I have been looking at auction theory and in the book Auction Theory by Krishna, there is one (seemingly simple) inequality that I just cannot follow.

Context: given a private valuation $x$, the optimal bidding strategy has been found $\beta(x)$. Now, the author wants to show that behaving and bidding as if you were of type $z$, $\beta(z)$ does not increase profits. Then, calculating the difference between the profit in the optimum and the profit if you would behave as if you were of type $z$ leads to the following inequality. $G(x)$ being a probability distriubtion: $$\pi(\beta(x),x) - \pi( \beta(z),x) = G(z)(z-x) - \int_x^zG(y)dy \geq 0$$

The profit functions were calculated from a first-price auction in case it helps anyone. My question is why the inequality holds. Why is $G(z)(z-x) - \int_x^zG(y)dy$ larger than 0?

I hope you can help me :)


2 Answers 2


$$ G(z) (z-x) = \int_x^z G(z) dy $$ and since $G$ is increasing on $[x,z]$, the right hand side is larger than $\int_x^z G(y) dy$.

  • $\begingroup$ Thank you very much! Do you perhaps also know why the equation hold for the case where z<x? That is for me the most confusing part of the inequality... @Giskard $\endgroup$
    – Snowrabbit
    Apr 21, 2021 at 21:40
  • $\begingroup$ @Snowrabbit Same reason. In that case $G(z)$ is smaller than $G(y)$ for all $y \in [z,x]$ and $(z-x)<0$. $\endgroup$
    – Giskard
    Apr 22, 2021 at 6:28
  • 1
    $\begingroup$ Now I see it! Thank you very much! $\endgroup$
    – Snowrabbit
    Apr 22, 2021 at 7:43

Although there already is an accepted answer, there is another way to see the global optimality - or rather the same way with a different formulation.

By construction, $$\frac{\partial \pi}{\partial b}(b,x) = - G((\beta)^{-1}(b)) + (x-b) \frac{G'((\beta)^{-1}(b))}{(\beta)'((\beta)^{-1}(b))}\Bigg{|}_{b=\beta(x)}= 0,$$ where $\frac{\partial \pi}{\partial b}(b,x)$ is increasing in $x$.

Now consider some bid $\widehat b<\beta(x)$. By continuity of $\beta$, there is a type $\widehat x<x$ such that $\beta(\widehat x)=\widehat b$. Hence, because $\widehat x<x$, $$\frac{\partial \Pi}{\partial b}(\widehat b,x) \geq \frac{\partial \Pi}{\partial b}(\widehat b, \widehat x) = \frac{\partial \Pi}{\partial b} (\beta(\widehat x),\widehat x) = 0. $$ Thus, the expected utility $\Pi( b,x)$ is increasing in $b$ for all $ b<\beta(x)$. Analogously, $\Pi(b,x)$ is decreasing for all $\widehat b'>\beta(x)$.


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