  Differentiating (2) with respect to d _H , we get

  $$ \frac{\partial V_{H}}{\partial d_{H}}=-\frac {d_{F}e^{m}}{(d_{H}+d_{F})^{2}}N_{H}a_{H}+ \frac {d_{F}e^{m}}{(d_{H}+d_{F})}\frac{\partial N_{H}}{\partial d_{H}}a_{H}+\gamma_{H}=0 $$

  (25)

  Differentiating (3) with respect to d _H , we get

  

  (26)
  $$\begin{aligned} \frac{\partial N_{H}}{\partial d_{H}} =&-\frac{k}{1-k}\frac {1}{(d_{H}+d_{F})}N_{H} \end{aligned}$$

  (27)

  Plugging (27) in (25), we get,

  $$ \frac{\partial V_{H}}{\partial d_{H}}=\frac {-d_{F}e^{m}N_{H}a_{H}}{(d_{H}+d_{F})^{2}}-\frac {d_{F}e^{m}a_{H}}{(d_{H}+d_{F})}\frac{kN_{H}}{(1-k)(d_{H}+d_{F})}- \gamma _{H}=0 $$

  (28)

  Or,

  $$ \frac{d_{F}e^{m}N_{H}a_{H}}{(d_{H}+d_{F})^{2}(1-k)}=\gamma _{H} $$

  (29)

  Or,

  $$ (d_{H}+d_{F})^{2}=\frac{1}{(1-k)\gamma _{H}}d_{F}e^{m}N_{H}a_{H} $$

  (30)

  Or,

  $$ (d_{H}+d_{F})=\bigl(B_{H}d_{F}e^{m}N_{H}a_{H} \bigr)^{\frac{1}{2}}, $$

  (31)

  where,

  $$ B_{H}=\frac{1}{(1-k)\gamma_{H}} $$

  (32)

  Or,

  $$ d_{H}^{*}=\bigl(B_{H}d_{F}^{*}e^{m}N_{H}a_{H} \bigr)^{1/2}-d_{F}^{*} $$

  (33)

  Similarly, for country F we have,

  $$ d_{F}^{*}=\bigl(B_{F}d_{H}^{*}e^{m}N_{F}a_{F} \bigr)^{1/2}-d_{H}^{*} $$

  (34)

  Plugging (34) in (33) we get,

  $$ d_{H}^{*}=\bigl(B_{H}e^{m}N_{H}a_{H} \bigr)^{1/2}\bigl(\bigl(B_{F}d_{H}^{*}e^{m}N_{F}a_{F} \bigr)^{1/2}-d_{H}^{*}\bigr)^{1/2}-\bigl( \bigl(B_{F}d_{H}^{*}e^{m}N_{F}a_{F} \bigr)^{1/2}-d_{H}^{*}\bigr) $$

  (35)

  Or,

  $$ \bigl(B_{F}d_{H}^{*}e^{m}N_{F}a_{F} \bigr)^{1/2}=\bigl(B_{H}e^{m}N_{H}a_{H} \bigr)^{1/2}\bigl(\bigl(B_{F}d_{H}^{*}e^{m}N_{F}a_{F} \bigr)^{1/2}-d_{H}^{*}\bigr)^{1/2} $$

  (36)

  Squaring both sides we get,

  $$ B_{F}d_{H}^{*}e^{m}N_{F}a_{F}= \bigl(B_{H}e^{m}N_{H}a_{H}\bigr) \bigl( \bigl(B_{F}d_{H}^{*}e^{m}N_{F}a_{F} \bigr)^{1/2}-d_{H}^{*}\bigr) $$

  (37)

  Or,

  $$ \bigl(B_{F}e^{m}N_{F}a_{F}+B_{H}e^{m}N_{H}a_{H} \bigr)d_{H}^{*}=B_{H}e^{m}N_{H}a_{H} \bigl(B_{F}d_{H}^{*}e^{m}N_{F}a_{F} \bigr)^{1/2} $$

  (38)

  Or,

  $$ \bigl(d_{H}^{*}\bigr)^{1/2}=\frac{(B_{F}e^{m}N_{F}a_{F})^{1/2}}{1+\frac {B_{F}e^{m}N_{F}a_{F}}{B_{H}e^{m}N_{H}a_{H}}}= \frac {(B_{F}e^{m}N_{F}a_{F})^{1/2}}{1+\frac{1}{T}\frac{N_{F}}{N_{H}}} $$

  (39)

  Where

  $$\begin{aligned} T =&\frac{B_{H}a_{H}}{B_{F}a_{F}} \end{aligned}$$

  (40)
  $$\begin{aligned} d_{H}^{*} =&\frac{(B_{F}e^{m}N_{F}a_{F})}{(1+\frac {N_{F}}{TN_{H}})^{2}}=\frac{(B_{F}e^{m}N_{F}a_{F})}{(1+\tau _{H})^{2}}= \frac{1}{p_{H2}X_{H}} \end{aligned}$$

  (41)

  Where,

  $$\begin{aligned} p_{H2} =&\frac{1}{(B_{F}e^{m}N_{F}a_{F})} \end{aligned}$$

  (42)
  $$\begin{aligned} X_{H} =&(1+\tau_{H})^{2} \end{aligned}$$

  (43)
  $$\begin{aligned} \tau_{H} =&\frac{N_{F}}{TN_{H}} \end{aligned}$$

  (44)

  Similarly,

  $$ d_{F}^{*}=\frac{B_{H}e^{m}N_{H}a_{H}}{X_{F}} $$

  (45)

  Where

  $$ X_{F}=\biggl(\frac{1+\tau_{H}}{\tau_{H}}\biggr)^{2} $$

  (46)

• (b) $\frac{\partial d_{H}^{*}}{\partial d_{F}^{*}}>0$.

  Differentiating (33) with respect to d _F , we get

  $$ \frac{\partial d_{H}^{*}}{\partial d_{F}^{*}}=\frac {1}{2}\bigl(B_{H}e^{m}N_{H}a_{H} \bigr)^{1/2}\bigl(d_{F}^{*}\bigr)^{-1/2}+ \bigl(B_{H}e^{m}a_{H}d_{F}^{*} \bigr)^{1/2}\frac {1}{2}N_{H}^{-1/2} \frac{\partial N_{H}}{\partial d_{F}^{*}}-1 $$

  (47)

  Differentiating (3) with respect to d _F , we get

  $$ \frac{\partial N_{H}}{\partial d_{F}^{*}}=\frac{kN_{H}}{(1-k)}\frac {d_{H}^{*}}{(d_{H}^{*}+d_{F}^{*})d_{F}^{*}} $$

  (48)

  Plugging (31), (33) and (34) into (48), we get

  $$ =\frac{kN_{H}}{(1-k)\tau _{H}(B_{H}e^{m}N_{H}a_{H}d_{F}^{*})^{1/2}} $$

  (49)

