<p>You question can be answered using a <em>revealed preference</em> argument.</p>
<p>Let <span class="math-container">$B = \{q \in \mathbb{R}^n_+| p' q \le m\}$</span> be some budget set of a consumer (i.e. <span class="math-container">$B$</span> gives all possible bundles that the consumer can choose).</p>
<p>Let <span class="math-container">$q^\ast$</span> be the optimal choice from <span class="math-container">$B$</span>, i.e. the bundle that optimizes the utility. Then for any other bundle <span class="math-container">$q \in B$</span>, it must be that <span class="math-container">$u(q^\ast) \ge u(q)$</span>.</p>
<blockquote>
<p><em>proof:</em> The proof is simple, if, towards a contradiction, we would have that <span class="math-container">$q \in B$</span> and  <span class="math-container">$u(q) &gt; u(q^\ast)$</span>, then this would contradict the assumption that <span class="math-container">$q^\ast$</span> was optimal in <span class="math-container">$B$</span>.</p>
</blockquote>
<p>Now, let's apply this to the two budget sets for your question.</p>
<p>The first budget set has a lump-sum tax equal to <span class="math-container">$R$</span>.
<span class="math-container">$$
B^1 = \{q \in \mathbb{R}^n_+ | p' q \le m - R\}.
$$</span>
The second budget set has an ad-valorem tax equal to <span class="math-container">$t$</span> for some good, say <span class="math-container">$i$</span>.
<span class="math-container">$$
B^2 = \{q \in \mathbb{R}^n_+| p' q + t q_i \le m\}.
$$</span>
Let <span class="math-container">$q^{1}$</span> be utility maximising, bundle in <span class="math-container">$B^1$</span> and let <span class="math-container">$q^2$</span> be the optimal, utility maximising, bundle in  <span class="math-container">$B^2$</span>.</p>
<p>In addition we assume that the revenue of the ad-valorem tax equals the revenue for the lump-sum tax:
<span class="math-container">$$
t q_i^2 = R.
$$</span>
We would like to show that <span class="math-container">$u(q^1) \ge u(q^2)$</span>, so the lump sum tax always gives at least as much utility as the ad-valorem tax.</p>
<p>Using the above <em>revealed preference</em> argument, it suffices to show that <span class="math-container">$q^2 \in B^1$</span>, which means that <span class="math-container">$q^2$</span> was a feasible option when <span class="math-container">$q^1$</span> was chosen.</p>
<p>Indeed, we have:
<span class="math-container">$$
q^2 \in B^2,\\
\iff p' q^2 + t q^2_i \le m,\\
\iff p'q^2 \le m - t q^2_i,\\
\iff p' q^2 \le m - R.\\
$$</span>
So <span class="math-container">$q^2 \in B^1$</span>.</p>
<p><strong>Remark</strong>: notice that we did not impose any restrictions on the utility function. The only thing that we assumed was that the consumer was able (from any budget set) to pick an optimal bundle.</p>

CC BY-SA 4.0
2021-05-02T12:27:58.210Z
Awesome. I would have preferred writing $q^*$ (or using letters) though because it might look like q squared otherwise, but that's just a minor thing. You certainly can teach math to econs.
j3141592653589793238
CC BY-SA 4.0
2021-05-02T07:57:21.437Z
2021-05-02T07:57:21.437Z
tdm
<p>I'm studying the optimal choice of consumers with regards to taxation.</p>
<p>I read that for consumers, income tax is generally (for Cobb-Douglas preferences) preferred compared to ad valorem tax:</p>
<p>If the budget constraint is generally <span class="math-container">$p_1x_1+p_2x_2=m$</span>, then an ad valorem tax <span class="math-container">$t$</span> will change it to <span class="math-container">$(p_1 + t)x_1+p_2x_2=m$</span>.</p>
<p>If the income tax is the same amount, we get <span class="math-container">$p_1x_1+p_2x_2=m-R*$</span>, where <span class="math-container">$R*=tx_1$</span> is the amount of taxes for both variants.</p>
<p>Now, the following figure demonstrates how an increased price of good 1 will lead to a preference of an income tax over ad valorem tax:
<a href="https://i.sstatic.net/va5U4.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/va5U4.png" alt="h. varian" /></a></p>
<p>I wonder if this is the case for all indifference curves or if there are certain indifference curves where consumers might prefer an ad valorem tax. I doubt this, but I think the chapter in H. Varian lacks a mathematical (algebraic or analytical) explanation for this which I feel would help me understand.</p>

