<p>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.</p>
<p>By construction,
<span class="math-container">$$\frac{\partial \pi}{\partial b}(b,x) = - G((\beta)^{-1}(b)) + (x-b)
\frac{G'((\beta)^{-1}(b))}{(\beta)'((\beta)^{-1}(b))}\Bigg{|}_{b=\beta(x)}= 0,$$</span>
where <span class="math-container">$\frac{\partial \pi}{\partial b}(b,x)$</span> is increasing in <span class="math-container">$x$</span>.</p>
<p>Now consider some bid <span class="math-container">$\widehat b&lt;\beta(x)$</span>.
By continuity of <span class="math-container">$\beta$</span>, there is a type <span class="math-container">$\widehat x&lt;x$</span> such that <span class="math-container">$\beta(\widehat x)=\widehat b$</span>.
Hence, because <span class="math-container">$\widehat  x&lt;x$</span>,
<span class="math-container">$$\frac{\partial \Pi}{\partial b}(\widehat b,x) \geq \frac{\partial \Pi}{\partial b}(\widehat b, \widehat x) = \frac{\partial \Pi}{\partial b} (\beta(\widehat x),\widehat x) = 0. $$</span>
Thus, the expected utility <span class="math-container">$\Pi( b,x)$</span> is increasing in <span class="math-container">$b$</span> for all <span class="math-container">$  b&lt;\beta(x)$</span>.
Analogously, <span class="math-container">$\Pi(b,x)$</span> is decreasing for all <span class="math-container">$\widehat b'&gt;\beta(x)$</span>.</p>

CC BY-SA 4.0
2021-04-30T13:48:24.980Z
2021-04-30T14:11:43.800Z
2021-04-30T14:11:43.800Z
Bayesian
<p><span class="math-container">$$
G(z) (z-x) =
\int_x^z G(z) dy
$$</span>
and since <span class="math-container">$G$</span> is increasing on <span class="math-container">$[x,z]$</span>, the right hand side is larger than
<span class="math-container">$\int_x^z G(y) dy$</span>.</p>

CC BY-SA 4.0
2021-04-21T21:40:18.280Z
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
Snowrabbit
CC BY-SA 4.0
2021-04-22T06:28:03.063Z
@Snowrabbit Same reason. In that case $G(z)$ is smaller than $G(y)$ for all $y \in [z,x]$ and $(z-x)<0$.
Giskard
CC BY-SA 4.0
2021-04-22T07:43:32.033Z
Now I see it! Thank you very much!
Snowrabbit
CC BY-SA 4.0
2021-04-21T21:33:34.927Z
2021-04-21T21:33:34.927Z
Giskard
<p>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.</p>
<p>Context: given a private valuation <span class="math-container">$x$</span>, the optimal bidding strategy has been found <span class="math-container">$\beta(x)$</span>. Now, the author wants to show that behaving and bidding as if you were of type <span class="math-container">$z$</span>, <span class="math-container">$\beta(z)$</span> 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 <span class="math-container">$z$</span> leads to the following inequality.
<span class="math-container">$G(x)$</span> being a probability distriubtion:
<span class="math-container">$$\pi(\beta(x),x) - \pi( \beta(z),x) = G(z)(z-x) - \int_x^zG(y)dy \geq 0$$</span></p>
<p>The profit functions were calculated from a first-price auction in case it helps anyone.
My question is why the inequality holds. Why is <span class="math-container">$G(z)(z-x) - \int_x^zG(y)dy$</span> larger than 0?</p>
<p>I hope you can help me :)</p>

<microeconomics><mathematical-economics><auctions>
Auction Theory: Proving that the found equilibrium is indeed optimal
Snowrabbit
# Auction Theory: Proving that the found equilibrium is indeed optimal

(Asked by: username_0 on 2021-04-21)

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 :)

**Tags:** microeconomics, mathematical-economics, auctions

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## 2 answers

### Answer by username_2 on 2021-04-21. Score: 5 | 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
$\int\_x^z G(y) dy$.

> **Comments:**
> **username_0**: 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  
> **username_2**: @Snowrabbit Same reason. In that case $G(z)$ is smaller than $G(y)$ for all $y \in [z,x]$ and $(z-x)<0$.  
> **username_0**: Now I see it! Thank you very much!  

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### Answer by username_1 on 2021-04-30. Score: 2

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 such that $\beta(\widehat x)=\widehat b$.
Hence, because $\widehat 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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