The impact of market innovations on the dissemination of social norms: the sustainability case

Abstract

The paper presents a model that links the diffusion process of a technological innovation and the dissemination of a social norm. Our approach introduces a new dimension to the interaction between markets and social norms beyond the interplay of monetary and non-monetary incentives: the innovation of material goods as a catalyst of norm evolution. We analyze how the dissemination of a social norm may be affected by product innovation, which adds to the variation of products with respect to their level of norm compliance. The market is linked to the process of norm adoption via psychological forces, endogenizing the rates of adoption and abandonment. We derive necessary and sufficient conditions for a) a positive impact of the innovation on the level of norm adoption and b) for multiplicity of norm equilibria. In concluding, we discuss several policy implications.

Appendix

Proof of Proposition 1: The demand system (19) in vector notion is given by $\left(\begin{array}{c}\hfill {p}^e\hfill \\ {}\hfill {p}^g\hfill \end{array}\right)= A\left(\begin{array}{c}\hfill {X}^e\hfill \\ {}\hfill {X}^g\hfill \end{array}\right)+ b$. According to Okuguchi and Szidarovszky (1990, p.34), given the linear structure of the model, negative definiteness of A + A ^T is sufficient for uniqueness of the Cournot equilibrium. Eigenvalues of A + A ^T are given by $-\frac{1}{\kappa -\lambda}\left(1\pm \sqrt{1-\frac{1}{\kappa -\lambda}}\right)$ and negative by inspection. QED

Proof of Proposition 2: $\frac{\partial {m}^{\ast }}{\partial q}\underset{<}{\overset{>}{=}}0\iff \frac{\partial {m}^{\ast }}{\partial q}\underset{<}{\overset{>}{=}}0.\mathrm{QED}$

Proof of Lemma 1: Claim 1 is obvious when X ^e and X ^g are strictly positive. Claim 2 follows from the fact that X ^e and X ^g are linear in q and thus solving Eq. (7) for q for any given value of $\dot{\tilde{q}}$ is tantamount to solving a polynomial of degree three. The first implication of Claim 3 follows from the fact that the denominator of the derivative $\frac{d\left({\tilde{X}}^e/{\tilde{X}}^g\right)}{ d q}$ is strictly positive and the numerator is given by:

$$ \begin{array}{l}\frac{{d\tilde{X}}^e}{ d q}{\tilde{X}}^g-\frac{{d\tilde{X}}^g}{ d q}{\tilde{X}}^e=\left({\Delta}^e\left(\frac{n}{n+1}\left({\theta}^g+{\Delta}^g q\right)-\frac{1}{n+1}\left({\theta}^e+{\Delta}^e q\right)\right)\frac{\lambda}{\kappa}\right)-\left(\frac{n}{n+1}{\Delta}^g-\frac{1}{n+1}{\Delta}^e\frac{\lambda}{\kappa}\right)\left({\theta}^e+{\Delta}^e q\right)\\ {}=\frac{n}{n+1}\left[{\theta}^g{\Delta}^e-{\theta}^e{\Delta}^g\right]\end{array} $$

(19)

The second implication of Claim 3 follows from the observation that all three terms summed up in

$$ \begin{array}{c}\frac{d\dot{\tilde{q}}}{ d q}=-\alpha \left({\sigma}_a\left(1+ CD\right)+{\sigma}_h\left(1- CD\right)\right)-\left(1-\alpha \right)\left({\sigma}_a\frac{{\tilde{X}}^e}{{\tilde{X}}^g}+{\sigma}_h\frac{{\tilde{X}}^g}{{\tilde{X}}^e}\right)\\ {}\kern1em +\left(1-\alpha \right)\left(\frac{\left(1- q\right){\sigma}_a}{{\left({\tilde{X}}^g\right)}^2}+\frac{q{\sigma}_h}{{\left({\tilde{X}}^e\right)}^2}\right)\left(\frac{{d\tilde{X}}^e}{ d q}{\tilde{X}}^g-\frac{{d\tilde{X}}^g}{ d q}{\tilde{X}}^e\right)\end{array} $$

(20)

are negative if $\frac{d\left({\tilde{X}}^e/{\tilde{X}}^g\right)}{ d q}\le 0.\mathrm{QED}$

1.1 Definition of MES and Approximation

In what follows, we will derive the vertices of the MES and reformulate the two differential equations $\dot{q}\left({q}^{Low}\right)<0$, $\dot{q}\left({q}^{High}\right)>0$ as differential equation for Δ^e(Δ^g).

At q=q ^ex such that ${\dot{q}}^{\prime }$(q ^ex.)=0, solving for the stationary points of (7) is equivalent to solving an equation in Δ^e,Δ^g: $\dot{q}$(q ^ex.(Δ^e,Δ^g))=0. Given strictly positive demand and a strictly positive conformity bias, $\dot{q}$(q)=0 is equivalent to:

$$ \left(1- q\right){\sigma}_a\frac{X^e}{X^g}- q{\sigma}_h\frac{X^g}{X^e}=\gamma +\beta q $$

(21)

where $\gamma =\frac{\alpha}{1-\alpha}{\sigma}_a\left(1+ CD\right),\beta =-\frac{\alpha}{1-\alpha}\left({\sigma}_a\left(1+ CD\right)+{\sigma}_h\left(1- CD\right)\right)$.

