This question stems from section, pages 76-78 of Bolton and Dewatripont "Contract Theory".

Suppose we have a single buyer and supplier. Buyers types are drawn from a continuum $\Theta = [\underline{\theta},\overline{\theta}]\owns \theta$ according to a CDF $F$ with density $f$ which is everywhere positive on $\Theta$. The supplier sells various goods which vary continuously with a parameter $q\in \mathbb{R}_+$ according to a price schedule $p(q)$. We can interpret $q$ directly as the quantity of the good supplied or as the quality of the good provided.

The buyer has utility function $u(q,p;\theta)$, given their type $\theta$, the $q$ they consume, and the price they pay, $p$.

The supplier has linear costs and their utility function is $p(q)-cq$. Assume the supplier has monopoly power and therefore solves the screening problem (applying the revelation principle), $$\max_{p(\cdot),q(\cdot)} \int_{[\underline{\theta},\overline{\theta}]}(p(\theta)-cq(\theta))f(\theta)d\theta$$ $$st. \begin{cases} u(q(\theta),p(\theta);\theta)\geq 0 & \forall \theta \in \Theta & (IR)_{\theta} \\ u(q(\theta),p(\theta);\theta)\geq u(q(\hat{\theta}),p(\hat{\theta});\theta) & \forall (\theta,\hat{\theta})\in \Theta^2 & (IC)_{\theta} \end{cases}$$

For this post, I ignore differentiability concerns and assume everything is smooth, including the optimal $p(\cdot)$ and $q(\cdot)$.

Now, in Bolton and Dewatripont, they state that the Spence-Mirrlees single crossing (SMSC) condition $$\frac{\partial }{\partial \theta} \left[-\frac{\partial u /\partial q}{\partial u / \partial \theta}\right]>0$$ allows us to replace the intractable IC constraints with a local adjacency condition and a monotonicity condition. However, in the textbook, they work with the specific functional form $u(q,p;\theta)=\theta v(q)-p$ where $v'>0$ and $v''<0$ and don't specify how the single crossing condition helps equate the IC constraints with their local variants in the more abstract setting.

I think the local IC conditions for a general $u$ will be $$\begin{align*} \frac{\partial u}{\partial q}(q(\theta),p(\theta);\theta)\frac{d q}{d \theta}(\theta)+ \frac{\partial u}{\partial p}(q(\theta),p(\theta);\theta)\frac{d p}{d \theta}(\theta) &=0 \\ \frac{d q}{d \theta}(\theta) &\geq 0 \end{align*}$$

I am stuck on proving these conditions are equivalent to the IC conditions when we assume SMSC and was hoping someone could point me in the right direction/outline how to prove this.

My Attempt

To prove the equivalence of the IC constraints and these conditions I first tried to prove the forward implication. That is, assume the IC conditions do hold. Define the function $$W(\theta,\hat{\theta})=u(q(\hat{\theta}),p(\hat{\theta}),\theta)$$ By the IC constraints for a fixed $\theta$, $W(\theta,\hat{\theta})$ is maximised at $\hat{\theta}=\theta$, so this gives the FOC and SOC $$\begin{align*} \frac{\partial u}{\partial q}\frac{d q}{d \theta}+ \frac{\partial u}{\partial p}\frac{d p}{d \theta}\Bigg|_{\hat{\theta}=\theta} &=0 \\ \frac{\partial^2 u}{\partial q^2}\left(\frac{d q}{d \theta}\right)^2 +\frac{\partial u}{\partial q}\frac{d^2 q}{d \theta^2}+\frac{\partial^2 u}{\partial p^2} \left(\frac{d p}{d \theta}\right)^2+\frac{\partial u}{\partial p}\frac{d^2 p}{d \theta^2}\Bigg|_{\hat{\theta}=\theta} &\leq 0\end{align*}$$

The FOC is our local adjacency condition so we just need to show the monotonicity condition. To do this in the specific case when $u(q,p;\theta)=\theta v(q)-p$, Bolton and Dewatripont differentiate the FOC with respect to $\theta$ and apply the SOC. Doing this (skipping the long algebra), I get that $$0\leq \frac{\partial^2 u}{\partial \theta \partial q}\frac{d q}{d \theta} +2 \frac{\partial^2 u}{\partial p \partial q}\frac{d q}{d \theta}\frac{d p}{d \theta}+\frac{\partial^2 u}{\partial \theta \partial p}\frac{d p}{d \theta} \quad (\dagger)$$ But I'm not sure how to use the SMSC condition to get the monotonicity condition. If I rearrange the FOC condition (and ignore potential division by 0) I get, $$-\frac{\partial u/\partial q}{\partial u /\partial p} =\frac{dp/d\theta}{dq/d\theta}$$ the SMSC condition gives me that $$0<\frac{\partial}{\partial \theta}\frac{dp/d\theta}{dq/d\theta} \quad (\ddagger)$$ which I think suggests that $\text{sgn}(dp/d\theta)=\text{sgn}(dq/d\theta)$ but I don't have a solid argument for this. Moreover, I can't see a way to use $(\dagger)$ and $(\ddagger)$ to show monotonicity.

