Bundling versus unbundling: asymmetric information on information externalities

Abstract

This paper addresses the benefits of bundling two sequential activities in the context of public–private partnerships (PPPs). The paper introduces a source of asymmetric information in the form of an externality parameter that links the building stage with subsequent operational activity. Within this framework, bundling allows the government to extract private information about the magnitude of the externality parameter. The framework also implies a higher degree of asymmetric information related to the operational stage than unbundling does when the contract is written. Our results indicate that the use of bundled contracts allows PPPs to be commitment devices that force governments to define ex-ante more coherent and informed plans, thereby improving investments and reducing unexpected cost overruns. However, because of the presence of asymmetric information, bundling makes any cost-reducing effort suboptimal during the operational phase.

Appendix

This Appendix contains the proofs and the the comparative statistical analysis.

1.1 Proof of Proposition 1

The expected function that is used to perform the welfare analysis is the following:

$$\begin{aligned} \int _{\theta ^{l}}^{\theta ^{h}}\lbrace S_{0}+s I-(1+\lambda )[T_{b}+T_{o}]+(T_{b}-C_{b}(I))+(T_{o}-C_o(\theta , e)-\phi (e))\rbrace f(\theta )d\theta \end{aligned}$$

1.1.1 Positive externality ($a=1$)

Using the new investment and effort functions, the first-order conditions in the bundling case become:

• $I^{B}=\frac{p}{1-hp}\frac{s+(1+\lambda )\theta }{1+\lambda }-\frac{p}{1-hp}\frac{\lambda }{1+\lambda }\frac{1-F(\theta )}{f(\theta )}$,

• $e^{B}=h I^B-\frac{\lambda }{1+\lambda }\frac{1-F(\theta )}{f(\theta )}$.

Substituting in the government’s objective formula, we obtain the value function under bundling:

$$\begin{aligned} V^{B}= & {} \int _{\theta ^{l}}^{\theta ^{h}}\left\{ S_{0}+s I^{B}-(1+\lambda )\left[ \frac{(I^{B})^{2}}{2p}+\frac{(e^{B})^{2}}{2h}+F-(\theta +e^{B})I^{B}\right] \right. \\&\quad \left. -\,\lambda \frac{1-F(\theta )}{f(\theta )}\frac{e^{B}}{h}\right\} f(\theta )d\theta \\= & {} \int _{\theta ^{l}}^{\theta ^{h}}\left\{ S_{0}+I^{B}(s+(1+\lambda )(\theta +e^B))\right. \\&\quad \left. -\,(1+\lambda )\left[ \frac{(I^{B})^{2}}{2p}+\frac{(I^{B})^{2} h}{2}+\frac{1}{2h}\left( \frac{\lambda }{1+\lambda }\frac{1-F(\theta )}{f(\theta )}\right) ^2 \right. \right. \\&\quad \left. \left. -\,\frac{\lambda }{1+\lambda }\frac{1-F(\theta )}{f(\theta )}I^B +F \right] -\lambda \frac{1-F(\theta )}{f(\theta )}\left( I^B-\frac{1}{h}\frac{\lambda }{1+\lambda }\frac{1-F(\theta )}{f(\theta )}\right) \right\} f(\theta )d\theta \\= & {} \int _{\theta ^{l}}^{\theta ^{h}}\left\{ S_{0}-(1+\lambda ) F+ \frac{(1+\lambda )(1-hp)}{2p}(I^{B})^2\right. \\&\quad \left. +\,(1+\lambda )\frac{1}{2h}\left( \frac{\lambda }{1+\lambda }\frac{1-F(\theta )}{f(\theta )}\right) ^2 \right\} f(\theta )d\theta \\= & {} \int _{\theta ^{l}}^{\theta ^{h}}\left\{ S_{0}-(1+\lambda ) F+ \frac{p (1+\lambda )}{2(1-hp)}\left( \frac{s+(1+\lambda )\theta }{1+\lambda }\right) ^2+\frac{p(1+\lambda )}{2(1-hp)}\left( \frac{\lambda }{1+\lambda }\frac{1-F(\theta )}{f(\theta )}\right) ^2\right. \\&\quad \left. -\,\frac{(1+\lambda )p}{1-hp}\frac{s+(1+\lambda )\theta }{1+\lambda }\frac{\lambda }{1+\lambda }\frac{1-F(\theta )}{f(\theta )}+(1+\lambda )\frac{1}{2h}\left( \frac{\lambda }{1+\lambda }\frac{1-F(\theta )}{f(\theta )}\right) ^2 \right\} f(\theta )d\theta \\= & {} S_{0}-(1+\lambda ) F+ \frac{p (1+\lambda )}{2(1-hp)}\left( \frac{s^2+(1+\lambda )^2({\overline{\theta }}+\sigma ^2_{\theta })+2s(1+\lambda ){\overline{\theta }}}{(1+\lambda )^2}\right) \\&\quad -\,\frac{\lambda }{h}\int _{\theta ^{l}}^{\theta ^{h}}\left\{ \frac{1-F(\theta )}{f(\theta )} \left[ \frac{hp}{(1-hp)}\frac{s+(1+\lambda )\theta }{1+\lambda }-\right. \right. \\&\quad \left. \left. -\,\frac{hp}{2(1-hp)}\left( \frac{\lambda }{1+\lambda }\frac{1-F(\theta )}{f(\theta )}\right) -\frac{1}{2}\left( \frac{\lambda }{1+\lambda }\frac{1-F(\theta )}{f(\theta )}\right) \right] \right\} f(\theta )d\theta \\= & {} S_{0}-(1+\lambda ) F+ \frac{p (1+\lambda )}{2(1-hp)}\left( \frac{s+(1+\lambda ){\overline{\theta }}}{1+\lambda }\right) ^2+ \frac{p(1+\lambda )}{2(1-hp)}\sigma ^2_{\theta }\\&\quad -\,\frac{\lambda }{2h}\int _{\theta ^{l}}^{\theta ^{h}}\left\{ \frac{1-F(\theta )}{f(\theta )}\left( e^B+\frac{hp}{1-hp}\frac{s+(1+\lambda )\theta }{1+\lambda }\right) \right\} f(\theta )d\theta \end{aligned}$$

