<p><strong>1)</strong> "To first order" appears to be a shorthand for "to a first-order approximation", meaning in turn, "to a linear approximation".</p>

<p><strong>2)</strong> It is not that "the one implies the other", it is that the one is consistent with the other.  </p>

<p>In attempting to arrive at a functional specification for the "tax distortion costs" function, say $C_D=h(T)$, first we reason that if tax revenue is zero (not due to inefficiency of tax collection and/or tax evasion, but because the tax rate is set to zero), then distortions due to tax will also be zero. This implies that $h(T)$ should not contain a constant term, so that $h(0) = 0$.  </p>

<p>Moreover <em>if</em> we reason (as D. Romer does) that "distortions by taxes are likely to increase more than proportionally with the amount of revenue raised", this implies that $h(T)$ should <em>not</em> be linear in $T$, but it should be convex (this is what it means to say "more than proportionally"). </p>

<p>The natural step then is to consider a quadratic distortion cost function, without a constant term:</p>

<p>$$C_D =  aT^2,\;\; a&gt;0$$</p>

<p>which reflects <em>both</em> our theoretical arguments (and in particular, $h(0) =0$).</p>

<p>But this also implies that "distortion costs are zero to a first-order approximation", because we have</p>

<p>$$h'(T) = 2aT \implies h'(0)=0$$</p>

<p>so a <em>first-order</em> approximation (Taylor expansion) around zero would give</p>

<p>$$h(T) \approx h(0) + h'(0)\cdot T = 0+0\cdot T =0$$</p>

CC BY-SA 3.0
2015-05-11T12:00:23.403Z
2015-05-11T12:00:23.403Z
Alecos Papadopoulos
<p>1) What does it mean that something (in this context the taxation) has (or doesn't) costs to first order?</p>

<p>So, below is an excerpt from Romer's <em>Advanced Macro</em> (p. 598) with such a statement that I've encountered and felt a bit lost. Altough the model developed later is pretty clear, I still don't get what the sentence with this 'costs to first order' really mean. </p>

<p>2) Also: how the last sentence is implied by the sentence before the last one? That is: what exactly (this first-order<em>ness</em>? how?) implies that $C \approx T^2$?</p>

<p><img src="https://i.sstatic.net/c6J4j.png" alt="enter image description here"></p>

<taxation><cost><adjustment-costs>
(tax with) costs to first order
Piotr Pieniążek
# (tax with) costs to first order

(Asked by: username_0 on 2015-05-10)

1) What does it mean that something (in this context the taxation) has (or doesn't) costs to first order?

So, below is an excerpt from Romer's *Advanced Macro* (p. 598) with such a statement that I've encountered and felt a bit lost. Altough the model developed later is pretty clear, I still don't get what the sentence with this 'costs to first order' really mean.

2) Also: how the last sentence is implied by the sentence before the last one? That is: what exactly (this first-order*ness*? how?) implies that $C \approx T^2$?

![enter image description here](https://i.sstatic.net/c6J4j.png)

**Tags:** taxation, cost, adjustment-costs


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## 1 answers

### Answer by username_1 on 2015-05-11. Score: 2 | Accepted answer

**1)** "To first order" appears to be a shorthand for "to a first-order approximation", meaning in turn, "to a linear approximation".

**2)** It is not that "the one implies the other", it is that the one is consistent with the other.

In attempting to arrive at a functional specification for the "tax distortion costs" function, say $C\_D=h(T)$, first we reason that if tax revenue is zero (not due to inefficiency of tax collection and/or tax evasion, but because the tax rate is set to zero), then distortions due to tax will also be zero. This implies that $h(T)$ should not contain a constant term, so that $h(0) = 0$.

Moreover *if* we reason (as D. Romer does) that "distortions by taxes are likely to increase more than proportionally with the amount of revenue raised", this implies that $h(T)$ should *not* be linear in $T$, but it should be convex (this is what it means to say "more than proportionally").

The natural step then is to consider a quadratic distortion cost function, without a constant term:

$$C\_D = aT^2,\;\; a>0$$

which reflects *both* our theoretical arguments (and in particular, $h(0) =0$).

But this also implies that "distortion costs are zero to a first-order approximation", because we have

$$h'(T) = 2aT \implies h'(0)=0$$

so a *first-order* approximation (Taylor expansion) around zero would give

$$h(T) \approx h(0) + h'(0)\cdot T = 0+0\cdot T =0$$


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