<p>I'm assuming by "at [the] second equality sign" you mean the translation of the right-hand side of the first equality sign to the right-hand side of the second equality sign. This can be handled by analyzing each integral separately.</p>

<hr>

<p>Notice that in the first (double) integral the left-hand side of the inner integral does not vary with $v_2$, so</p>

<p>$$\int_{a_1}^{\min\{v_2,b_1\}} \left[ v_2 f_2\left( v_2 \right) + F_2\left( v_2 \right) - 1 \right] f_1\left( v_1 \right) dv_1 \\\hspace{3cm}= \left[ v_2 f_2\left( v_2 \right) + F_2\left( v_2 \right) - 1 \right] \int_{a_1}^{\min\left\{ v_2, b_1 \right\}} f_1\left( v_1 \right) dv_1.$$</p>

<p>The upper bound of the right-hand integral is just there to cap the CDF, so $\int_{a_1}^{\min\{ v_2, b_1 \}} f_1( v_1 ) dv_1 = F_1( v_2 )$, taking as given that $F_1( x ) = 1$ for all $x \geq b_1$. In terms of the original equation this gives</p>

<p>$$ \int_{a_2}^{b_2} \int_{a_1}^{\min\{v_2,b_1\}} \left[ v_2 f_2\left( v_2 \right) + F_2\left( v_2 \right) - 1 \right] f_1\left( v_1 \right) dv_1 dv_2 \\\hspace{3cm}= \int_{a_2}^{b_2} \left[ v_2 f_2\left( v_2 \right) + F_2\left( v_2 \right) - 1 \right] F_1\left( v_2 \right) dv_2. \hspace{2cm} \text{(Term L)}$$</p>

<hr>

<p>Notice that the inner integrand in the second (double) integral can be written as</p>

<p>$$v_1 f_1\left( v_1 \right) + F_1\left( v_1 \right) = \frac{d}{dv} \left[ v F_1\left( v \right) \right]_{v = v_1}.$$</p>

<p>Since $a_1 F_1( a_1 ) = 0$ (because $F_1( a_1 ) = 0$) and $x F_1( x ) = x$ for all $x \geq b_1$, this gives</p>

<p>$$\int_{a_2}^{b_2} \int_{a_1}^{\min\left\{ v_2, b_1 \right\}} \left[ v_1 f_1\left( v_1 \right) + F_1\left( v_1 \right) \right] dv_1 f_2\left( v_2 \right) dv_2 \\ \hspace{3cm} = \int_{a_2}^{b_2} \min\left\{ v_2 F_1\left( v_2 \right), b_1 \right\} f_2\left( v_2 \right) dv_2. \hspace{2cm} \text{(Term R)}$$</p>

<hr>

<p>The second equality is $\text{(Term L)} - \text{(Term R)}$.</p>

CC BY-SA 4.0
2018-09-16T20:36:55.477Z
2018-09-16T20:36:55.477Z
kyle
<p>The following image is an excerpt taken from Myerson and Satterthwaite's 1983 paper <em>Efficient Mechanisms for Bilateral Trading</em>:
<a href="https://i.sstatic.net/iFxMP.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/iFxMP.jpg" alt="enter image description here"></a></p>

<p>I know what the theorem is all about, all I need is some help with the integrals 
 (especially the one at the second equality sign).</p>

CC BY-SA 4.0
2018-09-13T07:43:29.797Z
I was able to convert this double integral into a single one. I am unable to process the further calculations.
superhulk
CC BY-SA 4.0
2018-09-13T18:54:18.963Z
I'm working on a slightly different proof of this theorem for my thesis and this last step has also got me stuck for a bit.
Pedro Cavalcante
CC BY-SA 4.0
2018-09-16T22:07:46.083Z
We ask that questions do not have large pictures of math text, but that it is converted into MathJax type for this site, thanks.
Kitsune Cavalry
<mathematical-economics><game-theory><mechanism-design>
Help needed to understand proof in Myerson and Satterthwaite (1983)
superhulk
# Help needed to understand proof in Myerson and Satterthwaite (1983)

(Asked by: username_0 on 2018-09-12)

The following image is an excerpt taken from Myerson and Satterthwaite's 1983 paper *Efficient Mechanisms for Bilateral Trading*:
[![enter image description here](https://i.sstatic.net/iFxMP.jpg)](https://i.sstatic.net/iFxMP.jpg)

I know what the theorem is all about, all I need is some help with the integrals
(especially the one at the second equality sign).

