# Optimization problem with a constrant

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.

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

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).

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