  Plugging (49) in (47), we get

  $$\begin{aligned} \frac{\partial d_{H}^{*}}{\partial d_{F}^{*}} =&\frac {1}{2}\bigl(B_{H}e^{m}N_{H}a_{H} \bigr)^{1/2}\bigl(d_{F}^{*}\bigr)^{-1/2}+ \frac{1}{2}\frac {(B_{H}e^{m}a_{H}d_{F}^{*}N_{H})^{1/2}}{(B_{H}e^{m}a_{H}d_{F}^{*}N_{H})^{1/2}}\frac {k}{1-k}\frac{1}{\tau_{H}}-1 \\ =&\frac{1}{2}\frac {(B_{H}e^{m}N_{H}a_{H})^{1/2}X_{F}^{1/2}}{(B_{H}e^{m}N_{H}a_{H})^{1/2}}+\frac {1}{2}\frac{k}{1-k} \frac{1}{\tau_{H}}-1 \\ =&\frac{1}{2}\frac{(1+\tau_{H})}{\tau_{H}}+\frac{1}{2}\frac{k}{1-k} \frac {1}{\tau_{H}}-1 \\ =&\frac{(1-k)(1+\tau_{H})+k-2(1-k)\tau_{H}}{2(1-k)\tau_{H}} \\ =&\frac{1-(1-k)\tau_{H}}{2(1-k)\tau_{H}} \end{aligned}$$

  (50)

  Therefore, after simplification we get, $\frac{\partial d_{H}^{*}}{\partial d_{F}^{*}}>0$ if $\tau_{H}< \frac {1}{1-k}$,

  =0:

  if $\tau_{H}=\frac{1}{1-k}$,

  <0:

  if $\tau_{H}>\frac{1}{1-k}$.

  Similarly,

  $$\frac{\partial d_{F}^{*}}{\partial d_{H}^{*}}>0 \quad\mbox{when } \tau _{H}>1-k $$

  Similarly, $\frac{\partial d_{F}^{*}}{\partial d_{H}^{*}}=0$ when τ _H =1−k and $\frac{\partial d_{F}^{*}}{\partial d_{H}^{*}}<0$ when τ _H <1−k.

  Therefore, the reaction functions are stable iff:

  $$\tau_{H}\in\biggl(1-k,\frac{1}{1-k}\biggr) $$

• (c) $\frac{dd_{H}^{*}}{dr_{H}}<0$ for τ _H ∈(τ ₀,τ ₁) and ≥0 elsewhere.

  Differentiating (41) with respect to r _H , we get

  $$ \frac{\partial d_{H}^{*}}{\partial r_{H}}=\frac {B_{F}e^{m}a_{F}}{X_{H}}\frac{\partial N_{F}}{\partial r_{H}}-\frac {B_{F}e^{m}N_{F}a_{F}}{X_{H}^{2}} \frac{\partial X_{H}}{\partial r_{H}} $$

  (51)

  Note that,

  

  (52)

  Since $\frac{\partial\tilde{t}_{F}^{*}}{\partial r_{H}}=\frac {\partial\tilde{t}_{F}^{*}}{\partial r_{H}}=0$.

  Differentiating (3) with respect to r _H , we get

  

  (53)

  Similarly, for country F we have,

  

  (54)

  Using Lemma 3(d) in (53), we get

  

  (55)
  $$\begin{aligned} =& -N_{H}\frac{k}{(1-k)}\frac{\alpha_{H}}{(1+\alpha_{H}r_{H})} \end{aligned}$$

  (56)

  Differentiating (43) with respect to r _H , we get

  $$\begin{aligned} \frac{\partial X_{H}}{\partial r_{H}} =&2(1+\tau_{H})\frac{\partial\tau _{H}}{\partial r_{H}} \\ =&2(1+\tau_{H}) \biggl[\frac{1}{TN_{H}}\frac{\partial N_{F}}{\partial r_{H}}- \frac{1}{T}\frac{N_{F}}{N_{H}^{2}}\frac{\partial N_{H}}{\partial r_{H}} \biggr] \end{aligned}$$

  (57)

  Using (52) and plugging (56) in (57) we get,

  $$\begin{aligned} =&\frac{2(1+\tau_{H})N_{F}}{TN_{H}}\frac{k}{1-k}\frac{\alpha _{H}}{1+\alpha_{H}r_{H}} \end{aligned}$$

  (58)
  $$\begin{aligned} =&\frac{2(1+\tau_{H})\tau_{H}k\alpha_{H}}{(1-k)(1+\alpha _{H}r_{H})} \end{aligned}$$

  (59)

  Using (52) and plugging (59) in (51) we get,

  $$\begin{aligned} \frac{\partial d_{H}^{*}}{\partial r_{H}} =&-\frac {B_{F}e^{m}N_{F}a_{F}}{(1+\tau_{H})^{4}}\frac{2(1+\tau_{H})k\alpha _{H}\tau_{H}}{(1-k)(1+\alpha_{H}r_{H})} \\ =&-2\frac{B_{F}e^{m}N_{F}a_{F}\tau_{H}k\alpha_{H}}{(1+\tau _{H})^{3}(1-k)(1+\alpha_{H}r_{H})}<0 \end{aligned}$$

  (60)
  $$\begin{aligned} \frac{\partial d_{F}^{*}}{\partial r_{H}} =&\frac {B_{H}e^{m}a_{H}}{X_{F}}\frac{\partial N_{H}}{\partial r_{H}}-\frac {B_{H}e^{m}N_{H}a_{H}}{X_{F}^{2}} \frac{\partial X_{F}}{\partial r_{H}} \end{aligned}$$

  (61)
  $$\begin{aligned} \frac{\partial X_{F}}{\partial r_{H}} =&2\frac{(1+\tau_{H})}{\tau _{H}} \biggl[\frac{1}{\tau_{H}}- \frac{(1+\tau_{H})}{\tau_{H}^{2}} \biggr]\frac{\partial\tau_{H}}{\partial r_{H}} \\ =&-2\frac{(1+\tau_{H})}{\tau_{H}}\frac{1}{\tau_{H}^{2}} \biggl[\tau _{H} \frac{k}{1-k}\frac{\alpha_{H}}{(1+\alpha_{H}r_{H})} \biggr] \end{aligned}$$

  (62)

  Plugging (62) in (61), we get

  $$\begin{aligned} \frac{\partial d_{F}^{*}}{\partial r_{H}} =&-\tau_{H}^{2}\frac {B_{H}e^{m}a_{H}}{(1+\tau_{H})^{2}} \frac{kN_{H}}{1-k}\frac{\alpha _{H}}{1+\alpha_{H}r_{H}}+\frac{B_{H}e^{m}N_{H}a_{H}\tau_{H}^{4}}{(1+\tau _{H})^{4}}\frac{2(1+\tau_{H})}{\tau_{H}^{2}} \frac{k}{1-k}\frac{\alpha _{H}}{1+\alpha_{H}r_{H}} \\ =&\frac{-\tau_{H}^{2}B_{H}N_{H}e^{m}a_{H}k\alpha_{H}}{(1-k)(1+\alpha _{H}r_{H})(1+\tau_{H})^{2}} \biggl[1-\frac{2}{1+\tau_{H}} \biggr] \end{aligned}$$

  (63)

  Therefore, after some simplification we get,

  $$\frac{\partial d_{F}^{*}}{\partial r_{H}}>0 \quad\mbox{when }\tau _{H}<1 $$

  Similarly, $\frac{\partial d_{F}^{*}}{\partial r_{H}}=0$ when τ _H =1 and $\frac{\partial d_{F}^{*}}{\partial r_{H}}<0$ when τ _H >1.

