<p>In general, it is well possible that <span class="math-container">$u(x^*)&gt;u^*$</span> for an expenditure minimizing <span class="math-container">$x^*$</span>. Typical assumptions that guarantee <span class="math-container">$u(x^*)=x^*$</span> are that <span class="math-container">$u$</span> has domain <span class="math-container">$\mathbb{R}^l_+$</span>, is continuous, <span class="math-container">$p\gg 0$</span>, and <span class="math-container">$u^*&gt;0$</span>.</p>
<p>Here is the argument: Let <span class="math-container">$x$</span> satisfy <span class="math-container">$u(x)&gt;u^*&gt;0$</span>. We must have <span class="math-container">$x\neq 0$</span> and <span class="math-container">$p\cdot x&gt;0$</span>. Continuity of <span class="math-container">$u$</span> implies that for <span class="math-container">$\alpha\in(0,1)$</span> small enough, <span class="math-container">$u\big(\alpha 0+(1-\alpha)x\big)&gt;u^*$</span>. Since <span class="math-container">$p\gg 0$</span>, we also have <span class="math-container">$p\cdot\big(\alpha 0+(1-\alpha)x\big)=p\cdot(1-\alpha)x=(1-\alpha) p\cdot x&lt;p\cdot x$</span>. So a bundle that gives higher utility than <span class="math-container">$u^*$</span> cannot be expenditure minimizing, and every expenditure minimizing bundle must give utility exactly <span class="math-container">$u^*$</span>.</p>

CC BY-SA 4.0
2023-02-06T14:52:21.153Z
2023-02-06T14:52:21.153Z
Michael Greinecker
<p>Consider the utility function <span class="math-container">$u:\mathbb{R}^2_+\rightarrow\mathbb{R}$</span>, given by <span class="math-container">$u(x, y) = \min(x, y) + 1$</span>. Consider <span class="math-container">$u^*=0.5$</span>, solution to the expenditure minimisation problem when <span class="math-container">$p_X&gt;0, p_Y&gt;0$</span> will be <span class="math-container">$(0,0)$</span> and it satisfy <span class="math-container">$u(0,0)=1&gt;0.5=u^*$</span>.</p>
<p>Another example is with discontinuous utility <span class="math-container">$u:\mathbb{R}^2_+\rightarrow\mathbb{R}$</span>, given by <span class="math-container">$u(x, y) = \lfloor x+y\rfloor$</span> i.e. greatest integer less than or equal to <span class="math-container">$x+y$</span>. Consider <span class="math-container">$u^*=4.2$</span>, <span class="math-container">$p_X = 1$</span>, <span class="math-container">$p_Y = 2$</span>. Solution to the expenditure minimisation problem is <span class="math-container">$(5,0)$</span> and it satisfy <span class="math-container">$u(5,0) = 5&gt;4.2=u^*$</span>.</p>
<p>Another example is when the two commodities are bad. Consider utility <span class="math-container">$u:\mathbb{R}^2_+\rightarrow\mathbb{R}$</span>, given by <span class="math-container">$u(x, y) = 1-x-y$</span>. Consider <span class="math-container">$u^*= 0.5$</span>, solution to the expenditure minimisation problem when <span class="math-container">$p_X&gt;0, p_Y&gt;0$</span> will be <span class="math-container">$(0,0)$</span> and it satisfy <span class="math-container">$u(0,0)=1&gt;0.5=u^*$</span>.</p>

CC BY-SA 4.0
2023-02-06T15:08:31.573Z
2023-02-06T15:14:28.253Z
2023-02-06T15:14:28.253Z
Amit
<p>If the preferences are continuous and monotone, if you want to get a higher utility level <span class="math-container">$U’&gt; U^\star$</span>, you need to spend more money than the optimal expenditure for the current utility level <span class="math-container">$U^\star$</span>.</p>
<p>In functional notation, <span class="math-container">$e(p_x,p_y,U’) &gt; e(p_x,p_y,U^\star) $</span>, so you couldn’t attain the minimum for <span class="math-container">$U’ &gt; U^\star$</span>.</p>
<p>Taking this into account helps simplify the problem because optimizing for <span class="math-container">$u(x^\star) = u^\star$</span> can be done with a simple Lagrangian, while for <span class="math-container">$u(x^\star) \geq u^\star$</span> you need to work with the Kuhn-Tucker conditions which are a bit more complex.</p>

