<p>A monopolist will always produce on the elastic part of the demand curve. The idea is the following: if the output level is on the inelastic pat of the demand curve, then increasing prices, by 1% will reduce quantity sold by less than 1%, so total revenue increases. At the same time, total cost decreases (from the decreased output), resulting in increased profits. As such, the monopolist will have an incentive to deviate by increasing prices.</p>
<p>If the demand curve is unit-inelastic as in your case, there is no elastic part, so the monopolist will have an incentive to keep on increasing prices (as long as marginal costs are positive) until output is zero and prices tend to infinity. Basically, the optimization problem is not well defined.</p>
<p>Total revenues are given by <span class="math-container">$P_i Y_i$</span>, so:
<span class="math-container">$$
\begin{align*}
MR &amp;= Y_i + P_i \frac{\partial Y_i}{\partial P_i},\\
&amp;= Y_i\left(1 - \varepsilon \right),
\end{align*}
$$</span>
where <span class="math-container">$\varepsilon$</span> is the absolute value of the demand elasticity.</p>
<p>From this, it is easy to see that <span class="math-container">$MR$</span> is positive only if <span class="math-container">$\varepsilon &lt; 1$</span>. If the demand elasticity equals 1, <span class="math-container">$MR$</span> is zero which effectively means that revenues are constant for all prices. This means that the optimal output level will be where marginal costs are equal to zero. If marginal costs are always positive, this means that the monopolist will always have the incentive to keep on increasing prices - resulting in decreased output - as revenues stay the same while costs will decrease.</p>
<p>The sole exception with an optimal positive price level is where marginal costs are zero over an initial range of output. In this case, any output level in this region (and associated price level) will be optimal.</p>

