3
$\begingroup$

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).

$\endgroup$
4
  • 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
2
$\begingroup$

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 $\text{(Term L)} - \text{(Term R)}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.