<p>An attempt with adding an extra assumption:</p>
<p>Let <span class="math-container">$0 \in V(0)$</span> and <span class="math-container">$0 \notin V(y), \forall y&gt;0$</span></p>
<p>Since <span class="math-container">$V(y)$</span> is set of all <span class="math-container">$x$</span> that can at least produce <span class="math-container">$y$</span>, we have that <span class="math-container">$V(y’) \subseteq V(y), \forall y’&gt;y$</span></p>
<p>Now let <span class="math-container">$x^*$</span> (as defined in question) <span class="math-container">$\in V(y’)$</span>.</p>
<p>From definition: <span class="math-container">$wx^* \leq wx, \forall x \in V(y)$</span>. Therefore, we also have that</p>
<p><span class="math-container">$$wx^* \leq wx, \forall x \in V(y’)$$</span></p>
<p>So, <span class="math-container">$$x^* = \arg \min wx$$</span> <span class="math-container">$$\text{such that }  x \in V(y’)$$</span></p>
<p>Now this can be extended backward as well. Say <span class="math-container">$y’’&lt;y$</span> and</p>
<p><span class="math-container">$$x’’ = \arg \min wx$$</span> <span class="math-container">$$\text{such that } x \in V(y’’)$$</span></p>
<p>Following the logic from above, <span class="math-container">$x’’$</span> also minimizes cost for <span class="math-container">$y, y’$</span>.</p>
<p>Taking it all the way back, it must be that:</p>
<p><span class="math-container">$$x^* = \arg \min wx \,\,\, \forall x \in V(0)$$</span></p>
<p>However, due to the additional assumption (<em>inaction</em>), minimum cost for producing 0 is 0. So <span class="math-container">$x^*=0$</span> (for <span class="math-container">$w&gt;0$</span>), which is also a contradiction to the assumption stated below.</p>