Let $z(q)\equiv \frac{X^e}{X^g},\sigma \equiv {\sigma}_h/{\sigma}_a$, then (21) can be written as:

$$ {\left( z(q)\right)}^2=\frac{q}{\left(1- q\right)}\sigma +\frac{\left(\gamma +\beta q\right)}{\left(1- q\right)} z(q). $$

(22)

Taking the total derivative of (22) with respect to Δ^e, Δ^g and applying the envelope theorem gives us:

$$ \left[2 z\left({q}^{e x.}\right){\left.\frac{\partial z}{\partial {\varDelta}^e}\right|}_{q={q}^{e x.}}-\frac{\left(\gamma +\beta q\right)}{\left(1- q\right)}{\left.\frac{\partial z}{\partial {\varDelta}^e}\right|}_{q={q}^{e x.}}\right] d{\varDelta}^e+\left[2 z\left({q}^{e x.}\right){\left.\frac{\partial z}{\partial {\varDelta}^g}\right|}_{q={q}^{e x.}}-\frac{\left(\gamma +\beta q\right)}{\left(1- q\right)}{\left.\frac{\partial z}{\partial {\varDelta}^g}\right|}_{q={q}^{e x.}}\right] d{\varDelta}^g=0 $$

(23)

Equation (23) amounts to $\frac{n+1}{n}\frac{n+1}{n}+\frac{n+1}{n}=\frac{n+1}{n}$, which is equivalent to:

$$ \frac{d{\varDelta}^g}{d{\varDelta}^e}=\frac{\varDelta^g{q}^{e xtr.}+{\theta}^g}{\varDelta^e{q}^{e xtr.}+{\theta}^e}\left(>0\right) $$

(24)

Together with initial conditions: (Δe,Δg)|$\dot{q}$(q ^Max(Δe,Δg))=0 and (Δ^e,Δ^g)|$\dot{q}$(q ^Min(Δ^e,Δ^g))=0 the differential eq. (24) gives rise to two boundary functions: Δ^e,Min(Δ^g), Δ^e,Max(Δ^g).

Definition: All (Δ^e,Δ^g) pairs that satisfy the following three conditions define a parameter region such that multiple equilibria exist: (1) ${\varDelta}^e<-{\theta}^e+\frac{n}{\lambda}\left({\varDelta}^g+{\theta}^g\right)$, (2) Δ^e >Δ^e,Min(Δ^g), (3) Δ^e <Δ^e,Max(Δ^g). We will refer to this set as the multiple equilibria set (MES).

Before we continue, we will state some observations based on $\frac{d{\varDelta}^e}{d{\varDelta}^g}=\frac{1}{\frac{n+1{X}^g}{n{ X}^e}+\frac{\lambda}{n}}$ that will be helpful in the course of our argument:

• (1) The slopes of Δ^e,Min(Δ^g),Δ^e,Max(Δ^g) are positive and smaller than the slope of the third constraint${\varDelta}^e<-{\theta}^e+\frac{n}{\lambda}\left({\varDelta}^g+{\theta}^g\right)$.

• (2) By corollary 4, $\frac{d{\Delta}^g}{d{\Delta}^e}=\frac{\Delta^g q+{\theta}^g}{\Delta^e q+{\theta}^e}$ is ceteris paribus decreasing in q

• (3) At point A, the relevant constraints have the same slope.

We are able to determine the coordinates for points A and B (see Fig. 2) analytically. For better readability, Table 1 presents the results for α=0. Note that there exist multiple equilibria if and only if (Δ^e)^B >(Δ^e)^A.

Table 1 Vertices of multiple equilibria set for α=0

Approximation for α=0:

The differential equations given by (25) cannot be solved analytically. In the following, we present our approximation strategy for α=0, such that we can state explicit sufficient conditions for multiple equilibria to exist.

Note that the values for q that corresponds to (Δ^e,Δ^g) pairs that are elements of the graph of Δ^e,Max(Δ^g) range from ${q}^C=\frac{\left( n{\theta}^g-{\lambda \theta}^e\right)}{3\left( n{\varDelta}^g-\lambda {\varDelta}^e\right)+4\left( n{\theta}^g-{\lambda \theta}^e\right)}$ to q ^A=1. We can use the (Δ^g)^B as a lower bound for Δ^g and by that, can give a lower bound for q independent of Δ^e and Δ^g, i.e. $\underset{\bar{\mkern6mu}}{q}=\frac{4{\left( n+1\right)}^2{\theta}^e}{4{\left( n+1\right)}^2{\theta}^e+3\lambda \left( n{\theta}^g-{\lambda \theta}^e\right)}$. The system $\dot{q}\left({q}^C\right)=0$, ${\dot{q}}^{\prime}\left({q}^C\right)=0$ can be solved for Δ^e and Δ^g as a function of q. If we plug in $\underset{\bar{\mkern6mu}}{q}$, we get as point D a (Δ^e,Δ^g) pair on the graph of Δ^e,Max(Δ^g) that corresponds to a maximum for the dynamics in (7) that equals $\underset{\bar{\mkern6mu}}{q}$.

$$ {\left(\begin{array}{l}{\varDelta}^e\\ {}{\varDelta}^g\end{array}\right)}^D=\left(\begin{array}{l}\frac{4\sqrt{3{\theta}^e{\lambda \tau}^3}-{\theta}^e\left(4{\left( n+1\right)}^2{\theta}^e+9\lambda \tau \right)}{4{\left( n+1\right)}^2{\theta}^e}\\ {}\frac{3\underset{\bar{\mkern6mu}}{q}{\left({\varDelta}^e\right)}^D-2\tau +\sqrt{-9{\underset{\bar{\mkern6mu}}{q}}^2+6\underset{\bar{\mkern6mu}}{q}{\left({\varDelta}^e\right)}^D\left({\left({\varDelta}^e\right)}^D-{\theta}^e\right)+6\left(1- q\right){\left({\varDelta}^e\right)}^D{\theta}^e-{\left( n+1\right)}^2{\left({\theta}^e\right)}^2+{\tau}^2}}{3 n\underset{\bar{\mkern6mu}}{q}}\end{array}\right) $$

We approximate the upper and lower boundaries by linear functions intersecting point B and D, respectively. Our observation above, that the slope Δ^e,Min(Δ^g) is decreasing in q, gives us a lower bound for the slope by $\frac{\theta^e}{\theta^g}$. Figure 3 illustrates our approximation procedure. Note that, under our approach, MES is not empty if and only if the area spanned by X ^g(1)>0 and the two approximating linear function is non-empty.

