<p>Usually the change in variables <span class="math-container">$X=x-\delta$</span> allows to write
<span class="math-container">$$ e(p,u) = \min_{x \geq 0}\{ p'x : U(x-\delta) \geq u \} =  \max_{X \geq -\delta} \{ p'X : U(X) \geq u \} + p'\delta $$</span>
and so <span class="math-container">$ e(p,u) = E(p,u) + p'\delta.$</span></p>

CC BY-SA 4.0
2024-10-26T18:13:14.327Z
minimize is what you mean right? not maximize.
Jesper Hybel
CC BY-SA 4.0
2024-10-27T21:13:16.477Z
@Jesper Hypel: yes indeed, thank you. The first occurrence of max should be replaced by a min, but not the second, while $p'x$ is non-negative, $p'X$ becomes non-positive. I updated the post.
Bertrand
CC BY-SA 4.0
2020-12-08T08:42:05.260Z
2024-10-27T21:07:11.443Z
2024-10-27T21:07:11.443Z
Bertrand
<p>I think there is some mistake in your calculations for Hicksian Demand. Here is what I have got for the two-commodity case:</p>
<p>Given that <span class="math-container">$u:\mathbb{R}^2_+\rightarrow\mathbb{R}$</span> defined as <span class="math-container">$u(x_1,x_2)=-(x_1-\delta_1)^2-(x_2-\delta_2)^2$</span>,
expenditure minimisation problem is defined as follows:
<span class="math-container">\begin{eqnarray*} \min_{(x_1,x_2)\in\mathbb{R}^2_+} &amp; p_1x_1+p_2x_2 \\ \text{s.t. } &amp; -(x_1-\delta_1)^2-(x_2-\delta_2)^2\geq \overline{u}\end{eqnarray*}</span>
where <span class="math-container">$p_1&gt; 0, p_2&gt; 0$</span>, <span class="math-container">$\delta_1&gt; 0$</span>, <span class="math-container">$\delta_2&gt;0$</span> and <span class="math-container">$\overline{u}\leq 0$</span> are given.
[Please note that the above problem has no feasible solution for <span class="math-container">$\overline{u}&gt; 0$</span>.]</p>
<p>Solving the problem, we obtain the Hicksian demand as follows:
<span class="math-container">\begin{eqnarray*} (x_1^h,x_2^h)(p_1,p_2,\overline{u})=\begin{cases} (\delta_1,\delta_2)&amp;\text{if } \overline{u}=0  \\ (0,0) &amp; \text{if } \overline{u} \leq -\delta_1^2-\delta_2^2 \\ \left(0,\delta_2-\sqrt{-\overline{u}-\delta_1^2}\right) &amp; \text{if } \dfrac{\delta_1}{\sqrt{-\overline{u}-\delta_1^2}} \leq \dfrac{p_1}{p_2} \text{ and } -\delta_1^2-\delta_2^2 &lt;\overline{u} \leq -\delta_1^2 \\ \left(\delta_1-\sqrt{-\overline{u}-\delta_2^2},0\right) &amp; \text{if } \dfrac{\sqrt{-\overline{u}-\delta_2^2}}{\delta_2} \geq \dfrac{p_1}{p_2} \text{ and } -\delta_1^2-\delta_2^2 &lt;\overline{u} \leq -\delta_2^2 \\ \left(\delta_1-\left(\dfrac{-\overline{u}p_1^2}{p_1^2+p_2^2}\right)^{\frac{1}{2}},\delta_2-\left(\dfrac{-\overline{u}p_2^2}{p_1^2+p_2^2}\right)^{\frac{1}{2}}\right) &amp;  \text{ otherwise } \end{cases}\end{eqnarray*}</span></p>
<p>Consequently, expenditure function is <span class="math-container">$e:\mathbb{R}_{++}\times\mathbb{R}_{++}\times(\mathbb{R}_-\cup\{0\})\rightarrow\mathbb{R}_+$</span> is</p>
<p><span class="math-container">\begin{eqnarray*} e(p_1,p_2,\overline{u})=\begin{cases} p_1\delta_1+p_2\delta_2 &amp; \text{if } \overline{u}=0 \\ 
0 &amp; \text{if } \overline{u} \leq -\delta_1^2-\delta_2^2 \\ p_2\left(\delta_2-\sqrt{-\overline{u}-\delta_1^2}\right) &amp; \text{if } \dfrac{\delta_1}{\sqrt{-\overline{u}-\delta_1^2}} \leq \dfrac{p_1}{p_2} \text{ and } -\delta_1^2-\delta_2^2 &lt;\overline{u} \leq -\delta_1^2 \\ p_1\left(\delta_1-\sqrt{-\overline{u}-\delta_2^2}\right) &amp; \text{if } \dfrac{\sqrt{-\overline{u}-\delta_2^2}}{\delta_2} \geq \dfrac{p_1}{p_2} \text{ and } -\delta_1^2-\delta_2^2 &lt;\overline{u} \leq -\delta_2^2 \\  p_1\left(\delta_1-\left(\dfrac{-\overline{u}p_1^2}{p_1^2+p_2^2}\right)^{\frac{1}{2}}\right)+p_2\left(\delta_2-\left(\dfrac{-\overline{u}p_2^2}{p_1^2+p_2^2}\right)^{\frac{1}{2}}\right) &amp;  \text{otherwise} \end{cases} \end{eqnarray*}</span></p>
<p>Here is an example to illustrate different types of solutions to this problem. Here we consider <span class="math-container">$u(x_1,x_2)=-(x_1-2)^2-(x_2-1)^2$</span> and show the Hicksian demand in four different situations:</p>
<p><a href="https://i.sstatic.net/ASyGvR8J.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/ASyGvR8J.png" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2024-10-27T00:39:13.410Z
Thanks for your looking into this question. I did catch an error, but my work does not align with yours. I know this is a free website and you are likely busy, but if you can spell out your answer a bit more in terms of derivations I would appreciate it.
EconJohn
CC BY-SA 4.0
2024-10-27T04:42:51.587Z
Thanks for the comment. I'll try and add more details today. Actually, there are possibilities of corner solutions, solution at the bliss point, solution at (0,0) and other kinds of solutions depending on the parameter values, prices and the target utility in this problem.
Amit
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2024-10-27T09:17:41.503Z
@EconJohn I have added the graph to give an idea about different kinds of solutions. I hope that helps.
Amit
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2024-10-27T20:01:57.470Z
While I appreciate the included visual, I wanted to see your math for your derivations as this answer is unintuitive for a fairly standard problem.
EconJohn
CC BY-SA 4.0
2024-10-28T07:15:58.477Z
I have just done standard slope comparisons at the solutions on the axis and in the interior to write the Hicksian demand. That is why I just posted the graph to illustrate different types of solutions.
Amit
CC BY-SA 4.0
2024-10-26T07:10:22.140Z
2024-10-27T09:16:46.967Z
2024-10-27T09:16:46.967Z
Amit
<p>Edit: My previous answer contained a mistake for the case where <span class="math-container">$x$</span> is restricted to <span class="math-container">$\mathbb{R}_{+}^{n}$</span>. I removed this case from my answer.</p>

