Optimists, Pessimists, and the Precautionary Principle

Abstract

The precautionary principle has emerged as a leading guide to public decision-making about environmental risks under irreversibility and uncertainty. In this paper, we adopt a two-period model with irreversibility and agents differentiated by their degree of optimism to characterize the conditions under which the precautionary principle applies. We show in particular that it is more often applied if the decision-maker has an intermediate optimism index, if scientific research is more effective at reducing uncertainty, and if the same decision-maker makes the decisions for both periods. Moreover, we show that socially optimal decisions lead to apply the precautionary principle more often than an elected decision-maker does.

Appendices

Appendix 1: Proofs Without Irreversibility

Proof of Proposition 1

If $D_{1}=I$, to determine $D_{2}^{*},$ it is necessary to compare the individual evaluations of (I, I) and (I, A), taking into account the fact that the evaluation of the pair (I, A) depends on the result of the research.

Combining Table 1 and Eq. (8) gives:

$$\begin{aligned} W_{\alpha ,R}(I,I)& = 2V_{I} \\ W_{\alpha ,R}(I,A)& = q_{R}(\alpha )W(B,I,A)+(1-q_{R}(\alpha ))W(G,I,A) \\& = q_{R}(\alpha )\left( V_{I}+V-K\right) +(1-q_{R}(\alpha ))(V_{I}+V) \\& = V_{I}+V-Kq_{R}(\alpha ) \end{aligned}$$

Consequently, $D_{2}^{*}(I)=A\Leftrightarrow q_{R}(\alpha )<\frac{\Delta V}{K}.$

If $D_{1}=A$, to determine $D_{2}^{*},$ it is necessary to compare the individual evaluations of (A, I) and (A, A), taking into account the fact that both evaluations depend on the research result:

$$\begin{aligned} W_{\alpha ,R}(A,I)& = q_{R}(\alpha )W(B,A,I)+(1-q_{R}(\alpha ))W(G,A,I) \\& = q_{R}(\alpha )\left( V-K+V_{I}\right) +(1-q_{R}(\alpha ))(V+V_{I}) \\& = V-Kq_{R}(\alpha )+V_{I}\\ W_{\alpha ,R}(A,A)& = q_{R}(\alpha )W(B,A,A)+(1-q_{R}(\alpha ))W(G,A,A) \\& = q_{R}(\alpha )\left( 2V-2K\right) +(1-q_{R}(\alpha ))(2V) \\& = 2V-2Kq_{R}(\alpha ) \end{aligned}$$

Consequently, $D_{2}^{*}(A)=A\Leftrightarrow q_{R}(\alpha )<\frac{\Delta V}{K}.$

Then $q_{1}(\alpha )>q_{0}(\alpha )$ implies the proposition results. $\square$

Proof of Proposition 2

According to Equation (4), the welfare at the beginning of Period 1 if decisions $D_{1}$ and $D_{2}$ are taken is:

$$\begin{aligned} W_{\alpha }(D_{1},D_{2})& = q(\alpha )W(B,D_{1},D_{2})+(1-q(\alpha ))W(G,D_{1},D_{2}) \\& = q(\alpha )\left( W_{1}(D_{1})+W_{2}(D_{2})-K\left[ 1_{D_{1}=A}+1_{D_{2}=A} \right] \right) +(1-q(\alpha ))\left( W_{1}(D_{1})+W_{2}(D_{2})\right) \\& = W_{1}(D_{1})+W_{2}(D_{2})-Kq(\alpha )\left[ 1_{D_{1}=A}+1_{D_{2}=A}\right] \end{aligned}$$

We want to determine the value $D_{1}^{*}$ of $D_{1}$ which maximizes $W_{\alpha }(D_{1},D_{2}^{*})$, where $D_{2}^{*}$ is given by Proposition 1. Since $D_{2}^{*}$ does not depend on $D_{1}$ when there is no irreversibility (Proposition 1), it is equivalent maximizing $h(D_{1})=W_{1}(D_{1})-Kq(\alpha )1_{D_{1}=A}$

• $h(A)=V-Kq(\alpha )$ and $h(I)=V_{I}$, thus:

• $h(A)>h(I)\Leftrightarrow V-Kq(\alpha )>V_{I}$

• $h(A)>h(I)\Leftrightarrow U>q(\alpha )$

We can conclude that $D_{1}^{*}=A$ if $U>q(\alpha )$, and $D_{1}^{*}=I$ if $U<q(\alpha )$. $\square$

Appendix 2: Proofs with Irreversibility

Proof of Proposition 3

• (a) If $D_{1}=A$ then$\;D_{2}^{*}=A$ by irreversibility.

• (b) If $D_{1}=I$ both $D_{2}^{*}=A$ and $D_{2}^{*}=I$ are possible. To determine the preferred one, it is necessary to compare the corresponding individual evaluations, taking into account the fact that the evaluation of the pair (I, A) depends on the result of the research.

  $$\begin{aligned} W_{\alpha ,R}(I,I)& = 2V_{I} \\ W_{\alpha ,R}(I,A)& = q_{R}(\alpha )W(B,I,A)+(1-q_{R}(\alpha ))W(G,I,A) \\& = V_{I}+V-Kq_{R}(\alpha ) \end{aligned}$$

  Consequently, $D_{2}^{*}(I)=A\Leftrightarrow q_{R}(\alpha )<\frac{\Delta V}{K}.$

Then $q_{1}(\alpha )>q_{0}(\alpha )$ implies the results. $\square$

Proof of Proposition 4

To determine $D_{1}^{*}$, we have to compare $W_{\alpha }(I,D_{2}^{*}(I))$ and $W_{\alpha }(A,D_{2}^{*}(A)).$

Due to irreversibility, $W_{\alpha }(A,D_{2}^{*}(A))=W_{\alpha }(A,A)=2(V-Kq(\alpha ))$ where $q(\alpha )=\alpha q_{0}(\alpha )+(1-\alpha )q_{1}(\alpha )$ is the first period (ex ante) perception of the probability of the product being harmful.