Proposition (sufficient conditions): If α=0 and $0<{\theta}^e<\frac{n}{\lambda}{\theta}^g$, $\dot{\tilde{q}}=0$ has three solutions if:

$$ \left({\Delta}^e,{\Delta}^g\right)\in \left\{\begin{array}{l}\left.\left({\Delta}^e,{\Delta}^g\right)\right|{\Delta}^e<-{\theta}^e+\frac{n}{\lambda}\left({\Delta}^g+{\theta}^g\right),{\Delta}^e<\frac{\theta^e}{\theta^g}{\Delta}^g+\frac{{\left( n{\theta}^g-{\lambda \theta}^e\right)}^3}{4 n{\left( n+1\right)}^2{\theta}^g{\theta}^e}\\ {}\kern2.04em {\Delta}^e>\frac{{\left({\Delta}^e\right)}^D\underset{\bar{\mkern6mu}}{q}-{\theta}^e}{{\left({\Delta}^g\right)}^D\underset{\bar{\mkern6mu}}{q}-{\theta}^g}{\Delta}^g+{\left({\Delta}^e\right)}^D-\frac{{\left({\Delta}^e\right)}^D\underset{\bar{\mkern6mu}}{q}-{\theta}^e}{{\left({\Delta}^g\right)}^D\underset{\bar{\mkern6mu}}{q}-{\theta}^g}{\left({\Delta}^g\right)}^D\end{array}\right\}, $$

where $\underset{\bar{\mkern6mu}}{q}=\frac{4{\left( n+1\right)}^2{\theta}^e}{4{\left( n+1\right)}^2{\theta}^e+3\lambda \left( n{\theta}^g-{\lambda \theta}^e\right)}$.

Approximation for α ≠ 0

Since we have an analytical solution for point A, we now focus the general solution for the tangent point D:

$$ \begin{array}{l}{\left({\varDelta}^e\right)}^D=\frac{1}{4{\left(1+ n\right)}^2\left(1+\alpha \left(-2+{CD}^2\alpha \right)\right){\delta}_a}\left(-\Omega +\sqrt{\frac{8{\left(1+ n\right)}^2\left(1+\alpha \left(-2+{CD}^2\alpha \right)\right){\delta}_a{\theta}^e}{q^3\left(1-\alpha \right){\delta}_h}\Psi +{\Omega}^2}\right)\\ {}{\left({\varDelta}^g\right)}^D=\frac{1}{2{nq}^2\left(1-\alpha \right){\delta}_h}\left(\begin{array}{l}-{q}^2{\left({\varDelta}^e\right)}^D\left(\begin{array}{l}\left(1+ n\right)\alpha \left(\left(1+ CD\right){\delta}_a+\left(1- CD\right){\delta}_h\right)\\ {}-2\left(1-\alpha \right){\delta}_h\lambda \end{array}\right)\\ {}+ q\left(\begin{array}{l}-2{\delta}_h\tau +\alpha \Big(\left(1+ CD\right)\left(1+ n\right){\delta}_a\left({\left({\varDelta}^e\right)}^D-{\theta}^e\right)\\ {}+{\delta}_h\left(\left(-1+ CD\right)\left(1+ n\right){\theta}^e+2\tau \right)\Big)+\left(1+ n\right){\left({\varDelta}^e\right)}^D\Sigma \end{array}\right)\\ {}+\left(1+ n\right){\theta}^e\left(\left(1+ CD\right){\alpha \delta}_a+\Sigma \right)\end{array}\right)\end{array} $$