<p>Take <span class="math-container">$\bar{U} &gt; 0$</span>. Let's denote <span class="math-container">$\delta = (\delta_{1}, \ldots, \delta_{n})$</span> and <span class="math-container">$p = (p_{1}, \ldots, p_{n})$</span>. Assume <span class="math-container">$p \neq 0$</span>. 
We want to solve
<span class="math-container">\begin{align*}
\min_{x\in\mathbb{R}^{n}} p \cdot x
\qquad 
\text{s.t.}\quad
- (x - \delta) \cdot (x - \delta) \geq - \bar{U}.
\end{align*}</span></p>

<p>Let's first note that a solution must satisfy <span class="math-container">$(x - \delta) \cdot (x - \delta) = \bar{U}$</span>. 
This is because the left side of the constraint is continuous in <span class="math-container">$x$</span>, and because every neighbourhood of <span class="math-container">$x$</span> in <span class="math-container">$\mathbb{R}^{n}$</span> contains a point <span class="math-container">$x^{\prime}$</span> such that <span class="math-container">$p\cdot x^{\prime} &lt; p \cdot x$</span> (since <span class="math-container">$p \neq 0$</span>).</p>

<p>Setting up a Lagrangian, we solve
<span class="math-container">\begin{align*}
\min_{x\in\mathbb{R}^{n}, \lambda\in\mathbb{R}} p \cdot x
+ \lambda\left(
(x - \delta) \cdot (x - \delta) - \bar{U}
\right).
\end{align*}</span>
The first order conditions are
<span class="math-container">$$
p + 2 \lambda (x - \delta) = (0, \ldots, 0)
$$</span>
and
<span class="math-container">$$
(x - \delta) \cdot (x - \delta) - \bar{U} = 0.
$$</span>
Note that <span class="math-container">$\lambda$</span> cannot equal zero as otherwise the first FOC is contradicted. The first FOC is therefore equivalent to <span class="math-container">$(x - \delta) = - p / (2\lambda)$</span>.
Plugging this expression for <span class="math-container">$(x - \delta)$</span> into the second FOC, 
<span class="math-container">$$
\frac{p \cdot p}{4 \lambda^{2}} = \bar{U}
$$</span>
and hence <span class="math-container">$\lambda = \sqrt{\frac{p\cdot p}{4 \bar{U}}}$</span>. 
Finally, 
<span class="math-container">$$
x = \delta + p \frac{1}{2\delta} = \delta + p \sqrt{\frac{\bar{U}}{p\cdot p}}
$$</span></p>

