<p>As for your first question, about the second inequality below:</p>
<p><a href="https://i.sstatic.net/DyR2u.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/DyR2u.png" alt="enter image description here" /></a></p>
<p>first notice that the parts underlined in green are the same, as <span class="math-container">$\bar F = F(y_1)$</span></p>
<p>Then the text simply decomposes one of the integrals after the first inequality  in two intervals of integration (as the integral is additive on intervals of integration). That is, it sets:</p>
<p><span class="math-container">$ \int\limits_0^{y_1} F(y)dy= \int\limits_0^{y^*_1} F(y)dy+ \int\limits_{y^*_1}^{y_1} F(y)dy$</span></p>
<p>Therefore, we have:</p>
<p><span class="math-container">$ -[\int\limits_0^{y_1} F(y)dy- \int\limits_0^{y^*_1} F^*(y)dy]=-[\int\limits_0^{y^*_1} F(y)dy+ \int\limits_{y^*_1}^{y_1} F(y)dy-\int\limits_0^{y^*_1} F^*(y)dy]=-\int\limits_0^{y^*_1} (F(y)-F^*(y))dy+\int\limits_{y_1}^{y^*_1} F(y)dy$</span>  .</p>
<p>(notice that in the last integral, the extremes of integration have been exchanged).</p>
<p><span class="math-container">$\Box$</span></p>
<p><strong>Step 2</strong> of the answer.</p>
<blockquote>
<p>I don't understand how first mean value theorem (for definite
integrals) would have been applied here.</p>
</blockquote>
<p>The text you quoted says:</p>
<blockquote>
<p><strong>Applying the first mean-value theorem, the second term is positive.</strong></p>
</blockquote>
<p>I suppose that it refers to the second term of the equality reported above from the text,  that is</p>
<p><span class="math-container">$\int\limits_{y_1}^{y^*_1} F(y)dy- (y^*_1-y_1) F(y_1).\;\;\;\;\; (1)$</span></p>
<p>For the theorem of the mean value we can write:</p>
<p><span class="math-container">$\int\limits_{y_1}^{y^*_1} F(y)dy=(y^*_1-y_1)F(\xi)$</span>,</p>
<p>where <span class="math-container">$\xi \in (y_1, y^*_1)$</span>.</p>
<p>As <span class="math-container">$F$</span> is increasing, <span class="math-container">$F(\xi)&gt;F(y_1)$</span>, so (1) is positive.</p>

CC BY-SA 4.0
2022-10-29T12:17:53.313Z
To decompose that integral you as you did, wouldn't you need to know that y*1 is smaller than y1? How do you know? I'm also not sure I understand your reasoning for why ()>(1). You now seem to be saying y*1 is larger than y1?
Nasan
CC BY-SA 4.0
2022-10-29T12:46:20.957Z
No, the decomposition of the integral holds regardless of whether $y^*_1<y_1$ or vice versa. Simply, the integral changes its sign. As for $\xi$, to apply the mean value theorem we have to refer to an interval, in this case $(y_1,y_1^*)$, but if we suppose $y_1>y_1^*$ the result is the same, just signs are changed in the equality.
BakerStreet
CC BY-SA 4.0
2022-10-27T14:14:19.887Z
2022-10-27T19:23:29.770Z
2022-10-27T19:23:29.770Z
BakerStreet
<p>I'm reading Atkinson's &quot;On the measurement of inequality&quot; and I'm having difficulties understanding how this paper is arriving at certain outcomes, using integration by parts. Let me first post what I have been able to understand, and which could be needed for later operations:</p>
<p><a href="https://i.sstatic.net/kUPBW.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/kUPBW.png" alt="enter image description here" /></a></p>
<p>(3) is still straightforward integration by parts.</p>
<p><a href="https://i.sstatic.net/rKwjz.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/rKwjz.png" alt="enter image description here" /></a></p>
<p>I can still follow what's on the 1st row: we're comparing the 2 distributions here. However, I don't at understand how the right-hand side on the 2nd and 3rd rows is arrived at. I made an attempt at reproducing it (with my basic command of integration), by putting a negative sign in front of <span class="math-container">$ \int\limits_0^{y_1} F(y)dy $</span>
so that it becomes <span class="math-container">$ -\int\limits_{y_1}^0 F(y)dy $</span>.</p>
<p>That gives me: <span class="math-container">$ -(-\int\limits_{y_1}^0 F(y)dy - \int\limits_0^{y*_1} F*(y)dy) $</span>,</p>
<p>which then becomes: <span class="math-container">$ \int\limits_{y_1}^0 F(y)dy + \int\limits_0^{y*_1} F*(y)dy $</span>,</p>
<p>which would equal: <span class="math-container">$ \int\limits_{y_1}^{y*_1} F(y)dy $</span>, which is the 1st within the square brackets on the 3rd row.
However, I'm fairly certain this operation isn't correction, as F(y) and F*(y) are different functions and you can't add them like I did there. Moreover, I still wouldn't know the 1st term and 2nd terms within square brackets have been arrived at.</p>
<p>The paper then continues to say:</p>
<p><a href="https://i.sstatic.net/IHQzw.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/IHQzw.png" alt="enter image description here" /></a></p>
<p>I don't understand how first mean value theorem (for definite integrals) would have been applied here. I don't understand the relation with condition (2) or the inference to the relation between f(y) and f*(y). I then also don't understand the next set of operations that are said to flow out of (3). In particular I don't understand how the definition of the lorenz curve would imply this:
<a href="https://i.sstatic.net/diyoU.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/diyoU.png" alt="enter image description here" /></a></p>

