The following image is an excerpt taken from Myerson and Satterthwaite's 1983 paper Efficient Mechanisms for Bilateral Trading: enter image description here

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).

  • 1
    $\begingroup$ could you elaborate as to what in particular you deem unclear? $\endgroup$ – user11305 Sep 12 '18 at 20:51
  • $\begingroup$ I was able to convert this double integral into a single one. I am unable to process the further calculations. $\endgroup$ – superhulk Sep 13 '18 at 7:43
  • $\begingroup$ 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. $\endgroup$ – Pedro Cavalcante Sep 13 '18 at 18:54
  • $\begingroup$ We ask that questions do not have large pictures of math text, but that it is converted into MathJax type for this site, thanks. $\endgroup$ – Kitsune Cavalry Sep 16 '18 at 22:07

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.

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

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

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

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

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


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