The new efforts functions applied to the unbundling case yield, respectively:

• $I^{U}=p\frac{s+(1+\lambda ){\overline{\theta }}}{(1+\lambda )(1-hp)}$,

• $e^{U}=h I^U$.

Substituting in the government’s objective formula, we obtain the value function under unbundling:

$$\begin{aligned} V^{U}= & {} \int _{\theta ^{l}}^{\theta ^{h}}\left\{ S_{0}+s I^{U}-(1+\lambda )\left[ \frac{(I^{U})^{2}}{2p}+\frac{(e^{U})^{2}}{2h}+F-I^{U}(\theta +e^{U})\right] \right\} f(\theta )d\theta \\= & {} \int _{\theta ^{l}}^{\theta ^{h}}\left\{ S_{0}+s I^{U}-(1+\lambda )\left[ \frac{(I^{U})^{2}}{2p}+\frac{(I^{U})^{2} h}{2}+F-I^{U}(\theta +I^{U} h)\right] \right\} f(\theta )d\theta \\= & {} \int _{\theta ^{l}}^{\theta ^{h}}\left\{ S_{0}-(1+\lambda ) F+ I^{U} (s+(1+\lambda )\theta )-(1+\lambda )\frac{(I^{U})^{2}}{2p}(1-hp)\right\} f(\theta )d\theta \\= & {} S_{0}-(1+\lambda ) F+ I^{U} (s+(1+\lambda ){\overline{\theta }})-(1+\lambda )\frac{(I^{U})^{2}}{2p}(1-hp)\\= & {} S_{0}-(1+\lambda ) F+ I^{U} \frac{I^U}{p}(1+\lambda )(1-hp)-(1+\lambda )\frac{(I^{U})^{2}}{2p}(1-hp)\\= & {} S_{0}-(1+\lambda ) F + (1+\lambda )\frac{(I^{U})^{2}}{2p}(1-hp)\\= & {} S_{0}-(1+\lambda ) F +\frac{p(1+\lambda )}{2(1-hp)}\left( \frac{s+(1+\lambda ){\overline{\theta }}}{1+\lambda }\right) ^2 \end{aligned}$$

The net welfare gain/loss of governments when using bundling is equal to:

$$\begin{aligned} V^{B}-V^{U}= & {} \frac{p(1+\lambda )}{2(1-hp)}\sigma ^2_{\theta }-\frac{\lambda }{2h}\int _{\theta ^{l}}^{\theta ^{h}}\left\{ \frac{1-F(\theta )}{f(\theta )}\left( e^B+\frac{hp}{1-hp}\frac{s+(1+\lambda )\theta }{1+\lambda }\right) \right\} f(\theta )d\theta \end{aligned}$$

where $e^B=\frac{hp}{1-hp}\frac{s+(1+\lambda )\theta }{1+\lambda }-\frac{hp}{1-hp}\frac{\lambda }{1+\lambda }\frac{1-F(\theta )}{f(\theta )}-\frac{\lambda }{1+\lambda }\frac{1-F(\theta )}{f(\theta )}$.