  Now consider,

  $$ \frac{dd_{H}^{*}}{dr_{H}}=\frac{\partial d_{H}^{*}}{\partial r_{H}}+\frac{\partial d_{H}^{*}}{\partial d_{F}^{*}}\frac{\partial d_{F}^{*}}{\partial r_{H}} $$

  (64)

  Plugging in (50), (60) and (63) in (64), we get

  $$\begin{aligned} =&\frac{-2B_{F}e^{m}N_{F}a_{F}\tau_{H}k\alpha_{H}}{(1-k)(1+\tau _{H})^{3}(1+\alpha_{H}r_{H})} \\ &{}-\frac{1-(1-k)\tau_{H}}{2(1-k)\tau _{H}}\frac{B_{H}e^{m}N_{H}a_{H}k\alpha_{H}\tau_{H}^{2}}{(1-k)(1+\tau _{H})^{2}(1+\alpha_{H}r_{H})}\biggl( \frac{\tau_{H}-1}{\tau_{H}+1}\biggr) \end{aligned}$$

  (65)
  $$\begin{aligned} =& \biggl(\frac{1-\tau_{H}}{1+\tau_{H}} \biggr)\frac {B_{H}e^{m}N_{H}a_{H}k\alpha_{H}\tau_{H}^{2}(1-(1-k)\tau _{H})}{2(1-k)^{2}\tau_{H}(1+\tau_{H})^{2}(1+\alpha_{H}r_{H})} \\ &{}-\frac {2B_{F}N_{F}e^{m}a_{F}\tau_{H}k\alpha_{H}}{(1-k)(1+\tau _{H})^{3}(1+\alpha_{H}r_{H})} \end{aligned}$$

  (66)
  $$\begin{aligned} =&\frac{B_{F}e^{m}N_{F}a_{F}\tau_{H}k\alpha_{H}}{(1-k)(1+\tau _{H})^{3}(1+\alpha_{H}r_{H})} \biggl[\frac{(1-\tau_{H})(1-(1-k)\tau _{H})}{2(1-k)\tau_{H}}-2 \biggr] \end{aligned}$$

  (67)
  $$\begin{aligned} =&\frac{B_{F}e^{m}N_{F}a_{F}\tau_{H}k\alpha_{H}}{(1-k)(1+\tau _{H})^{3}(1+\alpha_{H}r_{H})} \\ &{}\times \biggl[\frac{(1-\tau_{H})(1-(1-k)\tau _{H})-4(1-k)\tau_{H}}{2(1-k)\tau_{H}} \biggr] \end{aligned}$$

  (68)
  $$\begin{aligned} =&\frac{B_{F}e^{m}N_{F}a_{F}\tau_{H}k\alpha_{H}}{(1-k)(1+\tau _{H})^{3}(1+\alpha_{H}r_{H})} \biggl[\frac{(1-k)\tau_{H}^{2}-(6-5k)\tau _{H}+1}{2(1-k)\tau_{H}} \biggr] \end{aligned}$$

  (69)
  $$\begin{aligned} \mbox{Let } y =&(1-k)\tau_{H}^{2}-(6-5k) \tau_{H}+1. \end{aligned}$$

  (70)

  Note that

  $$\operatorname{sign} \biggl(\frac{dd_{H}^{*}}{dr_{H}} \biggr) = \operatorname{sign}(y). $$

  Differentiating (70) with respect with τ _H we get,

  $$ \frac{dy}{d\tau_{H}}=2(1-k)\tau_{H}-(6-5k) $$

  (71)

  Therefore,

  $$\begin{aligned} \frac{dy}{d\tau_{H}} >&0 \quad\mbox{if } \tau_{H}>\frac {(6-5k)}{2(1-k)} \end{aligned}$$

  (72)
  $$\begin{aligned} \frac{dy}{d\tau_{H}} =&0 \quad\mbox{if } \tau_{H}=\frac {(6-5k)}{2(1-k)} \end{aligned}$$

  (73)
  $$\begin{aligned} \frac{dy}{d\tau_{H}} <&0\quad\mbox{if } \tau_{H}<\frac {(6-5k)}{2(1-k)} \end{aligned}$$

  (74)
  $$\begin{aligned} \frac{dy^{2}}{d\tau_{H}} =&2(1-k)>0 \end{aligned}$$

  (75)

  Therefore, y is convex, y(0)=1. Let τ ₀ and τ ₁ be the roots of y where τ ₀<τ ₁.

  Therefore, ∀τ _H ∈(τ ₀,τ ₁), y<0 and ≥0 elsewhere (see Fig. 2). Therefore, $\frac{dd_{H}^{*}}{dr_{H}}<0\ \forall\tau_{H}\in(\tau_{0},\tau_{1})$ and ≥0 elsewhere.

  Fig. 2

  

  y=f(τ _H ) (not drawn to scale)

• (d) $\frac{dd_{H}^{*}}{de_{H}}<0$.

  Differentiating (41) with respect to e _H , we get

  $$ \frac{\partial d_{H}^{*}}{\partial e_{H}}=\frac {mB_{F}N_{F}a_{F}e^{m-1}}{X_{H}}+\frac{B_{F}e^{m}a_{F}}{X_{H}}\frac {\partial N_{F}}{\partial e_{H}}- \frac {B_{F}e^{m}N_{F}a_{F}}{X_{H}^{2}}\frac{\partial X_{H}}{\partial e_{H}} $$

  (76)

  Differentiating (43) with respect to e _H , we note that

  $$ \frac{\partial X_{H}}{\partial e_{H}}=0 $$

  (77)

  Using (77) in (76) and Lemma 3(b), we get

  $$\begin{aligned} =&\frac{mB_{F}N_{F}a_{F}e^{m-1}}{X_{H}}+\frac {B_{F}a_{F}N_{F}mke^{m-1}}{X_{H}(1-k)} \end{aligned}$$

  (78)
  $$\begin{aligned} =&\frac{mB_{F}N_{F}a_{F}e^{m-1}}{X_{H}} \biggl[1+\frac{k}{1-k} \biggr] \end{aligned}$$

  (79)
  $$\begin{aligned} =&\frac{mB_{F}N_{F}a_{F}e^{m-1}}{(1-k)X_{H}}<0 \end{aligned}$$

  (80)

  Differentiating (45) with respect to e _H , we get

  $$\begin{aligned} \frac{\partial d_{F}^{*}}{\partial e_{H}} =&\frac {mB_{H}N_{H}a_{H}e^{m-1}}{X_{F}}+\frac {B_{H}a_{H}N_{H}mke^{m-1}}{X_{F}(1-k)}\quad \biggl( \mbox{Since }\frac{\partial X_{F}}{\partial e_{H}}=0\biggr) \\ =&\frac{mB_{H}N_{H}a_{H}e^{m-1}}{(1-k)X_{F}}<0 \end{aligned}$$

