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Consoder the following maximization problem $$\max_{\{\tau(\cdot),q(\cdot)\}}\int_{\underline{\theta}}^{\bar{\theta}}\left(\theta q(\theta)-\dfrac{\gamma\sigma^{2}}{2}q^2(\theta)-\tau(\theta)\right)f(\theta)d\theta$$ subject to $$\int_{\underline{\theta}}^{\bar{\theta}}\left(\tau(\theta)-v(\theta)q(\theta)\right)f(\theta)d\theta\geq\underline{\pi}$$ where $\theta=s-\gamma\sigma^2 I$ and has a bounded support, $[\underline{\theta},\bar{\theta}]$, $\gamma\sigma^2>0$ and $s\sim N(\bar{s},\sigma_1^{2})$ and $I\in\mathbb{R}$. The functions $u(\cdot)$, $\tau(\cdot)$ and $q(\cdot)$ are linear with respect to $\theta$, $\underline{\pi}$ is a constant and $f(\theta)$ is the pdf of the normal distribution.

This is a problem of the Biais, Rochet and Martimont paper in 2000 problem in subsection $3.5$. I am a little confused with the constraint and I can not understand how to solve it. It is not obvious to me. Thank you in advance!

$\underline{Hint:}$ They do not explicitly assumme that the $\theta$ variable follows a normal distribution, but this has nothing to do with the optimization problem.

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  • $\begingroup$ Would it be helpful to re-state my problem? Is something that it is not clear? What should I do? $\endgroup$ – Hunger Learn Mar 30 at 10:13
  • $\begingroup$ Can you include a link to the paper? People will be able to consult to help. $\endgroup$ – Daneel Olivaw Apr 5 at 12:17
  • $\begingroup$ @Daneel Olivaw of course I can!! $\endgroup$ – Hunger Learn Apr 5 at 12:19
  • $\begingroup$ good luck with the rest of the paper.. $\endgroup$ – Konstantin Apr 6 at 21:01
  • $\begingroup$ Although it starts with a classic and simple model it becomes more and more tricky...but it is a nice paper! $\endgroup$ – Hunger Learn Apr 6 at 21:02
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You should think of the integral as you would of a sum. Then the usual Lagrangian approach seems very natural.

\begin{align} \mathcal{L} &= \int_\underline{\theta}^\overline{\theta} \left( \theta q(\theta) - \frac{\gamma\sigma^2}{2}q^2(\theta) - \tau(\theta) \right) f(\theta) d \theta + \lambda \left(\int_{\underline{\theta}}^\overline{\theta} (\tau(\theta) - v(\theta) q(\theta)) f(\theta) d\theta \right)\\ &= \int_\underline{\theta}^\overline{\theta} \left( \theta q(\theta) - \frac{\gamma\sigma^2}{2}q^2(\theta) - \tau(\theta) + \lambda(\tau(\theta) - v(\theta) q(\theta)) \right) f(\theta) d\theta - \lambda \underline{\pi}\\ &= \int_\underline{\theta}^\overline{\theta} \left( \theta q(\theta) - \frac{\gamma\sigma^2}{2}q^2(\theta) - \lambda v(\theta) q(\theta) + (\lambda-1)\tau(\theta) \right) f(\theta) d\theta - \lambda \underline{\pi} \end{align}

Now, simply treat $\theta$ as a summation index and write the first-order conditions case-by-case (for each $\theta$):

\begin{align} \frac{\partial}{\partial q(\theta)} \mathcal{L} & = \theta - \gamma\sigma^2 q(\theta) - \lambda v(\theta) = 0 \\ \frac{\partial}{\partial \tau(\theta)} \mathcal{L} & = \lambda - 1 = 0 \end{align}

In the footnote 16 the authors make the assumption that the participation constraint is binding, and therefore $\lambda^* > 0$ and therefore $$ \lambda^* = 1, $$ according to the second first-order condition. Substituting the value of $\lambda^*$ into the first condition you get \begin{equation} q^*(\theta) = \frac{\theta - v(\theta)}{\gamma \sigma^2}. \end{equation}

The assumptions made in the last paragraph of the section ensure that $\tau^*(\theta)$ is non-zero almost everywhere (for all possible $\theta$ except for one arbitrary value, which they denote $\theta_0$ - it must depend on the exact functional form of $v$ should you choose one).

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  • $\begingroup$ thank you very much! $\endgroup$ – Hunger Learn Apr 6 at 21:01
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    $\begingroup$ It was my pleasure :) $\endgroup$ – Konstantin Apr 6 at 21:02
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    $\begingroup$ I hope someday, I can be so good to help someone also! $\endgroup$ – Hunger Learn Apr 6 at 21:03

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