<p>I couldn't think of a good hint. If you are having trouble with using the information given, (as the application of Walras + IVT is fairly straightforward), then there are not many tips that can help. I've opted to put out each step explicitly. Let me know if you don't understand a part, and I'll try to edit to make it clearer.</p>

<p>In the future, it is better if you try and explain what part you are having trouble with, so that people can better help you (and also to prevent you from getting your questions closed).</p>

<hr>

<p>We have:</p>

<p>$$Z(\vec p) = \bigg(Z_1(p_1, p_2), Z_2(p_1, p_2) \bigg)$$</p>

<p>and can substitute $p_1 = 1 - p_2$ as Herr K. has pointed out.</p>

<p>$$Z(\vec p) = \bigg(Z_1(p_1, 1 - p_1), Z_2(p_1, 1-p_1) \bigg)$$</p>

<p>$Z_1$ can be expressed as a function of one variable, $Z_1(p_1)$, so that the Intermediate Value Theorem can be applied.</p>

<p>On the interval $I = (0, 1) \in \mathbb{R}$, we have $Z_1: I \rightarrow \mathbb{R}$, a continuous (and bounded) function where given</p>

<p>$$\infty &gt; Z_1(p_1 = 0) &gt; 0 &gt; Z_1(p_1 = 1) &gt; -\infty$$</p>

<p>there exists $p_1^* \in (0, 1)$ such that $Z_1(p_1^*) = 0$</p>

<p>And by Walras' Law,</p>

<p>$$p_1^* Z_1(p_1^*, 1-p_1^*) + (1 - p_1^*) Z_2(p_1^*, 1-p_1^*) = 0$$</p>

<p>$$p_1^* \cdot 0 + (1 - p_1^*) Z_2(p_1^*, 1-p_1^*) = 0$$</p>

<p>and because $(1-p_1^*) &gt; 0$</p>

<p>$$Z_2(p_1^*, 1-p_1^*) = 0$$</p>

<p>meaning $Z(p_1^*) := Z(p^*_1, p^*_2) = (0, 0)$</p>

CC BY-SA 3.0
2016-11-22T08:13:19.213Z
2016-11-22T08:40:04.263Z
2016-11-22T08:40:04.263Z
Kitsune Cavalry
<p>A stronger hint: Write $p_2=1-p_1$, so that $Z(p)=Z(p_1,1-p_1)$. Use the conditions $Z_1(0,1)&gt;0,Z_1(1,0)&lt;0$ etc. and the intermediate value theorem to argue that there exists a $p_1^*\in(0,1)$ such that $Z_1(p_1^*,1-p_1^*)=0$. </p>

CC BY-SA 3.0
2016-11-21T23:54:40.073Z
2016-11-21T23:54:40.073Z
Herr K.
<p>Here's a problem restated from Ross Starr's <em>General Equilibrium Theory</em>.</p>

<blockquote>
  <p>Consider a two-commodity economy with an excess demand function $Z(p)=(Z_1(p),Z_2(p))$. The price space is $p \in P = \left \{ p | p \in \mathbb{R}^{2}, p \geq 0, p_1 + p_2 =1 \right \}$. Let $Z(p)$ be continuous, bounded and fulfill Walras' Law as an equality, that is $p_1Z_1(p)+p_2Z_2(p)=0$. Assume $Z_1(0,1)&gt;0$, $Z_1(1,0)&lt;0$, $Z_2(0,1)&lt;0$, $Z_2(1,0)&gt;0$.
  Use the <strong>intermediate value theorem</strong> and <strong>Walras' Law</strong> to show that the economy has a competive equilibrium. That is, demonstrate that there is a price vector $p* \in P$ so that $Z(p*)=(0,0)$.</p>
</blockquote>

<p>And I have a hint: Characterize $Z(p)$ as $Z(\alpha , 1- \alpha)$ for $0 \leq \alpha \leq 1$. Use the intermediate value theorem to find $0 \leq \alpha \leq 1$ so that $Z_1(\alpha , 1- \alpha)=0$. Then apply Walras' Law.</p>

<p>I'm having problem finding $\alpha$, how could I find it?</p>

CC BY-SA 3.0
2016-11-22T18:54:45.633Z
$\alpha$ is the $p_1^*$ in Kitsune's answer. No explicit value can be found because that would require the explicit functional form of $Z_1(\cdot)$ be given.
Herr K.
CC BY-SA 3.0
2016-11-22T19:45:11.640Z
What Herr K. said. Also why have you deleted the whole question OP?
Kitsune Cavalry
CC BY-SA 3.0
2016-11-23T00:58:28.053Z
The question as it stands is incomprehensible. The OP edited out almost all of the original question. As it stands I am voting to close.
Alecos Papadopoulos
CC BY-SA 3.0
2016-11-24T08:03:39.330Z
I rolled back the edit. If OP insists on changing it though, then we can react accordingly.
Kitsune Cavalry
<microeconomics><general-equilibrium>
Application of Intermediate Value Theorem for General Equilibrium
Renzo Mauricio Guzmán Anaya
# Application of Intermediate Value Theorem for General Equilibrium

(Asked by: username_0 on 2016-11-21)

Here's a problem restated from Ross Starr's *General Equilibrium Theory*.

