<p>As said in the comments, the Lemma you mentioned cannot be used to establish the homogeneity of degree <span class="math-container">$0$</span>  in prices and income of the demand functions, because it is only a necessary condition.</p>
<p>And the obviously straightforward way to show homogeneity is multiplying the variables in the budget constraint.</p>
<p>However, your question about whether it is possible to prove homogeneity by Euler's Theorem makes sense. Indeed, Euler's Theorem is a necessary and sufficient condition, and if demand is homogeneous it must hold, and the converse, if it holds the demand is homogeneous.
The question is how, practically, by calculations, to show this.</p>
<p>One has to proceed to find  the derivatives of the demand function with respect to prices and income and verifying that they satisfy the Euler's Theorem.</p>
<p>How to compute these derivatives?</p>
<p>We can resort to the <strong>implicit function theorem</strong>,  applying it to the <strong>first order conditions</strong> of the maximizing problem of the consumer, from which demand functions are derived.</p>
<p>I limit myself to the case of two goods, as for more goods the calculations are cumbersome, and also for two goods they  require a bit of patience.</p>
<p><span class="math-container">$$***$$</span></p>
<p>Assuming the utility function differentiable as needed, and assuming local non satiation, so that the budget constraint is fulfilled as equality, the <strong>optimization problem</strong> is given by</p>
<p><span class="math-container">$$u(x_1,x_2)\;\; \;\;\;max$$</span>
<span class="math-container">$$ s. t. p_1x_1+p_2x_2=m$$</span>,</p>
<p>where the variables have the usual meaning, <span class="math-container">$m$</span> is income.</p>
<p>Using Lagrange multipliers, <span class="math-container">$\lambda$</span>, the first order conditions can be written as :</p>
<p><span class="math-container">$$ m - p_1 x_1(p_1,p_2, m)- p_2 x_2(p_1,p_2, m) = 0. $$</span></p>
<p><span class="math-container">$$ \frac{\partial U( x_1(p_1,p_2, m), x_2(p_1,p_2, m))}{\partial x_1} - \lambda p_1 = 0 $$</span></p>
<p><span class="math-container">$$ \frac{\partial U( x_1(p_1,p_2, m), x_2(p_1,p_2, m))}{\partial x_2} - \lambda p_2 = 0 $$</span></p>
<p>The solutions of this system of equations are the demand functions <span class="math-container">$x_i(p_1,p_2, m)$</span>, <span class="math-container">$i=1,2$</span>.
Of course, it cannot be solved explicitly, but via the implicit function theorem (if its assumptions hold) the derivatives of the demand functions can be calculated.</p>
<p><span class="math-container">$$***$$</span></p>
<p>Let's calculate the <strong>derivatives of <span class="math-container">$x_1(p_1,p_2, m)$</span> with respect to <span class="math-container">$p_1, p_2, m$</span></strong>. This can made  as follows<span class="math-container">$^1$</span>.</p>
<p>Differentiating the first order conditions with respect to <span class="math-container">$p_1$</span> and writing it in matrix form we have:</p>
<p><span class="math-container">$$\begin{bmatrix} 0  &amp; -p_1  &amp; -p_2 \\ -p_1  &amp; u_{11}  &amp; u_{12} \\ -p_2  &amp; u_{21}  &amp; u_{22} \end{bmatrix} \begin {bmatrix} \frac{\partial \lambda}{\partial p_1}\\\frac{\partial x_1}{\partial p_1}\\\frac{\partial x_2}{\partial p_1} \end{bmatrix}= \begin {bmatrix} x_1\\\lambda\\0\end {bmatrix}.\;\;\;\;\;\;\;\;\;\;(1)$$</span></p>
<p>Using Cramer's rule for systems we have:</p>
<p><span class="math-container">$$\frac{\partial x_1}{\partial p_1}= \frac { \begin {vmatrix} 0 &amp; x_1 &amp; -p_2\\ -p_1&amp; \lambda &amp; u_{12}\\  -p_2 &amp; 0&amp; u_{22}\end {vmatrix} }{H}  \;\;\;\;\;  \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(2)  $$</span></p>
<p>where <span class="math-container">$H\neq 0$</span> is the determinant of the bordered hessian (the determinant of the first matrix in <span class="math-container">$(1)$</span>).</p>
<p>Expanding the determinant by the second column we have</p>
<p><span class="math-container">$$\frac{\partial x_1}{\partial p_1}= \lambda \frac { \begin {vmatrix} 0 &amp; -p_2 \\ -p_2 &amp; u_{22} \end{vmatrix}}{H}  -x_1  \frac {\begin {vmatrix} -p_1 &amp; u_{12}\\-p_2 &amp; u_{22}\end {vmatrix}} {H}.    \;\;\;\;\;\;\;\;\;(3)              $$</span></p>
<p>In the same way, differentiating the first order conditions with respect to <span class="math-container">$p_2$</span>, and using Cramer’s rule, we obtain the derivative of <span class="math-container">$x_1$</span> with respect to <span class="math-container">$p_2$</span>, that is</p>
<p><span class="math-container">$$\frac {\partial x_1} {\partial p_2}=- \lambda \frac {\begin {vmatrix}  0&amp;-p_2\\-p_1 &amp; u_{12}\end {vmatrix}}{H}   -x_2   \frac {\begin {vmatrix}  -p_1&amp; u_{12} \\-p_2 &amp; u_{22}\end {vmatrix}}{H}   \;\;\;\;\;(4)         $$</span></p>
<p>The third derivative, differentiating the first order conditions with respect to <span class="math-container">$m$</span> , is</p>
<p><span class="math-container">$$\frac {\partial x_1} {\partial m}= \frac {\begin {vmatrix} -p_1&amp; u_{12}\\-p_2&amp; u_{22}\end {vmatrix}}{H} \;\;\;\;\;\; \;\;\;\;\;\;\;\;\;\;\;\;\;\; \;\;\;\;\;\;\;\;\;\;\;\;\;(5)$$</span></p>
<p><span class="math-container">$$***$$</span>
If you have the patience to make calculations (I consider a miracle I haven't mistaken some sign), you can verify that <strong>Euler's Theorem holds</strong>.</p>
<p>In fact, we have to verify that:</p>
<p><span class="math-container">$$\frac {\partial x_1} {\partial p_1} p_1 +\frac {\partial x_1} {\partial p_2} p_2 +\frac {\partial x_1} {\partial m} m=0.    \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(6)$$</span></p>
<p>Using expressions <span class="math-container">$(3$</span>), <span class="math-container">$(4)$</span> and <span class="math-container">$(5$</span>) for the derivatives, and neglecting <span class="math-container">$H$</span> we have:</p>
<p><span class="math-container">$$(\lambda \begin {vmatrix} 0&amp; -p_2\\-p_2&amp; u_{22}\end {vmatrix} -x_1 \begin {vmatrix} -p_1&amp; u_{12}\\-p_2&amp; u_{22}\end {vmatrix}) p_1+ (-\lambda\begin {vmatrix} 0&amp; -p_2\\-p_1&amp; u_{12}\end {vmatrix}-x_2 \begin {vmatrix} -p_1&amp; u_{12}\\-p_2&amp; u_{22}\end {vmatrix}) p_2+ (\begin {vmatrix} -p_1&amp; u_{12}\\-p_2&amp; u_{22}\end {vmatrix}) m =0. \;\;\;\; (7)$$</span></p>
<p>Replacing  <span class="math-container">$m$</span> with <span class="math-container">$p_1x_1+p_2x_2$</span> and making products we can verify that <span class="math-container">$(7)$</span>, and therefore <span class="math-container">$(6)$</span>,  holds.</p>
<p>We have seen this way that Euler’s theorem holds for the demand of goods 1, <span class="math-container">$x_1(p_1, p_2, m)$</span></p>
<p>The same  procedure can show that it holds for the demand of good 2, <span class="math-container">$x_2 (p_1,p_2, m)$</span>.</p>
<p><span class="math-container">$$***$$</span></p>
<p>Let's say again this is not the most comfortable  way to show that the demands are homogeneous of degree <span class="math-container">$0$</span>. But it could be a useful exercise in calculating derivatives by the implicit functions theorem, and verifying the validity of Euler's Theorem in this case.</p>
<p>The use of the implicit functions theorem to calculate derivatives is important in economics, as it is a tool of comparative statics, both in microeconomics, as for demand functions,  and in macroeconomics, for instance to calculate the multipliers.</p>
<hr />
<p><span class="math-container">$^1$$\scriptsize\text {When using 'differentials' in the implicit functions theorem, a warning must be given, to not convey a wrong message: the use}$</span>
<span class="math-container">$\scriptsize\text {of 'differentials' as above is not rigorous, it is just a shortcut, a useful practical and mnemonic device to make calculations. But}$</span> <span class="math-container">$\scriptsize\text {if one looks at the implicit function theorem and its proof, there aren't differentials.}$</span></p>