$D_{2}^{*}(I)$ depends on the values of the parameters as shown in Proposition 3. Thus,

• 1. If $q_{0}(\alpha )>U$ then $D_{2}^{*}(I)=I$ for any R and $W_{\alpha }(I,D_{2}^{*}(I))=W_{\alpha }(I,I)=2V_{I};$

• 2. If $q_{1}(\alpha )<U$ then $D_{2}^{*}(I)=A$ for any R and $W_{\alpha }(I,D_{2}^{*}(I))=W_{\alpha }(I,A)=V_{I}+V-Kq(\alpha );$

• 3. If $q_{0}(\alpha )<U<q_{1}(\alpha )$ then $D_{2}^{*}(I)=I$ if $R=1$ and $D_{2}^{*}(I)=A$ if $R=0$ and according to Eq. (11) we have:

  $$\begin{aligned} W_{\alpha }(D_{1},D_{2}^{*}(D_{1}))& = W_{\alpha }(I,D_{2}^{*}(I))=\alpha W_{\alpha ,0}(I,A)+(1-\alpha )W_{\alpha ,1}(I,I)\\& = \alpha \left[ V_{I}+V-Kq_{0}(\alpha )\right] +(1-\alpha )\left[ 2V_{I} \right] =(2-\alpha )V_{I}+\alpha (V-Kq_{0}(\alpha )). \end{aligned}$$

Comparing $W_{\alpha }(A,A)$ and $W_{\alpha }(I,D_{2}^{*}(I))$ in each of the previous cases, we obtain:

• 1. If $q_{0}(\alpha )>U$ then $D_{1}^{*}=A\Leftrightarrow 2V-2Kq(\alpha )>2V_{I}$, i.e., $\frac{\Delta V}{K}>q(\alpha );$

• 2. If $q_{1}(\alpha )<U$ then $D_{1}^{*}=A\Leftrightarrow 2V-2Kq(\alpha )>V_{I}+V-Kq(\alpha )$

  i.e., $D_{1}^{*}=A\Leftrightarrow \frac{\Delta V}{K}>q(\alpha );$

• 3. If $q_{0}(\alpha )<U<q_{1}(\alpha )$ then $D_{1}^{*}=A\Leftrightarrow 2V-2Kq(\alpha )>(2-\alpha )V_{I}+\alpha (V-Kq_{0}(\alpha ))$,

  i.e., $D_{1}^{*}=A\Leftrightarrow \frac{\Delta V}{K}>\frac{2q(\alpha )-\alpha q_{0}(\alpha )}{2-\alpha }=M(\alpha )$

Since $q(\alpha )=\alpha q_{0}(\alpha )+(1-\alpha )q_{1}(\alpha )$, and $q_{0}(\alpha )\le q_{1}(\alpha )$,we have $q(\alpha )\in \left[ q_{0}(\alpha ),q_{1}(\alpha )\right]$.

This implies that $\left[ q_{0}(\alpha )>\frac{\Delta V}{K}\text { and }\frac{ \Delta V}{K}>q(\alpha )\right]$ is impossible, and that $q_{1}(\alpha )< \frac{\Delta V}{K}$ implies $\frac{\Delta V}{K}>q(\alpha )$.

Consequently,

• 1. If $q_{0}(\alpha )>U$ then $D_{1}^{*}=I;$

• 2. If $q_{1}(\alpha )<U$ then $D_{1}^{*}=A;$

• 3. If $q_{0}(\alpha )<U<q_{1}(\alpha )$ then $D_{1}^{*}=A$$\Leftrightarrow \frac{\Delta V}{K}>M(\alpha ).$

It is easy to show that $M(\alpha )=\frac{2q(\alpha )-\alpha q_{0}(\alpha )}{ 2-\alpha }\in \left[ q_{0}(\alpha ),q_{1}(\alpha )\right]$ which gives the first two results of the proposition.

Moreover,

$$\begin{aligned} M(\alpha )& = \frac{(2-\alpha )q(\alpha )+\alpha \left( q(\alpha )-q_{0}(\alpha )\right) }{2-\alpha }=q(\alpha )+\frac{\alpha }{2-\alpha } \left( q(\alpha )-q_{0}(\alpha )\right) \\& = q(\alpha )+\frac{\alpha (1-\alpha )}{2-\alpha }\left( q_{1}(\alpha )-q_{0}(\alpha )\right) =q(\alpha )+\frac{\alpha (1-\alpha )}{2-\alpha } \left( \delta -\delta ^{\prime }\right) \end{aligned}$$

$\square$

Appendix 3: Political Decision

Proof of Proposition 5

We have to compare $W_{\alpha _{1}}(I,D_{2}^{*}(\alpha _{2},I))$ and $W_{\alpha _{1}}(A,D_{2}^{*}(\alpha _{2},A)).$

Due to irreversibility, $W_{\alpha _{1}}(A,D_{2}^{*}(\alpha _{2},A))=W_{\alpha _{1}}(A,A)=2(V-Kq(\alpha _{1}))$ where $q(\alpha _{1})=\alpha _{1}q_{0}(\alpha _{1})+(1-\alpha _{1})q_{1}(\alpha _{1})$ is the first period (ex ante) perception of the probability of the product being harmful.

