<p>First let's look at the specific tax. The profit is
$$\pi_s=[P(q)-\tau_s]q-C(q).$$
Differentiating to establish the first-order condition:
$$P'(q)q+P(q)-\tau_s-C'(q)=0.$$
If we write $A=\tau_s q$ for the tax revenue then we can rewrite the FOC thus:
$$P'(q)q+P(q)-C'(q)=\frac{A}{q}.$$</p>

<hr>

<p>Now for ad valorem:
$$\pi_a=(1-\tau_a)P(q)q-C(q)$$
First-order condition:
$$(1-\tau_a)[P'(q)q+P(q)]-C'(q)=0.$$
We have $A=\tau_a q P(q)$, which can can use as follows:
$$[P'(q)q+P(q)]-\tau_aP'(q)q-\tau_aP(q)-C'(q)=0.$$
$$[P'(q)q+P(q)]-\frac{A}{P(q)}P'(q)-\frac{A}{q}-C'(q)=0.$$
$$P'(q)q+P(q)-C'(q)=\frac{A}{q}+\frac{A}{P(q)}P'(q).$$</p>

<hr>

<p>Note the the left-hand side of the last line in both sections is the same. Moreover, this expression ($P'(q)q+P(q)-C'(q)$) is the derivative of profits assuming zero tax. If profits are concave then this derivative is decreasing in $q$.</p>

<p>We can see that, for any value of $q$, the right-hand side is smaller for the ad valorem tax (because it has an extra term that is negative--remember that $P'(q)&lt;0$). Thus, we have something that looks like this:</p>

<p><a href="https://i.sstatic.net/pA01q.png" rel="noreferrer"><img src="https://i.sstatic.net/pA01q.png" alt="Intersection at a higher quantity for ad valorem tax"></a></p>

CC BY-SA 3.0
2017-03-13T09:12:59.740Z
2017-03-13T09:12:59.740Z
Ubiquitous
<p><a href="https://i.sstatic.net/80pPR.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/80pPR.jpg" alt="enter image description here"></a></p>

<p><strong>Graphical representation</strong>: in the pre-tax situation:- demand curve is D and marginal revenue curve is MR, optimum is $e_1$ , output is $Q_1$ and price is $P_1$.</p>

<p><strong>Case 1:</strong> Specific tax (τ):- After tax demand curve is represented by the dotted demand curve $D^s$ and the corresponding marginal revenue curve $MR^s$. Intersection with MC gives the optimum $e_2$ where price is $P_2$ and quantity is $Q_2$. Revenue is area A=τ$Q_2$.</p>

<p><strong>Case 2:</strong> Ad valorem tax (α) :- After tax demand curve is represented by the demand curve $D^a$ and the corresponding marginal revenue curve $MR^a$. Let α be so chosen that the corresponding marginal revenue curve intersects the cost curve at $Q_2$. Observe that although it reduces output by same amount, in this case, it raises higher revenue, i.e, area A+B=α$P_2$$Q_2$.</p>

<p>Clearly, from here we can conclude that had we imposed an ad valorem tax that collects the same revenue as the specific tax, reduction in output would be less.</p>

CC BY-SA 3.0
2018-02-26T04:34:43.577Z
2018-02-26T04:34:43.577Z
Rayee
<p>Hello I'm working through <strong>Microeconomic Theory : Basic Principles and Extensions</strong> of Nicholson and Snyder 10e, for an exam and I fail to get how to answer this question (p.517) :</p>

<blockquote>
  <p>A specific tax is a fixed amount per unit of output. If the tax rate
  is t per unit, total tax collections are tQ . Show that the imposition
  of a specific tax on a monopoly will reduce output more (and increase
  price more) than will the imposition of an ad valorem tax that
  collects the same tax revenue.</p>
</blockquote>

<p>So I state that both tax collects the same (s for specific and a for ad valorem)
$$q_s^m \times \tau_s =A$$
$$q_a^m \times p^m_a \times \tau_a=A  $$</p>

<p>Then I have tried some identities of $q^m$ and $p^m$ but I can't get to the proof. </p>

<p>$$q_s^m=\dfrac{C'(q_s^m)+\tau_s-p^m_s}{P'_s} $$
$$q_a^m=\dfrac{C'(q_a^m)}{(1-\tau_a)P'_a}-\dfrac{p_a^m}{P'_a}$$</p>

<p>where $P'_i$ is the derivative of the inverse demand function at the equilibrium point of the monopolist under the tax type $i$</p>

<p>I tried replacing $\tau$ as $\tau_s=\dfrac{q_a^m \times p_a^m \times \tau_a }{q_s^m}$ and then replacing it in $q_s^m$ but I get a quadratic form of no use...</p>

