Given a direct mechanism, we define a buyer's expected utility $u(\theta)$ conditional on her type being $\theta$ by $u(\theta)=\theta q(\theta)-t(\theta)$, where $q:[\underline{\theta},\bar{\theta}]\to[0,1]$ and $t:[\underline{\theta},\bar{\theta}]\to\mathbb{R}$.

We also define that a direct mechanism is incentive-compatible if truth-telling is optimal for every $\theta\in[\underline{\theta},\bar{\theta}]$, i.e., $$u(\theta)\geq \theta q(\theta')-t(\theta'),\quad\forall\theta,\theta'\in[\underline{\theta},\bar{\theta}].$$

LEMMA: For an incentive-compatible direct mechanism, we want to show that for all $\theta$ that $u$ is differentiable, we have $u'(\theta)=q(\theta)$.

PROOF: Consider any $\theta$ for which $u$ is differentiable. Let $\delta>0$. Then by incentive compatibility, we have the following:

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I don't understand how from incentive compatibility it follows the first and second inequality 2.6 and 2.8. How can we use the same $\theta$ in the proof while the definition clearly states $u(\theta)\geq \theta q(\theta')-t(\theta'),\forall\theta,\theta'$?

  • 2
    $\begingroup$ @HerrK. "An Introduction to the Theory of Mechanism Design" by Tilman Börgers - page 12, lemma 2.2 $\endgroup$
    – johnny09
    Commented Feb 27, 2019 at 5:51

2 Answers 2


By definition we have $$ \begin{align*} u(\theta) & = \theta q(\theta)-t(\theta) \\ \\ u(\theta + \delta) & = (\theta + \delta) q(\theta + \delta)-t(\theta + \delta) \end{align*} $$ By incentive compatibility (where $\theta + \delta$ is the true type, $\theta$ is the false type) we have $$ u(\theta + \delta) \geq (\theta + \delta) q(\theta)-t(\theta) $$ Using these (I used square brackets for clearer notation, there is no mathematical function to them) $$ \begin{align*} u(\theta + \delta) - u(\theta) & = \left[(\theta + \delta) q(\theta + \delta)-t(\theta + \delta)\right] - \left[\theta q(\theta)-t(\theta)\right] \\ \\ u(\theta + \delta) - u(\theta) & \geq \left[(\theta + \delta) q(\theta)-t(\theta)\right] - \left[\theta q(\theta)-t(\theta)\right] \end{align*} $$ This also holds if you divide by $\delta > 0$.

Similar argument for 2.8., using true type $\theta - \delta$ instead of $\theta + \delta$.


Let $u(x)=xq(x)-t(x)$. Incentive compatibility dictates that $xq(x)-t(x)\geq xq(z)-t(z)$, when $x$ is the observed private value. Using a little algebraic manipulation, it can be shown that,

$xq(x)-t(x)+zq(z)\geq xq(z)-t(z)+zq(z)$, or

$u(x)+zq(z)\geq u(z)+xq(z)$, or

$u(x)\geq u(z) +(x-z)q(z)$. Let this be $(1)$.

Similarly, $zq(z)-t(z)\geq zq(x)-t(x)$, when $z$ is the observed private value. Thus, from here as well, we have $u(z)\geq u(x) +(z-x)q(x)$. Let this expression be $(2)$.

From $(1)$, we have $\cfrac{u(x)-u(z)}{x-z} \geq q(z)$. Similarly, from $(2)$, we have $\cfrac{u(x)-u(z)}{x-z} \leq q(x)$. Now, we have the expression, \begin{align*} q(z) \leq \cfrac{u(x)-u(z)}{x-z} \leq q(x) \\ -(3) \end{align*}

EDIT As correctly pointed out by TheoreticalEconomist, if $x > z$, expression $(3)$ tells us that $q(x)$ is monotone. Also, as $u$ is a convex function, it is absolutely continuous. This tells us that $u$ is differentiable almost everywhere. Thus, wherever $u$ is differentiable, we have $u'(x) = q(x)$.

  • $\begingroup$ I’m less fond of this way of presenting the argument — it seems to implicitly assume that $q$ is continuous, since you want that $q(z+\delta) \to q(z)$ as $\delta \to 0$. $\endgroup$ Commented Feb 28, 2019 at 8:10
  • $\begingroup$ @TheoreticalEconomist You're right, this differentiability criteria holds almost every. $\endgroup$
    – superhulk
    Commented Feb 28, 2019 at 10:44
  • $\begingroup$ don’t you need $q$ to be continuous a.e. to say that? $\endgroup$ Commented Feb 28, 2019 at 11:12
  • $\begingroup$ @TheoreticalEconomist $u'(x)=q(x)$ whenever $u$ is differentiable. Also, $u$ is absolutely continuous, so it is differentiable almost everywhere, except for a set of countable points at the most. The left and right derivatives, although, exist for all $x$. $\endgroup$
    – superhulk
    Commented Feb 28, 2019 at 12:45
  • $\begingroup$ I know that. I don’t think you understand my point. From what you’ve written in your answer, one can only conclude that $u^\prime = q$ only if one assumes that $q$ is continuous a.e. Suppose $q$ where discontinuous a.e. The conclusion you assert at the end of your answer is no longer valid. Of course, this is ruled out by incentive compatibility (since $q$ must be monotone), but this is not at all clear in your answer. $\endgroup$ Commented Feb 28, 2019 at 12:54

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