CC BY-SA 4.0
2023-10-07T06:05:09.463Z
I have a question. Shouldn't the equation for the budget constraint be this:
(1 + τ ) p1x1 + p2x2 = m? The one that you have stated would be for quantity tax, even the picture you have attached says so.
Amit Daultani
<microeconomics><utility><consumer-theory><taxation><cobb-douglas>
Is an income tax always more favourable for consumers compared to ad valorem/quantity tax?
j3141592653589793238
# Is an income tax always more favourable for consumers compared to ad valorem/quantity tax?

(Asked by: username_0 on 2021-05-01)

I'm studying the optimal choice of consumers with regards to taxation.

I read that for consumers, income tax is generally (for Cobb-Douglas preferences) preferred compared to ad valorem tax:

If the budget constraint is generally $p\_1x\_1+p\_2x\_2=m$, then an ad valorem tax $t$ will change it to $(p\_1 + t)x\_1+p\_2x\_2=m$.

If the income tax is the same amount, we get $p\_1x\_1+p\_2x\_2=m-R\*$, where $R\*=tx\_1$ is the amount of taxes for both variants.

Now, the following figure demonstrates how an increased price of good 1 will lead to a preference of an income tax over ad valorem tax:
[![h. varian](https://i.sstatic.net/va5U4.png)](https://i.sstatic.net/va5U4.png)

I wonder if this is the case for all indifference curves or if there are certain indifference curves where consumers might prefer an ad valorem tax. I doubt this, but I think the chapter in H. Varian lacks a mathematical (algebraic or analytical) explanation for this which I feel would help me understand.

**Tags:** microeconomics, utility, consumer-theory, taxation, cobb-douglas

> **Comments:**
> **username_1**: I have a question. Shouldn't the equation for the budget constraint be this: (1 + τ ) p1x1 + p2x2 = m? The one that you have stated would be for quantity tax, even the picture you have attached says so.  

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## 1 answers

### Answer by username_2 on 2021-05-02. Score: 3 | Accepted answer

You question can be answered using a *revealed preference* argument.

Let $B = \{q \in \mathbb{R}^n\_+| p' q \le m\}$ be some budget set of a consumer (i.e. $B$ gives all possible bundles that the consumer can choose).

Let $q^\ast$ be the optimal choice from $B$, i.e. the bundle that optimizes the utility. Then for any other bundle $q \in B$, it must be that $u(q^\ast) \ge u(q)$.

> *proof:* The proof is simple, if, towards a contradiction, we would have that $q \in B$ and $u(q) > u(q^\ast)$, then this would contradict the assumption that $q^\ast$ was optimal in $B$.

Now, let's apply this to the two budget sets for your question.

The first budget set has a lump-sum tax equal to $R$.
$$
B^1 = \{q \in \mathbb{R}^n\_+ | p' q \le m - R\}.
$$
The second budget set has an ad-valorem tax equal to $t$ for some good, say $i$.
$$
B^2 = \{q \in \mathbb{R}^n\_+| p' q + t q\_i \le m\}.
$$
Let $q^{1}$ be utility maximising, bundle in $B^1$ and let $q^2$ be the optimal, utility maximising, bundle in $B^2$.

In addition we assume that the revenue of the ad-valorem tax equals the revenue for the lump-sum tax:
$$
t q\_i^2 = R.
$$
We would like to show that $u(q^1) \ge u(q^2)$, so the lump sum tax always gives at least as much utility as the ad-valorem tax.

Using the above *revealed preference* argument, it suffices to show that $q^2 \in B^1$, which means that $q^2$ was a feasible option when $q^1$ was chosen.

Indeed, we have:
$$
q^2 \in B^2,\\
\iff p' q^2 + t q^2\_i \le m,\\
\iff p'q^2 \le m - t q^2\_i,\\
\iff p' q^2 \le m - R.\\
$$
So $q^2 \in B^1$.

**Remark**: notice that we did not impose any restrictions on the utility function. The only thing that we assumed was that the consumer was able (from any budget set) to pick an optimal bundle.

> **Comments:**
> **username_0**: Awesome. I would have preferred writing $q^*$ (or using letters) though because it might look like q squared otherwise, but that's just a minor thing. You certainly can teach math to econs.  

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