<p>As Amit has commented, there is a mistake in the above expression. This however is not the answer I got as I have reworked through the math. So the answer which <em>I think is correct</em> along with my work is the following:</p>
<p>we are looking to solve
<span class="math-container">$$\arg \min_{x_1,x_2} p_1x_1+p_2x_2$$</span>
subject to:
<span class="math-container">$$\bar{U}\geq-(x_1-\delta_1)^2-(x_2-\delta _2)^2$$</span>
where <span class="math-container">$\bar{U}\leq0$</span>.</p>
<p><strong>Step 1: Set up Lagrangian</strong>
<span class="math-container">$$\mathcal{L}:p_1x_1+p_2x_2+\lambda[\bar{U}+(x_1-\delta_1)^2+(x_2-\delta _2)]$$</span></p>
<p><strong>Step 2: Take FOCs and derive optimality condition</strong>
<span class="math-container">$$(1)\ \ \ \ \ \ \ \frac{\partial\mathcal{L}}{\partial x_1}:p_1+2\lambda(x_1-\delta_1)=0$$</span>
<span class="math-container">$$(2)\ \ \ \ \ \ \ \frac{\partial\mathcal{L}}{\partial x_2}: p_2+2\lambda(x_2-\delta_2)=0$$</span></p>
<p>Rearranging the above we get:</p>
<p><span class="math-container">$$(1a)\ \ \ \ \ \ \ p_1=-2\lambda(x_1-\delta_1)$$</span>
<span class="math-container">$$(2a)\ \ \ \ \ \ \ p_2=-2\lambda(x_2-\delta_2)$$</span></p>
<p>Dividing (2a) by (1a) we get our optimality condition (where MRS equals our price ratio).</p>
<p><span class="math-container">$$\frac{p_2}{p_1}=\frac{x_2-\delta_2}{x_1-\delta_1}$$</span></p>
<p><strong>Step 3: Derive &quot;intermediate&quot; demand equation</strong></p>
<p>Rearranging our optimality condition we get:</p>
<p><span class="math-container">$$\hat{x_2}=\delta_2+\frac{p_2}{p_1}(x_1-\delta_1)$$</span></p>
<p><strong>Step 4: substitute intermediate demand equation into constraint to obtain hicksian demand</strong></p>
<p>the following algebra below helps give our answer (apologies for the messiness).</p>
<p><span class="math-container">$$\bar{U}=-(x_1-\delta_1)^2-(\hat{x_2}-\delta _2)^2$$</span></p>
<p><span class="math-container">$$\bar{U}=-(x_1-\delta_1)^2-\left(\delta_2+\frac{p_2}{p_1}(x_1-\delta_1)-\delta _2\right)^2$$</span>
<span class="math-container">$$\bar{U}=-(x_1-\delta_1)^2-\left(\frac{p_2}{p_1}(x_1-\delta_1)\right)^2$$</span>
<span class="math-container">$$\bar{U}=-(x_1-\delta_1)^2-\left(\frac{p_2}{p_1}(x_1-\delta_1)\right)^2$$</span></p>
<p><span class="math-container">$$\bar{U}=-(x_1-\delta_1)^2\left(1+\left(\frac{p_2}{p_1}\right)^2\right)$$</span></p>
<p>Heres the tricky part, recall that <span class="math-container">$\bar{U}\leq 0$</span> this allows us to do the following:</p>
<p><span class="math-container">$$(x_1-\delta_1)^2=\frac{-\bar{U}}{1+\left(\frac{p_2}{p_1}\right)^2}$$</span>
<span class="math-container">$$x_1-\delta_1=\left(\frac{-\bar{U}}{1+\left(\frac{p_2}{p_1}\right)^2}\right)^{\frac{1}{2}}$$</span></p>
<p>This is allowed because the term in the brackets is a positive number.</p>
<p>Moving over <span class="math-container">$\delta_1$</span> we have our hicksian demand for <span class="math-container">$x_1$</span>:</p>
<p><span class="math-container">$$x^h_1=\delta_1+\left(\frac{-\bar{U}}{1+\left(\frac{p_2}{p_1}\right)^2}\right)^{\frac{1}{2}}$$</span>
by symmetry the hicksian demand for <span class="math-container">$x_2$</span> is:</p>
<p><span class="math-container">$$x^h_2=\delta_2+\left(\frac{-\bar{U}}{1+\left(\frac{p_1}{p_2}\right)^2}\right)^{\frac{1}{2}}$$</span></p>
<p>I hope this helps.</p>