**Tags:** mathematical-economics, game-theory, mechanism-design

> **Comments:**
> **username_0**: I was able to convert this double integral into a single one. I am unable to process the further calculations.  
> **username_1**: I'm working on a slightly different proof of this theorem for my thesis and this last step has also got me stuck for a bit.  
> **username_2**: We ask that questions do not have large pictures of math text, but that it is converted into MathJax type for this site, thanks.  

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## 1 answers

### Answer by username_3 on 2018-09-16. Score: 2

I'm assuming by "at [the] second equality sign" you mean the translation of the right-hand side of the first equality sign to the right-hand side of the second equality sign. This can be handled by analyzing each integral separately.

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Notice that in the first (double) integral the left-hand side of the inner integral does not vary with $v\_2$, so

$$\int\_{a\_1}^{\min\{v\_2,b\_1\}} \left[ v\_2 f\_2\left( v\_2 \right) + F\_2\left( v\_2 \right) - 1 \right] f\_1\left( v\_1 \right) dv\_1 \\\hspace{3cm}= \left[ v\_2 f\_2\left( v\_2 \right) + F\_2\left( v\_2 \right) - 1 \right] \int\_{a\_1}^{\min\left\{ v\_2, b\_1 \right\}} f\_1\left( v\_1 \right) dv\_1.$$

The upper bound of the right-hand integral is just there to cap the CDF, so $\int\_{a\_1}^{\min\{ v\_2, b\_1 \}} f\_1( v\_1 ) dv\_1 = F\_1( v\_2 )$, taking as given that $F\_1( x ) = 1$ for all $x \geq b\_1$. In terms of the original equation this gives

$$ \int\_{a\_2}^{b\_2} \int\_{a\_1}^{\min\{v\_2,b\_1\}} \left[ v\_2 f\_2\left( v\_2 \right) + F\_2\left( v\_2 \right) - 1 \right] f\_1\left( v\_1 \right) dv\_1 dv\_2 \\\hspace{3cm}= \int\_{a\_2}^{b\_2} \left[ v\_2 f\_2\left( v\_2 \right) + F\_2\left( v\_2 \right) - 1 \right] F\_1\left( v\_2 \right) dv\_2. \hspace{2cm} \text{(Term L)}$$

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Notice that the inner integrand in the second (double) integral can be written as

$$v\_1 f\_1\left( v\_1 \right) + F\_1\left( v\_1 \right) = \frac{d}{dv} \left[ v F\_1\left( v \right) \right]\_{v = v\_1}.$$

Since $a\_1 F\_1( a\_1 ) = 0$ (because $F\_1( a\_1 ) = 0$) and $x F\_1( x ) = x$ for all $x \geq b\_1$, this gives

$$\int\_{a\_2}^{b\_2} \int\_{a\_1}^{\min\left\{ v\_2, b\_1 \right\}} \left[ v\_1 f\_1\left( v\_1 \right) + F\_1\left( v\_1 \right) \right] dv\_1 f\_2\left( v\_2 \right) dv\_2 \\ \hspace{3cm} = \int\_{a\_2}^{b\_2} \min\left\{ v\_2 F\_1\left( v\_2 \right), b\_1 \right\} f\_2\left( v\_2 \right) dv\_2. \hspace{2cm} \text{(Term R)}$$

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The second equality is $\text{(Term L)} - \text{(Term R)}$.


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