<p>Your attempt is almost correct. </p>

<p>Let the demand function be <span class="math-container">$D(p) = p^{-\epsilon}$</span> and the cost function some <span class="math-container">$C(D(p))$</span> so that <span class="math-container">$R(p) = D(p)p = p^{1-\epsilon}$</span>, right?</p>

<p>Then, I'd rewrite your first and second order conditions - as you only provided the expressions but it would be great to write them in a form of equations (or inequalities). So that:</p>

<p>First order condition: <span class="math-container">$R'(p) - C'(D(p))D'(p) =0$</span></p>

<p>And the second order: <span class="math-container">$R''(p) - C''\left(D(p)\right)\cdot D'(p)^2 - C'(D(p))D''(p) &lt; 0$</span></p>

<p>This is the moment when we kind of split our approaches. I also do agree that the second term in second order condition is negative. However, it is not necessarily <span class="math-container">$R''(p)$</span> that is negative - rather the whole relation <span class="math-container">$$R''(p) - C'(D(p))D''(p)&lt;0 \;(*)$$</span> Now, we want the second order condition to hold for every <span class="math-container">$p$</span> that satisfies the first order condition. Thus, I would substitute the <span class="math-container">$C'(D(p))$</span> from the first condition to the <span class="math-container">$(*)$</span> and hence, I end up with: <span class="math-container">$$R''(p) - \frac{R'(p)D''(p)}{D'(p)} &lt;0 \;(**)$$</span></p>

<p>Now it's time for the brute force. Calculate all derivatives in <span class="math-container">$(**)$</span>.</p>

<p><span class="math-container">$$ -\epsilon(1-\epsilon)p^{-1-\epsilon} - \frac{(1-\epsilon)p^{-\epsilon} \cdot -\epsilon(-1-\epsilon)p^{-2-\epsilon}}{-\epsilon p^{-1-\epsilon}} &lt;0 \implies \epsilon &gt; 1$$</span> for both, <span class="math-container">$p$</span> and <span class="math-container">$\epsilon$</span> positive.</p>

<p>P.S. If you think about it, letting <span class="math-container">$\epsilon&lt;1$</span> would imply that <span class="math-container">$R'(p)&gt;0$</span> and thus, the first order condition would have no solutions. Why? </p>

<p>P.P.S. I believe a similar exercise is in <em>Industrial Organization: Markets and Strategies</em> by Belleflamme and Peitz in their chapter about "monopoly pricing strategies" but with the inverse demands <span class="math-container">$P(q)$</span>.</p>

CC BY-SA 4.0
2020-04-12T15:04:07.840Z
Thank you very much! This one is from _The theory of industrial organization_ by Tirole. The chapters on monopoly pricing are quite similar to one another. I had seen the one you mention too when looking for answers!
econ86
CC BY-SA 4.0
2020-04-12T13:13:11.940Z
2020-04-12T13:13:11.940Z
bajun65537
<p>I'm doing a simple exercise from my textbook: suppose the revenue function is <span class="math-container">$R(p)=p^{1-\epsilon}$</span> with <span class="math-container">$\epsilon &gt; 0$</span>. Suppose the cost function is convex. Show that the profit function is quasi-concave if <span class="math-container">$\epsilon &gt; 1$</span>.</p>

<p>My attempt: Let the demand function then be <span class="math-container">$q(p)=p^{-\epsilon}$</span> and the cost function some <span class="math-container">$c(q(p))$</span>.</p>

<p>The first-order condition is
<span class="math-container">$(1-\epsilon)p^{-\epsilon}-c'(q(p))q'(p)$</span> and the second order condition <span class="math-container">$-\epsilon(1-\epsilon)p^{-\epsilon-1}-c''(q(p))q'(p)^2-c'(q(p))q''(p)$</span>.</p>

<p>The second term <span class="math-container">$-c''(q(p))q'(p)^2$</span> and third terms <span class="math-container">$-c'(q(p))q''(p)$</span> are negative under convexity of the cost function. Hence the second order condition is unambiguously negative if the first term is negative. But that happens for <span class="math-container">$\epsilon&lt;1$</span>, the opposite as conjectured.</p>