CC BY-SA 4.0
2024-10-27T04:58:03.567Z
This cannot be correct because in your expressions $x_1^h>\delta_1$ and $x_2^h>\delta_2$, but $(\delta_1,\delta_2)$ is a Bliss point so $u(\delta_1,\delta_2)>u(x_1^h,x_2^h)\geq \overline{U}$, and $(\delta_1,\delta_2)$ also costs less than $(x_1^h, x_2^h)$ because $x_1^h>\delta_1$ and $x_2^h>\delta_2$, contradicting that $(x_1^h, x_2^h)$ is expenditure minimising for $(p_1,p_2, \overline{U})$.
Amit
CC BY-SA 4.0
2024-10-27T05:06:17.017Z
Mistake is in the expression that you have written after this: $(x_1-\delta_1)^2=\frac{-\bar{U}}{1+\left(\frac{p_2}{p_1}\right)^2}$. Taking square root both sides yields LHS $x_1-\delta_1$ (which is negative) and RHS $\left(\frac{-\bar{U}}{1+\left(\frac{p_2}{p_1}\right)^2}\right)^{\frac{1}{2}}$ (which is positive). If you take positive square roots both sides, you'll get $\delta_1-x_1=\left(\frac{-\bar{U}}{1+\left(\frac{p_2}{p_1}\right)^2}\right)^{\frac{1}{2}}$.
Amit
CC BY-SA 4.0
2024-10-27T19:59:02.153Z
@Amit theres a bit to unpack in both of your comments. First note that $x_1^h$ and $x_2^h$ need not be bigger than $\delta_1$ and $\delta_2$ respectively. the square root term can be a positive number or negative number.

Your second comment I dont understand at all.

For these reasons Im keeping my answer as is for now.
EconJohn
CC BY-SA 4.0
2024-11-25T10:21:07.250Z
Thank you for both your answers! This has been very helpful!
Madriel Azère
CC BY-SA 4.0
2024-10-27T03:22:02.493Z
2024-10-27T03:41:00.617Z
2024-10-27T03:41:00.617Z
EconJohn
<p>As I've come across two very interesting posts (1 about generating an expenditure function for the bliss point utility hicksian demand and one about the topic in general) concerning Hicksian demands under a demand system with &quot;bliss point&quot; utility functions, I've tried to show that:</p>
<p><a href="https://i.sstatic.net/xbQdu3iI.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/xbQdu3iI.png" alt="enter image description here" /></a></p>
<p>But I never succeed and locked after simplifying the CPOs! Could someone help me on this one please?</p>
<p>Thank you a lot!</p>

CC BY-SA 4.0
2024-10-25T17:35:55.897Z
Welcome! Please consider taking the time to read the [faq] to familiarize yourself with some of our common practices. In addition, please use [MathJax](http://meta.math.stackexchange.com/q/5020/8348) to typeset mathematics instead of pasting images. Lastly, it would be helpful if you could include a short summary of (and links to) the two posts mentioned so that there is some background of your questions.
Herr K.
CC BY-SA 4.0
2024-10-26T10:07:04.473Z
I think there is mistake in the above expression. Please see this for the correct expression: https://economics.stackexchange.com/a/59381/11824
Amit
<microeconomics><demand>
Proof of Hicksian demand with Bliss Point utility function
Madriel Azère
# Proof of Hicksian demand with Bliss Point utility function

(Asked by: username_0 on 2024-10-25)

As I've come across two very interesting posts (1 about generating an expenditure function for the bliss point utility hicksian demand and one about the topic in general) concerning Hicksian demands under a demand system with "bliss point" utility functions, I've tried to show that:

[![enter image description here](https://i.sstatic.net/xbQdu3iI.png)](https://i.sstatic.net/xbQdu3iI.png)

But I never succeed and locked after simplifying the CPOs! Could someone help me on this one please?

Thank you a lot!

**Tags:** microeconomics, demand
> **Comments:**
> **username_1**: Welcome! Please consider taking the time to read the [faq] to familiarize yourself with some of our common practices. In addition, please use [MathJax](http://meta.math.stackexchange.com/q/5020/8348) to typeset mathematics instead of pasting images. Lastly, it would be helpful if you could include a short summary of (and links to) the two posts mentioned so that there is some background of your questions.  
> **username_2**: I think there is mistake in the above expression. Please see this for the correct expression: https://economics.stackexchange.com/a/59381/11824  

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## 1 answers

### Answer by username_3 on 2024-10-27. Score: 1

As Amit has commented, there is a mistake in the above expression. This however is not the answer I got as I have reworked through the math. So the answer which *I think is correct* along with my work is the following:

we are looking to solve
$$\arg \min\_{x\_1,x\_2} p\_1x\_1+p\_2x\_2$$
subject to:
$$\bar{U}\geq-(x\_1-\delta\_1)^2-(x\_2-\delta \_2)^2$$
where $\bar{U}\leq0$.