<p>To gain a geometric intuition, let's denote by <span class="math-container">$\Vert\cdot\Vert$</span> the Eucliden norm on <span class="math-container">$\mathbb{R}^{n}$</span> and observe that the set of feasible choices is
<span class="math-container">$$
\left\lbrace x \in\mathbb{R}^{n}\colon \Vert\delta - x \Vert \leq \sqrt{\bar{U}}\right\rbrace.
$$</span>
This is nothing but the closed ball of radiuc <span class="math-container">$\sqrt{\bar{U}}$</span> around <span class="math-container">$\delta$</span>. 
Minimizing the linear function <span class="math-container">$x\mapsto p \cdot x$</span> involves finding the point on the boundary of the ball where a level set of <span class="math-container">$x\mapsto p \cdot x$</span> is tangent to the ball.
We find this point by traveling from the ball's center at <span class="math-container">$\delta$</span> in the direction of <span class="math-container">$-p$</span>, i.e. orthogonal to the hyperplane, until we reach the boundary.</p>

<p>Here's an alternate proof that doesn't use any stuff involving Lagrangians:
As before, we may restrict attention to points <span class="math-container">$y$</span> satisfying <span class="math-container">$\Vert\delta - y \Vert = \sqrt{\bar{U}}$</span>. 
Any such point may be written as <span class="math-container">$y = \delta -  (\delta - y)\frac{\sqrt{\bar{U}}}{\Vert \delta - y \Vert}$</span>.
Consider the point <span class="math-container">$x = \delta - p \frac{\sqrt{\bar{U}}}{\Vert p \Vert}$</span>.
We claim that <span class="math-container">$p\cdot x \leq p \cdot y$</span> for any such point <span class="math-container">$y$</span> on the boundary. This is the case if and only if
<span class="math-container">\begin{align*}
p\cdot (\delta - y) \frac{1}{\Vert \delta - y \Vert}
\leq 
p\cdot p \frac{1}{\Vert p \Vert}.
\end{align*}</span>
The right side equals <span class="math-container">$\Vert p \Vert$</span>.
As for the left side, by the Cauchy-Schwarz inequality, 
<span class="math-container">\begin{align*}
p\cdot (\delta - y)  \frac{1}{\Vert \delta - y \Vert} \leq 
\Vert p \Vert \Vert \delta - y \Vert \frac{1}{\Vert \delta - y \Vert} = \Vert p\Vert,
\end{align*}</span>
and so we're done. Moreover, the Cauchy-Schwarz inequality holds strictly unless <span class="math-container">$y - \delta$</span> is a multiple of <span class="math-container">$p$</span>. In that case, however, either <span class="math-container">$p = \delta - y$</span> or <span class="math-container">$y$</span> does not lie on the boundary. So the point <span class="math-container">$x$</span> is in fact the unique minimizer.</p>

CC BY-SA 4.0
2019-11-13T20:44:34.957Z
Hi, looks like you're on to something. but it seems that you made a mistake.your solved FOC should be $(x-\delta)=-\frac{p}{2\lambda}$.
EconJohn
CC BY-SA 4.0
2019-11-13T21:02:42.707Z
You're right, that sign is wrong, but the rest of the argument is unchanged. I'll edit the answer
induction_is_a_laddah
CC BY-SA 4.0
2019-11-13T15:01:34.043Z
2019-11-13T21:03:49.510Z
2019-11-13T21:03:49.510Z
induction_is_a_laddah
<p><strong>EDIT:</strong> My answer has been modified almost entirely in response to the feedback in comments.</p>
<p><strong>TL;DR:</strong> The 2 good case is quirky and does not yield to a neat solution. Amit has posted a more complete solution for such systems, I will simply highlight why the solution in the original post is incomplete.</p>
<p><strong>Non-Graphical Argument</strong></p>
<p>A consumer will never demand any bundle outside <span class="math-container">$\{(x,y)\in\mathbb{R^2_+} \mid x \leq \delta_1, y \leq \delta_2\} $</span>. So it must be the case that <span class="math-container">$e(p_1, p_2, \bar U) \leq p_1\delta_1 + p_2\delta_2 $</span> - which is in contradiction the expenditure function stated above.</p>
<p>The 'correct' solution that the Lagrangian will give you (as discovered in the comments to this post) is</p>
<p><span class="math-container">$(x^c, y^c) = \left(\delta_1 - \left[\dfrac{- \bar U}{1 + (\frac{p_2}{p_1})^2}\right]^\frac{1}{2}, \delta_2 - \left[\dfrac{- \bar U}{1 + (\frac{p_1}{p_2})^2}\right]^\frac{1}{2}\right)$</span></p>
<p>However, even this is not complete.</p>
<p><strong>Graphical Argument</strong></p>
<p>The below arguments can also be seen succinctly in <a href="https://www.desmos.com/calculator/gm6svongyc" rel="nofollow noreferrer">this graph</a>:</p>
<p>The <span class="math-container">$(x^c, y^c)$</span> stated above is the point at which the budget constraint is tangent to the indifference curve.</p>
<ol>
<li><strong>Corner points are an issue</strong>: Here, <span class="math-container">$y^c &lt; 0$</span>. It isn't enough to set <span class="math-container">$y^c = 0$</span> and keep <span class="math-container">$x^c$</span> as is - you can see that in the graphic below, that would actually put you on a higher IC than you are targeting, even though there is a cheaper bundle available for the utility level that you are targeting.
<a href="https://i.sstatic.net/ASMRfD8J.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/ASMRfD8J.png" alt="enter image description here" /></a></li>
</ol>
<hr />
<ol start="2">
<li><strong>Target utility has a lower bound</strong> at/below which all other possible solution bundles are inferior to <span class="math-container">$(0,0)$</span> - <span class="math-container">$(x^c, y^c) = (0,0)$</span> for such cases.
<a href="https://i.sstatic.net/oRZxP2A4.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/oRZxP2A4.png" alt="enter image description here" /></a></li>
</ol>
<p><strike><a href="https://www.desmos.com/calculator/mqbydjksfd" rel="nofollow noreferrer">Here is a system</a> that tries to account for these, but you'll see that even this system fails..</strike></p>
<p>I think <a href="https://www.desmos.com/calculator/py7lbmtgbh" rel="nofollow noreferrer">this interactive demo</a> covers all edge cases</p>