where

$$ \begin{array}{l}\tau \equiv \left( n{\theta}^g-{\lambda \theta}^e\right)\hfill \\ {}\Sigma \equiv \sqrt{{\left(1+ CD\right)}^2{\left(-1+ q\right)}^2{\alpha}^2{\delta_a}^2-2\left(-1+ q\right) q\left(2+\alpha \left(-4+\alpha +{CD}^2\alpha \right)\right){\delta}_a{\delta}_h+{\left(-1+ CD\right)}^2{q}^2{\alpha}^2{\delta_h}^2}\hfill \\ {}\Psi \equiv \left(1+ n\right)\Big(2 q{\alpha \delta}_a{\delta}_h\left(-3\left(1+ n\right)\left(-3+2 q\right){\theta}^e+2\left(1+ CD\right) n\left(-1+ q\right){\theta}^g-2\left(1+ CD\right)\left(-1+ q\right){\theta}^e\lambda \right)+\hfill \\ {}{\alpha}^3\Big(-2 q{\delta}_a{\delta}_h\left(\left(1+ n\right)\left(-1+{CD}^2\left(-2+ q\right)+ q\right){\theta}^e-\left(1+ CD\right)\left(1+{CD}^2\right) n\left(-1+ q\right){\theta}^g+\left(1+ CD\right)\left(1+{CD}^2\right)\left(-1+ q\right){\theta}^e\lambda \right)+\hfill \\ {}{\left(1+ CD\right)}^2{\left(-1+ q\right)}^2{\delta_a}^2\left(-\left(1+ CD\right) n{\theta}^g+{\theta}^e\left(1+ n+\lambda + CD\lambda \right)\right)+{\left(-1+ CD\right)}^2{q}^2{\delta_h}^2\left(-\left(1+ CD\right) n{\theta}^g+{\theta}^e\left(1+ n+\lambda + CD\lambda \right)\right)\Big)+\hfill \\ {}\alpha \left(-\left(1+ CD\right)\left(1+ n\right)\left(-1+ q\right){\delta}_a{\theta}^e+ q{\delta}_h\left(-\left(-1+ CD\right)\left(1+ n\right){\theta}^e-4 n{\theta}^g+4{\theta}^e\lambda \right)\right)\Sigma +2 q{\delta}_h\left(\left(1+ n\right)\left(-3+2 q\right){\delta}_a{\theta}^e+\left( n{\theta}^g-{\theta}^e\lambda \right)\Sigma \right)-\hfill \\ {}{\alpha}^2\left({\left(1+ CD\right)}^2\left(1+ n\right){\left(-1+ q\right)}^2{\delta_a}^2{\theta}^e+{\delta}_a\right(2 q{\delta}_h\left(-\left(1+ n\right)\left(-7+{CD}^2\left(-2+ q\right)+5 q\right){\theta}^e+4\left(1+ CD\right) n\left(-1+ q\right){\theta}^g-4\left(1+ CD\right)\left(-1+ q\right){\theta}^e\lambda \right)-\hfill \\ {}\left(1+ CD\right)\left(-1+ q\right)\left(-\left(1+ CD\right) n{\theta}^g+{\theta}^e\left(1+ n+\lambda + CD\lambda \right)\right)\Sigma \left)+ q{\delta}_h\left({\left(-1+ CD\right)}^2\left(1+ n\right) q{\delta}_h{\theta}^e+\left(-\left(1+{CB}^2\right) n{\theta}^g+{\theta}^e\left(1+ n+\lambda + CD\left(-1- n+ CD\lambda \right)\right)\right)\Sigma \right)\right)\Big)\hfill \\ {}\Omega \equiv \frac{1}{q^2\left(-1+\alpha \right){\delta}_h}\left(\Psi +2{\left(1+ n\right)}^2 q\left(-1+\alpha \right)\left(1+\alpha \left(-2+{CD}^2\alpha \right)\right){\delta}_a{\delta}_h{\theta}^e\right)\hfill \end{array} $$

To find such a point, we apply the following approach: First, we express two of the conditions for the inflection point C, $\dot{q}$(q)=0, ${\dot{q}}^{\hbox{'}}$(q)=0 in terms of Δ^g(q). With these two conditions, we can solve for Δ^e. However, we still have to find a q that will be greater than q ^IP and independent of Δ^e and Δ^g.

For the general case α ≠ 0, we again choose q such that we can be sure that it will correspond to a point on the graph of Δ^e,Max(Δ^g). This can be achieved by choosing (Δ^g)^B as a lower bound for Δ^g and ‑θ ^e as a lower bound for Δ^e.

$$ \begin{array}{l}\underset{\bar{\mkern6mu}}{q}={q}^{IP}\left({\Delta}^g={\left({\Delta}^g\right)}^{\mathrm{B}},{\varDelta}^e=-{\theta}^e\right)=\\ {}\frac{\left(\begin{array}{l}\ {\delta}_h\left[\left(1- CD\right)\alpha \left(1+ n\right) n\left({\Delta}^g{\theta}^e+{\theta}^g{\varDelta}^e\right)+2 n\left(1-\alpha \right)\left( n{\Delta}^g-\lambda {\varDelta}^e\right)\left( n{\theta}^g-{\theta}^e\lambda \right)-2\lambda \left(\left(\left(1- CD\right)\left(1+ n\right)\alpha {\varDelta}^e\right){\theta}^e\right)\right]\\ {}+\left(1+ n\right){\delta}_a\left[\left(1+ CD\right) n\alpha \left({\Delta}^g{\theta}^e+{\theta}^g{\varDelta}^e\right)-{\left({\varDelta}^e\right)}^2\left(\left(1+ n\right)\left(1-\alpha \right)-\left(1+ CD\right)\alpha \lambda \right)+{\varDelta}^e\left( n\left(2{\theta}^e-\alpha \left(\left(1+ CD\right){\Delta}^g+2{\theta}^e\right)\right)+2{\theta}^e\left(1-\alpha -\left(1+ CD\right)\alpha \lambda \right)\right)\right]\end{array}\right)}{3\left(\left(1+ n\right){\delta}_a{\varDelta}^e\left(\left(1+ n\right)\left(-1+\alpha \right){\varDelta}^e-\left(1+ CD\right) n\alpha {\Delta}^g+\left(1+ CD\right)\alpha {\varDelta}^e\lambda \right)-{\delta}_h\left( n{\Delta}^g-{\varDelta}^e\lambda \right)\left( n{\Delta}^g-{\varDelta}^e\lambda +\alpha \left(\left(1- CD\right)\left(1+ n\right){\varDelta}^e+ n{\Delta}^g-{\varDelta}^e\lambda \right)\right)\right)}\end{array} $$

This gives us a lower bound for the maxima that correspond to Δ^e,Max(Δ^g) independent of Δ^e and Δ^g. We can then calculate the slope at point D: $\frac{d{\varDelta}^e}{d{\varDelta}^e}=\frac{d{\varDelta}^e}{d{\varDelta}^e}$.