CC BY-SA 4.0
2020-04-11T11:31:27.020Z
Perhaps edit your calculations into your question, that way people can see what goes wrong.
Giskard
CC BY-SA 4.0
2020-04-11T12:39:56.133Z
Yes, the second derivative may be negative, but so what? E.g., for $\epsilon = 1/2$ and $C(y) = 0$ you get $\Pi(p) = p^{1/2}$. Is this quasi-concave?
Giskard
CC BY-SA 4.0
2020-04-11T12:48:37.737Z
Well, I was trying to show that the objective is concave in prices, hence quasi-concave. How would you proceed instead to get that $\epsilon > 1$ result?
econ86
CC BY-SA 4.0
2020-04-11T13:57:01.663Z
Do you know the definition of quasi-concavity? Have you tried applying it?
Giskard
CC BY-SA 4.0
2020-04-11T14:26:07.690Z
I know the definition in terms of the $max$ operator and that of contour sets, but honestly don't know how to apply them there. A hint or an answer would be appreciated.
econ86
<microeconomics><mathematical-economics><industrial-organisation>
Quasi-concavity of profit function
econ86
# Quasi-concavity of profit function

(Asked by: username_0 on 2020-04-11)

I'm doing a simple exercise from my textbook: suppose the revenue function is $R(p)=p^{1-\epsilon}$ with $\epsilon > 0$. Suppose the cost function is convex. Show that the profit function is quasi-concave if $\epsilon > 1$.

My attempt: Let the demand function then be $q(p)=p^{-\epsilon}$ and the cost function some $c(q(p))$.

The first-order condition is
$(1-\epsilon)p^{-\epsilon}-c'(q(p))q'(p)$ and the second order condition $-\epsilon(1-\epsilon)p^{-\epsilon-1}-c''(q(p))q'(p)^2-c'(q(p))q''(p)$.

The second term $-c''(q(p))q'(p)^2$ and third terms $-c'(q(p))q''(p)$ are negative under convexity of the cost function. Hence the second order condition is unambiguously negative if the first term is negative. But that happens for $\epsilon<1$, the opposite as conjectured.

**Tags:** microeconomics, mathematical-economics, industrial-organisation
> **Comments:**
> **username_1**: Perhaps edit your calculations into your question, that way people can see what goes wrong.  
> **username_1**: Yes, the second derivative may be negative, but so what? E.g., for $\epsilon = 1/2$ and $C(y) = 0$ you get $\Pi(p) = p^{1/2}$. Is this quasi-concave?  
> **username_0**: Well, I was trying to show that the objective is concave in prices, hence quasi-concave. How would you proceed instead to get that $\epsilon > 1$ result?  
> **username_1**: Do you know the definition of quasi-concavity? Have you tried applying it?  
> **username_0**: I know the definition in terms of the $max$ operator and that of contour sets, but honestly don't know how to apply them there. A hint or an answer would be appreciated.  

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## 1 answers

### Answer by username_2 on 2020-04-12. Score: 2 | Accepted answer

Your attempt is almost correct.

Let the demand function be $D(p) = p^{-\epsilon}$ and the cost function some $C(D(p))$ so that $R(p) = D(p)p = p^{1-\epsilon}$, right?

Then, I'd rewrite your first and second order conditions - as you only provided the expressions but it would be great to write them in a form of equations (or inequalities). So that:

First order condition: $R'(p) - C'(D(p))D'(p) =0$

And the second order: $R''(p) - C''\left(D(p)\right)\cdot D'(p)^2 - C'(D(p))D''(p) < 0$

This is the moment when we kind of split our approaches. I also do agree that the second term in second order condition is negative. However, it is not necessarily $R''(p)$ that is negative - rather the whole relation $$R''(p) - C'(D(p))D''(p)<0 \;(\*)$$ Now, we want the second order condition to hold for every $p$ that satisfies the first order condition. Thus, I would substitute the $C'(D(p))$ from the first condition to the $(\*)$ and hence, I end up with: $$R''(p) - \frac{R'(p)D''(p)}{D'(p)} <0 \;(\*\*)$$

Now it's time for the brute force. Calculate all derivatives in $(\*\*)$.

$$ -\epsilon(1-\epsilon)p^{-1-\epsilon} - \frac{(1-\epsilon)p^{-\epsilon} \cdot -\epsilon(-1-\epsilon)p^{-2-\epsilon}}{-\epsilon p^{-1-\epsilon}} <0 \implies \epsilon > 1$$ for both, $p$ and $\epsilon$ positive.

P.S. If you think about it, letting $\epsilon<1$ would imply that $R'(p)>0$ and thus, the first order condition would have no solutions. Why?

P.P.S. I believe a similar exercise is in *Industrial Organization: Markets and Strategies* by Belleflamme and Peitz in their chapter about "monopoly pricing strategies" but with the inverse demands $P(q)$.

> **Comments:**
> **username_0**: Thank you very much! This one is from _The theory of industrial organization_ by Tirole. The chapters on monopoly pricing are quite similar to one another. I had seen the one you mention too when looking for answers!  

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