# Practice question on Correspondences and maximization

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

phew.

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.