For the reverse implication, I want to use the argument Bolton and Dewatripont use, but I don't understand how it works completely. Suppose that the IC constraint is violated for some $\theta$ but we have the local conditions holding. That is $\exists \hat{\theta}\in \Theta$ such that $\hat{\theta}\neq \theta$ and $$u(q(\theta),p(\theta);\theta)<u(q(\hat{\theta}),p(\hat{\theta});\theta)$$ Following the argument in the textbook, first assume $\hat{\theta}>\theta$. I want to integrate the inequality to get $$\int_{[\theta,\hat{\theta}]}\frac{\partial u}{\partial q}(q(x),p(x);\theta)\frac{d q}{d \theta}(x)+ \frac{\partial u}{\partial p}(q(x),p(x);\theta)\frac{d p}{d \theta}(x)dx>0 \quad (*)$$ But I don't understand why this would hold since I don't understand why the analogous condition in Bolton and Dewatripont holds. I have a feeling to derive the result I need to assume some additional structure on $u(q,p;\theta)$. What I want is to get $$\frac{\partial u}{\partial q}(q(x),p(x);\theta)<\frac{\partial u}{\partial q}(q(x),p(x);x) \quad x>\theta$$ and $$\frac{\partial u}{\partial p}(q(x),p(x);\theta)<\frac{\partial u}{\partial p}(q(x),p(x);x) \quad x>\theta$$ Then since by the monotonicity condition and $(\ddagger)$ $$\begin{align*} \frac{dq}{d\theta} &\geq 0 \\ \frac{dp}{d\theta} &\geq 0\end{align*}$$ so we would have that $$\frac{\partial u}{\partial q}(q(x),p(x);\theta)\frac{d q}{d \theta}(x)+ \frac{\partial u}{\partial p}(q(x),p(x);\theta)\frac{d p}{d \theta}(x)$$ $$<\frac{\partial u}{\partial q}(q(x),p(x);x)\frac{d q}{d \theta}(x)+ \frac{\partial u}{\partial p}(q(x),p(x);x)\frac{d p}{d \theta}(x)=0$$ Which means that $$\int_{[\theta,\hat{\theta}]}\frac{\partial u}{\partial q}(q(x),p(x);\theta)\frac{d q}{d \theta}(x)+ \frac{\partial u}{\partial p}(q(x),p(x);\theta)\frac{d p}{d \theta}(x)dx<0$$ which is a contradiction to $(*)$. We can use a similar argument, mutatis mutandis, for the case when $\hat{\theta}<\theta$.

Can someone help me understand how to use the SMSC condition properly and point out how to complete the proof or provide a reference where a proof is given?


1 Answer 1


Is it possible that you have a typo in the SMSC condition. I think it should be: $$ \frac{\partial}{\partial \theta} \left[-\frac{\partial u/\partial q}{\partial u/\partial p}\right] > 0. $$

If so, let us first show that the local IC conditions and the SMSC implies the monotonicity.

The local IC conditions require: $$ \begin{align*} &u_q q_\theta + u_p p_\theta = 0\\ &u_{qq} (q_\theta)^2 + u_q q_{\theta \theta} + 2u_{qp}q_\theta p_\theta + u_{pp} (p_\theta)^2 + u_p p_{\theta \theta} < 0 \end{align*} $$

Taking the derivative of the first order condition with respect to $\theta$ produces: $$ u_{qq} (q_{\theta})^2 + u_q q_{\theta \theta} + 2 u_{qp} q_{\theta} p_\theta + u_{pp} (p_\theta)^2 + u_p p_{\theta \theta} + u_{q\theta} q_\theta + u_{p\theta}p_\theta = 0. $$