CC BY-SA 4.0
2021-07-14T05:07:28.297Z
I'm not sure your argument is correct. Negating the statement in the question does not allow you to find an $x^\ast$ for **all** $y'$ and $y$ with $y' > y$ (but just for one of them). As a counterexample, let $V(0) = \mathbb{R}^n_+$ and let for all $y > 0$, $V(y) = \{x\in \mathbb{R}^n_+| x \ge 1\}$. Then your assumption is satisfied, but the condition in the question is not as for all $y' > y (>0)$ we have $V(y) = V(y')$.
tdm
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2021-07-14T16:18:55.700Z
Fair point. We basically need an assumption that ensures that the lower boundary for V(y) is excluded in V(y') which is what you did very explicitly. In texts I think this assumption is introduced, indirectly, through production function.
Dayne
CC BY-SA 4.0
2021-07-13T10:04:53.927Z
2021-07-13T10:04:53.927Z
Dayne
<p>I don't think you need convexity. However, I think you do need to assume some monotonicity condition. The following should work (but might not be the minimal set of assumptions that provides the result).</p>
<p>Consider the production possibility set <span class="math-container">$V(.)$</span>.
<span class="math-container">$$
V(y) = \{x \in \mathbb{R}^n_+| x \text{ can produce } y\}.
$$</span>
We assume that <span class="math-container">$V(y)$</span> is a closed non-empty subset of <span class="math-container">$\mathbb{R}^n_{+}$</span>. Let <span class="math-container">$w \in \mathbb{R}^n_{++}$</span> and define:
<span class="math-container">$$
c(y) = \arg\min_x wx \text{ s.t. } x \in V(y).
$$</span>
As <span class="math-container">$V(y)$</span> is closed, this problem is well defined. The value of <span class="math-container">$c(y)$</span> gives the minimal cost to produce <span class="math-container">$y$</span>. Define <span class="math-container">$X(y) = \{x \in V(y)| w x = c(y)\}$</span> as the set of all optimal solutions.</p>
<p><strong>Assumption 1:</strong> If  <span class="math-container">$y' &gt;y$</span>, then <span class="math-container">$V(y') \subseteq V(y)^\circ$</span> where <span class="math-container">$A^\circ$</span> is the interior of the set <span class="math-container">$A$</span> (relative to <span class="math-container">$\mathbb{R}^n_+$</span>).</p>
<p><strong>Assumption 2:</strong> if <span class="math-container">$y &gt; 0$</span> then <span class="math-container">$0 \notin V(y)$</span>.</p>
<p>Assumption 1 requires the production possibility sets to be strictly nested. Assumption 2 requires that we can not produce something from nothing.</p>
<p><strong>Lemma 1:</strong> If assumptions 1 and 2 are satisfied, then  <span class="math-container">$y' &gt; y$</span> implies <span class="math-container">$c(y') &gt; c(y)$</span>.</p>
<p><em>Proof:</em> Let <span class="math-container">$y' &gt; y$</span>. And let <span class="math-container">$x^\ast \in X(y')$</span>.  Then as <span class="math-container">$x^\ast \in V(y')$</span> we have by Assumption 1 that <span class="math-container">$x^\ast \in V(y)^\circ$</span>. Then we know there is a <span class="math-container">$\varepsilon &gt; 0$</span> such that <span class="math-container">$B_\varepsilon(x^\ast) \cap \mathbb{R}^n_{+} \subseteq V(y)$</span>. As <span class="math-container">$x^\ast \ne 0$</span> (by Assumption 2), we can find an <span class="math-container">$x' \in B_\varepsilon(x^\ast) \cap \mathbb{R}^n_+$</span> such that <span class="math-container">$x' &lt; x$</span>  and <span class="math-container">$x' \in V(y)$</span>. Then:
<span class="math-container">$$
c(y) \le w x' &lt; w x^\ast = c(y')
$$</span>
which demonstrates the proof.</p>
<p><strong>Theorem 2:</strong> Let <span class="math-container">$y' &gt; y$</span> and <span class="math-container">$x^\ast \in X(y)$</span> then <span class="math-container">$x^\ast \notin V(y)$</span>.</p>
<p><em>Proof:</em> Towards a contradiction assume that <span class="math-container">$y' &gt; y$</span>,  <span class="math-container">$x^\ast \in X(y)$</span> and <span class="math-container">$x^\ast \in V(y)$</span>. Then <span class="math-container">$c(y) = w x^\ast$</span> and <span class="math-container">$c(y') \le w x^\ast$</span>. However, this implies that:
<span class="math-container">$$
c(y') \le w x^\ast = c(y),
$$</span>
which contradicts Lemma 1.</p>

CC BY-SA 4.0
2021-07-12T17:40:42.670Z
:). So first, I didn’t mention monotonicity because i mentioned that V(y) is the set to *at least* produce y. So this should capture assumption 1. And as a nice surprise, just before you posted the answer I commented assumption 2. Thanks so much with assumption 2 I also proved this way :)
Dayne
CC BY-SA 4.0
2021-07-12T18:01:30.107Z
Also I think rather than assuming interior condition in assumption 1, we can add in assumption 2 that $0 \in V(0)$. That would suffice.
Dayne
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2021-07-13T04:56:58.137Z
@Dayne I'm not convinced, but please try if you think it works.
tdm
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2021-07-13T08:21:21.367Z
Didn't work through the proof, but some ' seems to be missing at the end.
VARulle
CC BY-SA 4.0
2021-07-13T10:05:58.340Z
@tdm: I have attempted the answer with the additional assumption of ‘inaction’ along with assumption 2 in your answer (relaxing assumption 1 in the process).
Dayne
CC BY-SA 4.0
2021-07-12T17:33:18.873Z
2021-07-12T17:33:18.873Z
tdm
<p>I am stuck at a very simple question. Let <span class="math-container">$V(y)$</span> be the set of all <span class="math-container">$x \in \mathbb{R}^n$</span> that can produce <em>at least</em> <span class="math-container">$y$</span>. We are given that <span class="math-container">$V(y)$</span> is convex set.</p>
<p>Given <span class="math-container">$w$</span>, factor prices, let</p>
<p><span class="math-container">$$x^* = \arg \min wx$$</span>
<span class="math-container">$$\text{such that } x \in V(y)$$</span></p>
<p>Show that <span class="math-container">$x^* \notin V(y’)$</span> for <span class="math-container">$y’&gt;y$</span></p>
<p>I tried using contradiction but only proved that if the above doesn’t hold then <span class="math-container">$x^*$</span> can produce infinite <span class="math-container">$y$</span>. Can I claim that as a contradiction (to what?) and say that proof is complete?</p>