  (81)
  $$\begin{aligned} \frac{dd_{H}^{*}}{de_{H}} =&\frac{\partial d_{H}^{*}}{\partial e_{H}}+\frac{\partial d_{H}^{*}}{\partial d_{F}^{*}}\frac{\partial d_{F}^{*}}{\partial e_{H}} \\ =&\frac{mB_{F}N_{F}a_{F}e^{m-1}}{X_{H}(1-k)}+\frac{(1-(1-k)\tau _{H})}{2(1-k)\tau_{H}}\frac{mB_{H}e^{m-1}N_{H}a_{H}}{X_{F}(1-k)} \\ =&\frac{mB_{F}N_{F}a_{F}e^{m-1}}{X_{H}(1-k)} \biggl[1+\frac{(1-(1-k)\tau _{H})}{2(1-k)\tau_{H}}\frac{1}{\tau_{H}} \tau_{H}^{2} \biggr] \\ =&\frac{mB_{F}N_{F}a_{F}e^{m-1}}{X_{H}(1-k)} \biggl[1+\frac{(1-(1-k)\tau _{H})}{2(1-k)} \biggr] \\ =&\frac{mB_{F}N_{F}a_{F}e^{m-1}}{X_{H}(1-k)} \biggl[\frac{(3-2k)-(1-k)\tau _{H}}{2(1-k)} \biggr] \end{aligned}$$

  (82)

  Since $\tau_{H}<\frac{1}{1-k}$, $\tau_{H}<\frac{(3-2k)}{(1-k)}$. Therefore, $\frac{dd_{H}^{*}}{de_{H}}<0$

  $$\begin{aligned} \frac{\partial d_{H}^{*}}{\partial e_{F}} =&\frac {mB_{F}N_{F}a_{F}e^{m-1}}{X_{H}}+\frac{B_{F}a_{F}e^{m}N_{F}}{X_{H}} \frac {mk}{1-k}e^{-1} \\ =&\frac{mB_{F}N_{F}a_{F}e^{m-1}}{X_{H}(1-k)}=\frac{\partial d_{H}^{*}}{\partial e_{H}} \end{aligned}$$

  (83)
  $$\begin{aligned} \frac{\partial d_{F}^{*}}{\partial e_{F}} =&\frac {mB_{H}N_{H}a_{H}e^{m-1}}{X_{F}}+\frac{B_{H}a_{H}e^{m}N_{H}}{X_{F}} \frac {mk}{1-k}e^{-1} \\ =&\frac{mB_{H}N_{H}a_{H}e^{m-1}}{X_{F}(1-k)}=\frac{\partial d_{F}^{*}}{\partial e_{H}} \end{aligned}$$

  (84)
  $$\begin{aligned} \frac{dd_{F}^{*}}{de_{F}} =&\frac{\partial d_{F}^{*}}{\partial e_{F}}+\frac{\partial d_{F}^{*}}{\partial d_{H}^{*}}\frac{\partial d_{H}^{*}}{\partial e_{F}} \\ =&\frac{mB_{H}N_{H}a_{H}e^{m-1}}{X_{F}(1-k)}+\frac{\tau _{H}-(1-k)}{2(1-k)}\frac{mB_{F}N_{F}a_{F}e^{m-1}}{X_{H}(1-k)} \\ =&\frac{mB_{H}N_{H}a_{H}e^{m-1}}{X_{F}(1-k)} \biggl[1+\frac{\tau _{H}-(1-k)}{2(1-k)\tau_{H}} \biggr] \\ =&\frac{mB_{H}N_{H}a_{H}e^{m-1}}{X_{F}(1-k)} \biggl[\frac{(3-2k)\tau _{H}-(1-k)}{2(1-k)\tau_{H}} \biggr] \end{aligned}$$

  (85)

  Consider (3−2k)τ _H −(1−k). This is increasing in τ _H . Note that $\tau_{H}\in(1-k,\frac{1}{1-k})$. Evaluating at τ _H =1−k, we get (3−2k)(1−k)−(1−k)>0. Since m<0, we have the following:

  $$ \frac{dd_{F}^{*}}{de_{F}}=\frac {mB_{H}N_{H}a_{H}e^{m-1}}{X_{F}(1-k)} \biggl[\frac{(3-2k)\tau _{H}-(1-k)}{2(1-k)\tau_{H}} \biggr]>0 $$

  (86)

• (e) $\frac{dd_{H}^{*}}{d\lambda}<0$.

  Differentiating (41) with respect to λ, we get

  $$ \frac{\partial d_{H}^{*}}{\partial\lambda}=\frac {B_{F}e^{m}a_{F}}{X_{H}}\frac{\partial N_{F}}{\partial\lambda}-\frac {B_{F}e^{m}N_{F}a_{F}}{X_{H}^{2}} \frac{\partial X_{H}}{\partial\lambda} $$

  (87)

  Differentiating (5) with respect to λ, we get

  

  (88)

  Since, , $\frac {\partial N_{F}}{\partial\lambda}<0$.

  Differentiating (43) with respect to λ, we get

  $$ \frac{\partial X_{H}}{\partial\lambda}=2(1+\tau_{H}){\frac{\partial \tau_{H}}{\partial\lambda}} $$

  (89)

  Differentiating (44) with respect to λ, we get

  

  (90)

  We next show that, .

  

  (91)

  Recall that,

  

  Therefore,

  

  (92)

  Similarly,

  

  (93)

  Therefore,

  

  (94)

  Therefore,

  $$\begin{aligned} =&\frac{N_{F}}{N_{H}}\frac{\tilde{t}_{H}^{*k}}{\tilde{t}_{F}^{*k}} \end{aligned}$$

  (95)
  

  (96)
  

  (97)

  Plugging (97) in (90) we get,

  

  (98)

  Therefore, $\frac{\partial\tau_{H}}{\partial\lambda}=0$ implies $\frac {\partial X_{H}}{\partial\lambda}=0$.

  Therefore,

  

  (99)

  Similarly,

  

  (100)
  $$\begin{aligned} &\frac{dd_{H}^{*}}{d\lambda}=\frac{\partial d_{H}^{*}}{\partial\lambda }+\frac{\partial d_{H}^{*}}{\partial d_{F}^{*}}\frac{\partial d_{F}^{*}}{\partial\lambda} \end{aligned}$$

  (101)
  

  (102)
  

  (103)
  

  (104)

  Now, using (92)–(94) we get,

  

  (105)

  Simplifying the expression within brackets and using the fact that, $\tau_{H}=\frac{1}{T}\frac{N_{F}}{N_{H}}=\frac{1}{T}\frac{\tilde{t}_{F}^{*k}}{\tilde{t}_{H}^{*k}}$, we get the following:

  

  (106)
  

  (107)

  Since, $\tau_{H}\in(1-k,\frac{1}{1-k}), \tau_{H}<\frac {3-2k}{1-k}$ (since k<1). Therefore, $\frac{dd_{H}^{*}}{d\lambda}<0$.

Appendix C: R&D effort

Lemma 6:

• (a) $r_{H}^{*}=\frac{1}{\alpha_{H}} [\frac{1}{\varepsilon _{H}}\Delta_{H}\delta_{H}R_{H}-1 ]$.