<microeconomics><taxation><monopoly><theory>
Monopoly and Taxes (Nicholson Exercise)
Alan K.
# Monopoly and Taxes (Nicholson Exercise)

(Asked by: username_0 on 2017-03-11)

Hello I'm working through **Microeconomic Theory : Basic Principles and Extensions** of Nicholson and Snyder 10e, for an exam and I fail to get how to answer this question (p.517) :

> A specific tax is a fixed amount per unit of output. If the tax rate
> is t per unit, total tax collections are tQ . Show that the imposition
> of a specific tax on a monopoly will reduce output more (and increase
> price more) than will the imposition of an ad valorem tax that
> collects the same tax revenue.

So I state that both tax collects the same (s for specific and a for ad valorem)
$$q\_s^m \times \tau\_s =A$$
$$q\_a^m \times p^m\_a \times \tau\_a=A $$

Then I have tried some identities of $q^m$ and $p^m$ but I can't get to the proof.

$$q\_s^m=\dfrac{C'(q\_s^m)+\tau\_s-p^m\_s}{P'\_s} $$
$$q\_a^m=\dfrac{C'(q\_a^m)}{(1-\tau\_a)P'\_a}-\dfrac{p\_a^m}{P'\_a}$$

where $P'\_i$ is the derivative of the inverse demand function at the equilibrium point of the monopolist under the tax type $i$

I tried replacing $\tau$ as $\tau\_s=\dfrac{q\_a^m \times p\_a^m \times \tau\_a }{q\_s^m}$ and then replacing it in $q\_s^m$ but I get a quadratic form of no use...

**Tags:** microeconomics, taxation, monopoly, theory


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## 2 answers

### Answer by username_1 on 2017-03-13. Score: 5

First let's look at the specific tax. The profit is
$$\pi\_s=[P(q)-\tau\_s]q-C(q).$$
Differentiating to establish the first-order condition:
$$P'(q)q+P(q)-\tau\_s-C'(q)=0.$$
If we write $A=\tau\_s q$ for the tax revenue then we can rewrite the FOC thus:
$$P'(q)q+P(q)-C'(q)=\frac{A}{q}.$$

---

Now for ad valorem:
$$\pi\_a=(1-\tau\_a)P(q)q-C(q)$$
First-order condition:
$$(1-\tau\_a)[P'(q)q+P(q)]-C'(q)=0.$$
We have $A=\tau\_a q P(q)$, which can can use as follows:
$$[P'(q)q+P(q)]-\tau\_aP'(q)q-\tau\_aP(q)-C'(q)=0.$$
$$[P'(q)q+P(q)]-\frac{A}{P(q)}P'(q)-\frac{A}{q}-C'(q)=0.$$
$$P'(q)q+P(q)-C'(q)=\frac{A}{q}+\frac{A}{P(q)}P'(q).$$

---

Note the the left-hand side of the last line in both sections is the same. Moreover, this expression ($P'(q)q+P(q)-C'(q)$) is the derivative of profits assuming zero tax. If profits are concave then this derivative is decreasing in $q$.

We can see that, for any value of $q$, the right-hand side is smaller for the ad valorem tax (because it has an extra term that is negative--remember that $P'(q)<0$). Thus, we have something that looks like this:

[![Intersection at a higher quantity for ad valorem tax](https://i.sstatic.net/pA01q.png)](https://i.sstatic.net/pA01q.png)


---

### Answer by username_2 on 2018-02-26. Score: 1

[![enter image description here](https://i.sstatic.net/80pPR.jpg)](https://i.sstatic.net/80pPR.jpg)

**Graphical representation**: in the pre-tax situation:- demand curve is D and marginal revenue curve is MR, optimum is $e\_1$ , output is $Q\_1$ and price is $P\_1$.

**Case 1:** Specific tax (τ):- After tax demand curve is represented by the dotted demand curve $D^s$ and the corresponding marginal revenue curve $MR^s$. Intersection with MC gives the optimum $e\_2$ where price is $P\_2$ and quantity is $Q\_2$. Revenue is area A=τ$Q\_2$.

**Case 2:** Ad valorem tax (α) :- After tax demand curve is represented by the demand curve $D^a$ and the corresponding marginal revenue curve $MR^a$. Let α be so chosen that the corresponding marginal revenue curve intersects the cost curve at $Q\_2$. Observe that although it reduces output by same amount, in this case, it raises higher revenue, i.e, area A+B=α$P\_2$$Q\_2$.

Clearly, from here we can conclude that had we imposed an ad valorem tax that collects the same revenue as the specific tax, reduction in output would be less.


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