> Consider a two-commodity economy with an excess demand function $Z(p)=(Z\_1(p),Z\_2(p))$. The price space is $p \in P = \left \{ p | p \in \mathbb{R}^{2}, p \geq 0, p\_1 + p\_2 =1 \right \}$. Let $Z(p)$ be continuous, bounded and fulfill Walras' Law as an equality, that is $p\_1Z\_1(p)+p\_2Z\_2(p)=0$. Assume $Z\_1(0,1)>0$, $Z\_1(1,0)<0$, $Z\_2(0,1)<0$, $Z\_2(1,0)>0$.
> Use the **intermediate value theorem** and **Walras' Law** to show that the economy has a competive equilibrium. That is, demonstrate that there is a price vector $p\* \in P$ so that $Z(p\*)=(0,0)$.

And I have a hint: Characterize $Z(p)$ as $Z(\alpha , 1- \alpha)$ for $0 \leq \alpha \leq 1$. Use the intermediate value theorem to find $0 \leq \alpha \leq 1$ so that $Z\_1(\alpha , 1- \alpha)=0$. Then apply Walras' Law.

I'm having problem finding $\alpha$, how could I find it?

**Tags:** microeconomics, general-equilibrium
> **Comments:**
> **username_1**: $\alpha$ is the $p_1^*$ in Kitsune's answer. No explicit value can be found because that would require the explicit functional form of $Z_1(\cdot)$ be given.  
> **username_2**: What Herr K. said. Also why have you deleted the whole question OP?  
> **username_3**: The question as it stands is incomprehensible. The OP edited out almost all of the original question. As it stands I am voting to close.  
> **username_2**: I rolled back the edit. If OP insists on changing it though, then we can react accordingly.  

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## 2 answers

### Answer by username_2 on 2016-11-22. Score: 4

I couldn't think of a good hint. If you are having trouble with using the information given, (as the application of Walras + IVT is fairly straightforward), then there are not many tips that can help. I've opted to put out each step explicitly. Let me know if you don't understand a part, and I'll try to edit to make it clearer.

In the future, it is better if you try and explain what part you are having trouble with, so that people can better help you (and also to prevent you from getting your questions closed).

---

We have:

$$Z(\vec p) = \bigg(Z\_1(p\_1, p\_2), Z\_2(p\_1, p\_2) \bigg)$$

and can substitute $p\_1 = 1 - p\_2$ as Herr K. has pointed out.

$$Z(\vec p) = \bigg(Z\_1(p\_1, 1 - p\_1), Z\_2(p\_1, 1-p\_1) \bigg)$$

$Z\_1$ can be expressed as a function of one variable, $Z\_1(p\_1)$, so that the Intermediate Value Theorem can be applied.

On the interval $I = (0, 1) \in \mathbb{R}$, we have $Z\_1: I \rightarrow \mathbb{R}$, a continuous (and bounded) function where given

$$\infty > Z\_1(p\_1 = 0) > 0 > Z\_1(p\_1 = 1) > -\infty$$

there exists $p\_1^\* \in (0, 1)$ such that $Z\_1(p\_1^\*) = 0$

And by Walras' Law,

$$p\_1^\* Z\_1(p\_1^\*, 1-p\_1^\*) + (1 - p\_1^\*) Z\_2(p\_1^\*, 1-p\_1^\*) = 0$$

$$p\_1^\* \cdot 0 + (1 - p\_1^\*) Z\_2(p\_1^\*, 1-p\_1^\*) = 0$$

and because $(1-p\_1^\*) > 0$

$$Z\_2(p\_1^\*, 1-p\_1^\*) = 0$$

meaning $Z(p\_1^\*) := Z(p^\*\_1, p^\*\_2) = (0, 0)$


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### Answer by username_1 on 2016-11-21. Score: 3

A stronger hint: Write $p\_2=1-p\_1$, so that $Z(p)=Z(p\_1,1-p\_1)$. Use the conditions $Z\_1(0,1)>0,Z\_1(1,0)<0$ etc. and the intermediate value theorem to argue that there exists a $p\_1^\*\in(0,1)$ such that $Z\_1(p\_1^\*,1-p\_1^\*)=0$.


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