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

<p>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 <span class="math-container">$(p',\hat\theta')=(p+f'(n\hat\theta)\Delta \hat{\theta},\hat\theta+\Delta \hat{\theta})$</span> does not satisfy (i).</p>

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

<p>Assume <span class="math-container">$f'(n\hat\theta)&lt;\alpha$</span>, and take <span class="math-container">$h$</span> within the interval <span class="math-container">$(f'(n\hat\theta),\alpha)$</span>.
For <span class="math-container">$\Delta \hat{\theta}$</span> small enough and <strong>negative</strong>, a firm that offers the package <span class="math-container">$(p',\hat\theta')=(p+h\Delta \hat{\theta},\hat\theta+\Delta \hat{\theta})$</span> satisfies (i) and (ii).</p>

CC BY-SA 4.0
2019-04-11T15:17:23.383Z
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
CC BY-SA 4.0
2019-04-11T15:25:05.290Z
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
CC BY-SA 4.0
2019-04-11T16:02:04.677Z
Yes, you are right. I edit my answer.
GuiWil
CC BY-SA 4.0
2019-04-11T17:35:34.860Z
Thank you! Your arguments make absolute sense and fixes the proof!
elbarto
CC BY-SA 4.0
2019-04-11T13:38:19.157Z
2019-04-11T16:39:03.473Z
2019-04-11T16:39:03.473Z
GuiWil
<p>I have a question regarding the proof of Proposition 1 in <a href="http://personal.lse.ac.uk/ghatak/csr.pdf" rel="nofollow noreferrer">Besley and Ghatak (2007)</a> 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 <span class="math-container">$\theta_c^*$</span> must satisfy <span class="math-container">$f'(n \theta_c^*) = \alpha$</span> in equilibrium. In their proof in the appendix they state:</p>

<p><a href="https://i.sstatic.net/3nfM5.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/3nfM5.png" alt="enter image description here"></a></p>

<p>However, I think there is a mistake because the package <span class="math-container">$(p', \widehat{\theta'})$</span> does <strong>not</strong> make each caring consumer strictly better off. </p>

<p>The new package <span class="math-container">$(p', \widehat{\theta'})$</span> gives utility:</p>

<p><span class="math-container">$$b-p'+f(\widehat{\theta'} + (n-1)\widehat{\theta}) $$</span></p>

<p>The old package <span class="math-container">$(p, \widehat{\theta})$</span> gives utility:</p>

<p><span class="math-container">$$b-p+f(n\widehat{\theta}) $$</span></p>

<p>The new package is strictly preferred to the old package iff</p>

<p><span class="math-container">$$b-p'+f(\widehat{\theta'} + (n-1)\widehat{\theta}) &gt; b-p+f(n\widehat{\theta})$$</span></p>

<p><span class="math-container">$$ f(n\widehat{\theta} + \Delta \widehat{\theta}) - f(n\widehat{\theta}) &gt; \Delta p$$</span></p>

<p>If we make the substitution as stated in the paper: <span class="math-container">$\Delta p = f'(n\widehat{\theta}) \Delta \widehat{\theta}$</span> and assuming we want a positive <span class="math-container">$\Delta \widehat{\theta}$</span>, then the above becomes:</p>

<p><span class="math-container">$$\frac{f(n\widehat{\theta} + \Delta \widehat{\theta}) - f(n\widehat{\theta})}{\Delta \widehat{\theta}} &gt; f'(n \widehat{\theta}) $$</span></p>

<p>If we pick a negative <span class="math-container">$\Delta \widehat{\theta}$</span>, then the above condition becomes </p>

<p><span class="math-container">$$\frac{f(n\widehat{\theta} + \Delta \widehat{\theta}) - f(n\widehat{\theta})}{\Delta \widehat{\theta}} &lt; f'(n \widehat{\theta}) $$</span></p>

<p>But <span class="math-container">$f$</span> is a increasing and strictly concave function, so the above inequalities are always false. So the new package is <strong>never</strong> 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 <span class="math-container">$\widehat{\theta}$</span> should be in equilibrium, if this is wrong then many other results in the paper are also wrong. </p>

<p>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?</p>

CC BY-SA 4.0
2019-04-08T14:54:24.610Z
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
CC BY-SA 4.0
2019-04-09T00:51:45.237Z
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
CC BY-SA 4.0
2019-04-09T00:55:17.843Z
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
<game-theory><nash-equilibrium><public-economics><public-goods>
Nash equilibrium - mistake in proof of paper?
elbarto
# Nash equilibrium - mistake in proof of paper?

(Asked by: username_0 on 2019-04-08)

I have a question regarding the proof of Proposition 1 in [Besley and Ghatak (2007)](http://personal.lse.ac.uk/ghatak/csr.pdf) 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:

[![enter image description here](https://i.sstatic.net/3nfM5.png)](https://i.sstatic.net/3nfM5.png)

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?

**Tags:** game-theory, nash-equilibrium, public-economics, public-goods

> **Comments:**
> **username_1**: 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?  
> **username_0**: 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.  
> **username_0**: 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.  

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## 1 answers

### Answer by username_2 on 2019-04-11. Score: 5 | Accepted 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).

> **Comments:**
> **username_0**: 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.  
> **username_0**: To avoid confusion, I have also just edited my original question to distinguish the cases when $\Delta \widehat{\theta}>0$ and $\Delta \widehat{\theta}<0$.  
> **username_2**: Yes, you are right. I edit my answer.  
> **username_0**: Thank you! Your arguments make absolute sense and fixes the proof!  

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