<p>I am reading a note on moral hazard and it shows that a problem like <span class="math-container">$\max _{s(x)} \mathbb{E}[V(x-s(x))]$</span> s.t. <span class="math-container">$\mathbb{E}[U(s(x))] \geq \bar{H}+c\left(a^{*}\right)$</span>  can be solved by setting up a simple Lagrangean <span class="math-container">$$\min _{\lambda} \max _{s(x)} \mathcal{L}=\mathbb{E}\left[V(x-s(x))\right]-\lambda\left[\bar{H}+c\left(a^{*}\right)-\mathbb{E}[U(s(x))]\right]$$</span>.</p>
<p>I guess this is a way to solve the non-equal constraint but I cannot understand the meaning of the minimax intuitively. And I wonder is there any difference of solving this minimax Lagrange comparing to solving the normal Lagrange FOCs and checking the complementary slackness conditions?</p>

CC BY-SA 4.0
2021-11-11T08:32:55.520Z
I am just guessing but I think they want to emphasize that they want to minimize value $\lambda$. If I remember correctly my graduate math classes, it is actually not proper to say that when you optimize the Lagrangian function you are taking the derivative wrt $\lambda$ when you use the constraint as a foc, maybe they want to emphasize that you should do that now
1muflon1
CC BY-SA 4.0
2021-11-12T06:31:25.000Z
This is the saddle point property of the Lagrangian---see any intro text on static optimization. Informally, $\lambda$ in the indirect marginal utility, and as usual, it should be minimized at the maximum (if the problem is well posed, say under convexity). If constraint is slack, $\lambda = 0$ which is the minimum value possible ($\lambda \geq 0$ is a constraint); this is just complementary slackness rephrased.
Michael
<mathematical-economics><optimization>
Constrained optimization problem written as minimax Lagrange
Alalalalaki
# Constrained optimization problem written as minimax Lagrange

(Asked by: username_0 on 2021-11-11)

I am reading a note on moral hazard and it shows that a problem like $\max \_{s(x)} \mathbb{E}[V(x-s(x))]$ s.t. $\mathbb{E}[U(s(x))] \geq \bar{H}+c\left(a^{\*}\right)$ can be solved by setting up a simple Lagrangean $$\min \_{\lambda} \max \_{s(x)} \mathcal{L}=\mathbb{E}\left[V(x-s(x))\right]-\lambda\left[\bar{H}+c\left(a^{\*}\right)-\mathbb{E}[U(s(x))]\right]$$.

I guess this is a way to solve the non-equal constraint but I cannot understand the meaning of the minimax intuitively. And I wonder is there any difference of solving this minimax Lagrange comparing to solving the normal Lagrange FOCs and checking the complementary slackness conditions?

**Tags:** mathematical-economics, optimization
> **Comments:**
> **username_1**: I am just guessing but I think they want to emphasize that they want to minimize value $\lambda$. If I remember correctly my graduate math classes, it is actually not proper to say that when you optimize the Lagrangian function you are taking the derivative wrt $\lambda$ when you use the constraint as a foc, maybe they want to emphasize that you should do that now  
> **username_2**: This is the saddle point property of the Lagrangian---see any intro text on static optimization. Informally, $\lambda$ in the indirect marginal utility, and as usual, it should be minimized at the maximum (if the problem is well posed, say under convexity). If constraint is slack, $\lambda = 0$ which is the minimum value possible ($\lambda \geq 0$ is a constraint); this is just complementary slackness rephrased.  

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## 0 answers