Proof of Proposition 4: The situation where MES is empty corresponds to the case point A, B and C are equal, i.e. where q ^Max=q ^Min≡q ^IP=1, $\dot{q}$(1)=0 and X ^g (1)=0. The latter two conditions provide a solution for Δ^g as a function of α: ${\varDelta}^g\left(\alpha \right)=\frac{{\alpha \lambda \delta}_h\left(1- CD\right)\left( n{\theta}^g-{\lambda \theta}^e\right)}{n\left( n+1\right)\left(1-\alpha \right){\delta}_a}-{\theta}^g$. The first condition amounts to a condition for α as a function of Δ^g:

$$ \begin{array}{l}\alpha \left({\varDelta}^g\right)=\hfill \\ {}\frac{2 n{\left( n+1\right)}^2{\delta}_a\left({\varDelta}^g+{\theta}^g\right)\left( n{\varDelta}^g+\tau \right)+{\delta}_h{\lambda}^2{\tau}^2}{2 n{\left( n+1\right)}^2{\delta}_a\left({\varDelta}^g+{\theta}^g\right)\left( n{\varDelta}^g+\tau \right)+{\delta}_h{\lambda}^2{\tau}^2+\left(1+ n\right)\lambda \tau \left( n\left(\left(1+ CD\right){\delta}_a+2\left(1- CD\right){\delta}_h\right)\left({\varDelta}^g+{\theta}^g\right)-\left(1- CD\right){\delta}_h{\theta}^e\lambda \right)}\hfill \end{array} $$

where τ=(nθ ^g ‑ λθ ^e). Solving these two equations for α yields the critical value:

$$ \begin{array}{l}{\alpha}^{c rit.}=\frac{\left(1- CD\right)\left( n+1\right){\theta}^e+2\left( n{\theta}^g-{\lambda \theta}^e\right)-\sqrt{{\left(\left(1- CD\right)\left( n+1\right){\theta}^e\right)}^2+4{\left( n{\theta}^g-{\lambda \theta}^e\right)}^2\left(1-{CD}^2\right)}}{2\left({CD}^2\left( n{\theta}^g-{\lambda \theta}^e\right)+\left( n+1\right)\left(1- CD\right){\theta}^e\right)}. Furthermore,\\ {}\frac{\partial {\alpha}^{c rit.}}{\partial {\theta}^g}=\frac{\partial {\alpha}^{c rit.}}{\partial n}=-\frac{\theta^e}{\theta^g}\frac{\partial {\alpha}^{c rit.}}{\partial {\theta}^e}=-\frac{n}{\theta^e}{\left.\frac{\partial {\alpha}^{c rit.}}{\partial \lambda}\right|}_{\begin{array}{l}{\theta}^e,{\theta}^g\\ {} fixed\end{array}}>0\ and\frac{\partial {\alpha}^{c rit.}}{\partial {\sigma}_a}=\frac{\partial {\alpha}^{c rit.}}{\partial {\sigma}_h}=0;\frac{\partial {\alpha}^{c rit.}}{\partial CD}>0\ which\\ {}\mathrm{implies}:\kern0.5em \frac{\partial {\alpha}^{c rit.}}{\partial {\chi}_h^e}<0;\frac{\partial {\alpha}^{c rit.}}{\partial {\chi}_h^g}>0;\frac{\partial {\alpha}^{c rit.}}{\partial {c}^e}>0;\frac{\partial {\alpha}^{c rit.}}{\partial {c}^g}<0;\frac{\partial {\alpha}^{c rit.}}{\partial k}>0. QED\end{array} $$

Derivation of the partial effects on the critical value α ^crit:

$$ \begin{array}{l}{\alpha}^{crit.}=\frac{\left(1- CD\right)\left( n+1\right){\theta}^e+2\left( n{\theta}^g-{\lambda \theta}^e\right)-\sqrt{{\left(\left(1- CD\right)\left( n+1\right){\theta}^e\right)}^2+4{\left( n{\theta}^g-{\lambda \theta}^e\right)}^2\left(1-{CD}^2\right)}}{2\left({CD}^2\left( n{\theta}^g-{\lambda \theta}^e\right)+\left( n+1\right)\left(1- CD\right){\theta}^e\right)}\\ {}\underset{y=2\left( n{\theta}^g-{\lambda \theta}^e\right)>0}{\overset{x=\left(1- CB\right)\left( n+1\right){\theta}^e>0}{=}}\kern1em \frac{x+ y-\sqrt{x^2+{y}^2\left(1-{CD}^2\right)}}{CD^2 y+2 x}\\ {}\end{array} $$

$$ \begin{array}{l}1.\mathrm{claim}:\frac{\partial {\alpha}^{crit.}}{\partial {\theta}^g}>0:\\ {}\kern2.16em \frac{\partial {\alpha}^{crit.}}{\partial {\theta}^g}=\frac{\left({y}^{\hbox{'}}-\frac{2{y y}^{\hbox{'}}\left(1-{CD}^2\right)}{2\sqrt{x^2+{y}^2\left(1-{CD}^2\right)}}\right)\left({CD}^2 y+2 x\right)-\left( x+ y-\sqrt{x^2+{y}^2\left(1-{CD}^2\right)}\right)\left({CD}^2{y}^{\hbox{'}}\right)}{{\left({CD}^2 y+2 x\right)}^2}\\ {}\kern2.28em =\frac{y^{\hbox{'}}\left({CD}^2 y+2 x\right)-\frac{2{y y}^{\hbox{'}}\left(1-{CD}^2\right)\left({CD}^2 y+2 x\right)}{2\sqrt{x^2+{y}^2\left(1-{CD}^2\right)}}-{ x CD}^2{y}^{\hbox{'}}-{ y CD}^2{y}^{\hbox{'}}+{CD}^2{y}^{\hbox{'}}\sqrt{x^2+{y}^2\left(1-{CD}^2\right)}}{{\left({CD}^2 y+2 x\right)}^2}\\ {}\kern2.28em =\frac{\left(2-{CD}^2\right){ x y}^{\hbox{'}}-\frac{2{y y}^{\hbox{'}}\left(1-{CD}^2\right)\left({CD}^2 y+2 x\right)}{2\sqrt{x^2+{y}^2\left(1-{CD}^2\right)}}+{CD}^2{y}^{\hbox{'}}\sqrt{x^2+{y}^2\left(1-{CD}^2\right)}}{{\left({CD}^2 y+2 x\right)}^2}>0\iff \end{array} $$