Using the second order conditions gives: $$ \begin{align*} &u_{q\theta} q_\theta + u_{p\theta} p_\theta > 0. \tag{1} \end{align*} $$ Next, the SMSC conditions require: $$ \partial_\theta(-u_q/u_p) > 0, $$ which is equivalent to (note that here the derivative of $\theta$ is only with respect to the third argument of $u_q$ and $u_p$): $$ \frac{-u_{q\theta} u_p + u_{p\theta} u_q}{(u_p)^2} > 0 $$ Multiplying by $(u_p)^2 > 0$ gives: $$ \begin{align*} -u_{q \theta} u_p + u_{p \theta} u_q> 0,\\ \end{align*} $$ Substituting the first order $u_p = - u_q \frac{q_\theta}{p_\theta}$ gives: $$ \begin{align*} &u_{q \theta}( u_q q_\theta/p_\theta) + u_{p \theta} u_q > 0,\\ \leftrightarrow &u_q \frac{1}{p_\theta} \left( u_{q \theta} q_\theta + u_{p \theta} p_\theta\right) > 0 \end{align*} $$ The second term is positive (equation $(1)$). This means that the first term $u_q/p_\theta$ is also positive.

Using the first order condition $u_q = - u_p\frac{p_\theta}{q_\theta}$ gives: $$ -u_p \frac{1}{q_\theta} > 0. $$ As $u_p < 0$ we need that $q_\theta > 0$.

Now for the second derivation, assume that the local IC conditions, the SMSC and the monotonicity conditions are satisfied, we need to show the IC conditions hold. As in your question, assume this is not true. Then there is a $\theta$ and $\hat \theta$ such that: $$ u(q(\theta), p(\theta), \theta) < u(q(\hat \theta), p(\hat \theta), \theta). $$ Assume that $\hat \theta > \theta$.

Note that (by integration): $$ \int_{\theta}^{\hat \theta} u_q(q(x), p(x), \theta) q_\theta(x) + u_p(q(x), p(x), \theta) p_\theta(x) dx = u(q(\hat \theta), p(\hat \theta), \theta) - u(q(\theta), p(\theta), \theta). $$ As such, we get that: $$ \int_{\theta}^{\hat \theta} u_q(q(x), p(x), \theta) q_\theta(x) + u_p(q(x), p(x), \theta) p_\theta(x) dx > 0. $$ By the first order condition, we have that $\frac{p_\theta(x)}{q_\theta(x)} = - \frac{u_q(q(x), p(x),x)}{u_p(q(x), p(x), x)}$. Substituting out gives: $$ \begin{align*} &\int_{\theta}^{\hat \theta} q_\theta(x)\left(u_q(q(x), p(x), \theta) - u_p(q(x), p(x), \theta) \frac{u_q(q(x), p(x),x)}{u_p(q(x), p(x),x)}\right) dx > 0.\\ &= \int_{\theta}^{\hat \theta} \frac{q_{\theta}(x)}{u_p(q(x), p(x), x)}\left(u_q(q(x), p(x), \theta) u_p(q(x), p(x), x) - u_p(q(x), p(x), \theta) u_q(q(x), p(x), x)\right) dx > 0 \end{align*} $$

The first term under the integration $\frac{q_\theta(x)}{u_p(q(x), p(x), x)}$ is negative as the numerator is positive (by the monotonicity condition) and the denominator negative. This means that there must be at least one $x > \theta$ such that: $$ u_q(q(x), p(x), \theta) u_p(q(x), p(x), x) - u_p(q(x), p(x), \theta) u_q(q(x), p(x), x) < 0. \tag{2} $$

Let us show that this is impossible. Consider the SMSC condition, which states that (for all $x$) the ratio $\frac{u_q(q(x), p(x), \theta)}{u_p(q(x), p(x), \theta)}$ is decreasing in $\theta$ so if $x > \theta$ we have that: $$ \frac{u_q(q(x), p(x), x)}{u_p(q(x), p(x), x)} < \frac{u_q(q(x), p(x), \theta)}{u_p(q(x), p(x), \theta)}. $$

This can be rewritten as (note that both $u_p(q(x), p(x), x)$ and $u_p(q(x), p(x), \theta)$ are negative): $$ u_q(q(x), p(x), \theta) u_p(q(x), p(x), x) - u_p(q(x), p(x), \theta) u_q(q(x), p(x), x) > 0. $$ This contradicts $(2)$. If $\hat \theta < \theta$ we get a similar reasoning but then we have values $x < \theta$.

  • $\begingroup$ Thank you for your answer! The SMSC in my question was indeed a typo. Your answer makes everything clear to me but I think there is a typo in the first half. In the expression immediately following "Next, the SMSC conditions require..." in the first half of your answer is it meant to be $\partial_{\theta}(-u_q/u_p) >0$ as in the original SMSC condition? I think this problem is resolved by flipping the inequality in expression (1) so that we get the same conclusion in the end. Thanks again! $\endgroup$ Commented Feb 3 at 17:26
  • $\begingroup$ Thanks, I corrected the mistakes. $\endgroup$
    – tdm
    Commented Feb 4 at 8:49

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