CC BY-SA 4.0
2024-10-29T14:41:21.863Z
This does not answer the question or point out any error. I looked at your desmos example. I like it but you need to remember that there are two solutions with the square root. See what I have edited here: https://www.desmos.com/calculator/spndtlroug
EconJohn
CC BY-SA 4.0
2024-10-29T19:24:56.093Z
The desmos graph strictly demonstrates the solution that you have claimed for the 2 good case in your original post. The "correct" demand graph that you have shared with me has the exact flaw that Amit has already pointed out - it doesn't account for extremes. See [this](https://www.desmos.com/calculator/gksiflsrdu), [this](https://www.desmos.com/calculator/erygybm5xi) and [this](https://www.desmos.com/calculator/qz9nwqtxo2) to see the specific cases where your corrected solution fails as well.
shgwani2804
CC BY-SA 4.0
2024-10-29T19:27:40.290Z
The answer also pretty clearly shows your error - the best available bundle costs $p_1 \delta_1 + p_2 \delta_2$. It is impossible to get a better bundle, by definition of bliss point. The expenditure function described in your original post is strictly higher than this for $\bar U < 0$. Using the other solution you have found, once you impose non-negativity constraints (again, I have already stated that the full answer would involve invoking the Kuhn-Tucker conditions), you will obtain the complete solution analytically.
shgwani2804
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2024-10-29T20:00:01.693Z
I see the corner solutions being possible, but I dont see how this invalidates the answer "partial marks" I have given so far. I understand how we can characterize the demands as $$x_1^c=\min \{\delta_1+\left[-\frac{\bar{U}}{1+(\frac{p_2}{p_1})^2}\right]^\frac{1}{2}, \delta_1-\left[-\frac{\bar{U}}{1+(\frac{p_2}{p_1})^2}\right]^\frac{1}{2}\}$$ and $$x_2^c=\min\{\delta_2+\left[-\frac{\bar{U}}{1+(\frac{p_1}{p_2})^2}\right]^\frac{1}{2}, \delta_2-\left[-\frac{\bar{U}}{1+(\frac{p_1}{p_2})^2}\right]^\frac{1}{2}\}$$ corners can come in a more parsimonious manner. I have some ideas I'll post later.
EconJohn
CC BY-SA 4.0
2024-10-29T20:02:09.280Z
To clarify, I agree that corners are an issue here. I dont think Amit's answer is clear though as it looks like there's possibility of imaginary solutions. Also I think we can come up with something prettier.
EconJohn
CC BY-SA 4.0
2024-10-30T04:29:35.043Z
The cleanest answer I can come up with for our compensated demands at this point is:

$$x_1^c=\max \{ \delta_1-\left[-\frac{\bar{U}}{1+(\frac{p_2}{p_1})^2}\right]^\frac{1}{2}, 0 \}$$

$$x_2^c=\max \{\delta_2-\left[-\frac{\bar{U}}{1+(\frac{p_1}{p_2})^2}\right]^\frac{1}{2}, 0 \}$$