$D_{2}^{*}(\alpha _{2},I)$ depends on the values of the parameters as shown in Proposition 3 with $\alpha =\alpha _{2}$. Thus,

• 1. If $q_{0}(\alpha _{2})>U$ then $D_{2}^{*}(\alpha _{2},I)=I$ for any R and $W_{\alpha _{1}}(I,D_{2}^{*}(\alpha _{2},I))=W_{\alpha _{1}}(I,I)=2V_{I};$

• 2. If $q_{1}(\alpha _{2})<U$ then $D_{2}^{*}(\alpha _{2},I)=A$ for any R and $W_{\alpha _{1}}(I,D_{2}^{*}(\alpha _{2},I))=W_{\alpha _{1}}(I,A)=V_{I}+V-Kq(\alpha _{1});$

• 3. If $q_{0}(\alpha _{2})<U<q_{1}(\alpha _{2})$ then $D_{2}^{*}(\alpha _{2},I)=I$ if $R=1$ and $D_{2}^{*}(\alpha _{2},I)=A$ if $R=0$ thus $W_{\alpha _{1}}(I,D_{2}^{*}(\alpha _{2},I))=\alpha _{1}W_{\alpha _{1},0}(I,A)+(1-\alpha _{1})W_{\alpha _{1},1}(I,I)$$=\alpha _{1}\left[ V_{I}+V-Kq_{0}(\alpha _{1})\right] +(1-\alpha _{1})\left[ 2V_{I}\right] =(2-\alpha _{1})V_{I}+\alpha _{1}(V-Kq_{0}(\alpha _{1})).$

Comparing $W_{\alpha _{1}}(A,A)$ and $W_{\alpha _{1}}(I,D_{2}^{*}(\alpha _{2},I))$ in each of the previous cases, we obtain:

• 1. If $q_{0}(\alpha _{2})>U$ then $D_{1}^{*}=A\Leftrightarrow 2V-2Kq(\alpha _{1})>2V_{I}$, i.e., $\frac{\Delta V}{K}>q(\alpha _{1});$

• 2. If $q_{1}(\alpha _{2})<U$ then $D_{1}^{*}=A\Leftrightarrow 2V-2Kq(\alpha _{1})>V_{I}+V-Kq(\alpha _{1})$,

  i.e., $D_{1}^{*}=A\Leftrightarrow \frac{\Delta V}{K}>q(\alpha _{1});$

• 3. If $q_{0}(\alpha _{2})<U<q_{1}(\alpha _{2})$ then

  $$\begin{aligned} D_{1}^{*}& = A\Leftrightarrow 2V-2Kq(\alpha _{1})>(2-\alpha _{1})V_{I}+\alpha _{1}(V-Kq_{0}(\alpha _{1})),~~\hbox {i.e.,}~~ \\ D_{1}^{*}& = A\Leftrightarrow \frac{\Delta V}{K}>\frac{2q(\alpha _{1})-\alpha _{1}q_{0}(\alpha _{1})}{2-\alpha _{1}}=M(\alpha _{1}) \end{aligned}$$

  As $q(\alpha _{1})<M(\alpha _{1})$, we can deduce then in the three cases:

$\frac{\Delta V}{K}<q(\alpha _{1})$ implies $D_{1}^{*}=I$

$\frac{\Delta V}{K}>M(\alpha _{1})$ implies $D_{1}^{*}=A$

If $q(\alpha _{1})<\frac{\Delta V}{K}<M(\alpha _{1})$, then:

$$\begin{aligned}&\frac{\Delta V}{K}\in \left( q_{0}(\alpha _{2}),q_{1}(\alpha _{2})\right) \Rightarrow D_{1}^{*}=I \\&\frac{\Delta V}{K}\notin \left[ q_{0}(\alpha _{2}),q_{1}(\alpha _{2})\right] \Rightarrow D_{1}^{*}=A \end{aligned}$$

which gives the result of the proposition. $\square$

Proof of Corollary 3

For $\alpha _{2}$ given, $0<\alpha _{2}<1$, we define $\alpha ^{\prime }$ and $\alpha ''$ by $q_{0}(\alpha _{2})=q(\alpha ^{\prime })=M(\alpha '')$

We know that $q_{0}$, q, and M are decreasing functions, with $q_{0}<q<M$ , so $\alpha _{2}<\alpha ^{\prime }<\alpha ''$.

Assume $\alpha _{1}\in \left( \alpha ^{\prime },\alpha ''\right)$; we have:

• If $U<q(\alpha _{1})$, then $D_{1}^{*}=I$ according to Proposition 5.

• If $U>M(\alpha _{1})$, then $D_{1}^{*}=A$ according to Proposition 5.

• Assume that $q(\alpha _{1})<U<M(\alpha _{1})$.

We have $q(\alpha _{1})<q_{0}(\alpha _{2})<M(\alpha _{1})$ since:

$$\alpha _{1}>\alpha ^{\prime }\Rightarrow q(\alpha _{1})<q(\alpha ^{\prime })=q_{0}(\alpha _{2}),$$

and

$$\alpha _{1}<\alpha \Rightarrow M(\alpha _{1})>M(\alpha '')=q_{0}(\alpha _{2})$$

We can then distinguish two sub-cases:

First sub-case: $q(\alpha _{1})<U<q_{0}(\alpha _{2})$, then $D_{1}^{*}=A$ according to Proposition 5.

Second sub-case: $q_{0}(\alpha _{2})<U<M(\alpha _{1})$, then $D_{1}^{*}=I$ according to Proposition 5. $\square$

Proof of Propositions 6 and 7

• (a) Majority voting on $D_{2}$

  For $D_{1}$ given, Proposition 3 gives the preferences of an agent $\alpha$ on $D_{2}$.