**Step 1: Set up Lagrangian**
$$\mathcal{L}:p\_1x\_1+p\_2x\_2+\lambda[\bar{U}+(x\_1-\delta\_1)^2+(x\_2-\delta \_2)]$$

**Step 2: Take FOCs and derive optimality condition**
$$(1)\ \ \ \ \ \ \ \frac{\partial\mathcal{L}}{\partial x\_1}:p\_1+2\lambda(x\_1-\delta\_1)=0$$
$$(2)\ \ \ \ \ \ \ \frac{\partial\mathcal{L}}{\partial x\_2}: p\_2+2\lambda(x\_2-\delta\_2)=0$$

Rearranging the above we get:

$$(1a)\ \ \ \ \ \ \ p\_1=-2\lambda(x\_1-\delta\_1)$$
$$(2a)\ \ \ \ \ \ \ p\_2=-2\lambda(x\_2-\delta\_2)$$

Dividing (2a) by (1a) we get our optimality condition (where MRS equals our price ratio).

$$\frac{p\_2}{p\_1}=\frac{x\_2-\delta\_2}{x\_1-\delta\_1}$$

**Step 3: Derive "intermediate" demand equation**

Rearranging our optimality condition we get:

$$\hat{x\_2}=\delta\_2+\frac{p\_2}{p\_1}(x\_1-\delta\_1)$$

**Step 4: substitute intermediate demand equation into constraint to obtain hicksian demand**

the following algebra below helps give our answer (apologies for the messiness).

$$\bar{U}=-(x\_1-\delta\_1)^2-(\hat{x\_2}-\delta \_2)^2$$

$$\bar{U}=-(x\_1-\delta\_1)^2-\left(\delta\_2+\frac{p\_2}{p\_1}(x\_1-\delta\_1)-\delta \_2\right)^2$$
$$\bar{U}=-(x\_1-\delta\_1)^2-\left(\frac{p\_2}{p\_1}(x\_1-\delta\_1)\right)^2$$
$$\bar{U}=-(x\_1-\delta\_1)^2-\left(\frac{p\_2}{p\_1}(x\_1-\delta\_1)\right)^2$$

$$\bar{U}=-(x\_1-\delta\_1)^2\left(1+\left(\frac{p\_2}{p\_1}\right)^2\right)$$

Heres the tricky part, recall that $\bar{U}\leq 0$ this allows us to do the following:

$$(x\_1-\delta\_1)^2=\frac{-\bar{U}}{1+\left(\frac{p\_2}{p\_1}\right)^2}$$
$$x\_1-\delta\_1=\left(\frac{-\bar{U}}{1+\left(\frac{p\_2}{p\_1}\right)^2}\right)^{\frac{1}{2}}$$

This is allowed because the term in the brackets is a positive number.

Moving over $\delta\_1$ we have our hicksian demand for $x\_1$:

$$x^h\_1=\delta\_1+\left(\frac{-\bar{U}}{1+\left(\frac{p\_2}{p\_1}\right)^2}\right)^{\frac{1}{2}}$$
by symmetry the hicksian demand for $x\_2$ is:

$$x^h\_2=\delta\_2+\left(\frac{-\bar{U}}{1+\left(\frac{p\_1}{p\_2}\right)^2}\right)^{\frac{1}{2}}$$

I hope this helps.

> **Comments:**
> **username_2**: This cannot be correct because in your expressions $x_1^h>\delta_1$ and $x_2^h>\delta_2$, but $(\delta_1,\delta_2)$ is a Bliss point so $u(\delta_1,\delta_2)>u(x_1^h,x_2^h)\geq \overline{U}$, and $(\delta_1,\delta_2)$ also costs less than $(x_1^h, x_2^h)$ because $x_1^h>\delta_1$ and $x_2^h>\delta_2$, contradicting that $(x_1^h, x_2^h)$ is expenditure minimising for $(p_1,p_2, \overline{U})$.  
> **username_2**: Mistake is in the expression that you have written after this: $(x_1-\delta_1)^2=\frac{-\bar{U}}{1+\left(\frac{p_2}{p_1}\right)^2}$. Taking square root both sides yields LHS $x_1-\delta_1$ (which is negative) and RHS $\left(\frac{-\bar{U}}{1+\left(\frac{p_2}{p_1}\right)^2}\right)^{\frac{1}{2}}$ (which is positive). If you take positive square roots both sides, you'll get $\delta_1-x_1=\left(\frac{-\bar{U}}{1+\left(\frac{p_2}{p_1}\right)^2}\right)^{\frac{1}{2}}$.  
> **username_3**: @Amit theres a bit to unpack in both of your comments. First note that $x_1^h$ and $x_2^h$ need not be bigger than $\delta_1$ and $\delta_2$ respectively. the square root term can be a positive number or negative number.  Your second comment I dont understand at all.  For these reasons Im keeping my answer as is for now.  
> **username_0**: Thank you for both your answers! This has been very helpful!  

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