$$ \begin{array}{l}\kern2.04em \left(2-{CD}^2\right){ x y}^{\hbox{'}}2\sqrt{x^2+{y}^2\left(1-{CD}^2\right)}-2{y y}^{\hbox{'}}\left(1-{CD}^2\right)\left({CD}^2 y+2 x\right)+{CD}^2{y}^{\hbox{'}}2\left({x}^2+{y}^2\left(1-{CD}^2\right)\right)>0\overset{y^{\hbox{'}}>0}{\iff}\\ {}\kern2.04em \left(2-{CD}^2\right) x\sqrt{x^2+{y}^2\left(1-{CD}^2\right)}- y\left(1-{CD}^2\right)\left({CD}^2 y+2 x\right)+{CD}^2\left({x}^2+{y}^2\left(1-{CD}^2\right)\right)>0\overset{x>0}{\iff}\\ {}\kern2.04em \left(2-{CD}^2\right)\sqrt{x^2+{y}^2\left(1-{CD}^2\right)}-2 y\left(1-{CD}^2\right)+{ x CD}^2>0\iff \\ {}\kern2.04em \left(2-{CD}^2\right)\sqrt{x^2+{y}^2\left(1-{CD}^2\right)}>2 y\left(1-{CD}^2\right)-{ x CD}^2\Leftarrow \\ {}\kern2.16em {\left(2-{CD}^2\right)}^2\left({x}^2+{y}^2\left(1-{CD}^2\right)\right)>{\left(2 y\left(1-{CD}^2\right)-{ x CD}^2\right)}^2\iff \\ {}\kern2.16em {\left(2-{CD}^2\right)}^2\left({x}^2+{y}^2\left(1-{CD}^2\right)\right)-{\left(2 y\left(1-{CD}^2\right)-{ x CD}^2\right)}^2=\left(1-{CD}^2\right){\left(2 x+{y CD}^2\right)}^2>0\end{array} $$

$$ \begin{array}{l}2.\mathrm{claim}:\frac{\partial {\alpha}^{crit.}}{\partial CD}>0\\ {}\frac{\partial {\alpha}^{crit.}}{\partial CD}=\frac{\left({x}^{\hbox{'}}-\frac{2{x x}^{\hbox{'}}-2{ CD y}^2}{2\sqrt{x^2+{y}^2\left(1-{CD}^2\right)}}\right)\left({CD}^2 y+2 x\right)-\left( x+ y-\sqrt{x^2+{y}^2\left(1-{CD}^2\right)}\right)\left(2 CDy+2{x}^{\hbox{'}}\right)}{{\left({CD}^2 y+2 x\right)}^2}\\ {}=\frac{x^{\hbox{'}}\left({CD}^2 y+2 x\right)-\frac{2{x x}^{\hbox{'}}-2{ CD y}^2}{2\sqrt{x^2+{y}^2\left(1-{CD}^2\right)}}\left({CD}^2 y+2 x\right)-\left( x+ y-\sqrt{x^2+{y}^2\left(1-{CD}^2\right)}\right)\left(2 CDy+2{x}^{\hbox{'}}\right)}{{\left({CD}^2 y+2 x\right)}^2}>0\\ {}\iff \\ {}{x}^{\hbox{'}}\left({CD}^2 y+2 x\right)-\frac{2{x x}^{\hbox{'}}-2{ CD y}^2}{2\sqrt{x^2+{y}^2\left(1-{CD}^2\right)}}\left({CD}^2 y+2 x\right)-\left(2 CDy+2{x}^{\hbox{'}}\right)\left( x+ y\right)+\left(2 CDy+2{x}^{\hbox{'}}\right)\sqrt{x^2+{y}^2\left(1-{CD}^2\right)}>0\\ {}\iff \\ {}-\frac{2{x x}^{\hbox{'}}-2{ CD y}^2}{2\sqrt{x^2+{y}^2\left(1-{CD}^2\right)}}\left({CD}^2 y+2 x\right)-\left(2 CDy\right)\left( x+ y\right)-\left(2-{CD}^2\right){x}^{\hbox{'}} y+\left(2 CDy+2{x}^{\hbox{'}}\right)\sqrt{x^2+{y}^2\left(1-{CD}^2\right)}>0\\ {}\overset{{}^{\ast }2\sqrt{x^2+{y}^2\left(1-{CD}^2\right)}}{\iff}\\ {}-\left(2{x x}^{\hbox{'}}-2{ CD y}^2\right)\left({CD}^2 y+2 x\right)-\left(2 CDy\left( x+ y\right)+\left(2-{CD}^2\right){x}^{\hbox{'}} y\right)2\sqrt{x^2+{y}^2\left(1-{CD}^2\right)}+\left(2 CDy+2{x}^{\hbox{'}}\right)2\left({x}^2+{y}^2\left(1-{CD}^2\right)\right)>0\end{array} $$