As a result, bundling dominates unbundling if and only if:

$$\begin{aligned} \sigma _{\theta }^{2}\ge & {} \frac{\lambda (1-hp)}{hp(1+\lambda )}\int _{\theta ^{l}}^{\theta ^{h}}\left\{ \frac{1-F(\theta )}{f(\theta )}\left( e^B+\frac{hp}{1-hp}\frac{s+(1+\lambda )\theta }{1+\lambda }\right) \right\} f(\theta )d\theta \\ \sigma _{\theta }^{2}\ge & {} \int _{\theta ^{l}}^{\theta ^{h}}\left\{ \frac{\lambda (1-hp)}{hp(1+\lambda )}\frac{1-F(\theta )}{f(\theta )}\left( \frac{hp}{1-hp}\frac{s+(1+\lambda )\theta }{1+\lambda }-\frac{hp}{1-hp}\frac{\lambda }{1+\lambda }\frac{1-F(\theta )}{f(\theta )}\right. \right. \\&\left. \left. -\,\frac{\lambda }{1+\lambda }\frac{1-F(\theta )}{f(\theta )}+\frac{hp}{1-hp}\frac{s+(1+\lambda )\theta }{1+\lambda }\right) \right\} f(\theta )d\theta \end{aligned}$$

1.1.2 Proof of Corollary 2—positive externality

To study the sign of the derivative of $\sigma _{\theta }^{w}$ varies with $\lambda$, it is sufficient to study the derivative of the integrand in the previous inequality with respect to $\lambda$ that is equal to:

$$\begin{aligned}= & {} \frac{1-hp}{hp}\frac{1}{(1+\lambda )^2}\frac{1-F(\theta )}{f(\theta )}\left( e^B+\frac{hp}{1-hp}\frac{s+(1+\lambda )\theta }{1+\lambda }\right) \\&\quad +\,\frac{1-hp}{hp}\frac{\lambda }{1+\lambda }\frac{1-F(\theta )}{f(\theta )}\left( -\frac{hp}{1-hp}\frac{2s}{(1+\lambda )^2}-\frac{hp}{1-hp}\frac{1}{(1+\lambda )^2}\frac{1-F(\theta )}{f(\theta )}\right. \\&\quad \left. -\,\frac{1}{(1+\lambda )^2}\frac{1-F(\theta )}{f(\theta )}\right) \\= & {} 2\frac{1-hp}{hp}\frac{1}{(1+\lambda )^2}\frac{1-F(\theta )}{f(\theta )}\left( \frac{hp}{1-hp}\frac{s(1-\lambda )+(1+\lambda )\theta }{1+\lambda }-\frac{hp}{1-hp}\frac{\lambda }{1+\lambda }\frac{1-F(\theta )}{f(\theta )}\right. \\&\quad \left. -\,\frac{\lambda }{1+\lambda }\frac{1-F(\theta )}{f(\theta )}\right) \\= & {} 2\frac{1-hp}{hp}\frac{1}{(1+\lambda )^2}\frac{1-F(\theta )}{f(\theta )}\left( e^B-\frac{hp}{1-hp}\frac{\lambda s}{1+\lambda }\right) \end{aligned}$$

As a conclusion, the value of $\sigma _{\theta }^{2}$ can increase or decrease with $\lambda$. Precisely, the derivative is positive if $\lambda =0$, while it may become negative as $\lambda$ increases. The second derivative is equal to:

$$\begin{aligned}= & {} 4\frac{1-hp}{hp}\frac{1}{(1+\lambda )^3}\frac{1-F(\theta )}{f(\theta )}\left( -e^B+\frac{hp}{1-hp}\frac{\lambda s}{1+\lambda }\right) + \\&\quad +\,2\frac{1-hp}{hp}\frac{1}{(1+\lambda )^2}\frac{1-F(\theta )}{f(\theta )}\left( -\frac{hp}{1-hp}\frac{2s}{(1+\lambda )^2}\right. \\&\quad \left. -\,\frac{hp}{1-hp}\frac{1}{(1+\lambda )^2}\frac{1-F(\theta )}{f(\theta )}-\frac{1}{(1+\lambda )^2}\frac{1-F(\theta )}{f(\theta )}\right) \\= & {} -2\frac{1-hp}{hp}\frac{1}{(1+\lambda )^3}\frac{1-F(\theta )}{f(\theta )}\left( 2e^B+2\frac{hp}{1-hp}\frac{s(1-\lambda )}{1+\lambda }\right. \\&\quad \left. +\,\frac{hp}{1-hp}\frac{1}{1+\lambda }\frac{1-F(\theta )}{f(\theta )}+\frac{1}{1+\lambda }\frac{1-F(\theta )}{f(\theta )}\right) \le 0 \end{aligned}$$

1.1.3 Negative externality ($a=-1$)

Using the new investment and effort functions, the first-order conditions in the bundling case become:

• $I^{B}=\frac{p}{1-hp}\frac{s-(1+\lambda )\theta }{1+\lambda }-\frac{p}{1-hp}\frac{\lambda }{1+\lambda }\frac{F(\theta )}{f(\theta )}$,

• $e^{B}=h I^B-\frac{\lambda }{1+\lambda }\frac{F(\theta )}{f(\theta )}$.

Substituting in the government’s objective formula, we obtain the value function under bundling:

$$\begin{aligned} V^{B}= & {} \int _{\theta ^{l}}^{\theta ^{h}}\left\{ S_{0}+s I^{B}-(1+\lambda )\left[ \frac{(I^{B})^{2}}{2p}+\frac{(e^{B})^{2}}{2h}+F+(\theta -e^{B})I^{B}\right] -\lambda \frac{F(\theta )}{f(\theta )}\frac{e^{B}}{h}\right\} f(\theta )d\theta \\= & {} \int _{\theta ^{l}}^{\theta ^{h}}\left\{ S_{0}+I^{B}(s-(1+\lambda )(\theta -e^B))\right. \\&\quad \left. -\,(1+\lambda )\left[ \frac{(I^{B})^{2}}{2p}+\frac{(I^{B})^{2} h}{2}+\frac{1}{2h}\left( \frac{\lambda }{1+\lambda }\frac{F(\theta )}{f(\theta )}\right) ^2 \right. \right. \\&\quad \left. \left. -\,\frac{\lambda }{1+\lambda }\frac{F(\theta )}{f(\theta )}I^B +F \right] -\lambda \frac{F(\theta )}{f(\theta )}\left( I^B-\frac{1}{h}\frac{\lambda }{1+\lambda }\frac{F(\theta )}{f(\theta )}\right) \right\} f(\theta )d\theta \\= & {} \int _{\theta ^{l}}^{\theta ^{h}}\left\{ S_{0}-(1+\lambda ) F+ \frac{(1+\lambda )(1-hp)}{2p}(I^{B})^2+(1+\lambda )\frac{1}{2h}\left( \frac{\lambda }{1+\lambda }\frac{F(\theta )}{f(\theta )}\right) ^2 \right\} f(\theta )d\theta \\= & {} \int _{\theta ^{l}}^{\theta ^{h}}\left\{ S_{0}-(1+\lambda ) F+ \frac{p (1+\lambda )}{2(1-hp)}\left( \frac{s-(1+\lambda )\theta }{1+\lambda }\right) ^2+\frac{p(1+\lambda )}{2(1-hp)}\left( \frac{\lambda }{1+\lambda }\frac{F(\theta )}{f(\theta )}\right) ^2\right. \\&\quad \left. -\,\frac{(1+\lambda )p}{1-hp}\frac{s-(1+\lambda )\theta }{1+\lambda }\frac{\lambda }{1+\lambda }\frac{F(\theta )}{f(\theta )}+(1+\lambda )\frac{1}{2h}\left( \frac{\lambda }{1+\lambda }\frac{F(\theta )}{f(\theta )}\right) ^2 \right\} f(\theta )d\theta \\= & {} S_{0}-(1+\lambda ) F+ \frac{p (1+\lambda )}{2(1-hp)}\left( \frac{s^2+(1+\lambda )^2({\overline{\theta }}+\sigma ^2_{\theta })-2s(1+\lambda ){\overline{\theta }}}{(1+\lambda )^2}\right) \\&\quad -\,\frac{\lambda }{h}\int _{\theta ^{l}}^{\theta ^{h}}\left\{ \frac{F(\theta )}{f(\theta )} \left[ \frac{hp}{(1-hp)}\frac{s-(1+\lambda )\theta }{1+\lambda }-\right. \right. \\&\quad \left. \left. -\,\frac{hp}{2(1-hp)}\left( \frac{\lambda }{1+\lambda }\frac{F(\theta )}{f(\theta )}\right) -\frac{1}{2}\left( \frac{\lambda }{1+\lambda }\frac{F(\theta )}{f(\theta )}\right) \right] \right\} f(\theta )d\theta \\= & {} S_{0}-(1+\lambda ) F+ \frac{p (1+\lambda )}{2(1-hp)}\left( \frac{s-(1+\lambda ){\overline{\theta }}}{1+\lambda }\right) ^2+ \frac{p(1+\lambda )}{2(1-hp)}\sigma ^2_{\theta }\\&\quad -\,\frac{\lambda }{2h}\int _{\theta ^{l}}^{\theta ^{h}}\left\{ \frac{F(\theta )}{f(\theta )}\left( e^B+\frac{hp}{1-hp}\frac{s-(1+\lambda )\theta }{1+\lambda }\right) \right\} f(\theta )d\theta \end{aligned}$$