CC BY-SA 4.0
2024-04-07T10:51:09.970Z
Why is $P = P_i$?
uninterestedacademic
CC BY-SA 4.0
2024-04-07T12:05:10.200Z
@tdm Thank you for your answer. In my class, we defined the markup as the constant $A$ such that $P_i=A\psi'(Y_i)$. This would give $\frac{P_i-\psi'(Y_i)}{P_i}=1+\frac{1}{A}$ so for this to be $1$ we would need $A=\infty$ which is where I am getting the infinite markup from. From $\frac{P_i-\psi'(Y_i)}{P_i}=1$ we get $\psi'(Y_i)=0$, could you explain why this would be the case here?
Joseph Basford
CC BY-SA 4.0
2024-04-07T12:11:41.980Z
I hadn't used that $P=P_i$ anywhere. This is true due to the symmetry of the intermediate monopolists' problems?
Joseph Basford
CC BY-SA 4.0
2024-04-08T05:33:30.707Z
I think you are right. I reformulated my answer. tldr: monopolists will always produce on the elastic part of the demand curve. You have an inelastic demand curve so your problem is not really well defined: there is no profit maximizing output that is strictly positive.
tdm
CC BY-SA 4.0
2024-04-08T06:43:47.407Z
@tdm I see! That's very clear now, thank you. That fits in well with the next part of the question which introduces two firms with heterogeneous costs which engage in Bertrand competition for each good intermediate good $i$. I think the Bertrand competition gets around the problem of an infinite markup since if the lowest cost firm prices too high the other firm will undercut them.
Joseph Basford
CC BY-SA 4.0
2024-04-06T18:58:36.983Z
2024-04-08T06:53:54.060Z
2024-04-08T06:53:54.060Z
tdm
<p>I'm going over a macro past paper and am stuck deriving and interpreting a result.</p>
<p>The question begins with the CES aggregator for aggregate output <span class="math-container">$Y$</span>, based on a continuum of intermediate goods <span class="math-container">$(y_i)$</span>, <span class="math-container">$$Y=\left[\int_0^1y_i^{\frac{\sigma-1}{\sigma}}di\right]^{\frac{\sigma}{\sigma-1}}$$</span> and looks at the limiting case when <span class="math-container">$\sigma \to 1$</span>. I can show that as <span class="math-container">$\sigma\to 1$</span>, we get <span class="math-container">$\ln Y = \int_0^1 \ln y_i di$</span> and the intermediate good demands are <span class="math-container">$$y_i=\left[\frac{P_{i}}{P}\right]^{-1}Y$$</span> Where <span class="math-container">$P_i$</span> is the price of the intermediate good, and <span class="math-container">$P$</span> is the aggregate price index determined by <span class="math-container">$$\ln P =\int_0^1 \ln P_i di$$</span></p>
<p>The question then asks to derive the markup of the intermediate market monopolists. To do this, I define <span class="math-container">$\psi(Y_i)$</span> as the minimal expenditure required to produce output <span class="math-container">$Y_i$</span>. Then, the intermediate good monopolists solve <span class="math-container">$$\max_{P_i}P_iY_i-\psi(Y_i)$$</span> <span class="math-container">$$st. \qquad Y_i=\left[\frac{P_i}{P}\right]^{-1}Y$$</span> This has first order condition <span class="math-container">$$0=P_i\frac{\partial Y_i}{\partial P_i}+Y_i-\psi'(Y_i)\frac{\partial Y_i}{\partial P_i}$$</span> Using from the condition that <span class="math-container">$\frac{\partial Y_i}{\partial P_i}=-\frac{P}{P_i^2}Y$</span> I derive that this FOC becomes <span class="math-container">$$0=\psi'(Y_i)\frac{P}{P_i^2}$$</span></p>
<p>I think from this, I can conclude that the markup must be infinite since either <span class="math-container">$P_i=\infty$</span> or <span class="math-container">$\psi'(Y_i)=0$</span>. However, I have no intuition for this result since agents can substitute away from a good as it becomes more expensive so I can't see why <span class="math-container">$P_i=\infty$</span> would hold. Additionally <span class="math-container">$\psi'(Y_i)=0$</span> implies 0 output of the intermediate good which I also think is meaningless. The usual markup is <span class="math-container">$\frac{\sigma}{\sigma-1}$</span> which does tend to infinity as <span class="math-container">$\sigma \to 1$</span>, but I am not sure why this is the case.</p>
<p>Can someone help me find where I've made a mistake or explain the intuition behind why the markup would be infinite when the elasticity of substitution is 1? I'm thinking I want to take logs somewhere in the intermediate monopolists problem but I'm not sure where.</p>

CC BY-SA 4.0
2024-04-03T05:35:27.797Z
"*I'm going over a macro past paper*" I am still confused why people are reluctant to give proper references. If you phrase it like this, a person querying the paper's title will not find your question, making it harder for them to benefit from the effort you and a future answerer put in.
Giskard
CC BY-SA 4.0
2024-04-03T07:41:20.290Z
@Giskard I agree, but my university does not release answers to past papers. Whilst it might just be professors’ time constraints which prohibit them from making answers, I didn’t want to become involved in hassle if they don’t want answers available as a course/university policy. I’ll ask my professor and edit the title if they are okay with a direct reference to the module and paper.
Joseph Basford
CC BY-SA 4.0
2024-04-03T08:44:44.043Z
Oh, so it's an exam/term paper, not a research paper?
Giskard
CC BY-SA 4.0
2024-04-03T09:13:31.880Z
@Giskard Yes sorry, British English confusion.
Joseph Basford
CC BY-SA 4.0
2024-04-05T04:23:05.813Z
I'm sorry I won't be of much help. Could you please tell me how $\displaystyle \lim_{\sigma \to 1} \frac{\ln \left(\int_{0}^{1} y_i^{\frac{\sigma - 1}{\sigma}} di\right)}{\frac{\sigma - 1}{\sigma}} = \int_0^1 \ln y_i di$? I hope that you're not taking the exponent $\frac{\sigma - 1}{\sigma}$ out of function inside the integral post-logarithmization since $\ln(\int) \neq \int (\ln)$.
uninterestedacademic
CC BY-SA 4.0
2024-04-05T23:26:09.953Z
@basicsubset It follows by L'Hopital's rule. Since $Y=\left[\int_0^1 y_i^{\frac{\sigma-1}{\sigma}} di\right]^{\frac{\sigma}{\sigma-1}}$ we have that $\frac{Y^{\frac{\sigma-1}{\sigma}}-1}{\sigma-1}=\int_0^1 \frac{y_i^{\frac{\sigma-1}{\sigma}}-1}{\sigma-1}di$. Appling L'Hopital then gives the result. A sufficient condition to interchange the limit and integral is that $y_i\in L^1$ which seems appropriate to assume.
Joseph Basford
<macroeconomics><elasticity-of-substitution><ces-function>
LSE EC417 2023: Markup as Elasticity Tends to Unity
Joseph Basford
# LSE EC417 2023: Markup as Elasticity Tends to Unity