CC BY-SA 4.0
2021-07-12T16:58:34.233Z
The result is not generally true for arbitrary production sets. Can you describe the setting in more detail? How is production modeled? Production sets? Production functions? What are the standing assumptions?
Michael Greinecker
CC BY-SA 4.0
2021-07-12T17:07:38.907Z
@MichaelGreinecker That sounds relieving. There is no set up really and is just a self created question. Could you tell me what minimum assumption would be required for this to be true?
Dayne
CC BY-SA 4.0
2021-07-12T17:30:55.260Z
@MichaelGreinecker: How about $0 \notin V(y)$ for $y>0$ as a necessary condition for the above to be true?
Dayne
<microeconomics><producer-theory>
Micro: proving that cost minimizing input vector for producing y cannot produce more than y
Dayne
# Micro: proving that cost minimizing input vector for producing y cannot produce more than y

(Asked by: username_0 on 2021-07-12)

I am stuck at a very simple question. Let $V(y)$ be the set of all $x \in \mathbb{R}^n$ that can produce *at least* $y$. We are given that $V(y)$ is convex set.

Given $w$, factor prices, let

$$x^\* = \arg \min wx$$
$$\text{such that } x \in V(y)$$

Show that $x^\* \notin V(y’)$ for $y’>y$

I tried using contradiction but only proved that if the above doesn’t hold then $x^\*$ can produce infinite $y$. Can I claim that as a contradiction (to what?) and say that proof is complete?

**Tags:** microeconomics, producer-theory

> **Comments:**
> **username_1**: The result is not generally true for arbitrary production sets. Can you describe the setting in more detail? How is production modeled? Production sets? Production functions? What are the standing assumptions?  
> **username_0**: @MichaelGreinecker That sounds relieving. There is no set up really and is just a self created question. Could you tell me what minimum assumption would be required for this to be true?  
> **username_0**: @MichaelGreinecker: How about $0 \notin V(y)$ for $y>0$ as a necessary condition for the above to be true?  

---

## 2 answers

### Answer by username_2 on 2021-07-12. Score: 3 | Accepted answer

I don't think you need convexity. However, I think you do need to assume some monotonicity condition. The following should work (but might not be the minimal set of assumptions that provides the result).

Consider the production possibility set $V(.)$.
$$
V(y) = \{x \in \mathbb{R}^n\_+| x \text{ can produce } y\}.
$$
We assume that $V(y)$ is a closed non-empty subset of $\mathbb{R}^n\_{+}$. Let $w \in \mathbb{R}^n\_{++}$ and define:
$$
c(y) = \arg\min\_x wx \text{ s.t. } x \in V(y).
$$
As $V(y)$ is closed, this problem is well defined. The value of $c(y)$ gives the minimal cost to produce $y$. Define $X(y) = \{x \in V(y)| w x = c(y)\}$ as the set of all optimal solutions.

**Assumption 1:** If $y' >y$, then $V(y') \subseteq V(y)^\circ$ where $A^\circ$ is the interior of the set $A$ (relative to $\mathbb{R}^n\_+$).

**Assumption 2:** if $y > 0$ then $0 \notin V(y)$.

Assumption 1 requires the production possibility sets to be strictly nested. Assumption 2 requires that we can not produce something from nothing.