CC BY-SA 4.0
2023-02-06T14:57:08.090Z
2023-02-06T15:10:32.433Z
2023-02-06T15:10:32.433Z
Nicolas Torres
<p>Under the assumption that the utility function is continuous and represents preference relations defined on the consumption set <span class="math-container">$X=R^{L}_+$</span>, and for <span class="math-container">$p &gt;&gt;0$</span>, the hicksian demand correspondence, <span class="math-container">$h(p,u)$</span>, possesses the property of non-excess utility.</p>
<p><span class="math-container">$h(p,u)$</span> is the set of optimal consumption bundles solving the expenditure minimization problem (EMP) at a price vector <span class="math-container">$p$</span> and a fixed level of utility <span class="math-container">$u$</span>.</p>
<p>I'm going to prove why a solution of the EMP, i.e., <span class="math-container">$x \in h(p,u)$</span>, is such that <span class="math-container">$u(x)=u$</span>, and <span class="math-container">$x$</span> is simply a consumption bundle solving the EMP.</p>
<p>Suppose there exists an <span class="math-container">$x' \in h(p,u^{\ast})$</span> such that <span class="math-container">$u(x')&gt;u^{\ast}$</span>, so <span class="math-container">$x'$</span> is a solution of the EMP but delivers a level of utility larger than <span class="math-container">$u^{\ast}$</span>. Take a scale-down version of <span class="math-container">$x'$</span>, say <span class="math-container">$x''=\alpha x'$</span>, where <span class="math-container">$\alpha \in (0,1)$</span>. For <span class="math-container">$\alpha$</span> sufficiently close to <span class="math-container">$1$</span>, by continuity of preferences, it must be that <span class="math-container">$u(x'') \ge u^{\ast}$</span>, but now <span class="math-container">$p \cdot x'' &lt; p x'$</span>, and this contraddicts <span class="math-container">$x'$</span> of being optimal in the EMP, and this is not possible. That's why <span class="math-container">$u(x')=u^{\ast}$</span> if <span class="math-container">$x' \in h(p,u^{\ast})$</span></p>

CC BY-SA 4.0
2023-02-06T14:55:02.657Z
2023-02-06T15:25:10.617Z
2023-02-06T15:25:10.617Z
Tony
<p>The typical expenditure min. problem wants to minimize expenditure under the constraint <span class="math-container">$u(x) \ge u^{\ast}$</span>.
Why the solution of this problem is such that <span class="math-container">$u(x^{\ast})=u^{\ast}$</span> and not <span class="math-container">$u(x^{\ast})&gt;u^{\ast}$</span>?</p>

CC BY-SA 4.0
2023-02-07T10:32:32.800Z
If $u(x^*)> u^*$ you might think you could reduce consumption and expenditure marginally to some $x^\prime$ and still have $u(x^\prime) \ge u^*$
Henry
<microeconomics><consumer-theory>
Expenditure min problem
Dimitru
# Expenditure min problem

(Asked by: username_0 on 2023-02-06)

The typical expenditure min. problem wants to minimize expenditure under the constraint $u(x) \ge u^{\ast}$.
Why the solution of this problem is such that $u(x^{\ast})=u^{\ast}$ and not $u(x^{\ast})>u^{\ast}$?

**Tags:** microeconomics, consumer-theory
> **Comments:**
> **username_1**: If $u(x^*)> u^*$ you might think you could reduce consumption and expenditure marginally to some $x^\prime$ and still have $u(x^\prime) \ge u^*$  

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## 4 answers

### Answer by username_5 on 2023-02-06. Score: 5 | Accepted answer

Under the assumption that the utility function is continuous and represents preference relations defined on the consumption set $X=R^{L}\_+$, and for $p >>0$, the hicksian demand correspondence, $h(p,u)$, possesses the property of non-excess utility.