  • If $D_{1}=A$, we have unanimity for $D_{2}^{*}(\alpha )=A$ since there is no other choice.

  • If $D_{1}=I$ and $R=0$, then $D_{2}^{*}(\alpha )=I$ iff $U<q_{0}(\alpha )$

  • If $D_{1}=I$ and $R=1$, then $D_{2}^{*}(\alpha )=I$ iff $U<q_{1}(\alpha )$

  $q_{0}$ and $q_{1}$ are affine functions, thus the median voter theorem applies, and $D_{2}^{**}=D_{2}^{*}(\alpha _{m_{2}})$

• (b) Majority voting on $D_{1}$

  Proposition 5 gives the preferences of an agent $\alpha _{1}$ on $D_{1}$ if $\alpha _{2}=\alpha _{m_{2}}$Three cases can be distinguished:

  • Case 1: If $U<q_{0}(\alpha _{2})$, then $D_{1}^{*}(\alpha _{1})=I$ iff $U<q(\alpha _{1})$

  • Case 2: If $U>q_{1}(\alpha _{2})$, then $D_{1}^{*}(\alpha _{1})=I$ iff $U<q(\alpha _{1})$

  • Case 3: If $q_{0}(\alpha _{2})<U<q_{1}(\alpha _{2})$, then $D_{1}^{*}(\alpha _{1})=I$ iff $U<M(\alpha _{1})$

In each of the three cases, the median voter theorem can be applied because q and M are continuous decreasing functions. Thus $D_{1}^{**}=D_{1}^{*}(\alpha _{m_{1}})$.

The median voter theorem applies for each period. Proposition 6 is then obtained by Corollary 2 with $\alpha =\alpha _{m}$. Proposition 7 is obtained by Proposition 3 with $\alpha =\alpha _{m_{2}}$ and Proposition 5 with $\alpha _{1}=\alpha _{m_{1}}$ and $\alpha _{2}=\alpha _{m_{2}}$. $\square$

Proof of Proposition 8

The political decision is given in Proposition 6 with $\alpha _{m}=\overline{ \alpha }$.

The socially optimal decisions are evaluated with a utilitarian criterion:

$${\mathcal {U}}(D_{1},D_{2})=\dfrac{1}{n}\sum _{i=1}^{n}W_{\alpha (i)}(D_{1},D_{2}).$$

• (a) Second-period socially optimal decisions

• If $D_{1}=A$ then $D_{2}^{opt}(A)=A$ due to irreversibility;

• If $D_{1}=I$, then to determine the optimal decision it is necessary to take into account the fact that the evaluation of the decision (I, A) depends on the result of the research.

  $$\begin{aligned} {\mathcal {U}}_{R}(I,I)& = \dfrac{1}{n}\sum _{i=1}^{n}W_{\alpha (i),R}(I,I)=2V_{I}=W_{{\overline{\alpha }},R}(I,I) \\ {\mathcal {U}}_{R}(I,A)& = \dfrac{1}{n}\sum _{i=1}^{n}W_{\alpha (i),R}(I,A)=\dfrac{ 1}{n}\sum _{i=1}^{n}\left[ V_{I}+V-Kq_{R}(\alpha (i))\right] \\& = V_{I}+V-Kq_{R}({\overline{\alpha }})=W_{{\overline{\alpha }} ,R}(I,A) \end{aligned}$$

due to the linearity of $q_{R}$ with respect to $\alpha .$

$${\mathcal {U}}_{R}(I,A)>{\mathcal {U}}_{R}(I,I)~\Leftrightarrow ~U>q_{R}( {\overline{\alpha }}).$$

It follows that given $D_{1}=I$, the utilitarian second-period decisions are the same as the political decisions when the median voter has an optimism index ${\overline{\alpha }}.$

From Proposition 3,

• 1. If $q_{0}({\overline{\alpha }})>U$ then $D_{2}^{opt}(I)=D_{2}^{**}(I)=I$ for any R; 

• 2. If $q_{1}({\overline{\alpha }})<U$ then $D_{2}^{opt}(I)=D_{2}^{**}(I)=A$ for any R; 

• 3. If $q_{0}({\overline{\alpha }})<U<q_{1}({\overline{\alpha }})$ then $D_{2}^{opt}(I)=D_{2}^{**}(I)=I$ if $R=1$ and $D_{2}^{opt}(I)=D_{2}^{ **}(I)=A$ if $R=0$

(b) First-period optimal decisions.

We have to compare ${\mathcal {U}}(A,D_{2}^{opt}(A))$ and ${\mathcal {U}} (I,D_{2}^{opt}(I)).$

Due to irreversibility,

$${\mathcal {U}}(A,D_{2}^{opt}(A))={\mathcal {U}}(A,A)=\dfrac{1}{n} \sum _{i=1}^{n}W_{\alpha (i)}(A,A)=\dfrac{1}{n}\sum _{i=1}^{n}2\left[ V-Kq(\alpha (i))\right] =2\left[ V-Kq({\overline{\alpha }})\right] =W_{ {\overline{\alpha }}}(A,A)$$

From (a), ${\mathcal {U}}(I,D_{2}^{opt}(I))$ depends on U, $q_{0}(\overline{ \alpha })$ and $q_{1}({\overline{\alpha }}).$

More specifically:

• 1. If $q_{0}({\overline{\alpha }})>U$ then $D_{2}^{opt}(I)=D_{2}^{**}(I)=I$ for any R and ${\mathcal {U}}(I,D_{2}^{opt}(I))={\mathcal {U}} (I,I)=2V_{I}=W_{{\overline{\alpha }}}(I,I)$

  ${\mathcal {U}}(A,D_{2}^{opt}(A))-{\mathcal {U}}(I,D_{2}^{opt}(I))=W_{\overline{ \alpha }}(A,A)-W_{{\overline{\alpha }}}(I,I)=2V-2Kq({\overline{\alpha }} )-2V_{I}<0$ if $q_{0}({\overline{\alpha }})>U$,

  thus $D_{1}^{opt}=D_{1}^{**}=I$; then $D^{opt}=D^{**}=(I,I)$.