<inequality><economic-measurement>
The application of integration by parts involving the lorenz curve
Nasan
# The application of integration by parts involving the lorenz curve

(Asked by: username_0 on 2022-10-27)

I'm reading Atkinson's "On the measurement of inequality" and I'm having difficulties understanding how this paper is arriving at certain outcomes, using integration by parts. Let me first post what I have been able to understand, and which could be needed for later operations:

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

(3) is still straightforward integration by parts.

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

I can still follow what's on the 1st row: we're comparing the 2 distributions here. However, I don't at understand how the right-hand side on the 2nd and 3rd rows is arrived at. I made an attempt at reproducing it (with my basic command of integration), by putting a negative sign in front of $ \int\limits\_0^{y\_1} F(y)dy $
so that it becomes $ -\int\limits\_{y\_1}^0 F(y)dy $.

That gives me: $ -(-\int\limits\_{y\_1}^0 F(y)dy - \int\limits\_0^{y\*\_1} F\*(y)dy) $,

which then becomes: $ \int\limits\_{y\_1}^0 F(y)dy + \int\limits\_0^{y\*\_1} F\*(y)dy $,

which would equal: $ \int\limits\_{y\_1}^{y\*\_1} F(y)dy $, which is the 1st within the square brackets on the 3rd row.
However, I'm fairly certain this operation isn't correction, as F(y) and F\*(y) are different functions and you can't add them like I did there. Moreover, I still wouldn't know the 1st term and 2nd terms within square brackets have been arrived at.

The paper then continues to say:

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

I don't understand how first mean value theorem (for definite integrals) would have been applied here. I don't understand the relation with condition (2) or the inference to the relation between f(y) and f\*(y). I then also don't understand the next set of operations that are said to flow out of (3). In particular I don't understand how the definition of the lorenz curve would imply this:
[![enter image description here](https://i.sstatic.net/diyoU.png)](https://i.sstatic.net/diyoU.png)

**Tags:** inequality, economic-measurement


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## 1 answers

### Answer by username_1 on 2022-10-27. Score: 1 | Accepted answer

As for your first question, about the second inequality below:

[![enter image description here](https://i.sstatic.net/DyR2u.png)](https://i.sstatic.net/DyR2u.png)

first notice that the parts underlined in green are the same, as $\bar F = F(y\_1)$

Then the text simply decomposes one of the integrals after the first inequality in two intervals of integration (as the integral is additive on intervals of integration). That is, it sets:

$ \int\limits\_0^{y\_1} F(y)dy= \int\limits\_0^{y^\*\_1} F(y)dy+ \int\limits\_{y^\*\_1}^{y\_1} F(y)dy$

Therefore, we have:

$ -[\int\limits\_0^{y\_1} F(y)dy- \int\limits\_0^{y^\*\_1} F^\*(y)dy]=-[\int\limits\_0^{y^\*\_1} F(y)dy+ \int\limits\_{y^\*\_1}^{y\_1} F(y)dy-\int\limits\_0^{y^\*\_1} F^\*(y)dy]=-\int\limits\_0^{y^\*\_1} (F(y)-F^\*(y))dy+\int\limits\_{y\_1}^{y^\*\_1} F(y)dy$ .

(notice that in the last integral, the extremes of integration have been exchanged).

$\Box$

**Step 2** of the answer.

> I don't understand how first mean value theorem (for definite
> integrals) would have been applied here.

The text you quoted says:

> **Applying the first mean-value theorem, the second term is positive.**

I suppose that it refers to the second term of the equality reported above from the text, that is

$\int\limits\_{y\_1}^{y^\*\_1} F(y)dy- (y^\*\_1-y\_1) F(y\_1).\;\;\;\;\; (1)$

For the theorem of the mean value we can write:

$\int\limits\_{y\_1}^{y^\*\_1} F(y)dy=(y^\*\_1-y\_1)F(\xi)$,

where $\xi \in (y\_1, y^\*\_1)$.

As $F$ is increasing, $F(\xi)>F(y\_1)$, so (1) is positive.

> **Comments:**
> **username_0**: To decompose that integral you as you did, wouldn't you need to know that y*1 is smaller than y1? How do you know? I'm also not sure I understand your reasoning for why ()>(1). You now seem to be saying y*1 is larger than y1?  
> **username_1**: No, the decomposition of the integral holds regardless of whether $y^*_1<y_1$ or vice versa. Simply, the integral changes its sign. As for $\xi$, to apply the mean value theorem we have to refer to an interval, in this case $(y_1,y_1^*)$, but if we suppose $y_1>y_1^*$ the result is the same, just signs are changed in the equality.  

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