(Asked by: username_0 on 2024-04-02)

I'm going over a macro past paper and am stuck deriving and interpreting a result.

The question begins with the CES aggregator for aggregate output $Y$, based on a continuum of intermediate goods $(y\_i)$, $$Y=\left[\int\_0^1y\_i^{\frac{\sigma-1}{\sigma}}di\right]^{\frac{\sigma}{\sigma-1}}$$ and looks at the limiting case when $\sigma \to 1$. I can show that as $\sigma\to 1$, we get $\ln Y = \int\_0^1 \ln y\_i di$ and the intermediate good demands are $$y\_i=\left[\frac{P\_{i}}{P}\right]^{-1}Y$$ Where $P\_i$ is the price of the intermediate good, and $P$ is the aggregate price index determined by $$\ln P =\int\_0^1 \ln P\_i di$$

The question then asks to derive the markup of the intermediate market monopolists. To do this, I define $\psi(Y\_i)$ as the minimal expenditure required to produce output $Y\_i$. Then, the intermediate good monopolists solve $$\max\_{P\_i}P\_iY\_i-\psi(Y\_i)$$ $$st. \qquad Y\_i=\left[\frac{P\_i}{P}\right]^{-1}Y$$ This has first order condition $$0=P\_i\frac{\partial Y\_i}{\partial P\_i}+Y\_i-\psi'(Y\_i)\frac{\partial Y\_i}{\partial P\_i}$$ Using from the condition that $\frac{\partial Y\_i}{\partial P\_i}=-\frac{P}{P\_i^2}Y$ I derive that this FOC becomes $$0=\psi'(Y\_i)\frac{P}{P\_i^2}$$

I think from this, I can conclude that the markup must be infinite since either $P\_i=\infty$ or $\psi'(Y\_i)=0$. However, I have no intuition for this result since agents can substitute away from a good as it becomes more expensive so I can't see why $P\_i=\infty$ would hold. Additionally $\psi'(Y\_i)=0$ implies 0 output of the intermediate good which I also think is meaningless. The usual markup is $\frac{\sigma}{\sigma-1}$ which does tend to infinity as $\sigma \to 1$, but I am not sure why this is the case.

Can someone help me find where I've made a mistake or explain the intuition behind why the markup would be infinite when the elasticity of substitution is 1? I'm thinking I want to take logs somewhere in the intermediate monopolists problem but I'm not sure where.