The new efforts functions applied to the unbundling case yield, respectively:

• $I^{U}=p\frac{s-(1+\lambda ){\overline{\theta }}}{(1+\lambda )(1-hp)}$,

• $e^{U}=h I^U$.

Substituting in the government’s objective formula, we obtain the value function under unbundling:

$$\begin{aligned} V^{U}= & {} \int _{\theta ^{l}}^{\theta ^{h}}\left\{ S_{0}+s I^{U}-(1+\lambda )\left[ \frac{(I^{U})^{2}}{2p}+\frac{(e^{U})^{2}}{2h}+F+I^{U}(\theta -e^{U})\right] \right\} f(\theta )d\theta \\= & {} \int _{\theta ^{l}}^{\theta ^{h}}\left\{ S_{0}+s I^{U}-(1+\lambda )\left[ \frac{(I^{U})^{2}}{2p}+\frac{(I^{U})^{2} h}{2}+F+I^{U}(\theta -I^{U} h)\right] \right\} f(\theta )d\theta \\= & {} \int _{\theta ^{l}}^{\theta ^{h}}\left\{ S_{0}-(1+\lambda ) F+ I^{U} (s-(1+\lambda )\theta )-(1+\lambda )\frac{(I^{U})^{2}}{2p}(1-hp)\right\} f(\theta )d\theta \\= & {} S_{0}-(1+\lambda ) F+ I^{U} (s-(1+\lambda ){\overline{\theta }})-(1+\lambda )\frac{(I^{U})^{2}}{2p}(1-hp)\\= & {} S_{0}-(1+\lambda ) F+ I^{U} \frac{I^U}{p}(1+\lambda )(1-hp)-(1+\lambda )\frac{(I^{U})^{2}}{2p}(1-hp)\\= & {} S_{0}-(1+\lambda ) F + (1+\lambda )\frac{(I^{U})^{2}}{2p}(1-hp)\\= & {} S_{0}-(1+\lambda ) F +\frac{p(1+\lambda )}{2(1-hp)}\left( \frac{s-(1+\lambda ){\overline{\theta }}}{1+\lambda }\right) ^2 \end{aligned}$$

The net welfare gain/loss of governments when using bundling is equal to:

$$\begin{aligned} V^{B}-V^{U}= & {} \frac{p(1+\lambda )}{2(1-hp)}\sigma ^2_{\theta }-\frac{\lambda }{2h}\int _{\theta ^{l}}^{\theta ^{h}}\left\{ \frac{F(\theta )}{f(\theta )}\left( e^B+\frac{hp}{1-hp}\frac{s-(1+\lambda )\theta }{1+\lambda }\right) \right\} f(\theta )d\theta \end{aligned}$$

where $e^B=\frac{hp}{1-hp}\frac{s-(1+\lambda )\theta }{1+\lambda }-\frac{hp}{1-hp}\frac{\lambda }{1+\lambda }\frac{F(\theta )}{f(\theta )}-\frac{\lambda }{1+\lambda }\frac{F(\theta )}{f(\theta )}$.