$h(p,u)$ is the set of optimal consumption bundles solving the expenditure minimization problem (EMP) at a price vector $p$ and a fixed level of utility $u$.

I'm going to prove why a solution of the EMP, i.e., $x \in h(p,u)$, is such that $u(x)=u$, and $x$ is simply a consumption bundle solving the EMP.

Suppose there exists an $x' \in h(p,u^{\ast})$ such that $u(x')>u^{\ast}$, so $x'$ is a solution of the EMP but delivers a level of utility larger than $u^{\ast}$. Take a scale-down version of $x'$, say $x''=\alpha x'$, where $\alpha \in (0,1)$. For $\alpha$ sufficiently close to $1$, by continuity of preferences, it must be that $u(x'') \ge u^{\ast}$, but now $p \cdot x'' < p x'$, and this contraddicts $x'$ of being optimal in the EMP, and this is not possible. That's why $u(x')=u^{\ast}$ if $x' \in h(p,u^{\ast})$


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### Answer by username_2 on 2023-02-06. Score: 3

In general, it is well possible that $u(x^\*)>u^\*$ for an expenditure minimizing $x^\*$. Typical assumptions that guarantee $u(x^\*)=x^\*$ are that $u$ has domain $\mathbb{R}^l\_+$, is continuous, $p\gg 0$, and $u^\*>0$.

Here is the argument: Let $x$ satisfy $u(x)>u^\*>0$. We must have $x\neq 0$ and $p\cdot x>0$. Continuity of $u$ implies that for $\alpha\in(0,1)$ small enough, $u\big(\alpha 0+(1-\alpha)x\big)>u^\*$. Since $p\gg 0$, we also have $p\cdot\big(\alpha 0+(1-\alpha)x\big)=p\cdot(1-\alpha)x=(1-\alpha) p\cdot x. So a bundle that gives higher utility than $u^\*$ cannot be expenditure minimizing, and every expenditure minimizing bundle must give utility exactly $u^\*$.


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### Answer by username_3 on 2023-02-06. Score: 2

Consider the utility function $u:\mathbb{R}^2\_+\rightarrow\mathbb{R}$, given by $u(x, y) = \min(x, y) + 1$. Consider $u^\*=0.5$, solution to the expenditure minimisation problem when $p\_X>0, p\_Y>0$ will be $(0,0)$ and it satisfy $u(0,0)=1>0.5=u^\*$.

Another example is with discontinuous utility $u:\mathbb{R}^2\_+\rightarrow\mathbb{R}$, given by $u(x, y) = \lfloor x+y\rfloor$ i.e. greatest integer less than or equal to $x+y$. Consider $u^\*=4.2$, $p\_X = 1$, $p\_Y = 2$. Solution to the expenditure minimisation problem is $(5,0)$ and it satisfy $u(5,0) = 5>4.2=u^\*$.

Another example is when the two commodities are bad. Consider utility $u:\mathbb{R}^2\_+\rightarrow\mathbb{R}$, given by $u(x, y) = 1-x-y$. Consider $u^\*= 0.5$, solution to the expenditure minimisation problem when $p\_X>0, p\_Y>0$ will be $(0,0)$ and it satisfy $u(0,0)=1>0.5=u^\*$.


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### Answer by username_4 on 2023-02-06. Score: 1

If the preferences are continuous and monotone, if you want to get a higher utility level $U’> U^\star$, you need to spend more money than the optimal expenditure for the current utility level $U^\star$.

In functional notation, $e(p\_x,p\_y,U’) > e(p\_x,p\_y,U^\star) $, so you couldn’t attain the minimum for $U’ > U^\star$.

Taking this into account helps simplify the problem because optimizing for $u(x^\star) = u^\star$ can be done with a simple Lagrangian, while for $u(x^\star) \geq u^\star$ you need to work with the Kuhn-Tucker conditions which are a bit more complex.


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