Certification and socially responsible production

Abstract

In many markets, consumers are unable to distinguish between goods that are produced in a socially responsible way and goods that are not. In such situations, socially responsible production is not a profit-maximizing strategy, even if the premium that consumers are willing to pay exceeds the costs. Only firms that are genuinely motivated by social responsibility would then produce in this way, and there would be too little socially responsible production. Improved opportunities for voluntary certification could potentially reduce this problem by allowing firms to signal their type. We examine how the possibility of certification affects the share of socially responsible production. Our main result is that increased certification may reduce the share of socially responsible production by reducing prices in the market for uncertified products and thus crowd out socially responsible producers who do not certify. This provide a mechanism through which certification might have adverse effects on socially responsible production, even when the certification process is perfect and when there is perfect competition among the producers.

Appendices

Appendix 1: Derivation of e(p, c)

Define the right-hand side of (8) as H(e; p, c), i.e.

$$\begin{aligned}{}[\epsilon G(p-m)-D(e;c) ]/[\epsilon G(p-m)-D(e,c)+(1-\epsilon ) G(p)] \equiv H(e;p,c). \end{aligned}$$

(15)

Equation (8) can then be written as

$$\begin{aligned} e=H(e;p,c). \end{aligned}$$

(16)

We focus on the case where there is a demand for certified products, i.e. where $c<\bar{z}$ such that $D(\cdot )>0$ for some $e>0$.² As D(e; c) is decreasing in e and c, it is straightforward to show that $H(\cdot )$ is increasing in e, c, and p. Assume that (i) $H_{ee}<0$ and (ii) $H_e(0;p;c)=- D_e(0,c)<1$ for the range of c and p we study.This ensures a unique solution to (16).³

Consider c-values such that $H(0;p,c)\ge 0$ for some p, and let p be sufficiently high to make $H(0;p,c)> 0$. When (i) holds, (ii) implies that $H_e(e;p,c)<1$ for all $e>0$. The $H(\cdot )$-curve then crosses the 45-degree line only once and from above. This implies that (15) gives a unique and stable solution with $e>0$. Equation (15) determines e as an increasing function of p for a given c, written as e(p; c). For each c, the e(p, c)-curve starts at the lowest p-value that yields $H(0;p,c)=0$. As $D(\cdot )>0$, it follows from (15) that $H(0;p,c)=0$ requires $G(p-m)>0$, which in turn requires $p>m$. Thus, all e(p; c)-curves start above m on the p-axis.

Differentiating (16) with respect to p yields

$$\begin{aligned} de/dp=\frac{1-H_e(e;p,c)}{ H_p(e;p,c)}. \end{aligned}$$

(17)

As $H_p(e;p,c)>0$ and $H_e(e;p,c)<1$, it follows that $de/dp>0$.

Differentiating (16) with respect to c yields

$$\begin{aligned} de/dc=\frac{1-H_e(e;p,c)}{ H_c(e;p,c)}. \end{aligned}$$

(18)

As $H_e(e;p,c)<1$ and $H_e(e;p,c)<1$, it follows that $de/dc>0$.

Appendix 2: Proof of Proposition 2

As shown in Sect. 2.3, Eq. (7) determines the price p as an increasing function of e, starting in v on the p-axis. For $c<\bar{z}$, the curve e(p; c) starts above m on the p-axis. As p goes to infinity, e(p, c) goes towards $\epsilon$, while $p(\epsilon )$ is a finite value. For c not too low, such that the e(p; c)-curve starts below v, it crosses the curve p(e) at least once from below. At the crossing, the e-curve is steeper than the p-curve, i.e. $1/e_p(p,c)>p(e)$, which means the solution is stable. Thus, there is at least one stable solution to (7) and (8) with $e>0$.⁴

Consider a stable solution with $e>0$. Differentiating (7) and (8) with respect to c yields

$$\begin{aligned} de/dc=\frac{e_c(p,c)}{1-e_p(p,c)p'(e)}. \end{aligned}$$

(19)

As $e_p(p,c)p'(e)<1$ in a stable solution, it follows that $de/dc>0$ , i.e. e goes down when c goes down. As $p'(e)>0$, p goes down when e goes down. With a lower c, the e(p; c)-curve shifts to the left and crosses the p(e)-curve at a lower value of e. Thus, a lower c gives rise to a lower fraction of ethical producers in the uncertified market.

Appendix 3: Proof of Proposition 3

Without access to certification, total welfare is given by (9). By using $e_0mx_0=\epsilon \int \limits _{0 }^{p_0-m } mdG(q)$, we can rewrite the welfare function as

$$\begin{aligned} W=v x+e_0\int \limits _{0 }^{x} [z(x)-m]dx-\left[ \epsilon \int \limits _{0 }^{p_0-m } qdG(q)+(1-\epsilon )\int \limits _{0 }^{p_0 } qdG(q) \right] . \end{aligned}$$

(20)

to show that $x_0$ is lower than optimal, we evaluate $W'(x)$ at $x_0$. Differentiating W with respect to x, taking into account that p and e are functions of x, and evaluating at $x_0$, yields

$$\begin{aligned} dW/dx_0=v+e_0z(x_0)-p_0 +(\lambda -e_0) \frac{1}{x_0} \int \limits _{0 }^{x_0 } [z(x)-m]dx, \end{aligned}$$

(21)

where $\lambda = \epsilon g(p_0-m)/[\epsilon g(p_0-m)+(1-\epsilon )g(p_0)$. As e is increasing in p, $\lambda - e_0>0$. From (5), $v+e_0z(x_0)=p_0$ in the no-certification equilibrium. Inserting this into (21) gives $dW/dx_0>0$, which means output is too low in the absence of certification.