• 2. If $q_{1}({\overline{\alpha }})<U$ then $D_{2}^{opt}(I)=D_{2}^{**}(I)=A$ for any R and ${\mathcal {U}}(I,D_{2}^{opt}(I))=\dfrac{1}{n} \sum _{i=1}^{n}W_{\alpha (i)}(I,A)=\dfrac{1}{n}\sum _{i=1}^{n}\left( V_{I}+V-Kq(\alpha (i))\right) =V_{I}+V-Kq({\overline{\alpha }})=W_{\overline{ \alpha }}(I,A)$

  ${\mathcal {U}}(A,D_{2}^{opt}(A))-{\mathcal {U}}(I,D_{2}^{opt}(I))=W_{\overline{ \alpha }}(A,A)-W_{{\overline{\alpha }}}(I,A)$

  $=2V-2Kq({\overline{\alpha }})-V_{I}-V+Kq({\overline{\alpha }})=V-V_{I}-Kq( {\overline{\alpha }})>0$ if $q_{1}({\overline{\alpha }})<U.$

  thus $D_{1}^{opt}=D_{1}^{**}=A$; then $D^{opt}=D^{**}=(A,A)$.

• 3. If $q_{0}({\overline{\alpha }})<U<q_{1}({\overline{\alpha }})$ then $D_{2}^{opt}(I)=D_{2}^{**}(I)=I$ if $R=1$ and $D_{2}^{opt}(I)=D_{2}^{ **}(I)=A$ if $R=0$. Moreover, according to Equation (11)

  $$\begin{aligned}&{\mathcal {U}}(I,D_{2}^{opt}(I))=\dfrac{1}{n}\sum _{i=1}^{n}W_{\alpha (i)}(I,D_{2}^{opt}(I)) \\&=\dfrac{1}{n}\sum _{i=1}^{n}\left[ \alpha (i)W_{\alpha (i),0}(I,D_{2}^{opt}(I))+(1-\alpha (i))W_{\alpha (i),1}(I,D_{2}^{opt}(I)) \right] \\&=\dfrac{1}{n}\sum _{i=1}^{n}\left[ \alpha (i)W_{\alpha (i),0}(I,A)+(1-\alpha (i))W_{\alpha (i),1}(I,I)\right] \\&=\dfrac{1}{n}\sum _{i=1}^{n}\left[ \alpha (i)\left( V_{I}+V-Kq_{0}(\alpha (i))\right) +(1-\alpha (i))\left( 2V_{I}\right) \right] \\&=\dfrac{1}{n}\sum _{i=1}^{n}\left[ \alpha (i)(V-V_{I})+2V_{I}-K\alpha (i)q_{0}(\alpha (i))\right] \\&={\overline{\alpha }}(V-V_{I})+2V_{I}-K\dfrac{1}{n}\sum _{i=1}^{n}\alpha (i)q_{0}(\alpha (i)) \\&{\mathcal {U}}(A,D_{2}^{opt}(A))={\mathcal {U}}(A,A)=2\left[ V-Kq(\overline{ \alpha })\right] =W_{{\overline{\alpha }}}(A,A) \end{aligned}$$

Moreover

$$\begin{aligned}&W_{{\overline{\alpha }}}(I,D_{2}^{opt}(I))={\overline{\alpha }}W_{\overline{ \alpha },0}(I,D_{2}^{opt}(I))+(1-{\overline{\alpha }})W_{{\overline{\alpha }} ,1}(I,D_{2}^{opt}(I)) \\&={\overline{\alpha }}W_{{\overline{\alpha }},0}(I,A)+(1-{\overline{\alpha }})W_{ {\overline{\alpha }},1}(I,I)={\overline{\alpha }}\left( V_{I}+V-Kq_{0}(\overline{ \alpha })\right) +(1-{\overline{\alpha }})(2V_{I}) \\&=2V_{I}+{\overline{\alpha }}(V-V_{I})-K{\overline{\alpha }}q_{0}(\overline{ \alpha }) \end{aligned}$$

Thus

$$\begin{aligned} {\mathcal {U}}(A,D_{2}^{opt}(A))-{\mathcal {U}}(I,D_{2}^{opt}(I))=(2-\overline{ \alpha })(V-V_{I})-2Kq({\overline{\alpha }})+K\dfrac{1}{n}\sum _{i=1}^{n}\alpha (i)q_{0}(\alpha (i)) \\ W_{{\overline{\alpha }}}(A,D_{2}^{opt}(A))-W_{{\overline{\alpha }} }(I,D_{2}^{opt}(I))=(2-{\overline{\alpha }})(V-V_{I})-2Kq({\overline{\alpha }})+K {\overline{\alpha }}q_{0}({\overline{\alpha }}) \end{aligned}$$