CC BY-SA 4.0
2022-11-20T18:44:07.227Z
Too long, I'll give it a read tomorrow. Thanks for your answer. I think it will help me for sure. (Accepting it for the time being; will ask you tomorrow if I have doubts.)
flower_bloom
CC BY-SA 4.0
2022-11-20T19:19:57.247Z
Thanks to you! I'm aware it is long, but it is the topic that is long! Anyway, good question!
BakerStreet
CC BY-SA 4.0
2022-11-20T17:05:10.327Z
2022-11-21T19:47:01.470Z
2022-11-21T19:47:01.470Z
BakerStreet
<p>Why is the demand function homogenous of degree <span class="math-container">$0$</span> in all prices and income?</p>
<p>If I know that the expenditure function is homogenous of degree <span class="math-container">$1$</span> in prices and income, how do I show, using the <em>lemma</em> stated below, that the Marshallian demand function is homogenous of degree <span class="math-container">$0$</span> in all prices and income?</p>
<p><em>Lemma</em>: If <span class="math-container">$f$</span> is homogenous of degree <span class="math-container">$1$</span>, then its partial dervatives <span class="math-container">$f_i$</span> are homogenous of degree <span class="math-container">$0$</span>.</p>
<p>I figured out that <span class="math-container">$x_i^{*}(tp,tI) = x_i^{*}(p,I)$</span> holds since <span class="math-container">$\max_x U(x) \text{ s.t. } p \cdot x \leq I$</span> and <span class="math-container">$\max_x U(x) \text{ s.t. } tp \cdot x \leq tI$</span> are the same maximization problems. But I am looking to show the same using Euler's theorem on homogenous functions.</p>