$$ \begin{array}{l}\iff \\ {}-2{x x}^{\hbox{'}}\left({CD}^2 y+2 x\right)+2{ CD y}^2\left({CD}^2 y+2 x\right)+\left(2 CD y+2{x}^{\hbox{'}}\right)2{x}^2+\left(2 CD y+2{x}^{\hbox{'}}\right)2{y}^2\left(1-{CD}^2\right)\\ {}-\left(2 CD y\left( x+ y\right)+\left(2-{CD}^2\right){x}^{\hbox{'}} y\right)2\sqrt{x^2+{y}^2\left(1-{CD}^2\right)}>0\\ {}\iff \\ {}-2{x x}^{\hbox{'}}{CD}^2 y+2{ CD y}^2\left({CD}^2 y+2 x\right)+2 CD y2{x}^2+\left(2 CD y+2{x}^{\hbox{'}}\right)2{y}^2-{CD}^2\left(2 CD y+2{x}^{\hbox{'}}\right)2{y}^2\\ {}-\left(2 CD y\left( x+ y\right)+\left(2-{CD}^2\right){x}^{\hbox{'}} y\right)2\sqrt{x^2+{y}^2\left(1-{CD}^2\right)}>0\\ {}\overset{:2 y, y>0}{\iff}\\ {}-{ x x}^{\hbox{'}}{CD}^2+ CD y\left({CD}^2 y+2 x\right)+ CD2{x}^2+\left(2 CD y+2{x}^{\hbox{'}}\right) y-{CD}^2\left(2 CD y+2{x}^{\hbox{'}}\right) y\\ {}-\left(2 CD\left( x+ y\right)+\left(2-{CD}^2\right){x}^{\hbox{'}}\right)\sqrt{x^2+{y}^2\left(1-{CD}^2\right)}>0\end{array} $$

$$ \begin{array}{l}\iff \\ {}-{ x x}^{\hbox{'}}{CD}^2+ CD y\left({CD}^2 y+2 x\right)+ CD2{x}^2+2{ CD y}^2-{CD}^22{ CD y}^2+2{x}^{\hbox{'}} y\left(\underset{=\left(1- CD\right)\left(1+ CD\right)}{\underbrace{1-{CD}^2}}\right)\\ {}-\left(2 CD\left( x+ y\right)+\left(2-{CD}^2\right){x}^{\hbox{'}}\right)\sqrt{x^2+{y}^2\left(1-{CD}^2\right)}>0\\ {}\overset{x^{\hbox{'}}=-\frac{x}{1- CB}}{\iff}\\ {}-{ x x}^{\hbox{'}}{CD}^2+ CD y\left({CD}^2 y+2 x\right)+ CD2{x}^2+2{ CD y}^2-{CD}^22{ CD y}^2-2 x y\left(1+ CD\right)\\ {}-\left(2 CD\left( x+ y\right)+\left(2-{CD}^2\right){x}^{\hbox{'}}\right)\sqrt{x^2+{y}^2\left(1-{CD}^2\right)}>0\\ {}\iff \\ {}\underset{>0,{x}^{\hbox{'}}<0}{\underbrace{-{ x x}^{\hbox{'}}{CD}^2}}+{ CD y}^2\underset{>0}{\underbrace{\left(2-{CD}^2\right)}}-2 x y+2{x}^2 CD-\left(2 CD\left( x+ y\right)+\left(2-{CD}^2\right){x}^{\hbox{'}}\right)\sqrt{x^2+{y}^2\left(1-{CD}^2\right)}>0\end{array} $$

In the next step, we first rearrange the term on the left-hand side of the last inequality and, second, we distinguish two cases to establish the strict positivity of the term.

$$ \begin{array}{l}-{ x x}^{\hbox{'}}{CD}^2+{CD y}^2\left(2-{CD}^2\right)-2 xy+2{x}^2 CD-\left(2 CD\left( x+ y\right)+\left(2-{CD}^2\right){x}^{\hbox{'}}\right)\sqrt{x^2+{y}^2\left(1-{CD}^2\right)}\overset{x^{\hbox{'}}=-\frac{x}{1- CD}}{=}\\ {}\\ {}{x}^2\frac{CD^2}{1- CD}+{CD y}^2\left(2-{CD}^2\right)-2 xy+2{x}^2 CD-\left(2 CD\left( x+ y\right)-\left(2-{CD}^2\right)\frac{x}{1- CD}\right)\sqrt{x^2+{y}^2\left(1-{CD}^2\right)}\overset{{}^{\ast}\left(1- CD\right)}{=}\\ {}{x}^2{CD}^2+{CD y}^2\left(2-{CD}^2\right)\left(1- CD\right)-2 xy\left(1- CD\right)+2{x}^2 CD\left(1- CD\right)-\left(2 CD\left(1- CD\right)\left( x+ y\right)-\left(2-{CD}^2\right) x\right)\sqrt{x^2+{y}^2\left(1-{CD}^2\right)}=\\ {}{x}^2 CD\left(2- CD\right)+{y}^2 CD\left(2-{CD}^2\right)\left(1- CD\right)-2 xy\left(1- CD\right)-\left(2 CD\left(1- CD\right)\left( x+ y\right)-\left(2-{CD}^2\right) x\right)\sqrt{x^2+{y}^2\left(1-{CD}^2\right)}=\\ {}{x}^2 CD\left(2- CD\right)+{y}^2 CD\left(2-{CD}^2\right)\left(1- CD\right)-2 xy\left(1- CD\right)-\left(2 CD\left(1- CD\right) y-\left(2+{CD}^2-2 CD\right) x\right)\sqrt{x^2+{y}^2\left(1-{CD}^2\right)}\\ {}\end{array} $$

$$ \begin{array}{l}\underset{\bar{\mkern6mu}}{1.\mathrm{case}:}\ \left(2 CD\left(1- CD\right) y-\left(2+{CD}^2-2 CD\right) x\right)>0\\ {}{x}^2 CD\left(2- CD\right)+{y}^2 CD\left(2-{CD}^2\right)\left(1- CD\right)-2 xy\left(1- CD\right)-\left(2 CD\left(1- CD\right) y-\left(2+{CD}^2-2 CD\right) x\right)\sqrt{x^2+{y}^2\left(1-{CD}^2\right)}\\ {}\overset{\sqrt{x^2+{y}^2\left(1-{CD}^2\right)}< x+ y\sqrt{\left(1-{CD}^2\right)}}{>}\\ {}{x}^2 CD\left(2- CD\right)+{y}^2 CD\left(2-{CD}^2\right)\left(1- CD\right)-2 xy\left(1- CD\right)-\left(2 CD\left(1- CD\right) y-\left(2+{CD}^2-2 CD\right) x\right)\left( x+ y\sqrt{\left(1-{CD}^2\right)}\right)>0\end{array} $$