theres still some kinks to work out, but its closer.
EconJohn
CC BY-SA 4.0
2024-10-30T06:22:29.913Z
@EconJohn that is closer and neater, yes. The only issue that I can see with that is [this](https://www.desmos.com/calculator/edfgappge8). The demand generated by the above functions is putting us on a higher IC than we're targeting, while bundles with the targeted IC are feasible and cheaper (look at the intercepts of the green IC in this graph). This problem arises when the price ratio is not 1
shgwani2804
CC BY-SA 4.0
2024-10-30T06:23:32.613Z
@EconJohn See [this completed system](https://www.desmos.com/calculator/rdmymc8j3y). It is not neat, but it recreates Amit's answer interactively. Perhaps we can find a way to make this neater.
shgwani2804
CC BY-SA 4.0
2024-10-30T14:54:39.717Z
Your desmos skills are excellent! I really like the pictures. Characterizing a solution seems quite akward. This is the most clear example of how standard optimization methods fail. if you update your answer to include these desmos charts (perhaps with some screenshots also since I dont know if the link would expire). I will accept your answer (even though it does not quite answer the initial question for an N good case, it turns out the 2 good case is quite tricky!).
EconJohn
CC BY-SA 4.0
2024-11-04T05:52:14.127Z
@EconJohn thank you. see the updated answer; I found a flaw with the 'completed system' I shared above in the comments, which I have mentioned as well in the updated answer. If I get time to recreate Amit's solution entirely, I will share that later.
shgwani2804
CC BY-SA 4.0
2024-10-29T09:48:23.117Z
2024-11-04T07:43:48.133Z
2024-11-04T07:43:48.133Z
shgwani2804
<p>I've been looking at expenditure systems and have been really interested in the behaviour of the demand system that underlies bliss points:</p>
<p>Consider the bliss point utility function of the following form:</p>
<p><span class="math-container">$$U(x_1,x_2)=-(x_1-\delta_1)^2-(x_2-\delta_2)^2$$</span></p>
<p>for two dimensions the corresponding hicksian demands are:</p>
<p><span class="math-container">$$x_1^c=\delta_1+\left[-\frac{\bar{U}}{1+(\frac{p_2}{p_1})^2}\right]^\frac{1}{2}$$</span>
<span class="math-container">$$x_2^c=\delta_2+\left[-\frac{\bar{U}}{1+(\frac{p_1}{p_2})^2}\right]^\frac{1}{2}$$</span></p>
<p>where <span class="math-container">$\bar{U}\leq 0$</span>.</p>
<p>It follows that the expenditure function is:
<span class="math-container">$$e(p_1,p_2,\bar{U})=p_1\delta_1+p_1\left[-\frac{\bar{U}}{1+(\frac{p_2}{p_1})^2}\right]^\frac{1}{2}+p_2\delta_2+p_2\left[-\frac{\bar{U}}{1+(\frac{p_1}{p_2})^2}\right]^\frac{1}{2}$$</span></p>
<p>Obviously expenditure functions can be much larger. however I'm having a hard time for generating a expenditure function for a number of <span class="math-container">$n$</span> goods.</p>
<p><strong>tldr</strong> What would the hicksian demands look like for the utility function:
<span class="math-container">$$U(\mathbf{x})=-\sum_{i=1}^n(x_i-\delta_i)^2$$</span></p>

<microeconomics><mathematical-economics>
Regarding the Expenditure Function Underlying a Bliss Point
EconJohn
# Regarding the Expenditure Function Underlying a Bliss Point

(Asked by: username_0 on 2019-11-08)

I've been looking at expenditure systems and have been really interested in the behaviour of the demand system that underlies bliss points:

Consider the bliss point utility function of the following form:

$$U(x\_1,x\_2)=-(x\_1-\delta\_1)^2-(x\_2-\delta\_2)^2$$

for two dimensions the corresponding hicksian demands are:

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

where $\bar{U}\leq 0$.

It follows that the expenditure function is:
$$e(p\_1,p\_2,\bar{U})=p\_1\delta\_1+p\_1\left[-\frac{\bar{U}}{1+(\frac{p\_2}{p\_1})^2}\right]^\frac{1}{2}+p\_2\delta\_2+p\_2\left[-\frac{\bar{U}}{1+(\frac{p\_1}{p\_2})^2}\right]^\frac{1}{2}$$

Obviously expenditure functions can be much larger. however I'm having a hard time for generating a expenditure function for a number of $n$ goods.

**tldr** What would the hicksian demands look like for the utility function:
$$U(\mathbf{x})=-\sum\_{i=1}^n(x\_i-\delta\_i)^2$$

**Tags:** microeconomics, mathematical-economics


---

## 4 answers

### Answer by username_5 on 2024-10-29. Score: 2 | Accepted answer

**EDIT:** My answer has been modified almost entirely in response to the feedback in comments.

**TL;DR:** The 2 good case is quirky and does not yield to a neat solution. username_3 has posted a more complete solution for such systems, I will simply highlight why the solution in the original post is incomplete.

**Non-Graphical Argument**

A consumer will never demand any bundle outside $\{(x,y)\in\mathbb{R^2\_+} \mid x \leq \delta\_1, y \leq \delta\_2\} $. So it must be the case that $e(p\_1, p\_2, \bar U) \leq p\_1\delta\_1 + p\_2\delta\_2 $ - which is in contradiction the expenditure function stated above.

The 'correct' solution that the Lagrangian will give you (as discovered in the comments to this post) is

$(x^c, y^c) = \left(\delta\_1 - \left[\dfrac{- \bar U}{1 + (\frac{p\_2}{p\_1})^2}\right]^\frac{1}{2}, \delta\_2 - \left[\dfrac{- \bar U}{1 + (\frac{p\_1}{p\_2})^2}\right]^\frac{1}{2}\right)$

However, even this is not complete.

**Graphical Argument**

The below arguments can also be seen succinctly in [this graph](https://www.desmos.com/calculator/gm6svongyc):

The $(x^c, y^c)$ stated above is the point at which the budget constraint is tangent to the indifference curve.