CC BY-SA 4.0
2022-11-14T13:36:22.887Z
Why do you think this can be shown using Euler's theorem?
Michael Greinecker
CC BY-SA 4.0
2022-11-15T05:47:49.383Z
@Julieawl9 I am looking for a solution using Euler's theorem.
flower_bloom
CC BY-SA 4.0
2022-11-15T05:49:34.017Z
@MichaelGreinecker Not exactly Euler, but using the fact that partial derivatives of homogenous functions of degree $n$ will be homogenous of degree $n-1$. The demands are partials of the indirect utility function (which is homogenous of degree $1$), so I think we can use the theorem. I have edited the question; hopefully, now it conveys what I wanted to say.
flower_bloom
CC BY-SA 4.0
2022-11-15T08:05:10.397Z
The indirect utility function is homogenous of degree 0 in prices and income. After all, it is simply the utility at Marshallian demand. So the premise of the question does not work.
Michael Greinecker
CC BY-SA 4.0
2022-11-15T08:38:40.127Z
@MichaelGreinecker Yes, agreed. Can we use the fact that expenditure function $E$ is homogenous of degree $1$? It's derivatives are Hicksian demands and not Marshallian demands, that's the problem. So maybe if we fix utility at $V(p,I)$ and write the expenditure function as $E(p, V(p, I))$ (where $V$ is the indirect utility), will that work?
flower_bloom
CC BY-SA 4.0
2022-11-15T08:45:22.957Z
I don't see why that should be possible. Hence my question.  And I don't understand why you want to do this.
Michael Greinecker
CC BY-SA 4.0
2022-11-15T15:35:31.773Z
Just another observation: the lemma you mentioned states a necessary condition, not a sufficient condition, so  it isn't possible to use it to prove homogeneity.
BakerStreet
<microeconomics><mathematical-economics><demand>
Why is the demand function homogenous of degree $0$ in all prices and income?
flower_bloom
# Why is the demand function homogenous of degree $0$ in all prices and income?

(Asked by: username_0 on 2022-11-13)

Why is the demand function homogenous of degree $0$ in all prices and income?

If I know that the expenditure function is homogenous of degree $1$ in prices and income, how do I show, using the *lemma* stated below, that the Marshallian demand function is homogenous of degree $0$ in all prices and income?

*Lemma*: If $f$ is homogenous of degree $1$, then its partial dervatives $f\_i$ are homogenous of degree $0$.

I figured out that $x\_i^{\*}(tp,tI) = x\_i^{\*}(p,I)$ holds since $\max\_x U(x) \text{ s.t. } p \cdot x \leq I$ and $\max\_x U(x) \text{ s.t. } tp \cdot x \leq tI$ are the same maximization problems. But I am looking to show the same using Euler's theorem on homogenous functions.