$$ \begin{array}{l}\overset{:\left(1- CD\right)}{\iff}\\ {}{x}^2\left(\frac{CD\left(2-{CD}^2\right)}{1- CD}+\frac{\left(2+{CD}^2-2 CD\right)}{1- CD}\right)+{CD y}^2\left(2-{CD}^2-2\sqrt{1-{CD}^2}\right)+2 xy\left(-1- CD+\left(2+{CD}^2-2 CD\right)\frac{\sqrt{\left(1-{CD}^2\right)}}{2\left(1- CD\right)}\right)=\\ {}={x}^2\left(\frac{2}{1- CD}\right)+{CD y}^2\underset{\ge 0}{\underbrace{\left(2-{CD}^2-2\sqrt{1-{CD}^2}\right)}}+2 xy\underset{\ge 0}{\underbrace{\left(-1- CD+\frac{1}{2\left(1- CD\right)}\left(1+{\left(1- CD\right)}^2\right)\right)}}>0\Rightarrow \mathrm{claim}\end{array} $$

$$ \begin{array}{l}\underset{\bar{\mkern6mu}}{2.\mathrm{case}:}\ \left(2 CD\left(1- CD\right) y-\left(2+{CD}^2-2 CD\right) x\right)<0\\ {}{x}^2 CD\left(2- CD\right)+{y}^2 CD\left(2-{CD}^2\right)\left(1- CD\right)-2 xy\left(1- CD\right)-\left(2 CD\left(1- CD\right) y-\left(2+{CD}^2-2 CD\right) x\right)\sqrt{x^2+{y}^2\left(1-{CD}^2\right)}=\\ {}{x}^2 CD\left(2- CD\right)+{y}^2 CD\left(2-{CD}^2\right)\left(1- CD\right)-2 xy\left(1- CD\right)+\underset{>0}{\underbrace{\left(\left(2+{CD}^2-2 CD\right) x-2 CD\left(1- CD\right) y\right)}}\sqrt{x^2+{y}^2\left(1-{CD}^2\right)}\\ {}\overset{\sqrt{x^2+{y}^2\left(1-{CD}^2\right)}> y\sqrt{\left(1-{CD}^2\right)}}{>}\\ {}{x}^2 CD\left(2- CD\right)+{y}^2 CD\left(2-{CD}^2\right)\left(1- CD\right)-2 xy\left(1- CD\right)+\left(\left(2+{CD}^2-2 CD\right) x-2 CD\left(1- CD\right) y\right) y\sqrt{\left(1-{CD}^2\right)}=\\ {}{x}^2 CD\left(2- CD\right)+{y}^2\left( CD\left(2-{CD}^2\right)\left(1- CD\right)-2 CD\left(1- CD\right)\sqrt{\left(1-{CD}^2\right)}\right)-2 xy\left(1- CD\right)+ xy\left(2+{CD}^2-2 CD\right)\sqrt{\left(1-{CD}^2\right)}=\\ {}{x}^2 CD\left(2- CD\right)+\left(1- CD\right){ CD y}^2\underset{\ge 0}{\underbrace{\left(2-{CD}^2-2\sqrt{\left(1-{CD}^2\right)}\right)}}+ xy\underset{\ge 0}{\underbrace{\left(\left(2+{CD}^2-2 CD\right)\sqrt{\left(1-{CD}^2\right)}-2\left(1- CD\right)\right)}}>0\Rightarrow \mathrm{claim}\end{array} $$

Proof of Proposition 5: Inserting equilibrium quantities given in (13) into (7) and evaluation at qº yields eq. (18). QED

Proof of Proposition 6: At the discontinuities we have m ^* =m ^eq and thus $\dot{q}(q)=\dot{\tilde{q}}(q)$. Otherwise, m ^* >m ^eq implies ${\tilde{X}}^e>{\widehat{X}}^e$ and ${\tilde{X}}^g<{\widehat{X}}^g$ due to $\frac{dX^{e\ast }}{dm}>0$ and $\frac{dX^{g\ast }}{dm}<0$. Hence, $\dot{q}(q)<\dot{\tilde{q}}(q)$ for all q in the intervals of continuity. For the second part of the claim, note that we can write $\dot{q}=\dot{q}\left(\frac{X^e}{X^g}\left( m(q), q\right), q\right)$ and thus, $\frac{d\dot{q}}{ d q}=\frac{\partial \dot{q}}{\partial \frac{X^e}{X^g}}\left(\frac{\partial \frac{X^e}{X^g}}{\partial m}\frac{d m}{ d q}+\frac{\partial \frac{X^e}{X^g}}{\partial q}\right)+\frac{\partial \dot{q}}{\partial q}$. Since $\frac{dm^{\ast }}{dq}=\frac{\Delta^e}{\sqrt{k\kappa}}$ and $\frac{dm^{eq}}{dq}=0$ for all q in the intervals of continuity, and the other terms in $\frac{d\dot{q}}{ d q}$ are the same for the discontinuous version of $\dot{q}$ and its continuous approximation $\dot{\tilde{q}}$, the observation $\frac{\partial \dot{q}}{\partial \frac{X^e}{X^g}}=\left(1-\alpha \right)\left(\left(1- q\right){\sigma}_a+\frac{q{\sigma}_h}{{\left(\frac{X^e}{X^g}\right)}^2}\right)>0$ implies the second claim of the Proposition. QED.