**Lemma 1:** If assumptions 1 and 2 are satisfied, then $y' > y$ implies $c(y') > c(y)$.

*Proof:* Let $y' > y$. And let $x^\ast \in X(y')$. Then as $x^\ast \in V(y')$ we have by Assumption 1 that $x^\ast \in V(y)^\circ$. Then we know there is a $\varepsilon > 0$ such that $B\_\varepsilon(x^\ast) \cap \mathbb{R}^n\_{+} \subseteq V(y)$. As $x^\ast \ne 0$ (by Assumption 2), we can find an $x' \in B\_\varepsilon(x^\ast) \cap \mathbb{R}^n\_+$ such that $x' < x$ and $x' \in V(y)$. Then:
$$
c(y) \le w x' < w x^\ast = c(y')
$$
which demonstrates the proof.

**Theorem 2:** Let $y' > y$ and $x^\ast \in X(y)$ then $x^\ast \notin V(y)$.

*Proof:* Towards a contradiction assume that $y' > y$, $x^\ast \in X(y)$ and $x^\ast \in V(y)$. Then $c(y) = w x^\ast$ and $c(y') \le w x^\ast$. However, this implies that:
$$
c(y') \le w x^\ast = c(y),
$$
which contradicts Lemma 1.

> **Comments:**
> **username_0**: :). So first, I didn’t mention monotonicity because i mentioned that V(y) is the set to *at least* produce y. So this should capture assumption 1. And as a nice surprise, just before you posted the answer I commented assumption 2. Thanks so much with assumption 2 I also proved this way :)  
> **username_0**: Also I think rather than assuming interior condition in assumption 1, we can add in assumption 2 that $0 \in V(0)$. That would suffice.  
> **username_2**: @username_0 I'm not convinced, but please try if you think it works.  
> **username_3**: Didn't work through the proof, but some ' seems to be missing at the end.  
> **username_0**: @username_2: I have attempted the answer with the additional assumption of ‘inaction’ along with assumption 2 in your answer (relaxing assumption 1 in the process).  

---

### Answer by username_0 on 2021-07-13. Score: 0

An attempt with adding an extra assumption:

Let $0 \in V(0)$ and $0 \notin V(y), \forall y>0$

Since $V(y)$ is set of all $x$ that can at least produce $y$, we have that $V(y’) \subseteq V(y), \forall y’>y$

Now let $x^\*$ (as defined in question) $\in V(y’)$.

From definition: $wx^\* \leq wx, \forall x \in V(y)$. Therefore, we also have that

$$wx^\* \leq wx, \forall x \in V(y’)$$

So, $$x^\* = \arg \min wx$$ $$\text{such that } x \in V(y’)$$

Now this can be extended backward as well. Say $y’’ and

$$x’’ = \arg \min wx$$ $$\text{such that } x \in V(y’’)$$

Following the logic from above, $x’’$ also minimizes cost for $y, y’$.

Taking it all the way back, it must be that:

$$x^\* = \arg \min wx \,\,\, \forall x \in V(0)$$

However, due to the additional assumption (*inaction*), minimum cost for producing 0 is 0. So $x^\*=0$ (for $w>0$), which is also a contradiction to the assumption stated below.

> **Comments:**
> **username_2**: I'm not sure your argument is correct. Negating the statement in the question does not allow you to find an $x^\ast$ for **all** $y'$ and $y$ with $y' > y$ (but just for one of them). As a counterexample, let $V(0) = \mathbb{R}^n_+$ and let for all $y > 0$, $V(y) = \{x\in \mathbb{R}^n_+| x \ge 1\}$. Then your assumption is satisfied, but the condition in the question is not as for all $y' > y (>0)$ we have $V(y) = V(y')$.  
> **username_0**: Fair point. We basically need an assumption that ensures that the lower boundary for V(y) is excluded in V(y') which is what you did very explicitly. In texts I think this assumption is introduced, indirectly, through production function.  

---