1. **Corner points are an issue**: Here, $y^c < 0$. It isn't enough to set $y^c = 0$ and keep $x^c$ as is - you can see that in the graphic below, that would actually put you on a higher IC than you are targeting, even though there is a cheaper bundle available for the utility level that you are targeting.
   [![enter image description here](https://i.sstatic.net/ASMRfD8J.png)](https://i.sstatic.net/ASMRfD8J.png)

---

2. **Target utility has a lower bound** at/below which all other possible solution bundles are inferior to $(0,0)$ - $(x^c, y^c) = (0,0)$ for such cases.
   [![enter image description here](https://i.sstatic.net/oRZxP2A4.png)](https://i.sstatic.net/oRZxP2A4.png)

[Here is a system](https://www.desmos.com/calculator/mqbydjksfd) that tries to account for these, but you'll see that even this system fails..

I think [this interactive demo](https://www.desmos.com/calculator/py7lbmtgbh) covers all edge cases

> **Comments:**
> **username_0**: This does not answer the question or point out any error. I looked at your desmos example. I like it but you need to remember that there are two solutions with the square root. See what I have edited here: https://www.desmos.com/calculator/spndtlroug  
> **username_5**: The desmos graph strictly demonstrates the solution that you have claimed for the 2 good case in your original post. The "correct" demand graph that you have shared with me has the exact flaw that username_3 has already pointed out - it doesn't account for extremes. See [this](https://www.desmos.com/calculator/gksiflsrdu), [this](https://www.desmos.com/calculator/erygybm5xi) and [this](https://www.desmos.com/calculator/qz9nwqtxo2) to see the specific cases where your corrected solution fails as well.  
> **username_5**: The answer also pretty clearly shows your error - the best available bundle costs $p_1 \delta_1 + p_2 \delta_2$. It is impossible to get a better bundle, by definition of bliss point. The expenditure function described in your original post is strictly higher than this for $\bar U < 0$. Using the other solution you have found, once you impose non-negativity constraints (again, I have already stated that the full answer would involve invoking the Kuhn-Tucker conditions), you will obtain the complete solution analytically.  
> **username_0**: I see the corner solutions being possible, but I dont see how this invalidates the answer "partial marks" I have given so far. I understand how we can characterize the demands as $$x_1^c=\min \{\delta_1+\left[-\frac{\bar{U}}{1+(\frac{p_2}{p_1})^2}\right]^\frac{1}{2}, \delta_1-\left[-\frac{\bar{U}}{1+(\frac{p_2}{p_1})^2}\right]^\frac{1}{2}\}$$ and $$x_2^c=\min\{\delta_2+\left[-\frac{\bar{U}}{1+(\frac{p_1}{p_2})^2}\right]^\frac{1}{2}, \delta_2-\left[-\frac{\bar{U}}{1+(\frac{p_1}{p_2})^2}\right]^\frac{1}{2}\}$$ corners can come in a more parsimonious manner. I have some ideas I'll post later.  
> **username_0**: To clarify, I agree that corners are an issue here. I dont think username_3's answer is clear though as it looks like there's possibility of imaginary solutions. Also I think we can come up with something prettier.  
> **username_0**: The cleanest answer I can come up with for our compensated demands at this point is:  $$x_1^c=\max \{ \delta_1-\left[-\frac{\bar{U}}{1+(\frac{p_2}{p_1})^2}\right]^\frac{1}{2}, 0 \}$$  $$x_2^c=\max \{\delta_2-\left[-\frac{\bar{U}}{1+(\frac{p_1}{p_2})^2}\right]^\frac{1}{2}, 0 \}$$  theres still some kinks to work out, but its closer.  
> **username_5**: @username_0 that is closer and neater, yes. The only issue that I can see with that is [this](https://www.desmos.com/calculator/edfgappge8). The demand generated by the above functions is putting us on a higher IC than we're targeting, while bundles with the targeted IC are feasible and cheaper (look at the intercepts of the green IC in this graph). This problem arises when the price ratio is not 1  
> **username_5**: @username_0 See [this completed system](https://www.desmos.com/calculator/rdmymc8j3y). It is not neat, but it recreates username_3's answer interactively. Perhaps we can find a way to make this neater.  
> **username_0**: Your desmos skills are excellent! I really like the pictures. Characterizing a solution seems quite akward. This is the most clear example of how standard optimization methods fail. if you update your answer to include these desmos charts (perhaps with some screenshots also since I dont know if the link would expire). I will accept your answer (even though it does not quite answer the initial question for an N good case, it turns out the 2 good case is quite tricky!).  
> **username_5**: @username_0 thank you. see the updated answer; I found a flaw with the 'completed system' I shared above in the comments, which I have mentioned as well in the updated answer. If I get time to recreate username_3's solution entirely, I will share that later.  

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### Answer by username_1 on 2020-12-08. Score: 3

Usually the change in variables $X=x-\delta$ allows to write
$$ e(p,u) = \min\_{x \geq 0}\{ p'x : U(x-\delta) \geq u \} = \max\_{X \geq -\delta} \{ p'X : U(X) \geq u \} + p'\delta $$
and so $ e(p,u) = E(p,u) + p'\delta.$

> **Comments:**
> **username_2**: minimize is what you mean right? not maximize.  
> **username_1**: @Jesper Hypel: yes indeed, thank you. The first occurrence of max should be replaced by a min, but not the second, while $p'x$ is non-negative, $p'X$ becomes non-positive. I updated the post.  