  Differentiating (2) with respect to r _H , we get

  $$\begin{aligned} \frac{dV_{H}}{dr_{H}} =&-\frac {d_{F}^{*}}{(d_{H}^{*}+d_{F}^{*})^{2}}e^{m}N_{H}a_{H} \frac {dd_{H}^{*}}{dr_{H}}+\gamma_{H}\frac{dd_{H}^{*}}{dr_{H}}+\varepsilon _{H}+\frac{d_{H}^{*}}{(d_{H}^{*}+d_{F}^{*})^{2}}e^{m}N_{H}a_{H} \frac {dd_{F}^{*}}{dr_{H}} \\ &{}+\frac{d_{F}^{*}}{(d_{H}^{*}+d_{F}^{*})}e^{m}a_{H}\frac {dN_{H}}{dt_{H}^{*}} \biggl(\frac{dt_{H}^{*}}{dd_{H}^{*}}\frac {dd_{H}^{*}}{dr_{H}}+\frac{dt_{H}^{*}}{dd_{F}^{*}} \frac {dd_{F}^{*}}{dr_{H}}+\frac{dt_{H}^{*}}{dr_{H}} \biggr)=0 \end{aligned}$$

  (108)

  Using (25) we get:

  Or,

  $$\begin{aligned} & \biggl[\frac {d_{H}^{*}}{(d_{H}^{*}+d_{F}^{*})^{2}}e^{m}N_{H}a_{H}+ \frac {d_{F}^{*}}{(d_{H}^{*}+d_{F}^{*})}e^{m}a_{H}N_{H} \frac{k}{(1-k)}\frac {d_{H}^{*}}{(d_{H}^{*}+d_{F}^{*})d_{F}^{*}} \biggr]\frac {dd_{F}^{*}}{dr_{H}} \\ &\quad{}-\frac{d_{F}^{*}}{(d_{H}^{*}+d_{F}^{*})}e^{m}a_{H}N_{H} \frac {k}{1-k}\frac{\alpha_{H}}{1+\alpha_{H}r_{H}}=-\varepsilon_{H} \end{aligned}$$

  (109)

  Or,

  $$ \frac{d_{H}^{*}}{(d_{H}^{*}+d_{F}^{*})^{2}}\frac {e^{m}N_{H}a_{H}}{1-k}\frac{dd_{F}^{*}}{dr_{H}}-\frac {d_{F}^{*}}{(d_{H}^{*}+d_{F}^{*})} \frac{e^{m}a_{H}N_{H}k\alpha _{H}}{(1-k)(1+\alpha_{H}r_{H})}=-\varepsilon_{H} $$

  (110)

  Using (31), (32), (41) and (45) and simplifying we get:

  Or,

  $$ \frac{1}{B_{H}\tau_{H}(1-k)}\frac{dd_{F}^{*}}{dr_{H}}-\frac {\tau_{H}}{(1-k)(1+\tau_{H})}\frac{e^{m}a_{H}N_{H}k\alpha_{H}}{(1+\alpha _{H}r_{H})}=- \varepsilon_{H} $$

  (111)

  Plugging the expression for $\frac{dd_{F}^{*}}{dr_{H}}$ in (111), we get

  Or,

  $$ \frac{B_{F}e^{m}N_{F}a_{F}k\alpha_{H}\tau _{H}(2(1-k)-(2-k)\tau_{H})}{B_{H}\tau_{H}(1-k)(1+\tau _{H})^{3}(1-k)^{2}(1+\alpha_{H}r_{H})}-\frac{\tau _{H}e^{m}a_{H}N_{H}\alpha_{H}k}{(1-k)(1+\tau_{H})(1+\alpha _{H}r_{H})}=-\varepsilon_{H} $$

  (112)

  Or,

  $$ \frac{k\alpha_{H}B_{F}e^{m}N_{F}a_{F}}{B_{H}(1-k)(1+\tau _{H})(1+\alpha_{H}r_{H})} \biggl[\frac{(2-k)\tau _{H}-2(1-k)}{(1-k)^{2}(1+\tau_{H})^{2}}+\frac{\tau _{H}B_{H}e^{m}a_{H}N_{H}}{B_{F}e^{m}a_{F}N_{F}} \biggr]= \varepsilon _{H} $$

  (113)

  Or,

  $$ \frac{k\alpha_{H}B_{F}e^{m}N_{F}a_{F}}{B_{H}(1-k)(1+\tau _{H})(1+\alpha_{H}r_{H})} \biggl[\frac{(2-k)\tau _{H}-2(1-k)}{(1-k)^{2}(1+\tau_{H})^{2}}+1 \biggr]=\varepsilon_{H} $$

  (114)

  Or,

  $$ \frac{k\alpha_{H}B_{F}a_{F}}{B_{H}(1-k)^{3}}\frac {e^{m}N_{F}}{(1+\tau_{H})^{3}} \bigl((2-k)\tau _{H}-2(1-k)+(1-k)^{2}(1+ \tau_{H})^{2} \bigr)=\varepsilon_{H}(1+\alpha _{H}r_{H}) $$

  (115)

  Or,

  $$ \frac{1}{\varepsilon_{H}}\Delta_{H}\delta_{H}R_{H}=1+ \alpha _{H}r_{H} $$

  (116)

  Where,

  $$\begin{aligned} \Delta_{H} =&\frac{k\alpha_{H}B_{F}a_{F}}{B_{H}(1-k)^{3}}>0 \end{aligned}$$

  (117)
  $$\begin{aligned} \delta_{H} =&e^{m}N_{F} \end{aligned}$$

  (118)
  $$\begin{aligned} R_{H} =&\frac{(2-k)\tau_{H}-2(1-k)+(1-k)^{2}(1+\tau_{H})^{2}}{(1+\tau _{H})^{3}} \end{aligned}$$

  (119)

  Therefore, we get,

  $$ r_{H}^{*}=\frac{1}{\alpha_{H}} \biggl[\frac{1}{\varepsilon_{H}} \Delta _{H}\delta_{H}R_{H}-1 \biggr] $$

  (120)

  Similarly, for country F we have,

  $$ r_{F}^{*}=\frac{1}{\alpha_{F}} \biggl[\frac{1}{\varepsilon_{F}} \Delta _{F}\delta_{F}R_{F}-1 \biggr] $$

  (121)

• (b) $\frac{\partial r_{H}^{*}}{\partial r_{F}^{*}}<0$.

  We first show that τ _H is a constant. Plugging (3) and (5) in (44), we get

  

  (122)

  Using (92) and (93) in (122), we get

  

  (123)
  

  (124)

  Plugging (10) and (11) and simplifying, we get

  $$ =\frac{1}{T} \biggl( \frac{\tilde{c}(1+\alpha _{F}r_{F}^{*})(d_{H}^{*}+d_{F}^{*})e^{-m}d_{F}^{*}}{\tilde{c}(1+\alpha _{H}r_{H}^{*})d_{H}^{*}e^{-m}(d_{H}^{*}+d_{F}^{*})} \biggr) ^{\frac {k}{k-1}} $$

  (125)

  Plugging (120) and (121) in (125) we get,

  $$ \tau_{H}=\frac{1}{T} \biggl(\frac{\tau_{H}\varepsilon_{H}\Delta _{F}R_{F}\delta_{F}}{\varepsilon_{F}\Delta_{H}R_{H}\delta_{H}} \biggr)^{\frac{k}{k-1}} $$

  (126)

  After some simplification, we get the following:

  $$\begin{aligned} \varPhi(\tau_{H}) =& \biggl(\frac{\varepsilon_{H}\Delta_{F}}{\varepsilon_{F}\Delta _{H}} \biggr)^{\frac{k-1}{k}}T^{\frac{2k-1}{k-1}} \end{aligned}$$