As a result, bundling dominates unbundling if and only if:

$$\begin{aligned} \sigma _{\theta }^{2}\ge & {} \frac{\lambda (1-hp)}{hp(1+\lambda )}\int _{\theta ^{l}}^{\theta ^{h}}\left\{ \frac{F(\theta )}{f(\theta )}\left( e^B+\frac{hp}{1-hp}\frac{s-(1+\lambda )\theta }{1+\lambda }\right) \right\} f(\theta )d\theta \\ \sigma _{\theta }^{2}\ge & {} \frac{\lambda (1-hp)}{hp(1+\lambda )}\int _{\theta ^{l}}^{\theta ^{h}}\left\{ \frac{F(\theta )}{f(\theta )}\left( \frac{hp}{1-hp}\frac{s-(1+\lambda )\theta }{1+\lambda }-\frac{hp}{1-hp}\frac{\lambda }{1+\lambda }\frac{F(\theta )}{f(\theta )}\right. \right. \\&\quad \left. \left. -\,\frac{\lambda }{1+\lambda }\frac{F(\theta )}{f(\theta )}+\frac{hp}{1-hp}\frac{s-(1+\lambda )\theta }{1+\lambda }\right) \right\} f(\theta )d\theta \end{aligned}$$

1.1.4 Proof of Corollary 2—negative externality

To study the sign of the derivative of $\sigma _{\theta }^{w}$ varies with $\lambda$, it is sufficient to study the derivative of the integrand in the previous inequality with respect to $\lambda$ that is equal to:

$$\begin{aligned}= & {} \frac{1-hp}{hp}\frac{1}{(1+\lambda )^2}\frac{F(\theta )}{f(\theta )}\left( e^B+\frac{hp}{1-hp}\frac{s-(1+\lambda )\theta }{1+\lambda }\right) \\&\quad +\,\frac{1-hp}{hp}\frac{\lambda }{1+\lambda }\frac{F(\theta )}{f(\theta )}\left( -\frac{hp}{1-hp}\frac{2s}{(1+\lambda )^2}\right. \\&\quad \left. -\,\frac{hp}{1-hp}\frac{1}{(1+\lambda )^2}\frac{F(\theta )}{f(\theta )}-\frac{1}{(1+\lambda )^2}\frac{F(\theta )}{f(\theta )}\right) \\= & {} 2\frac{1-hp}{hp}\frac{1}{(1+\lambda )^2}\frac{F(\theta )}{f(\theta )}\left( \frac{hp}{1-hp}\frac{s(1-\lambda )-(1+\lambda )\theta }{1+\lambda }\right. \\&\quad \left. -\,\frac{hp}{1-hp}\frac{\lambda }{1+\lambda }\frac{F(\theta )}{f(\theta )}-\frac{\lambda }{1+\lambda }\frac{F(\theta )}{f(\theta )}\right) \\= & {} 2\frac{1-hp}{hp}\frac{1}{(1+\lambda )^2}\frac{F(\theta )}{f(\theta )}\left( e^B-\frac{hp}{1-hp}\frac{\lambda s}{1+\lambda }\right) \end{aligned}$$

As a conclusion, the value of $\sigma _{\theta }^{2}$ can increase or decrease with $\lambda$. Precisely, the derivative is positive if $\lambda =0$, while it is negative if $\lambda =1$. The second derivative is equal to:

$$\begin{aligned}= & {} 4\frac{1-hp}{hp}\frac{1}{(1+\lambda )^3}\frac{F(\theta )}{f(\theta )}\left( -e^B+\frac{hp}{1-hp}\frac{\lambda s}{1+\lambda }\right) + \\&\quad +\,2\frac{1-hp}{hp}\frac{1}{(1+\lambda )^2}\frac{F(\theta )}{f(\theta )}\left( -\frac{hp}{1-hp}\frac{2s}{(1+\lambda )^2}\right. \\&\quad \left. -\,\frac{hp}{1-hp}\frac{1}{(1+\lambda )^2}\frac{F(\theta )}{f(\theta )}-\frac{1}{(1+\lambda )^2}\frac{F(\theta )}{f(\theta )}\right) \\= & {} -2\frac{1-hp}{hp}\frac{1}{(1+\lambda )^3}\frac{F(\theta )}{f(\theta )}\left( 2e^B+2\frac{hp}{1-hp}\frac{s(1-\lambda )}{1+\lambda }\right. \\&\quad \left. +\,\frac{hp}{1-hp}\frac{1}{1+\lambda }\frac{F(\theta )}{f(\theta )}+\frac{1}{1+\lambda }\frac{F(\theta )}{f(\theta )}\right) \le 0 \end{aligned}$$