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### Answer by username_3 on 2024-10-26. Score: 2

I think there is some mistake in your calculations for Hicksian Demand. Here is what I have got for the two-commodity case:

Given that $u:\mathbb{R}^2\_+\rightarrow\mathbb{R}$ defined as $u(x\_1,x\_2)=-(x\_1-\delta\_1)^2-(x\_2-\delta\_2)^2$,
expenditure minimisation problem is defined as follows:
\begin{eqnarray\*} \min\_{(x\_1,x\_2)\in\mathbb{R}^2\_+} & p\_1x\_1+p\_2x\_2 \\ \text{s.t. } & -(x\_1-\delta\_1)^2-(x\_2-\delta\_2)^2\geq \overline{u}\end{eqnarray\*}
where $p\_1> 0, p\_2> 0$, $\delta\_1> 0$, $\delta\_2>0$ and $\overline{u}\leq 0$ are given.
[Please note that the above problem has no feasible solution for $\overline{u}> 0$.]

Solving the problem, we obtain the Hicksian demand as follows:
\begin{eqnarray\*} (x\_1^h,x\_2^h)(p\_1,p\_2,\overline{u})=\begin{cases} (\delta\_1,\delta\_2)&\text{if } \overline{u}=0 \\ (0,0) & \text{if } \overline{u} \leq -\delta\_1^2-\delta\_2^2 \\ \left(0,\delta\_2-\sqrt{-\overline{u}-\delta\_1^2}\right) & \text{if } \dfrac{\delta\_1}{\sqrt{-\overline{u}-\delta\_1^2}} \leq \dfrac{p\_1}{p\_2} \text{ and } -\delta\_1^2-\delta\_2^2 <\overline{u} \leq -\delta\_1^2 \\ \left(\delta\_1-\sqrt{-\overline{u}-\delta\_2^2},0\right) & \text{if } \dfrac{\sqrt{-\overline{u}-\delta\_2^2}}{\delta\_2} \geq \dfrac{p\_1}{p\_2} \text{ and } -\delta\_1^2-\delta\_2^2 <\overline{u} \leq -\delta\_2^2 \\ \left(\delta\_1-\left(\dfrac{-\overline{u}p\_1^2}{p\_1^2+p\_2^2}\right)^{\frac{1}{2}},\delta\_2-\left(\dfrac{-\overline{u}p\_2^2}{p\_1^2+p\_2^2}\right)^{\frac{1}{2}}\right) & \text{ otherwise } \end{cases}\end{eqnarray\*}

Consequently, expenditure function is $e:\mathbb{R}\_{++}\times\mathbb{R}\_{++}\times(\mathbb{R}\_-\cup\{0\})\rightarrow\mathbb{R}\_+$ is

\begin{eqnarray\*} e(p\_1,p\_2,\overline{u})=\begin{cases} p\_1\delta\_1+p\_2\delta\_2 & \text{if } \overline{u}=0 \\
0 & \text{if } \overline{u} \leq -\delta\_1^2-\delta\_2^2 \\ p\_2\left(\delta\_2-\sqrt{-\overline{u}-\delta\_1^2}\right) & \text{if } \dfrac{\delta\_1}{\sqrt{-\overline{u}-\delta\_1^2}} \leq \dfrac{p\_1}{p\_2} \text{ and } -\delta\_1^2-\delta\_2^2 <\overline{u} \leq -\delta\_1^2 \\ p\_1\left(\delta\_1-\sqrt{-\overline{u}-\delta\_2^2}\right) & \text{if } \dfrac{\sqrt{-\overline{u}-\delta\_2^2}}{\delta\_2} \geq \dfrac{p\_1}{p\_2} \text{ and } -\delta\_1^2-\delta\_2^2 <\overline{u} \leq -\delta\_2^2 \\ p\_1\left(\delta\_1-\left(\dfrac{-\overline{u}p\_1^2}{p\_1^2+p\_2^2}\right)^{\frac{1}{2}}\right)+p\_2\left(\delta\_2-\left(\dfrac{-\overline{u}p\_2^2}{p\_1^2+p\_2^2}\right)^{\frac{1}{2}}\right) & \text{otherwise} \end{cases} \end{eqnarray\*}

Here is an example to illustrate different types of solutions to this problem. Here we consider $u(x\_1,x\_2)=-(x\_1-2)^2-(x\_2-1)^2$ and show the Hicksian demand in four different situations:

[![enter image description here](https://i.sstatic.net/ASyGvR8J.png)](https://i.sstatic.net/ASyGvR8J.png)

> **Comments:**
> **username_0**: Thanks for your looking into this question. I did catch an error, but my work does not align with yours. I know this is a free website and you are likely busy, but if you can spell out your answer a bit more in terms of derivations I would appreciate it.  
> **username_3**: Thanks for the comment. I'll try and add more details today. Actually, there are possibilities of corner solutions, solution at the bliss point, solution at (0,0) and other kinds of solutions depending on the parameter values, prices and the target utility in this problem.  
> **username_3**: @username_0 I have added the graph to give an idea about different kinds of solutions. I hope that helps.  
> **username_0**: While I appreciate the included visual, I wanted to see your math for your derivations as this answer is unintuitive for a fairly standard problem.  
> **username_3**: I have just done standard slope comparisons at the solutions on the axis and in the interior to write the Hicksian demand. That is why I just posted the graph to illustrate different types of solutions.  

