Auction Theory: Proving that the found equilibrium is indeed optimal

Question

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 :)

Giskard · Accepted Answer

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

Bayesian · Answer

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 $widehat b'>beta(x)$.

Auction Theory: Proving that the found equilibrium is indeed optimal

2 Answers

Add your own answers!

Ask a Question