Since $U=\frac{V-V_{I}}{K}$, and setting $S=\frac{1}{2-{\overline{\alpha }}} \left[ 2q({\overline{\alpha }})-\dfrac{1}{n}\sum _{i=1}^{n}\alpha (i)q_{0}(\alpha (i))\right]$, we have:

$$\begin{aligned}&W_{{\overline{\alpha }}}(A,D_{2}^{opt}(A))-W_{{\overline{\alpha }} }(I,D_{2}^{opt}(I))>0\Leftrightarrow U>\frac{1}{2-{\overline{\alpha }}}\left[ 2q({\overline{\alpha }})-{\overline{\alpha }}q_{0}({\overline{\alpha }})\right] =M( {\overline{\alpha }}) \\&{\mathcal {U}}(A,D_{2}^{opt}(A))-{\mathcal {U}}(I,D_{2}^{opt}(I))>0 \Leftrightarrow U>\frac{1}{2-{\overline{\alpha }}}\left[ 2q({\overline{\alpha }} )-\dfrac{1}{n}\sum _{i=1}^{n}\alpha (i)q_{0}(\alpha (i))\right] =S \end{aligned}$$

To achieve the proof of Proposition 8, it is sufficient to set ${\widehat{M}} =\min (S;q_{1}({\overline{\alpha }}))$ and to prove that $S>M(\overline{\alpha })$.

$$\begin{aligned}&S-M({\overline{\alpha }})=\frac{1}{2-{\overline{\alpha }}}\left[ 2q(\overline{ \alpha })-\dfrac{1}{n}\sum _{i=1}^{n}\alpha (i)q_{0}(\alpha (i))\right] - \frac{1}{2-{\overline{\alpha }}}\left[ 2q({\overline{\alpha }})-\overline{\alpha }q_{0}({\overline{\alpha }})\right] \\&S-M({\overline{\alpha }})=\frac{1}{2-{\overline{\alpha }}}\left[ \overline{ \alpha }q_{0}({\overline{\alpha }})-\dfrac{1}{n}\sum _{i=1}^{n}\alpha (i)q_{0}(\alpha (i))\right] =\frac{\delta ^{\prime }}{2-{\overline{\alpha }}} \left[ g({\overline{\alpha }})-\dfrac{1}{n}\sum _{i=1}^{n}g(\alpha (i))\right] \end{aligned}$$

Setting $g(\alpha )=\alpha (1-\alpha )$, because $q_{0}(\alpha )=(1-\delta )p+\delta ^{\prime }(1-\alpha )$,

and $\left[ g({\overline{\alpha }})-\dfrac{1}{n}\sum _{i=1}^{n}g(\alpha (i)) \right] >0$ since g is a concave function. $\square$

Appendix 4: Proofs with Costly Research

Proof of Proposition 9

• a. If $D_{1}=A$ then $D_{2}^{*}=A$ by irreversibility of authorization.

• b. If $D_{1}=I^{res}$, to determine $D_{2}^{*}$ we compare $W_{\alpha ,R}(I^{res},I)$ and $W_{\alpha ,R}(I^{res},A)$.

  $$\begin{aligned}&W_{\alpha ,R}(I^{res},I)=2V_{I}-C \\&W_{\alpha ,R}(I^{res},A)=q_{R}(\alpha )(V_{I}+V-K-C)+(1-q_{R}(\alpha ))(V_{I}+V-C) \\&=V_{I}+V-C-Kq_{R}(\alpha )\\&D_{2}^{*}(I^{res})=A\Leftrightarrow V_{I}+V-C-Kq_{R}(\alpha )>2V_{I}-C\\&D_{2}^{*}(I^{res})=A\Leftrightarrow U>q_{R}(\alpha )\,{\rm since} \,U=\frac{ V-V_{I}}{K} \end{aligned}.$$

• c. If $D_{1}=I^{no}$, to determine $D_{2}^{*}$ we compare $W_{\alpha }(I^{no},I)$ and $W_{\alpha }(I^{no},A)$.

  $$\begin{aligned}&W_{\alpha }(I^{no},I)=2V_{I} \\&W_{\alpha }(I^{no},A)=q(\alpha )(V_{I}+V-K)+(1-q(\alpha ))(V_{I}+V)=V_{I}+V-Kq(\alpha ) \\&D_{2}^{*}(I^{no})=A\Leftrightarrow V_{I}+V-Kq(\alpha )>2V_{I} \\&D_{2}^{*}(I^{no})=A\Leftrightarrow U>q(\alpha ) \end{aligned}$$

  $\square$

Proof of Proposition 10

We set $w_{\alpha }(D_{1})=W_{\alpha }(D_{1},D_{2}^{*}(D_{1})$. To determine $D_{1}^{*}$, we have to compare $w_{\alpha }(A)$, $w_{\alpha }(I^{res})$ and $w_{\alpha }(I^{no})$.

$$w_{\alpha }(A)=W_{\alpha }(A,A)=2V-2Kq(\alpha )$$

According to Proposition 9,

$$\begin{aligned} w_{\alpha }(I^{no})& = W_{\alpha }(I^{no},I)=2V_{I}\hbox { if }U<q(\alpha )\\ w_{\alpha }(I^{no})& = W_{\alpha }(I^{no},A)=V_{I}+V-Kq(\alpha )\,if \,U>q(\alpha )\end{aligned}$$

Moreover, $w_{\alpha }(I^{res})=W_{\alpha }(I^{res},D_{2}^{*}(I^{res}))$ thus:

• If $U<q_{0}(\alpha )$, then $w_{\alpha }(I^{res})=W_{\alpha }(I^{res},I)=2V_{I}-C$

• If $U>q_{1}(\alpha )$, then $w_{\alpha }(I^{res})=W_{\alpha }(I^{res},A)=V_{I}+V-Kq(\alpha )-C$

• If $q_{0}(\alpha )<U<q_{1}(\alpha )$, then $w_{\alpha }(I^{res})=\alpha W_{\alpha ,0}(I^{res},A)+(1-\alpha )W_{\alpha ,1}(I^{res},I)$$=\alpha \left[ V_{I}+V-C-Kq_{0}(\alpha )\right] +(1-\alpha )\left[ 2V_{I}-C \right] =(2-\alpha )V_{I}+\alpha (V-Kq_{0}(\alpha ))-C$

We have to study the three cases $U<q_{0}(\alpha )$, $U>q_{1}(\alpha )$ and $q_{0}(\alpha )<U<q_{1}(\alpha )$.