**Tags:** microeconomics, mathematical-economics, demand
> **Comments:**
> **username_1**: Why do you think this can be shown using Euler's theorem?  
> **username_0**: @Julieawl9 I am looking for a solution using Euler's theorem.  
> **username_0**: @MichaelGreinecker Not exactly Euler, but using the fact that partial derivatives of homogenous functions of degree $n$ will be homogenous of degree $n-1$. The demands are partials of the indirect utility function (which is homogenous of degree $1$), so I think we can use the theorem. I have edited the question; hopefully, now it conveys what I wanted to say.  
> **username_1**: The indirect utility function is homogenous of degree 0 in prices and income. After all, it is simply the utility at Marshallian demand. So the premise of the question does not work.  
> **username_0**: @MichaelGreinecker Yes, agreed. Can we use the fact that expenditure function $E$ is homogenous of degree $1$? It's derivatives are Hicksian demands and not Marshallian demands, that's the problem. So maybe if we fix utility at $V(p,I)$ and write the expenditure function as $E(p, V(p, I))$ (where $V$ is the indirect utility), will that work?  
> **username_1**: I don't see why that should be possible. Hence my question.  And I don't understand why you want to do this.  
> **username_2**: Just another observation: the lemma you mentioned states a necessary condition, not a sufficient condition, so  it isn't possible to use it to prove homogeneity.  

---

## 1 answers

### Answer by username_2 on 2022-11-20. Score: 2 | Accepted answer

As said in the comments, the Lemma you mentioned cannot be used to establish the homogeneity of degree $0$ in prices and income of the demand functions, because it is only a necessary condition.

And the obviously straightforward way to show homogeneity is multiplying the variables in the budget constraint.

However, your question about whether it is possible to prove homogeneity by Euler's Theorem makes sense. Indeed, Euler's Theorem is a necessary and sufficient condition, and if demand is homogeneous it must hold, and the converse, if it holds the demand is homogeneous.
The question is how, practically, by calculations, to show this.

One has to proceed to find the derivatives of the demand function with respect to prices and income and verifying that they satisfy the Euler's Theorem.

How to compute these derivatives?

We can resort to the **implicit function theorem**, applying it to the **first order conditions** of the maximizing problem of the consumer, from which demand functions are derived.

I limit myself to the case of two goods, as for more goods the calculations are cumbersome, and also for two goods they require a bit of patience.

$$\*\*\*$$

Assuming the utility function differentiable as needed, and assuming local non satiation, so that the budget constraint is fulfilled as equality, the **optimization problem** is given by

$$u(x\_1,x\_2)\;\; \;\;\;max$$
$$ s. t. p\_1x\_1+p\_2x\_2=m$$,

where the variables have the usual meaning, $m$ is income.

Using Lagrange multipliers, $\lambda$, the first order conditions can be written as :

$$ m - p\_1 x\_1(p\_1,p\_2, m)- p\_2 x\_2(p\_1,p\_2, m) = 0. $$

$$ \frac{\partial U( x\_1(p\_1,p\_2, m), x\_2(p\_1,p\_2, m))}{\partial x\_1} - \lambda p\_1 = 0 $$

$$ \frac{\partial U( x\_1(p\_1,p\_2, m), x\_2(p\_1,p\_2, m))}{\partial x\_2} - \lambda p\_2 = 0 $$

The solutions of this system of equations are the demand functions $x\_i(p\_1,p\_2, m)$, $i=1,2$.
Of course, it cannot be solved explicitly, but via the implicit function theorem (if its assumptions hold) the derivatives of the demand functions can be calculated.

$$\*\*\*$$

Let's calculate the **derivatives of $x\_1(p\_1,p\_2, m)$ with respect to $p\_1, p\_2, m$**. This can made as follows$^1$.

Differentiating the first order conditions with respect to $p\_1$ and writing it in matrix form we have:

$$\begin{bmatrix} 0 & -p\_1 & -p\_2 \\ -p\_1 & u\_{11} & u\_{12} \\ -p\_2 & u\_{21} & u\_{22} \end{bmatrix} \begin {bmatrix} \frac{\partial \lambda}{\partial p\_1}\\\frac{\partial x\_1}{\partial p\_1}\\\frac{\partial x\_2}{\partial p\_1} \end{bmatrix}= \begin {bmatrix} x\_1\\\lambda\\0\end {bmatrix}.\;\;\;\;\;\;\;\;\;\;(1)$$

Using Cramer's rule for systems we have:

$$\frac{\partial x\_1}{\partial p\_1}= \frac { \begin {vmatrix} 0 & x\_1 & -p\_2\\ -p\_1& \lambda & u\_{12}\\ -p\_2 & 0& u\_{22}\end {vmatrix} }{H} \;\;\;\;\; \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(2) $$

where $H\neq 0$ is the determinant of the bordered hessian (the determinant of the first matrix in $(1)$).

