We're learning about Theory of the Maximum. I tend to struggle with correspondences in this context, so I'm trying to work through some practice questions. I will start with some general notation of a canonical maximation problem (which can be found from Rajiv Sethi's lecture here, but reposted below so you don't have to go look).

Parameter set: $\Theta$

Choice set: $X$

Objective function: $f: X \times \Theta \to \mathbb{R}$

Constraint correspondence: $\Gamma: \Theta \rightrightarrows X$

Solution Correspondence: $\Gamma^*(\theta):= argmax_{x \in \Gamma(\theta)} f(x,\theta)$

The maximized value of the objective function: $f^*(x, \theta) = \max_{x \in \Gamma(\theta)} f(x,\theta)$


Ok, now consider the following maximization problem parametrized by $p \in [0,1]$:

$\max_{(x_1, x_2) \in \mathbb{R}_+^2} x_1 + 5x_2 $

s.t. $px_1 + x_2 \leq 1$

I know that we can write this in the form: $f(x,p) = x_1 + 5x_2$ and $\Gamma(p) = \{(x_1,x_2) \in \mathbb{R}^2_+: px_1 + x_2 \leq 1 \}$. I also know that at $\Gamma(0) = \{(x_1,x_2) \in \mathbb{R}^2_+: x_2 \leq 1 \}$ is not compact-valued, and thus we cannot apply the theorem of the maximum.

In the solution to this question, I see that the optimal policy correspondence is

$\Gamma^*(p) = \begin{cases} \emptyset & \text{if} \: p = 0 \\ \{(1/p,0)\} & \text{if} \: p = (0, 0.2) \\ \{ (x_1, x_2) \in \mathbb{R}_+^2: 0.2x_1 + x_2 = 1 \} & \text{if} \: p = 0.2 \\ \{(0,1)\} & \text{if} \: p = (0.2, 1] \end{cases}$

At $p = 0$, $\Gamma^*$ is empty-valued. For $p>0$ it is compact-valued and upper hemicontinuous. It fails to be lower hemicontinuous at $p=0.2$. Substituting $\Gamma^*(p)$ into the objective function, the value function is $f^*(p) = \max \{1/p , 5 \}$.

I'm not sure how, mechanically, to get to the optimal policy correspondence, as we didn't do anything like this in class, and I'm finding reading materials scarce. I would really appreciate if someone could walk me through the steps as if I'm a 5 year old.


I think the easiest way is to notice that since the problem is increasing in both arguments, we can assume $px_1 + x_2 = 1$ the budget binds (at least for $p \not =0$).

Substituting our constraint into the objective function, we have: $$ \max_{x_1} x_1 + 5(1 - p x_1) = x_1(1 - 5p) + 5 $$

If $1 - 5p<0$, we choose the smallest possible $x_1$, so $x_1 = 0$ and thus $x_2 = 1$.

If $1 - 5p = 0$, any $x_1$ would maximise the above.

If $1 - 5 p > 0$, you choose the biggest $x_1$ possible, so $x_2 = 0$ and $x_1 = \frac{1}{p}$.

If $p = 0$, there is no constraint on $x_1$, and your objective function is unbounded in $x_1$, so you would choose $x_1 = \infty$, so no solution exists.


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