<p>I think the easiest way is to notice that since the problem is increasing in both arguments, we can assume <span class="math-container">$px_1 + x_2 = 1$</span> the budget binds (at least for <span class="math-container">$p \not =0$</span>).</p>
<p>Substituting our constraint into the objective function, we have:
<span class="math-container">$$
\max_{x_1} x_1 + 5(1 - p x_1) = x_1(1 - 5p) + 5
$$</span></p>
<p>If <span class="math-container">$1 - 5p&lt;0$</span>, we choose the smallest possible <span class="math-container">$x_1$</span>, so <span class="math-container">$x_1 = 0$</span> and thus <span class="math-container">$x_2 = 1$</span>.</p>
<p>If <span class="math-container">$1 - 5p = 0$</span>, any <span class="math-container">$x_1$</span> would maximise the above.</p>
<p>If <span class="math-container">$1 - 5 p &gt; 0$</span>, you choose the biggest <span class="math-container">$x_1$</span> possible, so <span class="math-container">$x_2 = 0$</span> and <span class="math-container">$x_1 = \frac{1}{p}$</span>.</p>
<p>If <span class="math-container">$p = 0$</span>, there is no constraint on <span class="math-container">$x_1$</span>, and your objective function is unbounded in <span class="math-container">$x_1$</span>, so you would choose <span class="math-container">$x_1 = \infty$</span>, so no solution exists.</p>

CC BY-SA 4.0
2020-11-05T02:58:20.983Z
2020-11-05T02:58:20.983Z
Walrasian Auctioneer
<p>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 <a href="https://www.youtube.com/watch?v=HZ_zn-1CQrs&amp;list=PL0E5qztLTUQ0tU0QRh14lsYaMvnmu2VvN&amp;index=15&amp;t=110s" rel="nofollow noreferrer">here</a>, but reposted below so you don't have to go look).</p>
<p>Parameter set: <span class="math-container">$\Theta$</span></p>
<p>Choice set: <span class="math-container">$X$</span></p>
<p>Objective function: <span class="math-container">$f: X \times \Theta \to \mathbb{R}$</span></p>
<p>Constraint correspondence: <span class="math-container">$\Gamma: \Theta \rightrightarrows  X$</span></p>
<p>Solution Correspondence: <span class="math-container">$\Gamma^*(\theta):= argmax_{x \in \Gamma(\theta)} f(x,\theta)$</span></p>
<p>The maximized value of the objective function: <span class="math-container">$f^*(x, \theta) = \max_{x \in \Gamma(\theta)} f(x,\theta)$</span></p>
<p>phew.</p>
<p>Ok, now consider the following maximization problem parametrized by <span class="math-container">$p \in [0,1]$</span>:</p>
<p><span class="math-container">$\max_{(x_1, x_2) \in \mathbb{R}_+^2} x_1 + 5x_2 $</span></p>
<p>s.t. <span class="math-container">$px_1 + x_2 \leq 1$</span></p>
<p>I know that we can write this in the form: <span class="math-container">$f(x,p) = x_1 + 5x_2$</span> and <span class="math-container">$\Gamma(p) = \{(x_1,x_2) \in \mathbb{R}^2_+: px_1 + x_2 \leq 1 \}$</span>. I also know that at <span class="math-container">$\Gamma(0) = \{(x_1,x_2) \in \mathbb{R}^2_+: x_2 \leq 1 \}$</span> is not compact-valued, and thus we cannot apply the theorem of the maximum.</p>
<p>In the solution to this question, I see that the optimal policy correspondence is</p>
<p><span class="math-container">$\Gamma^*(p) = \begin{cases} 
\emptyset &amp; \text{if} \: p = 0 \\
\{(1/p,0)\} &amp; \text{if} \: p = (0, 0.2) \\
\{ (x_1, x_2) \in \mathbb{R}_+^2: 0.2x_1 + x_2 = 1 \} &amp; \text{if} \: p = 0.2 \\
\{(0,1)\} &amp; \text{if} \: p = (0.2, 1] 
\end{cases}$</span></p>
<p>At <span class="math-container">$p = 0$</span>, <span class="math-container">$\Gamma^*$</span> is empty-valued. For <span class="math-container">$p&gt;0$</span> it is compact-valued and upper hemicontinuous. It fails to be lower hemicontinuous at <span class="math-container">$p=0.2$</span>. Substituting <span class="math-container">$\Gamma^*(p)$</span> into the objective function, the value function is <span class="math-container">$f^*(p) = \max \{1/p , 5 \}$</span>.</p>
<p>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.</p>

<mathematical-economics><optimization>
Practice question on Correspondences and maximization
oracle_randy
# Practice question on Correspondences and maximization

(Asked by: username_0 on 2020-11-04)

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](https://www.youtube.com/watch?v=HZ_zn-1CQrs&list=PL0E5qztLTUQ0tU0QRh14lsYaMvnmu2VvN&index=15&t=110s), 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.

**Tags:** mathematical-economics, optimization

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## 1 answers

### Answer by username_1 on 2020-11-05. Score: 4

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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