# Nash equilibrium - mistake in proof of paper?

I have a question regarding the proof of Proposition 1 in Besley and Ghatak (2007) in Appendix A of their paper. It is a quite highly cited paper but I believe there is a mistake in the proof of their Proposition 1. The model is described in Section 2. of their paper. It is a very simple model that seeks to derive the Nash equilibrium of the market. In particular, they show that $$\theta_c^*$$ must satisfy $$f'(n \theta_c^*) = \alpha$$ in equilibrium. In their proof in the appendix they state: However, I think there is a mistake because the package $$(p', \widehat{\theta'})$$ does not make each caring consumer strictly better off.

The new package $$(p', \widehat{\theta'})$$ gives utility:

$$b-p'+f(\widehat{\theta'} + (n-1)\widehat{\theta})$$

The old package $$(p, \widehat{\theta})$$ gives utility:

$$b-p+f(n\widehat{\theta})$$

The new package is strictly preferred to the old package iff

$$b-p'+f(\widehat{\theta'} + (n-1)\widehat{\theta}) > b-p+f(n\widehat{\theta})$$

$$f(n\widehat{\theta} + \Delta \widehat{\theta}) - f(n\widehat{\theta}) > \Delta p$$

If we make the substitution as stated in the paper: $$\Delta p = f'(n\widehat{\theta}) \Delta \widehat{\theta}$$ and assuming we want a positive $$\Delta \widehat{\theta}$$, then the above becomes:

$$\frac{f(n\widehat{\theta} + \Delta \widehat{\theta}) - f(n\widehat{\theta})}{\Delta \widehat{\theta}} > f'(n \widehat{\theta})$$

If we pick a negative $$\Delta \widehat{\theta}$$, then the above condition becomes

$$\frac{f(n\widehat{\theta} + \Delta \widehat{\theta}) - f(n\widehat{\theta})}{\Delta \widehat{\theta}} < f'(n \widehat{\theta})$$

But $$f$$ is a increasing and strictly concave function, so the above inequalities are always false. So the new package is never preferred to the old package. Everything else in the proof is fine apart from this step and this is the most crucial step of the proof as it characterizes what $$\widehat{\theta}$$ should be in equilibrium, if this is wrong then many other results in the paper are also wrong.

So, is there a mistake in this part of the proof of the paper? And if so, is there a way to correct it so that the results still hold?

• I just skimmed the paper, so perhaps I missed something. But I noticed that your final inequality is different from theirs. While you're comparing with $f'(n\hat{\theta})$, they are comparing with $f'(n\hat{\theta}+ \Delta \hat{\theta})$. Isn't this the problem or I missed something? – Marcelo Gelati Apr 8 '19 at 14:54
• Yes, the inequality would be correct if it's $f'(n\widehat{\theta}+\Delta\widehat{\theta})$ on the RHS, but in their paper, they define $\Delta p = f'(n\widehat{\theta}) \Delta \widehat{\theta}$. In particular, I am confused on page 1661 of the paper where I quote: "We show that there exists a package of this form which will make each caring consumer strictly better off. The payoff of a caring consumer who accepts this package is $b-p'+f(\widehat{\theta'} + (n-1)\widehat{\theta})$" But as I just showed above, the new package cannot be strictly preferred. – elbarto Apr 9 '19 at 0:51
• Also, on page 1661 of the paper, their final inequality is perfectly fine, I am not referring to that inequality as that is talking about a different part of the proof. I am referring to the section of the proof above that where they are comparing the "old" package $(p, \widehat{\theta})$ to the "new" package $(p', \widehat{\theta'})$. This is a crucial step of the proof since everything else relies on this result and I believe there is a mistake here. – elbarto Apr 9 '19 at 0:55

## 1 Answer

The objective is to show that, as long as $$f'(n\hat\theta)\ne \alpha$$, a firm can always engineer a package $$(p',\hat\theta')=(p+\Delta p,\hat\theta+\Delta \hat{\theta})$$ such that (i) a caring consumer strictly prefers this package, and (ii) the firm makes positive profits.

1) Your maths are correct: the inequalities are always false when f is increasing and strictly concave. In that case, the proof in the paper is unclear. The package $$(p',\hat\theta')=(p+f'(n\hat\theta)\Delta \hat{\theta},\hat\theta+\Delta \hat{\theta})$$ does not satisfy (i).

2) You can show the proposition with the following reasoning. Assume $$f'(n\hat\theta)>\alpha$$. Take any $$h$$ within the interval $$(\alpha,f'(n\hat\theta))$$. For $$\Delta \hat{\theta}$$ small enough and positive, a firm that offers the package $$(p',\hat\theta')=(p+h\Delta \hat{\theta},\hat\theta+\Delta \hat{\theta})$$ satisfies (i) and (ii). With a first-order appoximation, we show (i): $$b-p'+f(\hat\theta'+(n-1)\hat\theta)>b-p+f(n \hat\theta),$$ and (ii): $$\Delta p - \alpha \Delta \theta>0.$$

Assume $$f'(n\hat\theta)<\alpha$$, and take $$h$$ within the interval $$(f'(n\hat\theta),\alpha)$$. For $$\Delta \hat{\theta}$$ small enough and negative, a firm that offers the package $$(p',\hat\theta')=(p+h\Delta \hat{\theta},\hat\theta+\Delta \hat{\theta})$$ satisfies (i) and (ii).

• Thanks so much for looking into this! However, I believe your argument above is not quite right because if we pick $\Delta \widehat{\theta}<0$, then the correct condition should be $$\frac{f(n\hat\theta+\Delta\hat\theta)-f(n\hat\theta)}{\Delta\hat\theta}<f'(n\hat\theta)$$ More generally, the package $(p', \widehat{\theta'})$ is preferred to $(p, \widehat{\theta})$ iff $f(n\hat\theta+\Delta\hat\theta)-f(n\hat\theta) > \Delta p$. In your example, we have $\ln(1) - \ln(2) \approx -0.6931$ while $\Delta p = f'(n\hat\theta)\Delta \widehat{\theta} =-0.5$, so the new package is not preferred. – elbarto Apr 11 '19 at 15:17
• To avoid confusion, I have also just edited my original question to distinguish the cases when $\Delta \widehat{\theta}>0$ and $\Delta \widehat{\theta}<0$. – elbarto Apr 11 '19 at 15:25
• Yes, you are right. I edit my answer. – GuiWil Apr 11 '19 at 16:02
• Thank you! Your arguments make absolute sense and fixes the proof! – elbarto Apr 11 '19 at 17:35