<p>If the utility is continuous and locally insatiable then the Hicksian demand equals the Walrasian demand.</p>

<p>So you'll need to look for a utility function which violates atleast one of these.</p>

<p>Lets try a simple violation of non-satiation <span class="math-container">$$
u(x,y) = \left\{
        \begin{array}{ll}
            x+y &amp; \quad x+y \leq 1 \\
            1 &amp; \quad \text{ otherwise}
        \end{array}
    \right.
$$</span></p>

<p>With <span class="math-container">$p=(1,1)$</span> and <span class="math-container">$w=2$</span> Then <span class="math-container">$x(p,w)=\{(x,y): 1\leq x+y\leq 2\}$</span>. This is becuse the consumer's problem is <span class="math-container">$$\max_{x,y} u(x,y)$$</span> <span class="math-container">$$\text{st }x+y\leq 2$$</span></p>

<p><span class="math-container">$u(x(p,w))=1$</span> so apply the EMP to this utility <span class="math-container">$$\min x+y$$</span> <span class="math-container">$$\text{st } x+y\geq 1$$</span>
This gives <span class="math-container">$h(p,1)=\{(x,y): x+y= 1\}$</span>. Hence, <span class="math-container">$h(p,v(p,w))\neq x(p,w)$</span>.</p>

<p>I think a violation of non-satiation is much easier to construct a counter example with. If we have a 'thick' indifference curve then the UMP will choose the entire curve but the EMP will choose at most the lower envelope, as why spend more to get the same utility? </p>

<p>Conversely, we can use continuity to find an example of where <span class="math-container">$h(p,v)$</span> is not a subset of <span class="math-container">$x(p,e(p,v))$</span>. Take <span class="math-container">$p$</span> and <span class="math-container">$w$</span> as above and </p>

<p><span class="math-container">$$
u(x,y) = \left\{
        \begin{array}{ll}
            x+y+1 &amp; \quad x+y \geq 2,\ x\geq y \\
            x+y &amp; \quad \text{ otherwise}
        \end{array}
    \right.
$$</span></p>

<p>The EMP with <span class="math-container">$v=2$</span> is</p>

<p><span class="math-container">$$\min x+y$$</span>
<span class="math-container">$$\text{st }u(x,y)\geq 2$$</span></p>

<p><span class="math-container">$h(p,2)=\{(x,y): x+y=2\}$</span>. Now look at the UMP at <span class="math-container">$p,w$</span></p>

<p><span class="math-container">$$\max u(x,y)$$</span></p>

<p><span class="math-container">$$x+y\leq 2$$</span></p>

<p><span class="math-container">$x(p,w)=\{(x,y):\ x+y=2, x\geq y\}$</span>. Hence, <span class="math-container">$x(p,e(p,v))$</span> is a proper subset of <span class="math-container">$h(p,v)$</span>.</p>

<p>I found it alot easier to come up with a counterexample without continuity (our <span class="math-container">$u$</span> is locally non-satiated). Perhaps with continuity you get <span class="math-container">$h\subset x$</span> and local non-satiation you get the reverse? Ill leave that to you.</p>

CC BY-SA 4.0
2019-09-08T14:44:11.297Z
2019-09-09T15:44:11.253Z
2019-09-09T15:44:11.253Z
guest12382
<p>I am looking for a preference relation which satisfies the above property. </p>

<microeconomics><utility>
A case where the solution to Expenditure Minimisation Problem is not a solution to Utility Maximistion Problem
DrStrangeLove
# A case where the solution to Expenditure Minimisation Problem is not a solution to Utility Maximistion Problem

(Asked by: username_0 on 2019-09-08)

I am looking for a preference relation which satisfies the above property.

**Tags:** microeconomics, utility

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## 1 answers

### Answer by username_1 on 2019-09-08. Score: 5 | Accepted answer

If the utility is continuous and locally insatiable then the Hicksian demand equals the Walrasian demand.

So you'll need to look for a utility function which violates atleast one of these.

Lets try a simple violation of non-satiation $$
u(x,y) = \left\{
\begin{array}{ll}
x+y & \quad x+y \leq 1 \\
1 & \quad \text{ otherwise}
\end{array}
\right.
$$

With $p=(1,1)$ and $w=2$ Then $x(p,w)=\{(x,y): 1\leq x+y\leq 2\}$. This is becuse the consumer's problem is $$\max\_{x,y} u(x,y)$$ $$\text{st }x+y\leq 2$$

$u(x(p,w))=1$ so apply the EMP to this utility $$\min x+y$$ $$\text{st } x+y\geq 1$$
This gives $h(p,1)=\{(x,y): x+y= 1\}$. Hence, $h(p,v(p,w))\neq x(p,w)$.

I think a violation of non-satiation is much easier to construct a counter example with. If we have a 'thick' indifference curve then the UMP will choose the entire curve but the EMP will choose at most the lower envelope, as why spend more to get the same utility?

Conversely, we can use continuity to find an example of where $h(p,v)$ is not a subset of $x(p,e(p,v))$. Take $p$ and $w$ as above and

$$
u(x,y) = \left\{
\begin{array}{ll}
x+y+1 & \quad x+y \geq 2,\ x\geq y \\
x+y & \quad \text{ otherwise}
\end{array}
\right.
$$

The EMP with $v=2$ is

$$\min x+y$$
$$\text{st }u(x,y)\geq 2$$

$h(p,2)=\{(x,y): x+y=2\}$. Now look at the UMP at $p,w$

$$\max u(x,y)$$

$$x+y\leq 2$$

$x(p,w)=\{(x,y):\ x+y=2, x\geq y\}$. Hence, $x(p,e(p,v))$ is a proper subset of $h(p,v)$.

I found it alot easier to come up with a counterexample without continuity (our $u$ is locally non-satiated). Perhaps with continuity you get $h\subset x$ and local non-satiation you get the reverse? Ill leave that to you.


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