I am working through Rochet and Stohle (2003) chapter on multi-dimensional screening, and I am struggling filling in the blanks between equation (2.1) p. 154, and its simplified form on page 155. In particular, my question concerns:

Consumers are of type $\theta$ $\in$ $\Theta$ = [$\underline\theta$, $\bar\theta$] with associated cdf $F(\theta)$, which is absolutely continuous, and associated density function $f(\theta)$ = $F'(\theta)$. This is the distribution of types.

There also is a preference relation for each consumer for $q \in Q = [0,\bar q]$: $u = v(q, \theta) - P$ with the single crossing property $v_{q\theta} > 0 $. I am adding this for context, but all information may not be necessary for my particular question.

This renders an indirect utility function defined by:

$u(\theta) = \max_{\rm {q\in Q}} \{v(q, \theta) - P(q)\}$

The monopoly firm is using a non-linear tariff, $P(q)$, and wants to maximize its expected payoff (I am skipping some information not needed for my question on what renders this equation):

$E(\pi) = \int^\bar\theta_\underline\theta [S(q(\theta), \theta) - u(\theta)] dF(\theta)$ equation (2.1) in text

subject to:

$du/d\theta = v_\theta(q(\theta), \theta)$

$dq(\theta)/d\theta\geq0 $

$IR\space constraint$

Note that the function $S(q(\theta), \theta)$ is not important for my question, hence, I am not including its definition here.

Now to my question. The authors makes a simplification of the problem:

$E(\pi) = \int^\bar\theta_\underline\theta [S(q(\theta), \theta) - \frac {1-F(\theta)}{f(\theta)}v_\theta(q(\theta), \theta) - u(\underline\theta)]dF(\theta)$

subject to:

$dq(\theta)/d\theta\geq0 $

$IR\space constraint$

Now, this to me is the same as saying:

$\int^\bar\theta_\underline\theta u(\theta) dF(\theta) = \int^\bar\theta_\underline\theta [\frac {1-F(\theta)}{f(\theta)}v_\theta(q(\theta), \theta) + u(\underline\theta)]dF(\theta) $

And here is where I need help to fill in the steps. As the authors explain, this is done by integration by parts, and using one of the constraints. I get something like this:

$\int^\bar\theta_\underline\theta u(\theta) dF(\theta) = u(\theta)F(\theta)|^\bar\theta_\underline\theta - \int^\bar\theta_\underline\theta F(\theta)v_\theta(q(\theta),\theta)d\theta$

I can simplify this a bit further, but it does not simplify to the authors' expression. Hence, can anybody help me fill in the blanks?

I hope that I have left sufficient information, otherwise, please let me know what I need to clarify.


1 Answer 1


I'll start from your last equation.

\begin{align*} \int_{\underline{\theta}}^{\bar{\theta}}{u(\theta) dF(\theta)} & = \Big[ u(\theta)F(\theta) \Big]_{\underline{\theta}}^{\bar{\theta}} - \int_{\underline{\theta}}^{\bar{\theta}}{F(\theta) v_{\theta}(q(\theta),\theta) d\theta} \\ & = u(\bar{\theta})F(\bar{\theta})-u(\underline{\theta})F(\underline{\theta})-\int_{\underline{\theta}}^{\bar{\theta}}{F(\theta) v_{\theta}(q(\theta),\theta) d\theta} \\ & = u(\bar{\theta})-\int_{\underline{\theta}}^{\bar{\theta}}{F(\theta) v_{\theta}(q(\theta),\theta) d\theta} \text{ since } F(\bar{\theta})=1, F(\underline{\theta})=0\\ & = u(\underline{\theta})+(u(\bar{\theta})-u(\underline{\theta}))-\int_{\underline{\theta}}^{\bar{\theta}}{F(\theta) v_{\theta}(q(\theta),\theta) d\theta} \\ & = u(\underline{\theta})+\int_{\underline{\theta}}^{\bar{\theta}}{u'(\theta) d\theta}-\int_{\underline{\theta}}^{\bar{\theta}}{F(\theta) v_{\theta}(q(\theta),\theta) d\theta} \\ & = u(\underline{\theta})+\int_{\underline{\theta}}^{\bar{\theta}}{v_{\theta}(q(\theta),\theta) d\theta}-\int_{\underline{\theta}}^{\bar{\theta}}{F(\theta) v_{\theta}(q(\theta),\theta) d\theta} \text{ since } u'=v_{\theta}\\ & = u(\underline{\theta})+\int_{\underline{\theta}}^{\bar{\theta}}{(1-F(\theta)) v_{\theta}(q(\theta),\theta) d\theta} \\ & = u(\underline{\theta})+\int_{\underline{\theta}}^{\bar{\theta}}{\dfrac{1-F(\theta)}{f(\theta)} v_{\theta}(q(\theta),\theta) dF(\theta)} \\ & = \int_{\underline{\theta}}^{\bar{\theta}}{u(\underline{\theta})dF(\theta)}+\int_{\underline{\theta}}^{\bar{\theta}}{\dfrac{1-F(\theta)}{f(\theta)} v_{\theta}(q(\theta),\theta) dF(\theta)} \\ & = \int_{\underline{\theta}}^{\bar{\theta}}{\Big[u(\underline{\theta})+\dfrac{1-F(\theta)}{f(\theta)} v_{\theta}(q(\theta),\theta) \Big]dF(\theta)} \end{align*}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.