• 1. If $U<q_{0}(\alpha )$, then $w_{\alpha }(I^{res})=2V_{I}-C<2V_{I}=w_{\alpha }(I^{no})$

  $$\begin{aligned}&D_{1}^{*}=I^{no}\Leftrightarrow w_{\alpha }(I^{no})>w_{\alpha }(A) \\&D_{1}^{*}=I^{no}\Leftrightarrow 2V_{I}>2V-2Kq(\alpha ) \\&D_{1}^{*}=I^{no}\Leftrightarrow q(\alpha )>U\,\,\hbox {true here since} \,\,q(\alpha )>q_{0}(\alpha ) \end{aligned}$$

  Conclusion: $D_{1}^{*}=I^{no}$ if $U<q_{0}(\alpha )$.

• 2. If $U>q_{1}(\alpha )$, then $w_{\alpha }(I^{res})=V_{I}+V-Kq(\alpha )-C<V_{I}+V-Kq(\alpha )=w_{\alpha }(I^{no})$

  $$\begin{aligned}&D_{1}^{*}=A\Leftrightarrow w_{\alpha }(A)>w_{\alpha }(I^{no}) \\&D_{1}^{*}=A\Leftrightarrow 2V-2Kq(\alpha )>V_{I}+V-Kq(\alpha ) \\&D_{1}^{*}=A\Leftrightarrow U>q(\alpha )\,\, \hbox {true here because} \,\, q_{1}(\alpha )>q(\alpha )\\ \end{aligned}$$

  Conclusion: $D_{1}^{*}=A$ if $U>q_{1}(\alpha )$.

• 3. If $q_{0}(\alpha )<U<q_{1}(\alpha )$, then:

  $$w_{\alpha }(A)>w_{\alpha }(I^{no})\Leftrightarrow U>q(\alpha )$$

  and

  $$\begin{aligned}&w_{\alpha }(A)>w_{\alpha }(I^{res})\Leftrightarrow 2V-2Kq(\alpha )>(2-\alpha )V_{I}+\alpha (V-Kq_{0}(\alpha ))-C \\&w_{\alpha }(A)>w_{\alpha }(I^{res})\Leftrightarrow (2-\alpha )(V-V_{I})>2Kq(\alpha )-\alpha Kq_{0}(\alpha )-C \\&w_{\alpha }(A)>w_{\alpha }(I^{res})\Leftrightarrow U>M_{C}(\alpha ), \end{aligned}$$

  setting $M_{C}(\alpha )=\frac{1}{(2-\alpha )}\left[ 2q(\alpha )-\alpha q_{0}(\alpha )-\frac{C}{K}\right] =M(\alpha )-\frac{C}{(2-\alpha )K} =q(\alpha )+\frac{\alpha (1-\alpha )(\delta -\delta ^{\prime })-\frac{C}{K}}{ 2-\alpha }$

and if $U<q(\alpha )$:

$$\begin{aligned}&w_{\alpha }(I^{res})>w_{\alpha }(I^{no})\Leftrightarrow (2-\alpha )V_{I}+\alpha (V-Kq_{0}(\alpha ))-C>2V_{I} \\&w_{\alpha }(I^{res})>w_{\alpha }(I^{no})\Leftrightarrow \alpha (V-V_{I})>\alpha Kq_{0}(\alpha )+C \\&w_{\alpha }(I^{res})>w_{\alpha }(I^{no})\Leftrightarrow U>q_{0}(\alpha )+ \frac{C}{\alpha K} \end{aligned}$$

This implies that if $q_{0}(\alpha )<U<q_{1}(\alpha )$:

$$\begin{aligned}&D_{1}^{*}=A\Leftrightarrow \left[ U>q(\alpha )\text { and } U>M_{C}(\alpha )\right] \\&D_{1}^{*}=I^{no}\Leftrightarrow \left[ U<q(\alpha )\text { and } U<q_{0}(\alpha )+\frac{C}{\alpha K}\right] \\&D_{1}^{*}=I^{res}\Leftrightarrow \left[ q_{0}(\alpha )+\frac{C}{\alpha K }<U<M_{C}(\alpha )\right] \end{aligned}$$

To find $D_{1}^{*}$, we just need to compare $M_{C}(\alpha )$, $q(\alpha )$ and $q_{0}(\alpha )+\frac{C}{\alpha K}$.

First Case. We assume that research is very expensive, i.e., $C>(\frac{ \delta -\delta ^{\prime }}{4})K.$

Comparing $q(\alpha )$ and $M_{C}(\alpha )$:

$$\begin{aligned}&M_{C}(\alpha )<q(\alpha )\Leftrightarrow q(\alpha )+\frac{\alpha (1-\alpha )(\delta -\delta ^{\prime })}{2-\alpha }-\frac{C}{(2-\alpha )K}<q(\alpha ) \\&M_{C}(\alpha )<q(\alpha )\Leftrightarrow \alpha (1-\alpha )(\delta -\delta ^{\prime })<\frac{C}{K} \end{aligned}$$

Let $h(\alpha )=\alpha (1-\alpha )(\delta -\delta ^{\prime })$ on $\alpha \in [0;1]$. Then $\max _{0\le \alpha \le 1}h(\alpha )=h(\frac{1}{2})= \frac{\delta -\delta ^{\prime }}{4}$

As we are in the first case, $\max _{0\le \alpha \le 1}h(\alpha )<\frac{C}{K }$, i.e., $h(\alpha )<\frac{C}{K}$, $\forall \alpha$.