  (127)
  $$\begin{aligned} \tau_{H} =&\varPhi^{-1} \biggl( \biggl(\frac{\varepsilon_{H}\Delta_{F}}{\varepsilon _{F}\Delta_{H}} \biggr)^{\frac{k-1}{k}}T^{\frac{2k-1}{k-1}} \biggr) \end{aligned}$$

  (128)

  Since R _H =f(τ _H ), R _H must also be a constant. Differentiating (120) with respect to r _F , we get

  $$ \frac{\partial r_{H}^{*}}{\partial r_{F}^{*}}=\frac{\Delta _{H}R_{H}}{\alpha_{H}\varepsilon_{H}}e^{m}\frac{\partial N_{F}}{\partial r_{F}^{*}} $$

  (129)

  Plugging (54) in (129), we get

  $$ =-\frac{\Delta_{H}R_{H}}{\alpha_{H}\varepsilon_{H}}\frac{e^{m}}{(1-k)}\frac {N_{F}k\alpha_{F}}{(1+\alpha_{F}r_{F}^{*})} $$

  (130)

  Plugging (121) in (130), we get

  $$ =-\frac{\Delta_{H}R_{H}}{\alpha_{H}\varepsilon_{H}}\frac{e^{m}}{(1-k)}\frac {N_{F}k\alpha_{F}}{\frac{\Delta_{F}\delta_{F}R_{F}}{\varepsilon _{F}}} $$

  (131)

  Let

  $$ \frac{\Delta_{H}R_{H}\varepsilon_{F}\alpha_{F}kN_{F}}{\Delta _{F}R_{F}\varepsilon_{H}(1-k)N_{H}\alpha_{H}}=\beta_{1} $$

  (132)

  Therefore,

  $$ \frac{\partial r_{H}^{*}}{\partial r_{F}^{*}}=-\beta_{1}<0 $$

  (133)

  Similarly, for country F we have,

  $$ \frac{\partial r_{F}^{*}}{\partial r_{H}^{*}}=-\frac{\Delta _{F}R_{F}\varepsilon_{H}\alpha_{H}kN_{H}}{\Delta_{H}R_{H}\varepsilon _{F}(1-k)N_{F}\alpha_{F}}=-\beta_{2}<0 $$

  (134)

  Where,

  $$ \beta_{2}=\frac{\Delta_{F}R_{F}\varepsilon_{H}\alpha_{H}kN_{H}}{\Delta _{H}R_{H}\varepsilon_{F}(1-k)N_{F}\alpha_{F}} $$

  (135)

  Note that,

  $$ \beta_{1}\beta_{2}= \biggl(\frac{k}{1-k} \biggr)^{2} $$

  (136)

• (c) $\frac{dr_{H}^{*}}{de_{H}}>0$ when $k>\frac{1}{2}$, <0 when $k<\frac{1}{2}$.

  Differentiating (120) with respect to e _H , we get

  $$ \frac{\partial r_{H}^{*}}{\partial e_{H}}=\frac{\Delta_{H}R_{H}}{\alpha _{H}\varepsilon_{H}} \biggl(me^{m-1}N_{F}+e^{m} \frac{\partial N_{F}}{\partial e_{H}} \biggr) $$

  (137)

  Differentiating (5) with respect to e _H , we get

  $$ \frac{\partial N_{F}}{\partial e_{H}}=\frac{N_{F}mke^{-1}}{1-k} $$

  (138)

  Plugging (138) in (137), we get

  $$\begin{aligned} \frac{\partial r_{H}^{*}}{\partial e_{H}} =&\frac{\Delta_{H}R_{H}}{\alpha _{H}\varepsilon_{H}}e^{m-1} \biggl(mN_{F}+mN_{F} \frac{k}{1-k} \biggr) \\ =&\frac{\Delta_{H}R_{H}}{\alpha_{H}\varepsilon_{H}}\frac {me^{m-1}N_{F}}{1-k}<0 \end{aligned}$$

  (139)

  Similarly, for country F we have,

  $$ \frac{\partial r_{F}^{*}}{\partial e_{H}}=\frac{\Delta_{F}R_{F}}{\alpha _{F}\varepsilon_{F}}\frac{me^{m-1}N_{H}}{1-k}<0 $$

  (140)

  Now consider,

  $$ \frac{dr_{H}^{*}}{de_{H}}=\frac{\partial r_{H}^{*}}{\partial e_{H}}+\frac{\partial r_{F}^{*}}{\partial r_{F}^{*}}\frac{\partial r_{F}^{*}}{\partial e_{H}} $$

  (141)

  Plugging (139) and (140) in (141), we get

  $$\begin{aligned} =&\frac{\Delta_{H}R_{H}}{\alpha_{H}\varepsilon_{H}}\frac {me^{m-1}N_{F}}{1-k}-\beta_{1} \frac{\Delta_{F}R_{F}me^{m-1}N_{H}}{\alpha _{F}\varepsilon_{F}(1-k)} \end{aligned}$$

  (142)
  $$\begin{aligned} =&\frac{\Delta_{H}R_{H}me^{m-1}}{\alpha_{H}\varepsilon_{H}(1-k)}N_{F} \biggl[1-\beta_{1} \frac{\Delta_{F}R_{F}N_{H}\alpha_{H}\varepsilon_{H}}{\Delta _{H}R_{H}\alpha_{F}\varepsilon_{F}N_{F}} \biggr] \end{aligned}$$

  (143)
  $$\begin{aligned} =&\frac{\Delta_{H}R_{H}me^{m-1}}{\alpha_{H}\varepsilon_{H}(1-k)}N_{F} \biggl[1-\beta_{1} \frac{(1-k)}{k}\beta_{2} \biggr] \end{aligned}$$

  (144)

  Since $\beta_{1}\beta{2}=(\frac{k}{1-k})^{2}$

  $$\begin{aligned} =&\frac{\Delta_{H}R_{H}me^{m-1}}{\alpha_{H}\varepsilon_{H}(1-k)}N_{F} \biggl[1-\frac{k}{(1-k)} \biggr] \end{aligned}$$

  (145)
  $$\begin{aligned} =&\frac{\Delta_{H}R_{H}me^{m-1}}{\alpha_{H}\varepsilon_{H}(1-k)}N_{F} \biggl[\frac{(1-2k)}{(1-k)} \biggr] \end{aligned}$$

  (146)

  Therefore, $\frac{dr_{H}^{*}}{de_{H}}<0$ when $k<\frac{1}{2}$. Similarly, the other case can be shown to hold true.

• (d) $\frac{dr_{H}^{*}}{d\lambda}>0$ when $k>\frac{1}{2}$, <0 when $k<\frac{1}{2}$.

  Differentiating (120) with respect to λ, we get

  

  (147)

  Differentiating (121) with respect to λ, we get

  

  (148)
  $$\begin{aligned} \frac{dr_{H}^{*}}{d\lambda} =&\frac{\partial r_{H}^{*}}{\partial\lambda }+\frac{\partial r_{H}^{*}}{\partial r_{F}^{*}}\frac{\partial r_{F}^{*}}{\partial\lambda} \end{aligned}$$

  (149)

  Using (133), (147) and (148), we get

  

  (150)
  

  (151)

  Recall that,

  

  Therefore,

  

  (152)

  Therefore, $\frac{dr_{H}^{*}}{d\lambda}<0$ if $k<\frac{1}{2}$. Similarly, the other case can be shown to hold true. Similar results hold for $\frac{dr_{F}^{*}}{d\lambda}$.