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### Answer by username_4 on 2019-11-13. Score: 1

Edit: My previous answer contained a mistake for the case where $x$ is restricted to $\mathbb{R}\_{+}^{n}$. I removed this case from my answer.

Take $\bar{U} > 0$. Let's denote $\delta = (\delta\_{1}, \ldots, \delta\_{n})$ and $p = (p\_{1}, \ldots, p\_{n})$. Assume $p \neq 0$.
We want to solve
\begin{align\*}
\min\_{x\in\mathbb{R}^{n}} p \cdot x
\qquad
\text{s.t.}\quad
- (x - \delta) \cdot (x - \delta) \geq - \bar{U}.
\end{align\*}

Let's first note that a solution must satisfy $(x - \delta) \cdot (x - \delta) = \bar{U}$.
This is because the left side of the constraint is continuous in $x$, and because every neighbourhood of $x$ in $\mathbb{R}^{n}$ contains a point $x^{\prime}$ such that $p\cdot x^{\prime} < p \cdot x$ (since $p \neq 0$).

Setting up a Lagrangian, we solve
\begin{align\*}
\min\_{x\in\mathbb{R}^{n}, \lambda\in\mathbb{R}} p \cdot x
+ \lambda\left(
(x - \delta) \cdot (x - \delta) - \bar{U}
\right).
\end{align\*}
The first order conditions are
$$
p + 2 \lambda (x - \delta) = (0, \ldots, 0)
$$
and
$$
(x - \delta) \cdot (x - \delta) - \bar{U} = 0.
$$
Note that $\lambda$ cannot equal zero as otherwise the first FOC is contradicted. The first FOC is therefore equivalent to $(x - \delta) = - p / (2\lambda)$.
Plugging this expression for $(x - \delta)$ into the second FOC,
$$
\frac{p \cdot p}{4 \lambda^{2}} = \bar{U}
$$
and hence $\lambda = \sqrt{\frac{p\cdot p}{4 \bar{U}}}$.
Finally,
$$
x = \delta + p \frac{1}{2\delta} = \delta + p \sqrt{\frac{\bar{U}}{p\cdot p}}
$$

To gain a geometric intuition, let's denote by $\Vert\cdot\Vert$ the Eucliden norm on $\mathbb{R}^{n}$ and observe that the set of feasible choices is
$$
\left\lbrace x \in\mathbb{R}^{n}\colon \Vert\delta - x \Vert \leq \sqrt{\bar{U}}\right\rbrace.
$$
This is nothing but the closed ball of radiuc $\sqrt{\bar{U}}$ around $\delta$.
Minimizing the linear function $x\mapsto p \cdot x$ involves finding the point on the boundary of the ball where a level set of $x\mapsto p \cdot x$ is tangent to the ball.
We find this point by traveling from the ball's center at $\delta$ in the direction of $-p$, i.e. orthogonal to the hyperplane, until we reach the boundary.

Here's an alternate proof that doesn't use any stuff involving Lagrangians:
As before, we may restrict attention to points $y$ satisfying $\Vert\delta - y \Vert = \sqrt{\bar{U}}$.
Any such point may be written as $y = \delta - (\delta - y)\frac{\sqrt{\bar{U}}}{\Vert \delta - y \Vert}$.
Consider the point $x = \delta - p \frac{\sqrt{\bar{U}}}{\Vert p \Vert}$.
We claim that $p\cdot x \leq p \cdot y$ for any such point $y$ on the boundary. This is the case if and only if
\begin{align\*}
p\cdot (\delta - y) \frac{1}{\Vert \delta - y \Vert}
\leq
p\cdot p \frac{1}{\Vert p \Vert}.
\end{align\*}
The right side equals $\Vert p \Vert$.
As for the left side, by the Cauchy-Schwarz inequality,
\begin{align\*}
p\cdot (\delta - y) \frac{1}{\Vert \delta - y \Vert} \leq
\Vert p \Vert \Vert \delta - y \Vert \frac{1}{\Vert \delta - y \Vert} = \Vert p\Vert,
\end{align\*}
and so we're done. Moreover, the Cauchy-Schwarz inequality holds strictly unless $y - \delta$ is a multiple of $p$. In that case, however, either $p = \delta - y$ or $y$ does not lie on the boundary. So the point $x$ is in fact the unique minimizer.

> **Comments:**
> **username_0**: Hi, looks like you're on to something. but it seems that you made a mistake.your solved FOC should be $(x-\delta)=-\frac{p}{2\lambda}$.  
> **username_4**: You're right, that sign is wrong, but the rest of the argument is unchanged. I'll edit the answer  

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