This implies that $M_{C}(\alpha )<q(\alpha )$, $\forall \alpha$.

Comparing $q(\alpha )$ and $q_{0}(\alpha )+\frac{C}{\alpha K}$:

$$\begin{aligned}&q(\alpha )<q_{0}(\alpha )+\frac{C}{\alpha K}\Leftrightarrow (1-\delta )p+\delta (1-\alpha )<(1-\delta )p+\delta ^{\prime }(1-\alpha )+\frac{C}{ \alpha K} \\&q(\alpha )<q_{0}(\alpha )+\frac{C}{\alpha K}\Leftrightarrow (\delta -\delta ^{\prime })(1-\alpha )<\frac{C}{\alpha K} \\&q(\alpha )<q_{0}(\alpha )+\frac{C}{\alpha K}\Leftrightarrow h(\alpha )< \frac{C}{K} \,always true since we are in the first case.\end{aligned}$$

This implies that $q(\alpha )<q_{0}(\alpha )+\frac{C}{\alpha K}$, $\forall \alpha$.

Conclusion: in the first case, $M_{C}(\alpha )<q(\alpha )<q_{0}(\alpha )+ \frac{C}{\alpha K}$, $\forall \alpha$.

Thus $D_{1}^{*}=A$ if $U>q(\alpha )$, and $D_{1}^{*}=I^{no}$ if $U<q(\alpha )$.

Second Case. We assume that research is not very expensive, i.e., $C\le ( \frac{\delta -\delta ^{\prime }}{4})K.$

Comparing $q(\alpha )$ and $M_{C}(\alpha )$:

$$\begin{aligned}&M_{C}(\alpha )\ge q(\alpha )\Leftrightarrow q(\alpha )+\frac{\alpha (1-\alpha )(\delta -\delta ^{\prime })}{2-\alpha }-\frac{C}{(2-\alpha )K} \ge q(\alpha ) \\&M_{C}(\alpha )\ge q(\alpha )\Leftrightarrow \alpha (1-\alpha )(\delta -\delta ^{\prime })\ge \frac{C}{K} \\&M_{C}(\alpha )\ge q(\alpha )\Leftrightarrow h(\alpha )\ge \frac{C}{K} \end{aligned}$$

Here $\max _{0\le \alpha \le 1}h(\alpha )=h(\frac{1}{2})=\frac{\delta -\delta ^{\prime }}{4}\ge \frac{C}{K}$.

Then there exists $\beta \in [0;\frac{1}{2}]$ with: $h(\alpha )\ge \frac{C}{K}$ if and only if $\alpha \in \left[\frac{1}{2}-\beta ;\frac{1}{2 }+\beta \right]$.

This means here that $M_{C}(\alpha )\ge q(\alpha )\Leftrightarrow \alpha \in \left[\frac{1}{2}-\beta ;\frac{1}{2}+\beta \right]$.

Comparing $q(\alpha )$ and $q_{0}(\alpha )+\frac{C}{\alpha K}$:

$$\begin{aligned}&q(\alpha )\ge q_{0}(\alpha )+\frac{C}{\alpha K}\Leftrightarrow (1-\delta )p+\delta (1-\alpha )\ge (1-\delta )p+\delta ^{\prime }(1-\alpha )+\frac{C}{ \alpha K} \\&q(\alpha )\ge q_{0}(\alpha )+\frac{C}{\alpha K}\Leftrightarrow (\delta -\delta ^{\prime })(1-\alpha )\ge \frac{C}{\alpha K} \\&q(\alpha )\ge q_{0}(\alpha )+\frac{C}{\alpha K}\Leftrightarrow h(\alpha )\ge \frac{C}{K}\,\, \hbox {true only for} \,\,\alpha \in \left[\frac{1}{2}-\beta ; \frac{1}{2}+\beta \right]. \end{aligned}$$

This means here that $q(\alpha )\ge q_{0}(\alpha )+\frac{C}{\alpha K} \Leftrightarrow \alpha \in \left[\frac{1}{2}-\beta ;\frac{1}{2}+\beta \right]$

Conclusion: in the second case,

$$\begin{aligned}&M_{C}(\alpha )<q(\alpha )<q_{0}(\alpha )+\frac{C}{\alpha K}, \forall \alpha \notin \left[\frac{1}{2}-\beta ;\frac{1}{2}+\beta \right]. \\&M_{C}(\alpha )\ge q(\alpha )\ge q_{0}(\alpha )+\frac{C}{\alpha K}, \forall \alpha \in \left[\frac{1}{2}-\beta ;\frac{1}{2}+\beta \right]. \end{aligned}$$

Thus for $\alpha \notin \left[\frac{1}{2}-\beta ;\frac{1}{2}+\beta \right]$, $D_{1}^{*}=A$ if $U>q(\alpha )$, and $D_{1}^{*}=I^{no}$ if $U<q(\alpha )$.

For $\alpha \in \left[\frac{1}{2}-\beta ;\frac{1}{2}+\beta \right]$:

$$\begin{aligned}&D_{1}^{*}=A\hbox { if }U>M_{C}(\alpha ) \\&D_{1}^{*}=I^{no}\hbox { if }U<q_{0}(\alpha )+\frac{C}{\alpha K} \\&D_{1}^{*}=I^{res}\hbox { if }q_{0}(\alpha )+\frac{C}{\alpha K}<U<M_{C}(\alpha ) \end{aligned}$$

$\square$