Appendix D: Pre-emption

Lemma 7:

• (a) $e_{H}^{*}= (\frac{-\mu_{H}}{mG_{H}N_{H}a_{H}} )^{\frac {1}{m-1}}-e_{F}^{*}$.

  Differentiating (2) with respect to e _H , we get

  $$\begin{aligned} \frac{dV_{H}}{de_{H}} =&-\frac {d_{F}^{*}}{(d_{H}^{*}+d_{F}^{*})^{2}}e^{m}N_{H}a_{H} \biggl(\frac {dd_{H}^{*}}{dr_{H}^{*}}\frac{dr_{H}^{*}}{de_{H}}+\frac {dd_{H}^{*}}{dr_{F}^{*}} \frac{dr_{F}^{*}}{de_{H}}+\frac {dd_{H}^{*}}{de_{H}} \biggr) \\ &{}+\frac{d_{H}^{*}}{(d_{H}^{*}+d_{F}^{*})^{2}}e^{m}N_{H}a_{H} \biggl( \frac {dd_{F}^{*}}{dr_{H}^{*}}\frac{dr_{H}^{*}}{de_{H}}+\frac {dd_{F}^{*}}{dr_{F}^{*}}\frac{dr_{F}^{*}}{de_{H}}+ \frac {dd_{F}^{*}}{de_{H}} \biggr) \\ &{}+\frac {d_{F}^{*}}{(d_{H}^{*}+d_{F}^{*})}N_{H}a_{H}me^{m-1} \\ &{}+\frac{d_{F}^{*}}{(d_{H}^{*}+d_{F}^{*})}e^{m}a_{H}\frac {dN_{H}}{dt_{H}^{*}} \biggl(\frac{dt_{H}^{*}}{dd_{H}^{*}}\frac {dd_{H}^{*}}{dr_{H}^{*}}\frac{dr_{H}^{*}}{de_{H}}+ \frac {dt_{H}^{*}}{dd_{F}^{*}}\frac{dd_{F}^{*}}{dr_{H}^{*}}\frac {dr_{H}^{*}}{de_{H}}+\frac{dt_{H}^{*}}{dd_{H}^{*}} \frac {dd_{H}^{*}}{dr_{F}^{*}}\frac{dr_{F}^{*}}{de_{H}} \biggr) \\ &{}+\frac{d_{F}^{*}}{(d_{H}^{*}+d_{F}^{*})}e^{m}a_{H}\frac {dN_{H}}{dt_{H}^{*}} \biggl(\frac{dt_{H}^{*}}{dd_{F}^{*}}\frac {dd_{F}^{*}}{dr_{F}^{*}}\frac{dr_{F}^{*}}{de_{H}}+ \frac {dt_{H}^{*}}{dd_{H}^{*}}\frac{dd_{H}^{*}}{de_{H}}+\frac {dt_{H}^{*}}{dd_{F}^{*}}\frac{dd_{F}^{*}}{de_{H}}+ \frac {dt_{H}^{*}}{dr_{H}^{*}}\frac{dr_{H}^{*}}{de_{H}} \\ &{}+\frac {dt_{H}^{*}}{de_{H}} \biggr) \\ &{}+\gamma_{H} \biggl(\frac{dd_{H}^{*}}{dr_{H}^{*}}\frac {dr_{H}^{*}}{de_{H}}+ \frac{dd_{H}^{*}}{dr_{F}^{*}}\frac {dr_{F}^{*}}{de_{H}}+\frac{dd_{H}^{*}}{de_{H}} \biggr)+ \varepsilon_{H}\frac {dr_{H}^{*}}{de_{H}}+\mu_{H}=0 \end{aligned}$$

  (153)

  Using (28) and (109) and simplifying we get the following:

  $$\begin{aligned} =&\frac{d_{H}^{*}}{(d_{H}^{*}+d_{F}^{*})^{2}}e^{m}N_{H}a_{H} \biggl( \frac {dd_{F}^{*}}{dr_{F}^{*}}\frac{dr_{F}^{*}}{de_{H}}+\frac {dd_{F}^{*}}{de_{H}} \biggr) \\ &{} +\frac {d_{F}^{*}}{(d_{H}^{*}+d_{F}^{*})}me^{m-1}N_{H}a_{H} \\ &{}+ \frac {d_{F}^{*}a_{H}e^{m}}{(d_{H}^{*}+d_{F}^{*})}\frac {dN_{H}}{dt_{H}^{*}} \biggl(\frac{dt_{H}^{*}}{dd_{F}^{*}} \frac {dd_{F}^{*}}{dr_{F}^{*}}\frac{dr_{F}^{*}}{de_{H}}+\frac {dt_{H}^{*}}{dd_{F}^{*}}\frac{dd_{F}^{*}}{de_{H}}+ \frac {dt_{H}^{*}}{de_{H}} \biggr) \\ =&-\mu_{H} \end{aligned}$$

  (154)

  Plugging in expressions for $\frac{dt_{H}^{*}}{dd_{F}^{*}}$, $\frac {dd_{F}^{*}}{dr_{F}^{*}}$, $\frac{dr_{F}^{*}}{de_{H}}$, $\frac {dd_{F}^{*}}{de_{H}}$ and $\frac{dt_{H}^{*}}{de_{H}}$, we get, or

  $$\begin{aligned} &\frac {d_{H}^{*}}{(d_{H}^{*}+d_{F}^{*})^{2}}e^{m}N_{H}a_{H}\biggl( \frac {B_{F}e^{m}N_{F}a_{F}k\alpha_{F}\chi_{2}}{(1+\tau_{H})^{3}(1-k)(1+\alpha _{F}r_{F}^{*})}\frac{R_{F}\Delta_{F}mN_{H}e^{m-1}(1-2k)}{\alpha _{F}\varepsilon_{F}(1-k)^{2}} \\ &\quad{}+\frac {mB_{F}N_{F}a_{F}e^{m-1}}{2X_{H}(1-k)^{2}}\bigl((3-2k)\tau _{H}-(1-k)\bigr) \\ &\quad{}+\frac{d_{F}^{*}}{(d_{H}^{*}+d_{F}^{*})}me^{m-1}N_{H}a_{H}- \frac {d_{F}^{*}kN_{H}}{(d_{H}^{*}+d_{F}^{*})^{2}(1-k)}e^{m}a_{H} \\ &\quad{}\times\biggl(\frac {d_{H}^{*}B_{F}e^{m}N_{F}a_{F}k\alpha_{F}\chi_{2}}{d_{F}^{*}(1+\tau _{H})^{3}(1-k)(1+\alpha_{F}r_{F}^{*})} \frac{R_{F}\Delta _{F}me^{m-1}N_{H}(1-2k)}{\alpha_{F}\varepsilon_{F}(1-k)^{2}}\biggr) \\ &\quad{}-\frac{d_{F}^{*}}{(d_{H}^{*}+d_{F}^{*})^{2}}e^{m}a_{H}\frac {N_{H}k}{(1-k)} \frac{d_{H}^{*}}{d_{F}^{*}}\frac {mB_{F}N_{F}a_{F}e^{m-1}}{2X_{H}(1-k)^{2}}\bigl((3-2k)\tau_{H}-(1-k)\bigr) \\ &\quad{}+ \frac {d_{F}^{*}}{(d_{H}^{*}+d_{F}^{*})}e^{m}a_{H}\frac {N_{H}mke^{-1}}{1-k}=- \mu_{H} \end{aligned}$$