Expanding the determinant by the second column we have

$$\frac{\partial x\_1}{\partial p\_1}= \lambda \frac { \begin {vmatrix} 0 & -p\_2 \\ -p\_2 & u\_{22} \end{vmatrix}}{H} -x\_1 \frac {\begin {vmatrix} -p\_1 & u\_{12}\\-p\_2 & u\_{22}\end {vmatrix}} {H}. \;\;\;\;\;\;\;\;\;(3) $$

In the same way, differentiating the first order conditions with respect to $p\_2$, and using Cramer’s rule, we obtain the derivative of $x\_1$ with respect to $p\_2$, that is

$$\frac {\partial x\_1} {\partial p\_2}=- \lambda \frac {\begin {vmatrix} 0&-p\_2\\-p\_1 & u\_{12}\end {vmatrix}}{H} -x\_2 \frac {\begin {vmatrix} -p\_1& u\_{12} \\-p\_2 & u\_{22}\end {vmatrix}}{H} \;\;\;\;\;(4) $$

The third derivative, differentiating the first order conditions with respect to $m$ , is

$$\frac {\partial x\_1} {\partial m}= \frac {\begin {vmatrix} -p\_1& u\_{12}\\-p\_2& u\_{22}\end {vmatrix}}{H} \;\;\;\;\;\; \;\;\;\;\;\;\;\;\;\;\;\;\;\; \;\;\;\;\;\;\;\;\;\;\;\;\;(5)$$

$$\*\*\*$$
If you have the patience to make calculations (I consider a miracle I haven't mistaken some sign), you can verify that **Euler's Theorem holds**.

In fact, we have to verify that:

$$\frac {\partial x\_1} {\partial p\_1} p\_1 +\frac {\partial x\_1} {\partial p\_2} p\_2 +\frac {\partial x\_1} {\partial m} m=0. \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(6)$$

Using expressions $(3$), $(4)$ and $(5$) for the derivatives, and neglecting $H$ we have:

$$(\lambda \begin {vmatrix} 0& -p\_2\\-p\_2& u\_{22}\end {vmatrix} -x\_1 \begin {vmatrix} -p\_1& u\_{12}\\-p\_2& u\_{22}\end {vmatrix}) p\_1+ (-\lambda\begin {vmatrix} 0& -p\_2\\-p\_1& u\_{12}\end {vmatrix}-x\_2 \begin {vmatrix} -p\_1& u\_{12}\\-p\_2& u\_{22}\end {vmatrix}) p\_2+ (\begin {vmatrix} -p\_1& u\_{12}\\-p\_2& u\_{22}\end {vmatrix}) m =0. \;\;\;\; (7)$$

Replacing $m$ with $p\_1x\_1+p\_2x\_2$ and making products we can verify that $(7)$, and therefore $(6)$, holds.

We have seen this way that Euler’s theorem holds for the demand of goods 1, $x\_1(p\_1, p\_2, m)$

The same procedure can show that it holds for the demand of good 2, $x\_2 (p\_1,p\_2, m)$.

$$\*\*\*$$

Let's say again this is not the most comfortable way to show that the demands are homogeneous of degree $0$. But it could be a useful exercise in calculating derivatives by the implicit functions theorem, and verifying the validity of Euler's Theorem in this case.

The use of the implicit functions theorem to calculate derivatives is important in economics, as it is a tool of comparative statics, both in microeconomics, as for demand functions, and in macroeconomics, for instance to calculate the multipliers.

---

$^1$$\scriptsize\text {When using 'differentials' in the implicit functions theorem, a warning must be given, to not convey a wrong message: the use}$
$\scriptsize\text {of 'differentials' as above is not rigorous, it is just a shortcut, a useful practical and mnemonic device to make calculations. But}$ $\scriptsize\text {if one looks at the implicit function theorem and its proof, there aren't differentials.}$

> **Comments:**
> **username_0**: Too long, I'll give it a read tomorrow. Thanks for your answer. I think it will help me for sure. (Accepting it for the time being; will ask you tomorrow if I have doubts.)  
> **username_2**: Thanks to you! I'm aware it is long, but it is the topic that is long! Anyway, good question!  

---