**Tags:** macroeconomics, elasticity-of-substitution, ces-function

> **Comments:**
> **username_1**: "*I'm going over a macro past paper*" I am still confused why people are reluctant to give proper references. If you phrase it like this, a person querying the paper's title will not find your question, making it harder for them to benefit from the effort you and a future answerer put in.  
> **username_0**: @username_1 I agree, but my university does not release answers to past papers. Whilst it might just be professors’ time constraints which prohibit them from making answers, I didn’t want to become involved in hassle if they don’t want answers available as a course/university policy. I’ll ask my professor and edit the title if they are okay with a direct reference to the module and paper.  
> **username_1**: Oh, so it's an exam/term paper, not a research paper?  
> **username_0**: @username_1 Yes sorry, British English confusion.  
> **username_2**: I'm sorry I won't be of much help. Could you please tell me how $\displaystyle \lim_{\sigma \to 1} \frac{\ln \left(\int_{0}^{1} y_i^{\frac{\sigma - 1}{\sigma}} di\right)}{\frac{\sigma - 1}{\sigma}} = \int_0^1 \ln y_i di$? I hope that you're not taking the exponent $\frac{\sigma - 1}{\sigma}$ out of function inside the integral post-logarithmization since $\ln(\int) \neq \int (\ln)$.  
> **username_0**: @basicsubset It follows by L'Hopital's rule. Since $Y=\left[\int_0^1 y_i^{\frac{\sigma-1}{\sigma}} di\right]^{\frac{\sigma}{\sigma-1}}$ we have that $\frac{Y^{\frac{\sigma-1}{\sigma}}-1}{\sigma-1}=\int_0^1 \frac{y_i^{\frac{\sigma-1}{\sigma}}-1}{\sigma-1}di$. Appling L'Hopital then gives the result. A sufficient condition to interchange the limit and integral is that $y_i\in L^1$ which seems appropriate to assume.  

---

## 1 answers

### Answer by username_3 on 2024-04-06. Score: 2 | Accepted answer

A monopolist will always produce on the elastic part of the demand curve. The idea is the following: if the output level is on the inelastic pat of the demand curve, then increasing prices, by 1% will reduce quantity sold by less than 1%, so total revenue increases. At the same time, total cost decreases (from the decreased output), resulting in increased profits. As such, the monopolist will have an incentive to deviate by increasing prices.

If the demand curve is unit-inelastic as in your case, there is no elastic part, so the monopolist will have an incentive to keep on increasing prices (as long as marginal costs are positive) until output is zero and prices tend to infinity. Basically, the optimization problem is not well defined.

Total revenues are given by $P\_i Y\_i$, so:
$$
\begin{align\*}
MR &= Y\_i + P\_i \frac{\partial Y\_i}{\partial P\_i},\\
&= Y\_i\left(1 - \varepsilon \right),
\end{align\*}
$$
where $\varepsilon$ is the absolute value of the demand elasticity.

From this, it is easy to see that $MR$ is positive only if $\varepsilon < 1$. If the demand elasticity equals 1, $MR$ is zero which effectively means that revenues are constant for all prices. This means that the optimal output level will be where marginal costs are equal to zero. If marginal costs are always positive, this means that the monopolist will always have the incentive to keep on increasing prices - resulting in decreased output - as revenues stay the same while costs will decrease.

The sole exception with an optimal positive price level is where marginal costs are zero over an initial range of output. In this case, any output level in this region (and associated price level) will be optimal.

> **Comments:**
> **username_2**: Why is $P = P_i$?  
> **username_0**: @username_3 Thank you for your answer. In my class, we defined the markup as the constant $A$ such that $P_i=A\psi'(Y_i)$. This would give $\frac{P_i-\psi'(Y_i)}{P_i}=1+\frac{1}{A}$ so for this to be $1$ we would need $A=\infty$ which is where I am getting the infinite markup from. From $\frac{P_i-\psi'(Y_i)}{P_i}=1$ we get $\psi'(Y_i)=0$, could you explain why this would be the case here?  
> **username_0**: I hadn't used that $P=P_i$ anywhere. This is true due to the symmetry of the intermediate monopolists' problems?  
> **username_3**: I think you are right. I reformulated my answer. tldr: monopolists will always produce on the elastic part of the demand curve. You have an inelastic demand curve so your problem is not really well defined: there is no profit maximizing output that is strictly positive.  
> **username_0**: @username_3 I see! That's very clear now, thank you. That fits in well with the next part of the question which introduces two firms with heterogeneous costs which engage in Bertrand competition for each good intermediate good $i$. I think the Bertrand competition gets around the problem of an infinite markup since if the lowest cost firm prices too high the other firm will undercut them.  

---