  (155)

  where,

  $$\chi_{2}=\frac{\tau_{H}^{2}-(4-3k)\tau_{H}+(1-k)}{2(1-k)} $$

  Simplifying further by plugging in (120) and (121), we get:

  Or,

  $$\begin{aligned} &\frac {d_{H}^{*}e^{m}N_{H}a_{H}}{(d_{H}^{*}+d_{F}^{*})^{2}}\frac {B_{F}e^{m}a_{F}N_{F}}{(1-k)^{2}} \\ &\quad{}\times\biggl(\frac{R_{F}\Delta_{F}N_{H}k\alpha _{F}me^{m-1}(1-2k)\chi_{2}}{(1+\tau_{H})^{3}(1-k)\alpha_{F}\varepsilon _{F}\frac{\Delta_{F}\delta_{F}R_{F}}{\varepsilon_{F}}}+ \frac{m((3-2k)\tau _{H}-(1-k))}{2X_{H}e} \biggr) \\ &\quad{}+\frac{d_{F}^{*}}{(d_{H}^{*}+d_{F}^{*})}me^{m-1}N_{H}a_{H} \\ &\quad{}- \frac {d_{F}^{*}}{(d_{H}^{*}+d_{F}^{*})^{2}}\frac {e^{m}a_{H}kN_{H}}{(1-k)} \biggl(\frac {d_{H}^{*}B_{F}e^{m}N_{F}a_{F}k\alpha_{F}R_{F}\Delta _{F}me^{m-1}N_{H}(1-2k)\chi_{2}}{d_{F}^{*}(1+\tau_{H})^{3}(1-k)\frac {\Delta_{F}R_{F}\delta_{F}\alpha_{F}\varepsilon_{F}(1-k)^{2}}{\varepsilon _{F}}} \biggr) \\ &\quad{}-\frac{d_{F}^{*}}{(d_{H}^{*}+d_{F}^{*})^{2}}\frac {e^{m}a_{H}N_{H}k}{1-k}\frac{d_{H}^{*}}{d_{F}^{*}} \frac {mB_{F}N_{F}a_{F}e^{m-1}}{2X_{H}(1-k)^{2}}\bigl((3-2k)\tau _{H}-(1-k)\bigr) \\ &\quad{}+\frac{d_{F}^{*}}{(d_{H}^{*}+d_{F}^{*})}\frac {e^{m-1}a_{H}kN_{H}m}{(1-k)}=-\mu_{H} \end{aligned}$$

  (156)

  Using (31), (41) and (45) and simplifying we get:

  Or,

  $$\begin{aligned} &\frac{e^{m-1}N_{H}a_{H}m}{(1-k)^{3}} \biggl(\frac{(1-2k)\chi _{2}}{(1+\tau_{H})^{3}(1-k)}+\frac{((3-2k)\tau_{H}-(1-k))}{2X_{H}} \biggr)+ \frac{me^{m-1}N_{H}a_{H}}{(1-k)X_{F}^{\frac{1}{2}}} \\ &\quad=-\mu_{H} \end{aligned}$$

  (157)

  Or,

  $$ mN_{H}a_{H}G_{H}e^{m-1}=- \mu_{H} $$

  (158)

  This gives the following expression for e

  $$ e^{*}= \biggl(\frac{-\mu_{H}}{mG_{H}N_{H}a_{H}} \biggr)^{\frac {1}{m-1}} $$

  (159)

  Where,

  $$G_{H}=\frac{1}{(1-k)^{3}} \biggl(\frac{(1-2k)}{(1+\tau _{H})^{3}(1-k)^{3}}+ \frac{((3-2k)\tau_{H}-(1-k))}{2X_{H}} \biggr)+\frac {1}{(1-k)X_{F}^{\frac{1}{2}}} $$

  Therefore,

  $$ e_{H}^{*}= \biggl(\frac{-\mu_{H}}{mG_{H}N_{H}a_{H}} \biggr)^{\frac {1}{m-1}}-e_{F}^{*} $$

  (160)

• (b) $\frac{\partial e_{H}^{*}}{\partial e_{F}^{*}}=\frac {km}{(1-k)(1-m)}-1$.

  First, using the same logic as we used for $e_{H}^{*}$, we get the following expression for country F

  $$ mN_{F}a_{F}G_{F}e^{m-1}=- \mu_{F} $$

  (161)

  This gives the following expression for e

  $$ e^{*}= \biggl(\frac{-\mu_{F}}{mG_{F}N_{F}a_{F}} \biggr)^{\frac {1}{m-1}} $$

  (162)

  Therefore,

  $$\begin{aligned} e_{F}^{*} =& \biggl(\frac{-\mu_{F}}{mG_{F}N_{F}a_{F}} \biggr)^{\frac {1}{m-1}}-e_{H}^{*} \end{aligned}$$

  (163)
  $$\begin{aligned} \frac{\partial e_{H}^{*}}{\partial e_{F}^{*}} =& \biggl(\frac{-\mu _{H}}{mG_{H}N_{H}a_{H}} \biggr)^{\frac{1}{m-1}} \frac{1}{1-m}N_{H}^{\frac {1}{1-m}-1}\frac{\partial N_{H}}{\partial e_{F}}-1 \\ =& \biggl(\frac{-\mu_{H}}{mG_{H}N_{H}a_{H}} \biggr)^{\frac{1}{m-1}}\frac {1}{1-m}N_{H}^{\frac{1}{1-m}-1} \frac{kN_{H}m}{(1-k)e}-1 \\ =& \biggl(\frac{-\mu_{H}}{mG_{H}N_{H}a_{H}} \biggr)^{\frac{1}{m-1}}\frac {km}{(1-k)e(1-m)}-1 \\ =&\frac{km}{(1-k)(1-m)}-1 \end{aligned}$$

  (164)

  This leads to the following results.

  $$ \frac{\partial e_{H}^{*}}{\partial e_{F}^{*}}=\frac {km}{(1-k)(1-m)}-1<0 $$

  (165)

  Similarly,

  $$ \frac{\partial e_{F}^{*}}{\partial e_{H}^{*}}=\frac {km}{(1-k)(1-m)}-1<0 $$

  (166)

• (c) $\frac{\partial e_{H}^{*}}{\partial\lambda}>0$.

  Differentiating (160) with respect to λ, we get

  

  (167)

  Similarly,

  

  (168)
  $$\begin{aligned} \frac{de_{H}^{*}}{d\lambda} =&\frac{\partial e_{H}^{*}}{\partial\lambda }+\frac{\partial e_{H}^{*}}{\partial e_{F}^{*}}\frac{\partial e_{F}^{*}}{\partial\lambda} \end{aligned}$$

  (169)

  Plugging (167) and (168) in (169), we get

  

  (170)
  

  (171)
  

  (172)

  Recall that,

  

  Therefore,

  

  (173)

  Similarly, $\frac{de_{H}^{*}}{d\lambda}>0$.